Mi oppenheim - signals and systems 2ed solucionario
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Apr 07, 2014
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About This Presentation
Oppenheim Solution Handbook
Slide Content
EL SOLUCIONARIO
DATE TT A RI.
DATE TT A RI.
CORRELTEB SoLuTIons
121 €) () x Gal + ju}
big ne
4 4)
ELD yy existe g)-c(t-a)] ent tog
6") > -0F yl Y=xl ea
121 €) () x Gal + ju}
big ne
4 4)
ELD yy existe g)-c(t-a)] ent tog
6") > -0F yl Y=xl ea
1a.
12
13.
14
sar to Cartesian coordinates
RE = Font) = +
Converting rom Cartesian to polar coordinates:
naj = ett
(0) Bao [Tora
(9249 = SOD, Joao] = 1. Terre Bo =f ont)?
u [oran à [= ja
(6) 2 = cat. Therefore, Ba = [jetée = J” coste = 20,
roe dian fotos ju ET) as
(2) ae) (al (D Ta Ba SS lt PUR
Pa=0, bem Be con
ci ie
[uno Pas
coin) There, Be = $e lei? = E own) os,
en 1 frota
rn o Ma aw rT À (ESS ni
(a) The signal xn) is shifted by 3 tothe right. The sifted signal willbe zero forn < 1
and n> 7
(b) The signal af is shifted by 4 to the le The shied signal wil be zero for n < =
and n> 0.
12
13.
14
sar to Cartesian coordinates
RE = Font) = +
Converting rom Cartesian to polar coordinates:
naj = ett
(0) Bao [Tora
(9249 = SOD, Joao] = 1. Terre Bo =f ont)?
u [oran à [= ja
(6) 2 = cat. Therefore, Ba = [jetée = J” coste = 20,
roe dian fotos ju ET) as
(2) ae) (al (D Ta Ba SS lt PUR
Pa=0, bem Be con
ci ie
[uno Pas
coin) There, Be = $e lei? = E own) os,
en 1 frota
rn o Ma aw rT À (ESS ni
(a) The signal xn) is shifted by 3 tothe right. The sifted signal willbe zero forn < 1
and n> 7
(b) The signal af is shifted by 4 to the le The shied signal wil be zero for n < =
and n> 0.
(€) The signa af is fippod. The Sipped signal wil be zero for n < —4 and > 2
(2) The signal zn) is Ripped and the ied signal i shifted by 2 to the right. This new
signal willbe zero farn < -2 and n > 4
(6) The signal zin] is flipped and the flipped signal is shifted by 2 Lo the let. This new
signal wil be zero form < ~6 and n > 0.
1.5. (a) 2(1=#) is obtained by Aipping 20) and shifting the flipped signal by 10 the right.
‘Therefore, 2(1 — 0) wil be zero for t > -2.
6) From (a), we know that 2(1=#) izo for € > 2. Similarly, (21) is zero for >
Therefore, 2(1 = 2) + 2(2~ 0) willbe ero for 4 > 2.
(©) 2(3) i obtained by linearly compresing 20) by a Factor 073. Therefore, (31) wil be
ero for #< 1
(a) 2(2/3) is obtained by linearly stretching €) by a factor of. ‘Therefore, (4/3) wil be
ero for <8
1.6. (a) z(t) is not periodic because itis zero for 4 < 0.
(0) za] = 1 for am. Therefore, it is periodic with a fundamental period of
(0) zul) is 8 shown inthe Figure SL,
Figure 81.6
“Therefore, its parodie with a fundamental period of 4
17 (o)
un)
Enter = oi ia Ja a +
Therefore, Sole) is eco for In > 3.
(0) Since 2a{) is an odd signal, Ev(za(0) 15 zero for al vales of
©
Eotealn
Kap di
Len + tn) = ¿lud
‘Therefore, Een} à zero when In] <3 and when [nj o.
@
oes} Bad) = Hertel 4)
Therefore, Ev{zul)) l ero only when | > co
2
(2) The signal zn) is Ripped and the ied signal i shifted by 2 to the right. This new
signal willbe zero farn < -2 and n > 4
(6) The signal zin] is flipped and the flipped signal is shifted by 2 Lo the let. This new
signal wil be zero form < ~6 and n > 0.
1.5. (a) 2(1=#) is obtained by Aipping 20) and shifting the flipped signal by 10 the right.
‘Therefore, 2(1 — 0) wil be zero for t > -2.
6) From (a), we know that 2(1=#) izo for € > 2. Similarly, (21) is zero for >
Therefore, 2(1 = 2) + 2(2~ 0) willbe ero for 4 > 2.
(©) 2(3) i obtained by linearly compresing 20) by a Factor 073. Therefore, (31) wil be
ero for #< 1
(a) 2(2/3) is obtained by linearly stretching €) by a factor of. ‘Therefore, (4/3) wil be
ero for <8
1.6. (a) z(t) is not periodic because itis zero for 4 < 0.
(0) za] = 1 for am. Therefore, it is periodic with a fundamental period of
(0) zul) is 8 shown inthe Figure SL,
Figure 81.6
“Therefore, its parodie with a fundamental period of 4
17 (o)
un)
Enter = oi ia Ja a +
Therefore, Sole) is eco for In > 3.
(0) Since 2a{) is an odd signal, Ev(za(0) 15 zero for al vales of
©
Eotealn
Kap di
Len + tn) = ¿lud
‘Therefore, Een} à zero when In] <3 and when [nj o.
@
oes} Bad) = Hertel 4)
Therefore, Ev{zul)) l ero only when | > co
2
right.
il be
il be
18,
140.
in
The signal alo a shown in Figure
(0) Retzıld) = -2 = 22 cos{ae +=)
(b) Retz} = VB one) cost +29) = cos($) = e costa +0)
(©) Reza) = etai) = et cos + $)
(a) Re} = = sn(100) = eZ sn(L00E + x) = 6" eua(LOD à 7)
(6) 210) sa periode complex exponent
a)
i nr)
‘he fundamental period of (0) is = #
D) za) is a complex exponential maltptied by a decaying exponential. Therefor, le)
is ot period
(6) za) ia periodic signal
af
zul] is a complex exponential with a fundamental period of 2 = 2.
(a) alo] i periodic signal. The fundamental period is given by N = mi Bj) = (i)
By choosing m = 3, we obtain the fundamental period to be 10
de) zu) wot pei. li complex exponential with uy = 3/5. We cannot nd
too m such that vs) e loa integer. Therefore, sf is wt perio.
xt
Period of fist term in RES = #5 = §
Period of second term in RS = 2 =$
Therefore, the overall signal is periode witha period which i the last common multiple
cf the period of the ist and second terms, This is equal 10 5.
2ecs(101+1) ~sn( at = 1)
om
aly] ae
Period ofthe frst torm in the RES = 1
Period ofthe second term in the RES mg) = 7 (when m = 2)
Period ofthe third term in the RS = mg) = 5 (when m
‘Therefore, the overall signal sf) is periodie with a period which is the least common
multiple of the periods ofthe three terms in xn). Tiss aqua to 35.
shifting the flipped signal by 3 to the right. Therefore, an]
M and no 3.
il be
il be
18,
140.
in
The signal alo a shown in Figure
(0) Retzıld) = -2 = 22 cos{ae +=)
(b) Retz} = VB one) cost +29) = cos($) = e costa +0)
(©) Reza) = etai) = et cos + $)
(a) Re} = = sn(100) = eZ sn(L00E + x) = 6" eua(LOD à 7)
(6) 210) sa periode complex exponent
a)
i nr)
‘he fundamental period of (0) is = #
D) za) is a complex exponential maltptied by a decaying exponential. Therefor, le)
is ot period
(6) za) ia periodic signal
af
zul] is a complex exponential with a fundamental period of 2 = 2.
(a) alo] i periodic signal. The fundamental period is given by N = mi Bj) = (i)
By choosing m = 3, we obtain the fundamental period to be 10
de) zu) wot pei. li complex exponential with uy = 3/5. We cannot nd
too m such that vs) e loa integer. Therefore, sf is wt perio.
xt
Period of fist term in RES = #5 = §
Period of second term in RS = 2 =$
Therefore, the overall signal is periode witha period which i the last common multiple
cf the period of the ist and second terms, This is equal 10 5.
2ecs(101+1) ~sn( at = 1)
om
aly] ae
Period ofthe frst torm in the RES = 1
Period ofthe second term in the RES mg) = 7 (when m = 2)
Period ofthe third term in the RS = mg) = 5 (when m
‘Therefore, the overall signal sf) is periodie with a period which is the least common
multiple of the periods ofthe three terms in xn). Tiss aqua to 35.
shifting the flipped signal by 3 to the right. Therefore, an]
M and no 3.
1.14. ‘The signal (1) and its derivative g(t) ae shown in Figure SI.
> 3
nie)
4
xy A
Figure $1.14
-2
‘Therefore,
a) =3 E des Y sean)
“This imp that Ay =, =0, 43 ==3, and ta
4.38, (a) The signal sl, which the input oS, the same a po) Therefore,
ii = mind don
= an 2 + qui
= tea) tante Mn + laa
= amfr 24 sen ind
"The input-output relationship or Sis
aie) = 220-2) + Sin
> 3
nie)
4
xy A
Figure $1.14
-2
‘Therefore,
a) =3 E des Y sean)
“This imp that Ay =, =0, 43 ==3, and ta
4.38, (a) The signal sl, which the input oS, the same a po) Therefore,
ii = mind don
= an 2 + qui
= tea) tante Mn + laa
= amfr 24 sen ind
"The input-output relationship or Sis
aie) = 220-2) + Sin
!
(The topar romo ds ac ne I rr nih 4 au ye
mans em. a ely ry ng ac & lors
Sy zn the Sgn a ai de de ptt Se e me a
Tacos
nie = ta dno
= abl anda
E)
= Zen 2] + Grain ~ 3) + 2safn dl
opa pt mini o Sc gin
bl = ln 2) 83 2
146, (a) The ets nt memory eon] epend on ps a f
(0) Ti ut af goa ly = Ss =
(€) From the resul of part (b), we may conclude that te system output is always zero for
inputs ofthe form Bin ~ A, € Z. Therefore, th system is not invertible.
1.17. (a) The system is not causal because the output y) at some time may depend on future
values ot 2(). Ror instance, y(n) = 20)
(0) Consider two arbitrary inputs (0) and 20.
alt) > al) =, (in)
O)
Let zul) be a linear combination of 8) and 22(). That,
aut) = oni +00)
‘where a and b ae arbitrary scalars. LE x(t isthe input to the given system, then the
corresponding opt zul) i
an)
= am (inf) + ba (inf)
= at) Hl)
mo
Therefore, the sytem near,
1.18. (8) Consider two arbitrary inputs sn) and af
all Y alt
(The topar romo ds ac ne I rr nih 4 au ye
mans em. a ely ry ng ac & lors
Sy zn the Sgn a ai de de ptt Se e me a
Tacos
nie = ta dno
= abl anda
E)
= Zen 2] + Grain ~ 3) + 2safn dl
opa pt mini o Sc gin
bl = ln 2) 83 2
146, (a) The ets nt memory eon] epend on ps a f
(0) Ti ut af goa ly = Ss =
(€) From the resul of part (b), we may conclude that te system output is always zero for
inputs ofthe form Bin ~ A, € Z. Therefore, th system is not invertible.
1.17. (a) The system is not causal because the output y) at some time may depend on future
values ot 2(). Ror instance, y(n) = 20)
(0) Consider two arbitrary inputs (0) and 20.
alt) > al) =, (in)
O)
Let zul) be a linear combination of 8) and 22(). That,
aut) = oni +00)
‘where a and b ae arbitrary scalars. LE x(t isthe input to the given system, then the
corresponding opt zul) i
an)
= am (inf) + ba (inf)
= at) Hl)
mo
Therefore, the sytem near,
1.18. (8) Consider two arbitrary inputs sn) and af
all Y alt
sean] mt Ye 218
Let za) be linear combination off) and so. That is,
sn) = eso] + ban]
hore and b ae abitraty scalar. I fn] the input tothe given system, then the
Corresponding output san] is
wi) = À ol
Fa E
= entr + bin)
Therefor, the system i linear.
(6) Consider an arbitrary input uf] Let
ante) = Y au
be the corresponding output, Considers second input zn) obtained by shifting fr]
in ime aber)
The output corresponding to this input is
ane aus E ales "I" oat
Ao ot tat u
mania Sat
Therion, |
va) = bn ml
“This rolls that the system is time-invariant,
(©) fein}. < B, then
sie) < (ne + D
‘Therefore, O £ (Amp + 1)B.
Let za) be linear combination off) and so. That is,
sn) = eso] + ban]
hore and b ae abitraty scalar. I fn] the input tothe given system, then the
Corresponding output san] is
wi) = À ol
Fa E
= entr + bin)
Therefor, the system i linear.
(6) Consider an arbitrary input uf] Let
ante) = Y au
be the corresponding output, Considers second input zn) obtained by shifting fr]
in ime aber)
The output corresponding to this input is
ane aus E ales "I" oat
Ao ot tat u
mania Sat
Therion, |
va) = bn ml
“This rolls that the system is time-invariant,
(©) fein}. < B, then
sie) < (ne + D
‘Therefore, O £ (Amp + 1)B.
in)
119. (@) (0
ON)
Consider tw arbitray inputs 211) and zul)
A) > m()=én(t=0
a(t) a = Pratt = 1)
Let s(t) be a ine combination of (0) and za). That i,
salt) = 021(2)+803(0)
wre à and bare arbitrary sealars. 123) ls the input tothe given system, then
the corresponding output y(t) ls
m0 = Pet)
= Plant 1) + bexlt—1))
= and +b)
‘Therefor, the system is Tinea.
Consideran arbitrary input zu() Let
HO = En)
bo the corresponding output. Consider a second input 21) obtained by shifting
(9 in time
a(t) = nl)
“The output corresponding to this input ie
POELE EEE EN)
Ao note that
ml to) = (= He 10) lO,
‘Therefor, he system is not time-invariant,
Consider two arbitrary inputs fn] and zn)
sale] alla
sale) — alo} ab
at af] be a ner combination of fn] and af
le} + tl
here a and bare arta scalar, Ii the opt to the gen system, en
{Be coresponding output fl
ati,
mp) = nz
= (ein 2) + tela 2)
= den] 42 aan?)
# ow + bale]
‘Therefore, the system is not linear.
1
119. (@) (0
ON)
Consider tw arbitray inputs 211) and zul)
A) > m()=én(t=0
a(t) a = Pratt = 1)
Let s(t) be a ine combination of (0) and za). That i,
salt) = 021(2)+803(0)
wre à and bare arbitrary sealars. 123) ls the input tothe given system, then
the corresponding output y(t) ls
m0 = Pet)
= Plant 1) + bexlt—1))
= and +b)
‘Therefor, the system is Tinea.
Consideran arbitrary input zu() Let
HO = En)
bo the corresponding output. Consider a second input 21) obtained by shifting
(9 in time
a(t) = nl)
“The output corresponding to this input ie
POELE EEE EN)
Ao note that
ml to) = (= He 10) lO,
‘Therefor, he system is not time-invariant,
Consider two arbitrary inputs fn] and zn)
sale] alla
sale) — alo} ab
at af] be a ner combination of fn] and af
le} + tl
here a and bare arta scalar, Ii the opt to the gen system, en
{Be coresponding output fl
ati,
mp) = nz
= (ein 2) + tela 2)
= den] 42 aan?)
# ow + bale]
‘Therefore, the system is not linear.
1
Ki) Consider an arbitary input zo. Let
be the corresponding output. Considera second input ala] obtained by sbifting
auf] in time
PORTES)
"Tho output corresponding o his input is
slo) ln 2] = sf 2 no]
to note hat
me ss —2= 0)
Therefore,
sale] = fn ral
‘his implies that the system is time-invariant,
(©) (0) Consider two arbitrary inputs fn] and af.
a) — ml ln al
ln + zu
Let za] be a linear combination off) and alo). That is,
cala] = exif) +: bool)
cell ale
the input tothe given system, then
whore and b ae arbitrar salas. If o
‘he corresponding output paf] is
PORT ETS
= entr} + bin + D-enin—
= alsin Yala 1) + Male +
= apie) + bale]
= tex
“an 1)
Therefore, the ape I linear.
(i) Consider an arbitrary input fa]. Let
nie] = fn +)
‘be the corresponding output. Consider «second input za] obtained by shifting
zul] in ime
zul] = en ve)
‘The output corresponding to this input is
nf) = zn + 1] = zn eilt 1 ml ala 1 0)
8
be the corresponding output. Considera second input ala] obtained by sbifting
auf] in time
PORTES)
"Tho output corresponding o his input is
slo) ln 2] = sf 2 no]
to note hat
me ss —2= 0)
Therefore,
sale] = fn ral
‘his implies that the system is time-invariant,
(©) (0) Consider two arbitrary inputs fn] and af.
a) — ml ln al
ln + zu
Let za] be a linear combination off) and alo). That is,
cala] = exif) +: bool)
cell ale
the input tothe given system, then
whore and b ae arbitrar salas. If o
‘he corresponding output paf] is
PORT ETS
= entr} + bin + D-enin—
= alsin Yala 1) + Male +
= apie) + bale]
= tex
“an 1)
Therefore, the ape I linear.
(i) Consider an arbitrary input fa]. Let
nie] = fn +)
‘be the corresponding output. Consider «second input za] obtained by shifting
zul] in ime
zul] = en ve)
‘The output corresponding to this input is
nf) = zn + 1] = zn eilt 1 ml ala 1 0)
8
ing
ing
ON]
0)
Also note that
la na] = la +1 ro] fn 1 0]
Therefore,
man) wn nal
“This plis that the system is time-invariant.
‘Consider two arbitrary inputs z() and zul)
au) nl = Ost)
alt) — mio = Od)
Let 2) be a linear combination of z(t) and 2). That i,
al) = ont) + alt)
where a and bare arbitrary salas, If s(t) is Ue input to the given system, then
‘he corresponding output ys) ie
ni) = Oda)
= Ola) Halt}
Per Er FRE
an + nl)
‘Therefore, the system is line.
Consider an arbitrary input 1. Let
ala)
w(t) = Ou} = BO
be the corresponding output. Consider a second input s(t) obtained by sifting
ln in time
al) =
‘The output corresponding o tis Input is
wl) = Oteo) = 20-260
= a) al
=4)
Also note that
alt= ta) (tro)
(t-te) And
‘Therefore, the system Is not time-invariant.
ing
ON]
0)
Also note that
la na] = la +1 ro] fn 1 0]
Therefore,
man) wn nal
“This plis that the system is time-invariant.
‘Consider two arbitrary inputs z() and zul)
au) nl = Ost)
alt) — mio = Od)
Let 2) be a linear combination of z(t) and 2). That i,
al) = ont) + alt)
where a and bare arbitrary salas, If s(t) is Ue input to the given system, then
‘he corresponding output ys) ie
ni) = Oda)
= Ola) Halt}
Per Er FRE
an + nl)
‘Therefore, the system is line.
Consider an arbitrary input 1. Let
ala)
w(t) = Ou} = BO
be the corresponding output. Consider a second input s(t) obtained by sifting
ln in time
al) =
‘The output corresponding o tis Input is
wl) = Oteo) = 20-260
= a) al
=4)
Also note that
alt= ta) (tro)
(t-te) And
‘Therefore, the system Is not time-invariant.
1.20. (a) Given
at)
EC)
uo
a yl) mer
Since the system linear,
Lt zer) yt
20 ern
Einen
0) Motor tt
ste (te) =
Using the inearity property, we may once again write
a= Het ald = Herat er mon)
Therefore,
zul) = lat 1/2)) — w(t) = cote 1)
(2 he signa are sktehed in Figure 51.21
re) KA a alte)
+} 4, =
HER STs
gta) cds
DE ga
+. It
Fi 47 4
Figure 51.21
The signals are sketched in Figure SL22
TS. The even and odd parts are sketched in Figure $1.23,
at)
EC)
uo
a yl) mer
Since the system linear,
Lt zer) yt
20 ern
Einen
0) Motor tt
ste (te) =
Using the inearity property, we may once again write
a= Het ald = Herat er mon)
Therefore,
zul) = lat 1/2)) — w(t) = cote 1)
(2 he signa are sktehed in Figure 51.21
re) KA a alte)
+} 4, =
HER STs
gta) cds
DE ga
+. It
Fi 47 4
Figure 51.21
The signals are sketched in Figure SL22
TS. The even and odd parts are sketched in Figure $1.23,
sb) Rn
Culte. lille, te. at.
aire, allie. „le. a]
y TT
X
a 10] ar (dy
G La + pera at]
Hd Pre ] wi 1 Il
À sn Tan + CRE: a ere ma
A ce)
l a 4 loo 122 a
pe] u.
N, et
@ me
«e, xt)
A Maz
e
Za
O E 4
a.
ro e 24
NT ote
D € =
©
Figure $1.23
Culte. lille, te. at.
aire, allie. „le. a]
y TT
X
a 10] ar (dy
G La + pera at]
Hd Pre ] wi 1 Il
À sn Tan + CRE: a ere ma
A ce)
l a 4 loo 122 a
pe] u.
N, et
@ me
«e, xt)
A Maz
e
Za
O E 4
a.
ro e 24
NT ote
D € =
©
Figure $1.23
126
1.24. ‘The even and odd parts are sketched in Figure S124
(0) Pei
(6) 2 = [1 + cos(dt—24/9)/2. Period
(a) 2) = cosltnt)/2. Period, pero
[SET
1.26. (a) Periode, period = 2+/(4) = 9/2
le, period = 2r/(e) = 2.
(Ar) = 1/2.
infärt)u(e) - sinart)ul-A)/2 Not periode
(9) Not parodie
(2) Periodic, period = 7.
(0) Not periodic
(e) Periodic, period = 8
/2)fcos(Sen/) + coslrn/4). Perdi period = 8
die, period = 16
N Ar) (a) Linear, stable
(b) Memory, linen, causa, stable
(6) Linear
(a) Linea,
causal, stable
(6) Time invariant, nar, causal, stable
(©) Linea,
(0) Tm
stable
avaro nar, casal
period = 2n/(4) = /2
1.24. ‘The even and odd parts are sketched in Figure S124
(0) Pei
(6) 2 = [1 + cos(dt—24/9)/2. Period
(a) 2) = cosltnt)/2. Period, pero
[SET
1.26. (a) Periode, period = 2+/(4) = 9/2
le, period = 2r/(e) = 2.
(Ar) = 1/2.
infärt)u(e) - sinart)ul-A)/2 Not periode
(9) Not parodie
(2) Periodic, period = 7.
(0) Not periodic
(e) Periodic, period = 8
/2)fcos(Sen/) + coslrn/4). Perdi period = 8
die, period = 16
N Ar) (a) Linear, stable
(b) Memory, linen, causa, stable
(6) Linear
(a) Linea,
causal, stable
(6) Time invariant, nar, causal, stable
(©) Linea,
(0) Tm
stable
avaro nar, casal
period = 2n/(4) = /2
ing
een ATAN
130.
131
132
(a) invertible, Imre tem y(t) = 2 + 0
(0) Non invertible, The signals (9 and 20) = (4) + 2 give the same output.
(©) Non ivertibl, fp] and 2} give the same output
(0) velo Inverse sytem: y) = 5/4
(6) Inverso eme sytem: yp) = +1) for 2 0 and lo] al form <0
(8) Non inertie, in ad ln] lve the same rn
(6) Inveribe verse system: fo) = al —
A) Tari. Inverse eaten y) = 20+ dee
aver, averno ate: yo = z = (1/23 — 1
(6) Non invertible I(t any constant, then u) =0.
(0) Non invertible, dp and 2] rent In in) =
(0) Tavo averse ay: yt) = 2/2)
{mon invertible uf] = + dn 1 nd ze) = A ve o)
(0) Incio. averse system: fe) = «o,
(0) Note that zuf) = zul) - (= 2). Therefore, using linearity we get pa) = vl) =
gilt = he This a shown in Figure SI
(b) Note that s(t) = 2u(t)+2:(¢ D. Therefore, using linearity we get y(t) = wll) +
AE 1) "This in as shown in Figure S132
ya) po
23 |
9 = to HAT =
Figure $1.31
All statements ae Ic
(1) 209 period with period T; vl) periodic, period 7/2.
(2) ult) perodie period 7; (1) periodic, period 27.
(9) x() periodie, period T; ya.) periodic, period 27.
(4) vt) perdi period T; (1) periodic, period 7/2.
. (1 True. fol = zfn + Malo) mín + Ma. La. periodic with No = N/2 if is even,
and with period Ny = ICN is dd.
5
een ATAN
130.
131
132
(a) invertible, Imre tem y(t) = 2 + 0
(0) Non invertible, The signals (9 and 20) = (4) + 2 give the same output.
(©) Non ivertibl, fp] and 2} give the same output
(0) velo Inverse sytem: y) = 5/4
(6) Inverso eme sytem: yp) = +1) for 2 0 and lo] al form <0
(8) Non inertie, in ad ln] lve the same rn
(6) Inveribe verse system: fo) = al —
A) Tari. Inverse eaten y) = 20+ dee
aver, averno ate: yo = z = (1/23 — 1
(6) Non invertible I(t any constant, then u) =0.
(0) Non invertible, dp and 2] rent In in) =
(0) Tavo averse ay: yt) = 2/2)
{mon invertible uf] = + dn 1 nd ze) = A ve o)
(0) Incio. averse system: fe) = «o,
(0) Note that zuf) = zul) - (= 2). Therefore, using linearity we get pa) = vl) =
gilt = he This a shown in Figure SI
(b) Note that s(t) = 2u(t)+2:(¢ D. Therefore, using linearity we get y(t) = wll) +
AE 1) "This in as shown in Figure S132
ya) po
23 |
9 = to HAT =
Figure $1.31
All statements ae Ic
(1) 209 period with period T; vl) periodic, period 7/2.
(2) ult) perodie period 7; (1) periodic, period 27.
(9) x() periodie, period T; ya.) periodic, period 27.
(4) vt) perdi period T; (1) periodic, period 7/2.
. (1 True. fol = zfn + Malo) mín + Ma. La. periodic with No = N/2 if is even,
and with period Ny = ICN is dd.
5
(2) Fase. sna} period does no imply af is periodic. Le let sn] = fn] + Ain] where
[A mem slo news
fr NU md Mt {ae neal
“Phen sn) = 22 perdio but zn is cleatly not period.
(9) Te, an MI = ed la + Nol = val] where Ne
(4) Tre. van + N] = aol si + Nol
—
À where No
1.34) (a) Consider
=
fio, sj + [a] = 0- Thee, the given summation evaluates to zo
(0) Dein) aia,
(0) Let glo] = ball. Then
tail =~
ven) = ana
spi hat po
(6) Consider
Faw car
Eta Ea
Using the resul of part (0), we know that [nx
the result of part (a) we may conclude tha
2% savia) =0.
DEC
is an odd signal. Therefore, using
Therefor, A o u
Y Pm Y sit Y seh)
(2) Consider
MO
[sous EOS [zer
Again, since ao) is odd,
(atbsstod -0.
Que [wa [avé
‘Therefore,
16
[A mem slo news
fr NU md Mt {ae neal
“Phen sn) = 22 perdio but zn is cleatly not period.
(9) Te, an MI = ed la + Nol = val] where Ne
(4) Tre. van + N] = aol si + Nol
—
À where No
1.34) (a) Consider
=
fio, sj + [a] = 0- Thee, the given summation evaluates to zo
(0) Dein) aia,
(0) Let glo] = ball. Then
tail =~
ven) = ana
spi hat po
(6) Consider
Faw car
Eta Ea
Using the resul of part (0), we know that [nx
the result of part (a) we may conclude tha
2% savia) =0.
DEC
is an odd signal. Therefore, using
Therefor, A o u
Y Pm Y sit Y seh)
(2) Consider
MO
[sous EOS [zer
Again, since ao) is odd,
(atbsstod -0.
Que [wa [avé
‘Therefore,
16
1.36. We want to And the smallest No such that (2/0) = 2 or No = kN/m, where isan
Intege. Tf No has to bean integer, then M must be multiple of m/ and m/k must be an
integer, This implies that m/f à à divisor of both m and. Also, we want the smallest
Possible No, then m/k should be the GCD of m and N. Therefore, No = M/gedm, N)
1.36. (a) Ha in periodic MONIT ge, where u =
2 x 7.
Eure + 7
This implis that
rational number
(6) 67/7 = pla then x(n} = er. Te fundamental period is 9/g0o(9,0) and the
fundamental frequency ls
= ar
cat = cdi
Bed, = “Testing
(©) p/ezdlo,a) periods of 2() are needed
137. (a) From the definition of day(s we have
Gal)
[lucerne
= tal?
(0) Note ram prt (a) at ba) = 0 Tis imps that fa) ven. herir,
the odd par offal) i uo
(0 Her, = at 7 and dy) = Bel
(2:38) (a) We know that 255 20 = day). Therefore,
[acer
ESO
jm a)
‘This implies that
sa = 40)
(0) The plots ae as shown in Figure SL,
1.39, Wet
meld) = ina = 0
ao,
ao = Jo
Intege. Tf No has to bean integer, then M must be multiple of m/ and m/k must be an
integer, This implies that m/f à à divisor of both m and. Also, we want the smallest
Possible No, then m/k should be the GCD of m and N. Therefore, No = M/gedm, N)
1.36. (a) Ha in periodic MONIT ge, where u =
2 x 7.
Eure + 7
This implis that
rational number
(6) 67/7 = pla then x(n} = er. Te fundamental period is 9/g0o(9,0) and the
fundamental frequency ls
= ar
cat = cdi
Bed, = “Testing
(©) p/ezdlo,a) periods of 2() are needed
137. (a) From the definition of day(s we have
Gal)
[lucerne
= tal?
(0) Note ram prt (a) at ba) = 0 Tis imps that fa) ven. herir,
the odd par offal) i uo
(0 Her, = at 7 and dy) = Bel
(2:38) (a) We know that 255 20 = day). Therefore,
[acer
ESO
jm a)
‘This implies that
sa = 40)
(0) The plots ae as shown in Figure SL,
1.39, Wet
meld) = ina = 0
ao,
ao = Jo
140.
uke
N
+ ma a +
whe
E
4 9 a Y
so [unser [Tura
tee teo
ei, En
(a) If system I ative, then
02 2) = #4) wol
Also, à system is homogeneous, then
0= 020) —+ 900.0 = 0
(0) ut) = 280) ie such a system.
(6) No. For example, consider y(
[rer wit 6 = (ule, Then (9 = 0
for > 1, but y(t) = for #> 1
6
uke
N
+ ma a +
whe
E
4 9 a Y
so [unser [Tura
tee teo
ei, En
(a) If system I ative, then
02 2) = #4) wol
Also, à system is homogeneous, then
0= 020) —+ 900.0 = 0
(0) ut) = 280) ie such a system.
(6) No. For example, consider y(
[rer wit 6 = (ule, Then (9 = 0
for > 1, but y(t) = for #> 1
6
!
i
1,
N
i
|
i
i
i
|
i
(6) Since ry > ra > O and —1 5 conf 09) <1,
Gall = ri+ri amm
S rh ré armee 0)
= nf
aod
(eatin? = + rar a tal
1.54. (9) For a =1, is fail obvious that
Foro 1, we may wie
car re
Therefore a
Foie
(0) Re jl <1,
Jno =
‘Therefore, from the result of the previous part,
(Ee) - 4)
(4) We may write
for jal <1.
1.55. (a) The desired sum i
i
1,
N
i
|
i
i
i
|
i
(6) Since ry > ra > O and —1 5 conf 09) <1,
Gall = ri+ri amm
S rh ré armee 0)
= nf
aod
(eatin? = + rar a tal
1.54. (9) For a =1, is fail obvious that
Foro 1, we may wie
car re
Therefore a
Foie
(0) Re jl <1,
Jno =
‘Therefore, from the result of the previous part,
(Ee) - 4)
(4) We may write
for jal <1.
1.55. (a) The desired sum i
Chapter 2 Answers
2.1. (a) We know that
nfo] = ato hin] = 3” Allein — 4) 21-1)
“The signals fr] and Ala) are as shown in Figure 82.1.
2 + =
\ ah) hh)
3
Sin =o
ia Figure $2.1
From this figure, we can easily see that the above convolution sum reduces to
al) = Atala +1] + Hein - 1)
= Dal 1] + 22{n— 1}
This gives
v
Bi +1] + al + D 1)+2óln — 2) —26(n 4)
(b) We know that
nfo] =2jn+2]e lo} = À Miele +2
Comparing with eq. (S2.1-1), we see that =>
vale] = El
(€) We may rewrite eg. (S2.1-1) as
ni = seh) = À ein
Similarly, we may write
ban À akin 2—
Comparing this with eq. ($2.1), we see that
al) = min +2]
30
2.1. (a) We know that
nfo] = ato hin] = 3” Allein — 4) 21-1)
“The signals fr] and Ala) are as shown in Figure 82.1.
2 + =
\ ah) hh)
3
Sin =o
ia Figure $2.1
From this figure, we can easily see that the above convolution sum reduces to
al) = Atala +1] + Hein - 1)
= Dal 1] + 22{n— 1}
This gives
v
Bi +1] + al + D 1)+2óln — 2) —26(n 4)
(b) We know that
nfo] =2jn+2]e lo} = À Miele +2
Comparing with eq. (S2.1-1), we see that =>
vale] = El
(€) We may rewrite eg. (S2.1-1) as
ni = seh) = À ein
Similarly, we may write
ban À akin 2—
Comparing this with eq. ($2.1), we see that
al) = min +2]
30
“JU ils
4 En
Figure 52.4
25. The signal yl is
Yin) ala) Afr] = CL CEE
In this caso, this summation reduces to
ule) = Satin - 4 = En
From this it is clear that yln] is a summation of shifted replicas of fín]. Since the last
replica will begin at n = 9 and Aln) is zero for n > N, yfn) is zero for n > N +9. Using
this and the fact that y[14] = 0, we may conclude that N can at most be 4. Furthermore,
‘we can conclude that Aja] has at least 5 non-zero points. The only value of
N which satiies both these conditions is 4.
2.6. From the given information, we have:
sn Ata) = À ati 4
wa)
$ dirt ink
E Gran k=
= bara)
Replacing k by p-1,
viol = Er un +ri (926-1)
=
For n> 0 the above equation reducto,
Sas 1
vb La =p
32
4 En
Figure 52.4
25. The signal yl is
Yin) ala) Afr] = CL CEE
In this caso, this summation reduces to
ule) = Satin - 4 = En
From this it is clear that yln] is a summation of shifted replicas of fín]. Since the last
replica will begin at n = 9 and Aln) is zero for n > N, yfn) is zero for n > N +9. Using
this and the fact that y[14] = 0, we may conclude that N can at most be 4. Furthermore,
‘we can conclude that Aja] has at least 5 non-zero points. The only value of
N which satiies both these conditions is 4.
2.6. From the given information, we have:
sn Ata) = À ati 4
wa)
$ dirt ink
E Gran k=
= bara)
Replacing k by p-1,
viol = Er un +ri (926-1)
=
For n> 0 the above equation reducto,
Sas 1
vb La =p
32
“Ue DL
erage RTE
Figure 82.7
i
] que
E
a
2.8. Using the convolution integral,
at) en) = [amena f° E Mr)zte = mdr.
Given that h(t) = 6(¢+2) +25(¢+ 1), he above integral reduce to
a(t) u(t) = a(t +2) + 2 1)
‘The signals 2(¢ +2) and 2e(¢ + 1) are plotted in Figure 82.8. u
A
2
a aro) A Al)
2 \
= 0 ı
+
= 8
+ 7 Figure 82.8
‘Using these plots, we can easily show that
+3,
alta -1<t50
WO=) 2-2, o<tst
o otherwise
2.9. Using the given definition for the signal h(t), we may write
et, 7>5
ered
o 4<r<b
Mr) = Pur + 4) + ele — 5)
‘Therefore,
Ore
Here e, >
o —i<r<-4
If we now shift the signal A(~r) by £ to the right, then the resultant signal h(t ~r) will be
em, rt
Mena} LES]
o (perc
34
erage RTE
Figure 82.7
i
] que
E
a
2.8. Using the convolution integral,
at) en) = [amena f° E Mr)zte = mdr.
Given that h(t) = 6(¢+2) +25(¢+ 1), he above integral reduce to
a(t) u(t) = a(t +2) + 2 1)
‘The signals 2(¢ +2) and 2e(¢ + 1) are plotted in Figure 82.8. u
A
2
a aro) A Al)
2 \
= 0 ı
+
= 8
+ 7 Figure 82.8
‘Using these plots, we can easily show that
+3,
alta -1<t50
WO=) 2-2, o<tst
o otherwise
2.9. Using the given definition for the signal h(t), we may write
et, 7>5
ered
o 4<r<b
Mr) = Pur + 4) + ele — 5)
‘Therefore,
Ore
Here e, >
o —i<r<-4
If we now shift the signal A(~r) by £ to the right, then the resultant signal h(t ~r) will be
em, rt
Mena} LES]
o (perc
34
‘Therefore,
er
Azt-5, B
2.10. From the given information, we may sketch 2(t) and h(t) as shown in Figure $2.10.
(a) With the aid of the plots in Figure $2.10, we can show that y(t) = 2(t) «A(t is as
shown in Figure $2.10.
12
2) . hte)
ot car +
JUN x ihe
Figure 82.10
Therefore,
‘ ostsa
a astsl
VOZ] tas 1<e<(i+e)
o cbarmioo
(6) From the plot of fe), ii clear that $42 has discontinuities at 0, a, 1, and 1 a. If
we want 42 to have only thre discontinuities, then we need to ensure that a = 1.
2.11. (a) From the given information, we se that A) non zero only for 0 < 1 € oo. Therefore,
2 +h = [krai
[ert-r-2-ut
We can easily show that (ul ~ 3) = ut r — 8) non sro only In the range
{= 5) <1 <(t— 3). Therefore, for ¢ < 3, the above integral evaluates to zero. For
3.< £5, the above integral is
FC] E
For £> 5, the integral is
ult)
a
= 5))ar
er
Azt-5, B
2.10. From the given information, we may sketch 2(t) and h(t) as shown in Figure $2.10.
(a) With the aid of the plots in Figure $2.10, we can show that y(t) = 2(t) «A(t is as
shown in Figure $2.10.
12
2) . hte)
ot car +
JUN x ihe
Figure 82.10
Therefore,
‘ ostsa
a astsl
VOZ] tas 1<e<(i+e)
o cbarmioo
(6) From the plot of fe), ii clear that $42 has discontinuities at 0, a, 1, and 1 a. If
we want 42 to have only thre discontinuities, then we need to ensure that a = 1.
2.11. (a) From the given information, we se that A) non zero only for 0 < 1 € oo. Therefore,
2 +h = [krai
[ert-r-2-ut
We can easily show that (ul ~ 3) = ut r — 8) non sro only In the range
{= 5) <1 <(t— 3). Therefore, for ¢ < 3, the above integral evaluates to zero. For
3.< £5, the above integral is
FC] E
For £> 5, the integral is
ult)
a
= 5))ar
2.12.
218.
‚Therefore, he result of this convolution may be expressed as
(9) By diferentiating 2() with respect 10 time we get
EN 5-3) 5-5)
‘Therefore,
a= u el) Due 3) — Hufe — 5)
(©) From tho result of part (a), we may computo the derivative of y(t) Lo be
o moots
del) _ 3) Peer
(einen, Beto
‘This is exactly equal to g(t). Therefore, gie) = “42.
‘The signal y(t) may be written as
U) =e Mult 46) +e Mult 2) re tue) re dr E EE
In the range 0 <¢ <3, we may write y(t) as
vo eMule + 6) +e AE)
Er.
mette 4e tte)
Therefore, À = yobs.
(a) We requir that
(3) wa Qs = 6h)
Putting n = 1 and solving for A gives A = }.
(b) From part (a), we know that
Men = ay
Ale = Le = an)
From the ét fa es st e may age that
218.
‚Therefore, he result of this convolution may be expressed as
(9) By diferentiating 2() with respect 10 time we get
EN 5-3) 5-5)
‘Therefore,
a= u el) Due 3) — Hufe — 5)
(©) From tho result of part (a), we may computo the derivative of y(t) Lo be
o moots
del) _ 3) Peer
(einen, Beto
‘This is exactly equal to g(t). Therefore, gie) = “42.
‘The signal y(t) may be written as
U) =e Mult 46) +e Mult 2) re tue) re dr E EE
In the range 0 <¢ <3, we may write y(t) as
vo eMule + 6) +e AE)
Er.
mette 4e tte)
Therefore, À = yobs.
(a) We requir that
(3) wa Qs = 6h)
Putting n = 1 and solving for A gives A = }.
(b) From part (a), we know that
Men = ay
Ale = Le = an)
From the ét fa es st e may age that
2.14.
215.
246.
(8) We first determine if (6) is absolutely integrable as follows
[mer feras
‘Therefore, hy (t) is the impulse response of a stable LITT system.
(b) We determine if a(t) is absolutely integrable as follows
[mcr = [7 etcosanae
This integral is clearly finite-valued because e~*|cos(2t)| is an exponentially decaying
funcion inthe range 0 < € = 00, "Therefore, a(t) isthe pulse response ofa stable
Unt system.
(a) We determine if hyn} is absolutely summable as follows
Y Il = cosa
te =
“This sum doesnot have a finite value because the function kcoa(A)] increases as che
value of ke increases. Therefore, hyn) cannot bo the impulse response of a stable LIT
system,
(b) We determine if a(n] is absolutely summable as follows
$ path= Etna
IA
‘Therefore, hr is the impulse response of a stable LTI system.
(a) True, This may be easily argued by noting that convolution may be viewed as the
process of carrying out the superposition of a number of echos of la], The fist such
«cho will occur at the location of the fist non zero sample of fr). In this case, the
first echo will occur at M. The echo of Ar] which occurs at y = Ni will have its Best
‘nom zero sample at he time location Ni +N. "Therefore, fr all values of which are
lesser that N + No, the output y] is ero.
(0) False. Consider
al = ele)
= Satie
From this,
sina) DiR
2 «= 3)
‘This shows thatthe given statement i fale.
El
215.
246.
(8) We first determine if (6) is absolutely integrable as follows
[mer feras
‘Therefore, hy (t) is the impulse response of a stable LITT system.
(b) We determine if a(t) is absolutely integrable as follows
[mcr = [7 etcosanae
This integral is clearly finite-valued because e~*|cos(2t)| is an exponentially decaying
funcion inthe range 0 < € = 00, "Therefore, a(t) isthe pulse response ofa stable
Unt system.
(a) We determine if hyn} is absolutely summable as follows
Y Il = cosa
te =
“This sum doesnot have a finite value because the function kcoa(A)] increases as che
value of ke increases. Therefore, hyn) cannot bo the impulse response of a stable LIT
system,
(b) We determine if a(n] is absolutely summable as follows
$ path= Etna
IA
‘Therefore, hr is the impulse response of a stable LTI system.
(a) True, This may be easily argued by noting that convolution may be viewed as the
process of carrying out the superposition of a number of echos of la], The fist such
«cho will occur at the location of the fist non zero sample of fr). In this case, the
first echo will occur at M. The echo of Ar] which occurs at y = Ni will have its Best
‘nom zero sample at he time location Ni +N. "Therefore, fr all values of which are
lesser that N + No, the output y] is ero.
(0) False. Consider
al = ele)
= Satie
From this,
sina) DiR
2 «= 3)
‘This shows thatthe given statement i fale.
El
(e) True. Consider
ult) = 210 A)
[ne - Der
From this,
un = [Teint ner
= Farmer
= Nenn)
‘This shows that the given statement is true.
(4) True. This may be argued by considering,
l= at) ent) = RECU
in Figure $2.16, we plot z{r) and Alt - 1) under the assumptions that (1) 2(¢) =
for t > Ty and (2) h(t) = 0 for £ > Ta. Clearly, the product 2(r)h(t = 7) is zero if
Figure $2.16
1-73 > Ty. Therefore, y(t) =0 for > Ti + D
2.17. (a) We know that y(t) is the sum of the particular and homogeneous soluti
given differential equation. Wo first determine the particular solution yl
the method specified in Example 2.14. Since we are given that the input is z{t) =
eH0u(t) for £> 0, we hypothesize that for £ > 0
Ke,
EC]
Substituting for z(t) and y(t) in the given differential equation,
LR ance H 2 Kran
38
ult) = 210 A)
[ne - Der
From this,
un = [Teint ner
= Farmer
= Nenn)
‘This shows that the given statement is true.
(4) True. This may be argued by considering,
l= at) ent) = RECU
in Figure $2.16, we plot z{r) and Alt - 1) under the assumptions that (1) 2(¢) =
for t > Ty and (2) h(t) = 0 for £ > Ta. Clearly, the product 2(r)h(t = 7) is zero if
Figure $2.16
1-73 > Ty. Therefore, y(t) =0 for > Ti + D
2.17. (a) We know that y(t) is the sum of the particular and homogeneous soluti
given differential equation. Wo first determine the particular solution yl
the method specified in Example 2.14. Since we are given that the input is z{t) =
eH0u(t) for £> 0, we hypothesize that for £ > 0
Ke,
EC]
Substituting for z(t) and y(t) in the given differential equation,
LR ance H 2 Kran
38
This gives
(Atak 4K =1, 3K
‘Therefore,
1
D
Wan y 1>0
In order to determine the homogeneous solution, we hypothesize that
w(t) =
w(t) = Act
Since the homogeneous solution has to satisfy the following differential equation
dnl)
HO La) =0,
we obtain
Ase! + 4Aet = Acts 44)
‘This implics that ¢ = 4 for any A. The overall solution to the differential equation
now becomes
y o,
e A, 150
WO) = ete es >
Now in order to determine the constant A, we use the fact thatthe system satisfies the
condition of initial rest. Given that y(0) = 0, we may conclude that
1
a
EEE]
A+
‘Therefore for ¢> 0,
1
EE]
Since the system satisfies the condition of initial rest, y(t) = U for £ < 0. Therefore,
ut) = eae], o> 0
ud)
tn] cy
(0) The output wil ow e the rel pr ofthe ane obtained in pat)
O
E)
(Atak 4K =1, 3K
‘Therefore,
1
D
Wan y 1>0
In order to determine the homogeneous solution, we hypothesize that
w(t) =
w(t) = Act
Since the homogeneous solution has to satisfy the following differential equation
dnl)
HO La) =0,
we obtain
Ase! + 4Aet = Acts 44)
‘This implics that ¢ = 4 for any A. The overall solution to the differential equation
now becomes
y o,
e A, 150
WO) = ete es >
Now in order to determine the constant A, we use the fact thatthe system satisfies the
condition of initial rest. Given that y(0) = 0, we may conclude that
1
a
EEE]
A+
‘Therefore for ¢> 0,
1
EE]
Since the system satisfies the condition of initial rest, y(t) = U for £ < 0. Therefore,
ut) = eae], o> 0
ud)
tn] cy
(0) The output wil ow e the rel pr ofthe ane obtained in pat)
O
E)
2.18. Since the system is causal, yn] = 0 for n-< 1. Now,
vit) = Intro + 1=1
Ye = jul +20] =j40=5
vi] = MR
Ñ Lyme
um) = GY
“Therefore,
in) = un = 1)
2.19. (a) Consider the dference equation relating ur] and ufn] for Sy
ol = aufn — 1) + Bu)
From this we may write 1
ei
inl- Gun —1)
and
soln -1] = Sule 1] Sun ~ 2)
‘Weighting the previous equation by 1/2 and subtracting from the one before, we obtain
1
juin 1
ule] = Gin ~ vin — again
Substituting this in he difeence equation relating uf] and zn] or Si,
Aa en pe D + Foul -2)= sb)
That i, ;
Yin) (a + hub — 1] — Fale — 2] + Bein)
‘Comparing with the given equation relating y(n] and z{n], we obtain
pai
1
a=}
vit) = Intro + 1=1
Ye = jul +20] =j40=5
vi] = MR
Ñ Lyme
um) = GY
“Therefore,
in) = un = 1)
2.19. (a) Consider the dference equation relating ur] and ufn] for Sy
ol = aufn — 1) + Bu)
From this we may write 1
ei
inl- Gun —1)
and
soln -1] = Sule 1] Sun ~ 2)
‘Weighting the previous equation by 1/2 and subtracting from the one before, we obtain
1
juin 1
ule] = Gin ~ vin — again
Substituting this in he difeence equation relating uf] and zn] or Si,
Aa en pe D + Foul -2)= sb)
That i, ;
Yin) (a + hub — 1] — Fale — 2] + Bein)
‘Comparing with the given equation relating y(n] and z{n], we obtain
pai
1
a=}
Now,
ax)
de
which is the desired integral. We now evaluate the valu ofthe integral us
delt)
dt les
ut = reed
= sine) = 0
2.21, (a) The desired convolution is
vl) = zin] + Alm)
$ stehen
a
- ran forn20
EEE un) fra 8
F
(6) From (a), .
velar » | sin] = + eat)
(© Porn <6,
e E
For n > 6,
Er E
“Therefore, j
à af eye, n<6
ES
(4) The desired convolution is
DE
= AO + at = 3) + en — 22h + hir — 4)
He = 1] +f 2] + Mn 3) + na
a)
‘This is as shown in Figure 82.21
ax)
de
which is the desired integral. We now evaluate the valu ofthe integral us
delt)
dt les
ut = reed
= sine) = 0
2.21, (a) The desired convolution is
vl) = zin] + Alm)
$ stehen
a
- ran forn20
EEE un) fra 8
F
(6) From (a), .
velar » | sin] = + eat)
(© Porn <6,
e E
For n > 6,
Er E
“Therefore, j
à af eye, n<6
ES
(4) The desired convolution is
DE
= AO + at = 3) + en — 22h + hir — 4)
He = 1] +f 2] + Mn 3) + na
a)
‘This is as shown in Figure 82.21
‚lin,
2.22. (a) The desired convolution is
u(t)
[ora
f ented, 120
Then
= { Cu) a 4
ter Phu(s) a=p
(b) The desired convolution is
u)
[stom ner
This may be written as
[eri faire, t<1
won| [ena Leena, rsess
A, 3stso
o, bce
‘Therefore,
uf!
(Der + el 2, <3
(ASE 38156
(aye = 2680-7) 4 M9),
| o bet
2.22. (a) The desired convolution is
u(t)
[ora
f ented, 120
Then
= { Cu) a 4
ter Phu(s) a=p
(b) The desired convolution is
u)
[stom ner
This may be written as
[eri faire, t<1
won| [ena Leena, rsess
A, 3stso
o, bce
‘Therefore,
uf!
(Der + el 2, <3
(ASE 38156
(aye = 2680-7) 4 M9),
| o bet
(©) The desired convolution is
wo = [Carmine
= Fame nee
Th gs us
o 1
won | Gent Leto
(2/=)[cos(m(t - 3)) ~ 1}, 3<t<5
o set
(d) Let
A) = hut) - 380-2,
where
mole sist
Now, :
D = At) +20 = fal #210 Gt 2)
We hare
iy i "
pat? Gale 1)? + be HE D)
‘Therefore,
vio = fat = La nt + ne lea 40) ai 20)
(e) 2(0) periodic implies y() periodic. determine 1 period only. We have
Coronas fa esmare jet, pete}
ve) = E fi u
[Laune freine
Parera, ]<t<d
‘The period of y(t) is 2
2.23. y(t) is sketched in Figure 82.23 for the different values of T.
2.24, (a) We are given that hafn) = öfn] + dfn — 1]. Therefore,
ar al = fn] + 25) = + fa — 2)
a
wo = [Carmine
= Fame nee
Th gs us
o 1
won | Gent Leto
(2/=)[cos(m(t - 3)) ~ 1}, 3<t<5
o set
(d) Let
A) = hut) - 380-2,
where
mole sist
Now, :
D = At) +20 = fal #210 Gt 2)
We hare
iy i "
pat? Gale 1)? + be HE D)
‘Therefore,
vio = fat = La nt + ne lea 40) ai 20)
(e) 2(0) periodic implies y() periodic. determine 1 period only. We have
Coronas fa esmare jet, pete}
ve) = E fi u
[Laune freine
Parera, ]<t<d
‘The period of y(t) is 2
2.23. y(t) is sketched in Figure 82.23 for the different values of T.
2.24, (a) We are given that hafn) = öfn] + dfn — 1]. Therefore,
ar al = fn] + 25) = + fa — 2)
a
Ta Teh
A
Gane RS
WO em Ii
AMARAL
HS + TZ
Figure 62:28
since
fn] = Aufn] + {hafn] + habra),
ag
An] = Par +2ha fe — 1] + Pan = 2)
Therefore,
io) = alo}
HU = ha + 2%)
HE = taf] + 2h] + ul)
M8] = taf + 2h + Ha]
HE] = al) + 2h(8) + Pa (2)
A] = Bal + 2 + a
CCE
NE
hyn] =0 for n < 0 and n> 5.
(b) In this case,
sn = 2h = Ar] Mn)
(9°
2.28. (a) We may write x(n) as
A
Gane RS
WO em Ii
AMARAL
HS + TZ
Figure 62:28
since
fn] = Aufn] + {hafn] + habra),
ag
An] = Par +2ha fe — 1] + Pan = 2)
Therefore,
io) = alo}
HU = ha + 2%)
HE = taf] + 2h] + ul)
M8] = taf + 2h + Ha]
HE] = al) + 2h(8) + Pa (2)
A] = Bal + 2 + a
CCE
NE
hyn] =0 for n < 0 and n> 5.
(b) In this case,
sn = 2h = Ar] Mn)
(9°
2.28. (a) We may write x(n) as
(b) Now,
fn) = sfr] + fn] = slr] bn = 1:
‘Therefore,
2{1- apart) +2{1- art} = art,
vin}=4 1
6,
‘Therefore, yn] = (1/2)"*"uln +3
(©) We have
valo] = zolo) + zafn] = ufn +3] un +2] = él)
(4) From the result of part (c), we get
vin) = wal] + lr
[voa
£ E lr) = ardt
[st frena
if rinde
= Ac
ln +3] = (1/2) un +9)
2.27. ‘The proof is as follows,
Ay
2.28. (4) Gansal because hn] =O fr <0. Stale cause DY)? = 5/4 < oe.
(b) Not causal because fín] #0 for n <0. Stable because > (0.8)" = 5 < co.
.
(6) Anti-causal because a] = 0 for n > 0. Unstable because. DS (1/2)" = 00
(4) Not causal because la] #0 for n <0. Stable becaute 2 = < oo
(0) Causal because ifn} = 0 forn <0, Unstable because the srond ter besomesinnte
(© Not causal because hf] 4 0 for n< 0. Stable because J Al = 305/3 < 00
a
fn) = sfr] + fn] = slr] bn = 1:
‘Therefore,
2{1- apart) +2{1- art} = art,
vin}=4 1
6,
‘Therefore, yn] = (1/2)"*"uln +3
(©) We have
valo] = zolo) + zafn] = ufn +3] un +2] = él)
(4) From the result of part (c), we get
vin) = wal] + lr
[voa
£ E lr) = ardt
[st frena
if rinde
= Ac
ln +3] = (1/2) un +9)
2.27. ‘The proof is as follows,
Ay
2.28. (4) Gansal because hn] =O fr <0. Stale cause DY)? = 5/4 < oe.
(b) Not causal because fín] #0 for n <0. Stable because > (0.8)" = 5 < co.
.
(6) Anti-causal because a] = 0 for n > 0. Unstable because. DS (1/2)" = 00
(4) Not causal because la] #0 for n <0. Stable becaute 2 = < oo
(0) Causal because ifn} = 0 forn <0, Unstable because the srond ter besomesinnte
(© Not causal because hf] 4 0 for n< 0. Stable because J Al = 305/3 < 00
a
2.20.
2.30.
231.
2.82.
(e) Gant bez i ben <0. Sala sem À =
(6) Causa baso A) = Or £<0. Stable because DIE =
4 < 00,
AS
(9 Ma cool bee A Orc abe ea | lt = €
(0) a ui eco) 40 10 Same |” a = 72 <
(0 a onl beans 40 <0 Sto tc | Bat = 1/8 < on
(0 Casal bes) =D rt <0, Sle mae |” Mt =<,
(9 ana ans) =O <0 Una ML
We need to find the output of the system when the input is 2{n] = dfn). Since we are asked
40 assume initial rest, we may conclude that yfa] = 0 for n < 0. Now,
vin] = sfr] - 2y{n - 2).
‘Therefore,
vl ==, oft
and so on. In closed form,
“This the impute response of he system.
Anti! rst implies that ya) =O form < -2. Now
ie) = 20) +200 2 in — 1
There,
da = 21 vi
ld] = 56,y15] = -10, yla] = -120(-2)"- form 25.
(a) HE yal
(1/2)", then we need to verify
OO
(Clearly this is true.
stl] =
2.30.
231.
2.82.
(e) Gant bez i ben <0. Sala sem À =
(6) Causa baso A) = Or £<0. Stable because DIE =
4 < 00,
AS
(9 Ma cool bee A Orc abe ea | lt = €
(0) a ui eco) 40 10 Same |” a = 72 <
(0 a onl beans 40 <0 Sto tc | Bat = 1/8 < on
(0 Casal bes) =D rt <0, Sle mae |” Mt =<,
(9 ana ans) =O <0 Una ML
We need to find the output of the system when the input is 2{n] = dfn). Since we are asked
40 assume initial rest, we may conclude that yfa] = 0 for n < 0. Now,
vin] = sfr] - 2y{n - 2).
‘Therefore,
vl ==, oft
and so on. In closed form,
“This the impute response of he system.
Anti! rst implies that ya) =O form < -2. Now
ie) = 20) +200 2 in — 1
There,
da = 21 vi
ld] = 56,y15] = -10, yla] = -120(-2)"- form 25.
(a) HE yal
(1/2)", then we need to verify
OO
(Clearly this is true.
stl] =
(0) rei ta rn 20
OOO
= 2
(€) From eq. (P2.32:), we know that yl0] = 2[0] + (1/24(-1) = x[0] = 1. Now we also
have
WO=4+B => A=1-B=3.
2.38. (a) () From Example 2.14, we know that
mit) | u(t).
(ji) We solve this along the lines of Example 2.14, First assume that yp(t) is of the
form Ke for € > 0. Then using eq. (P2.5%1), we get for £ > 0
1
RIAL HK
Wo now know that 3,6) = fe for > 0. We may hypothesize the homogeneous
solution o be of the form
une) = Acc“,
Therefor,
Mi) = At te, EU
Ascuming initial rst, we can conclude that y(t) = 0 for <0. Therefor,
MO =0=4+] +
Then,
1,2]
mo [jojo
(ii) Let the input be za(() = acHu(t) + Aetu(t). Assume that the particular solution
vale) is of the form
vet) = Kia! + Kafe
for + > 0. Using eq, (P2.33-1), we get
SK yae™ + 2K Be + 2Kyae™ + 2KaBe = a + pe
Equating the coefficients of e and ¢ on both sides, we get
4
OOO
= 2
(€) From eq. (P2.32:), we know that yl0] = 2[0] + (1/24(-1) = x[0] = 1. Now we also
have
WO=4+B => A=1-B=3.
2.38. (a) () From Example 2.14, we know that
mit) | u(t).
(ji) We solve this along the lines of Example 2.14, First assume that yp(t) is of the
form Ke for € > 0. Then using eq. (P2.5%1), we get for £ > 0
1
RIAL HK
Wo now know that 3,6) = fe for > 0. We may hypothesize the homogeneous
solution o be of the form
une) = Acc“,
Therefor,
Mi) = At te, EU
Ascuming initial rst, we can conclude that y(t) = 0 for <0. Therefor,
MO =0=4+] +
Then,
1,2]
mo [jojo
(ii) Let the input be za(() = acHu(t) + Aetu(t). Assume that the particular solution
vale) is of the form
vet) = Kia! + Kafe
for + > 0. Using eq, (P2.33-1), we get
SK yae™ + 2K Be + 2Kyae™ + 2KaBe = a + pe
Equating the coefficients of e and ¢ on both sides, we get
4
2.37. Let us consider two inputs
m) =0, forall
and
a) = efult) — ult 1)
Since the system is linear, the response yı{!) = 0 for all
Now let us ind the output ya(t) when the input is x(t). The particular solute
the form
were, 0<t<i
Substituting in eq. (P2.88-1), we get
3
Now, including the homogencous solution whi
overall solution:
of the form ya(t) = Ae, we get the
nz le, ocre >
Using this we got À = 64/8, Therefore,
Assuming final rest, we have y(1) =
n=-femede, <tc (6237-1)
For £ <0, we note that z(t) = 0. ‘Thus the particular solution is zero in this range and
e, t<0. (8237-2)
ne) =
Since the two pieces of he solution for ya(£) in eqs. (82.37-1) and (S2.37-2) must match at
= 0, we can determine B from the equ
1_1
LS da
which yields
wio=(f-fe)o% ta
Now note that since z(t) = z(t) for £ < 0, it must be true that for a causal system
lt) = alt) fort < 0. However, the results of obtained above show that this is not true.
‘Therefore, the system is not causal.
2.88. The block diagrams are as shown in Figure $2.38,
2.39. ‘The block diagrams are as shown in Figure $2.39,
(a) Note that
ult) a.
[cosas
‘Therefore,
A(t) = eMule — 2).
ss
m) =0, forall
and
a) = efult) — ult 1)
Since the system is linear, the response yı{!) = 0 for all
Now let us ind the output ya(t) when the input is x(t). The particular solute
the form
were, 0<t<i
Substituting in eq. (P2.88-1), we get
3
Now, including the homogencous solution whi
overall solution:
of the form ya(t) = Ae, we get the
nz le, ocre >
Using this we got À = 64/8, Therefore,
Assuming final rest, we have y(1) =
n=-femede, <tc (6237-1)
For £ <0, we note that z(t) = 0. ‘Thus the particular solution is zero in this range and
e, t<0. (8237-2)
ne) =
Since the two pieces of he solution for ya(£) in eqs. (82.37-1) and (S2.37-2) must match at
= 0, we can determine B from the equ
1_1
LS da
which yields
wio=(f-fe)o% ta
Now note that since z(t) = z(t) for £ < 0, it must be true that for a causal system
lt) = alt) fort < 0. However, the results of obtained above show that this is not true.
‘Therefore, the system is not causal.
2.88. The block diagrams are as shown in Figure $2.38,
2.39. ‘The block diagrams are as shown in Figure $2.39,
(a) Note that
ult) a.
[cosas
‘Therefore,
A(t) = eMule — 2).
ss
Ma
Un)
OT a 27 O Yo
Figure $2.38
Figure $2.30
(0) We have
u) = [unan
= Fer r+1)-a(t=r-2)
A(x) and 2(¢~) are as shown in the figure below.
Using this igure, we may write
o ter
Bar 1er, 1<t<a
ult) = de
dar Op 1>4
2.41. (a) We may write
sin] asín — 1]
ruf) = rum]
ain
an
56
Un)
OT a 27 O Yo
Figure $2.38
Figure $2.30
(0) We have
u) = [unan
= Fer r+1)-a(t=r-2)
A(x) and 2(¢~) are as shown in the figure below.
Using this igure, we may write
o ter
Bar 1er, 1<t<a
ult) = de
dar Op 1>4
2.41. (a) We may write
sin] asín — 1]
ruf) = rum]
ain
an
56
4,
“The signals for all parts of this problem are plotted in the Figure $2.47.
ya YU)
2
A 1
° E
0]
%
1 4
: a E
or 3
co |
“a RCE S247
.. (a) True. TEL) periodic and nonzero, then
SS
‘Therefore, h(t is unstable.
(b) False, For example, inverse of fr] = dfn — A] is gi] dfn + A] which is noncansal
(e) False. For example Afr] = ul] implies that
> IAr}} = oo.
‘This is an unstable system.
(a) True. Assuming tha hf) is bounded and nonzero in the range mi < n <a,
Y mini < co.
a
bis implies that the system i stable
(€) False. For example, h(t) = e'u(t) is causal but not stable.
(©) False. For example, the cascade of a causal system with impulso response hıln] =
ln 1) and a noweausal system with impulse response Aal] = 0 + 1] leads to a
Syotem with overall impulse response given y Af] = fn) « hal] = ei
e
“The signals for all parts of this problem are plotted in the Figure $2.47.
ya YU)
2
A 1
° E
0]
%
1 4
: a E
or 3
co |
“a RCE S247
.. (a) True. TEL) periodic and nonzero, then
SS
‘Therefore, h(t is unstable.
(b) False, For example, inverse of fr] = dfn — A] is gi] dfn + A] which is noncansal
(e) False. For example Afr] = ul] implies that
> IAr}} = oo.
‘This is an unstable system.
(a) True. Assuming tha hf) is bounded and nonzero in the range mi < n <a,
Y mini < co.
a
bis implies that the system i stable
(€) False. For example, h(t) = e'u(t) is causal but not stable.
(©) False. For example, the cascade of a causal system with impulso response hıln] =
ln 1) and a noweausal system with impulse response Aal] = 0 + 1] leads to a
Syotem with overall impulse response given y Af] = fn) « hal] = ei
e
(6) False. For example, if h(t)
u(t), then sl) =
à = e"tu(t) and
I late t+ et] = 00.
Although the sytem is stable the step response i not absolutely integrable
Cu) True. We may write uf] = Sn ~ A. Therefore,
=
sin) ns
If af = 0 for n < 0, then h(n] = 0 for n < D and the system is causal,
2.40. (0) It is a bowided input. jz{n| < 1 = Be for all n.
(b) Consider
0 E
$ Ibal] 0
ie
‘Therefore, the output is not bounded. Thus, the system is not stable and absolute
summability is necessary.
(©) Let
o irn) <0
“= { od, iene 20
Now, [z(6)| < 1 forall. Therefore, z(t) is a bounded input Now,
vo = [Cami
_ fre
= Lin
f "moi =0
Therefore, the system is unstable if the impulse response is not absolutely integrable
2.50. (a) The output wil be anit) + bal
(6) The output willbe 21(¢~ 7)
6
u(t), then sl) =
à = e"tu(t) and
I late t+ et] = 00.
Although the sytem is stable the step response i not absolutely integrable
Cu) True. We may write uf] = Sn ~ A. Therefore,
=
sin) ns
If af = 0 for n < 0, then h(n] = 0 for n < D and the system is causal,
2.40. (0) It is a bowided input. jz{n| < 1 = Be for all n.
(b) Consider
0 E
$ Ibal] 0
ie
‘Therefore, the output is not bounded. Thus, the system is not stable and absolute
summability is necessary.
(©) Let
o irn) <0
“= { od, iene 20
Now, [z(6)| < 1 forall. Therefore, z(t) is a bounded input Now,
vo = [Cami
_ fre
= Lin
f "moi =0
Therefore, the system is unstable if the impulse response is not absolutely integrable
2.50. (a) The output wil be anit) + bal
(6) The output willbe 21(¢~ 7)
6
(4)
Figure $2.57
(b) The figures corresponding to the remaining parts of this problem are shown in the
Figure $2.57.
2.58. (a) Realizing that z2{n] = yıln), we may eliminate these from the two given difference
‘equations. ‘This would give us
2yaln] = val — 1] + valo = = ln] ~ Si ln ~ 4).
‘This is the same as the overall difference equation.
(b) The figures corresponding to the remaining parts of this problem are shown in Figure
$258.
2.50. (a) Ie th gen iowa eqn oc nd pin. gt
sf nero ft series da
Therefore, A = ~ao/ay, B =01/a1, C = bo/a
(b) Realizing that za(t) = yx(t), we may eliminate these from the two given integral equa
tions. This would give us
w=
n= Af mine] an + Ca.
n
Figure $2.57
(b) The figures corresponding to the remaining parts of this problem are shown in the
Figure $2.57.
2.58. (a) Realizing that z2{n] = yıln), we may eliminate these from the two given difference
‘equations. ‘This would give us
2yaln] = val — 1] + valo = = ln] ~ Si ln ~ 4).
‘This is the same as the overall difference equation.
(b) The figures corresponding to the remaining parts of this problem are shown in Figure
$258.
2.50. (a) Ie th gen iowa eqn oc nd pin. gt
sf nero ft series da
Therefore, A = ~ao/ay, B =01/a1, C = bo/a
(b) Realizing that za(t) = yx(t), we may eliminate these from the two given integral equa
tions. This would give us
w=
n= Af mine] an + Ca.
n
400
e Figure 52.58 ®
(6) The figures corresponding to (he remaining part of this problem are shown in Figure
$250.
2.60. (a) Integrating the given diffrenil eqation once and simplifying, we get
+ Bf f soi]
‘Therefore, A =—01/07, B ==09/02, C = dafur, D = bi/02, E = iofar.
(b) Realiing that 22(¢) = yn(t), we may eliminate these from the two given integral oque.
tions.
(©) The figures corresponding to the remaining parts of this problem are shown in Figure
$2.60,
vo = ae ever ff senos
ried Baty
2.61. (a) (i) From Kirchof's voltage law, we know that the input voltage must equal the sum
‘of the voltages across the inductor and capacitor. Therefore,
ay = 10220 à y
2
Ber
e Figure 52.58 ®
(6) The figures corresponding to (he remaining part of this problem are shown in Figure
$250.
2.60. (a) Integrating the given diffrenil eqation once and simplifying, we get
+ Bf f soi]
‘Therefore, A =—01/07, B ==09/02, C = dafur, D = bi/02, E = iofar.
(b) Realiing that 22(¢) = yn(t), we may eliminate these from the two given integral oque.
tions.
(©) The figures corresponding to the remaining parts of this problem are shown in Figure
$2.60,
vo = ae ever ff senos
ried Baty
2.61. (a) (i) From Kirchof's voltage law, we know that the input voltage must equal the sum
‘of the voltages across the inductor and capacitor. Therefore,
ay = 10220 à y
2
Ber
Chapter 3 Answers
Using the Fourier series synthesis eq, (38),
a.
32.
34.
2)
= à qe HT y que y a, y POST
= EN y 2e On GATO: je
= 400470 Bent
= oo ++ El
Using the Fourier series synthesis eq. (396).
A A 4. ge gen
A eke gnu
HN y 2 Han
FE: Be,
1+ 2eotEn + 7) + doos( “En +
142
‘The given signal is
=)
From this, we may conclude that the fundamental frequency of z{) is 2x/6
Lee à Lee _ 2 jen 9 age
24 pote + Leu ajetreo
2 24 Sen Bein os ay HO
ae, St Br,
np Asin Ene
non-zero Fourier series coefcients of 2() are:
a =2,
3
ir,
)
Since wy = r, T = 25/49 = 2. Therefore,
Now,
and for k #0
Lf get,
Luca
3
lc
Der
ir
Y acta
[isa tf ramo
Lodge
Using the Fourier series synthesis eq, (38),
a.
32.
34.
2)
= à qe HT y que y a, y POST
= EN y 2e On GATO: je
= 400470 Bent
= oo ++ El
Using the Fourier series synthesis eq. (396).
A A 4. ge gen
A eke gnu
HN y 2 Han
FE: Be,
1+ 2eotEn + 7) + doos( “En +
142
‘The given signal is
=)
From this, we may conclude that the fundamental frequency of z{) is 2x/6
Lee à Lee _ 2 jen 9 age
24 pote + Leu ajetreo
2 24 Sen Bein os ay HO
ae, St Br,
np Asin Ene
non-zero Fourier series coefcients of 2() are:
a =2,
3
ir,
)
Since wy = r, T = 25/49 = 2. Therefore,
Now,
and for k #0
Lf get,
Luca
3
lc
Der
ir
Y acta
[isa tf ramo
Lodge
3.
3.
87.
Both si ) and ft 1) ae parodie wich fundemental period Ti = 2. Since vt is
tinea combination of 21 (1-1) and 21(¢~ D), itis ale periodic wit fundemental period
Ma = 2. Therefore, u = an.
since u) ay, wing th ets bl 3 ve have
ant) ES at)
alt ES ave IMO ap y(t 1) ES ane HOH
Tree,
alan E ete y a yor HHO)
ay + 0-4)
(a) Comparing zu(t) with the Fourier series synthesis oq, (3.8), wo obtain the Fourier
series coccion of z(t) to be
LW, oss 100
Pa
From Table 3.1 we know that if zı(£) is real, then az has to be conjugate-symmetric,
Prom ae Sucede ot to fr 21), Le sgl not real valued.
Sri the Fourier series celia ol (0) are
cos(kr), 100 << k S100
A ps
rom Tabl 3.1 we kaow hat i lt is eal hen os bas tobe omlugatesynmetri,
ie, ay = 0". Since this is true for 22(), the signal is real valued.
Sari he Fourier sin ces of 0) are
: _ { jsnthr/2) 100€ < 100
.. La otherwise
From Table 3.1 we know tat iz) i real, then as has to be conjugatesymmnetric,
he ap = a? y. Sine this is true for za), the signal is real valued.
(6) For a signal to e even ts Fourier seis coeicents must be even. This is true only
for ale)
Given that
20 Ban
weine
of) = ES, ty stos.
Theatre, a
pa
AT wee
sr
3.
87.
Both si ) and ft 1) ae parodie wich fundemental period Ti = 2. Since vt is
tinea combination of 21 (1-1) and 21(¢~ D), itis ale periodic wit fundemental period
Ma = 2. Therefore, u = an.
since u) ay, wing th ets bl 3 ve have
ant) ES at)
alt ES ave IMO ap y(t 1) ES ane HOH
Tree,
alan E ete y a yor HHO)
ay + 0-4)
(a) Comparing zu(t) with the Fourier series synthesis oq, (3.8), wo obtain the Fourier
series coccion of z(t) to be
LW, oss 100
Pa
From Table 3.1 we know that if zı(£) is real, then az has to be conjugate-symmetric,
Prom ae Sucede ot to fr 21), Le sgl not real valued.
Sri the Fourier series celia ol (0) are
cos(kr), 100 << k S100
A ps
rom Tabl 3.1 we kaow hat i lt is eal hen os bas tobe omlugatesynmetri,
ie, ay = 0". Since this is true for 22(), the signal is real valued.
Sari he Fourier sin ces of 0) are
: _ { jsnthr/2) 100€ < 100
.. La otherwise
From Table 3.1 we know tat iz) i real, then as has to be conjugatesymmnetric,
he ap = a? y. Sine this is true for za), the signal is real valued.
(6) For a signal to e even ts Fourier seis coeicents must be even. This is true only
for ale)
Given that
20 Ban
weine
of) = ES, ty stos.
Theatre, a
pa
AT wee
sr
38.
30.
Since 2(t is real and odd (eue 1), its Fourier sores coefficients az aro purely imaginary and
odd (See Table 3.1). Therefore, ag = -a-4 and ao = 0. Also, since itis given that ax = 0
for [E] > 1, the only unknown Fourier series cooficients are ay and a. Using Parseval's
ratio, :
7, ore Zur
for tho given signal we have
Using the informa
given in che (4) along with the above equation,
E
‘Therefore,
ot ar
vi ET
The two possible signals which satisfy the given information are
a= me > a A e VB sin(nt)
20 ST eee HOME sit)
‘The period of the given signal is 4. Therefore,
a
ae ¿Dei
2
Lect
> sin
‘This gives
a
as=1+2
30.
Since 2(t is real and odd (eue 1), its Fourier sores coefficients az aro purely imaginary and
odd (See Table 3.1). Therefore, ag = -a-4 and ao = 0. Also, since itis given that ax = 0
for [E] > 1, the only unknown Fourier series cooficients are ay and a. Using Parseval's
ratio, :
7, ore Zur
for tho given signal we have
Using the informa
given in che (4) along with the above equation,
E
‘Therefore,
ot ar
vi ET
The two possible signals which satisfy the given information are
a= me > a A e VB sin(nt)
20 ST eee HOME sit)
‘The period of the given signal is 4. Therefore,
a
ae ¿Dei
2
Lect
> sin
‘This gives
a
as=1+2
3.10. Since the Fourier series coefficients repeat every N, we have
= aso asar ‚ud mar
Furthermore, since the signal is real and od, the Fourier series ooefheiens az will be purely
imaginary and odd. Therefore, ao = 0 and
Finally,
+ aah ad
3.11. Since the Fourier serie coefficients repeat every N = 10, we have aı
more, since z[r] is ral and even, o is also real and even. Therefore, 01
alo given that
oe
tr.
ing Parseval’s relation,
Y mf = 50
>
Dia = 5
fail? + loi? 408+ lu = 50
E
+ Yolo? = 0
=
Therefore, ay = 0 for £=2,--- 8. Now using the synthesis 0q.(3.4), we have
die D ach at
ES
3.12. Using the multiplication property (ee Table 32), we have
aie) FS ad Sabet
were ro
Ende aba aden + abs
Ey be + Moya + a +
89
= aso asar ‚ud mar
Furthermore, since the signal is real and od, the Fourier series ooefheiens az will be purely
imaginary and odd. Therefore, ao = 0 and
Finally,
+ aah ad
3.11. Since the Fourier serie coefficients repeat every N = 10, we have aı
more, since z[r] is ral and even, o is also real and even. Therefore, 01
alo given that
oe
tr.
ing Parseval’s relation,
Y mf = 50
>
Dia = 5
fail? + loi? 408+ lu = 50
E
+ Yolo? = 0
=
Therefore, ay = 0 for £=2,--- 8. Now using the synthesis 0q.(3.4), we have
die D ach at
ES
3.12. Using the multiplication property (ee Table 32), we have
aie) FS ad Sabet
were ro
Ende aba aden + abs
Ey be + Moya + a +
89
318.
is1 for all values of E, lt is clear that by + Du + 26-9 + 204.3 will be 6 for all
values of k. Therefore,
zıln)zaln) #56, for all&
Let us first evaluate the Fourier series coeicionts of z(t). Cleary, since 2() is real and
oat, sis purely imaginary and odd. Therefore, 0. Now,
7
a geo,
¿Lo a
1 fee 1 [has
When z(t) is passed through an LTI system with froqueney response H(ju), the output
(8 is given by (see Section 38)
ult) = Y alla
where un = 37 = 3. Since ay is non gero only for odd values off, we ned to evaluate the
bore summation only for odd À. Furthermore, note that
sin(én)
Kr)
H (gk) = HÜkLR/M) =
is always zero for odd values of k. Therefore,
u(t) =0.
8.14. ‘The signal z[n) is periodic with period N 4. Its Fourier series coefficients are
From the results presented in Section 38, we know that the output yin] is given by
à = Eo
à ons semi
HA 4 LN)
(s3.4-1)
go
is1 for all values of E, lt is clear that by + Du + 26-9 + 204.3 will be 6 for all
values of k. Therefore,
zıln)zaln) #56, for all&
Let us first evaluate the Fourier series coeicionts of z(t). Cleary, since 2() is real and
oat, sis purely imaginary and odd. Therefore, 0. Now,
7
a geo,
¿Lo a
1 fee 1 [has
When z(t) is passed through an LTI system with froqueney response H(ju), the output
(8 is given by (see Section 38)
ult) = Y alla
where un = 37 = 3. Since ay is non gero only for odd values off, we ned to evaluate the
bore summation only for odd À. Furthermore, note that
sin(én)
Kr)
H (gk) = HÜkLR/M) =
is always zero for odd values of k. Therefore,
u(t) =0.
8.14. ‘The signal z[n) is periodic with period N 4. Its Fourier series coefficients are
From the results presented in Section 38, we know that the output yin] is given by
à = Eo
à ons semi
HA 4 LN)
(s3.4-1)
go
From the given information, we know that uf is
un) = ln D
= ont
jeans Lae
= jeune Yon
“Comparing this with eq, (63.41), we have
Me) = HEN =0
and
meh=2et, md = BF
3.45. From the results of Section 23,
w= E a
where an = 3 = 12 since Hu) is ero fr lol > 10, the largest value of I} for which
Ue is nonzero should be such that
Ver < 100
«ps implies that Yl <8. Therefore, for I > 8, a guaran 1 De ze.
(a) The given signal ln] is
sap) = Car = = de
Therefore, x(n is periodic with period N = 2 and it’s Fourier series coeficions in the
range 0-S# £ 1 are
a =0, and a=)
Using the result drived in Section 3.6, the output uf is ven I
ui) AAA
ro
Dean
=0
(6) The signa zl) i period wth period N = 16. The sins! an] may be written as.
ale) = genen. pe D + G ayer HM aan
= Seem AO O tana
a
un) = ln D
= ont
jeans Lae
= jeune Yon
“Comparing this with eq, (63.41), we have
Me) = HEN =0
and
meh=2et, md = BF
3.45. From the results of Section 23,
w= E a
where an = 3 = 12 since Hu) is ero fr lol > 10, the largest value of I} for which
Ue is nonzero should be such that
Ver < 100
«ps implies that Yl <8. Therefore, for I > 8, a guaran 1 De ze.
(a) The given signal ln] is
sap) = Car = = de
Therefore, x(n is periodic with period N = 2 and it’s Fourier series coeficions in the
range 0-S# £ 1 are
a =0, and a=)
Using the result drived in Section 3.6, the output uf is ven I
ui) AAA
ro
Dean
=0
(6) The signa zl) i period wth period N = 16. The sins! an] may be written as.
ale) = genen. pe D + G ayer HM aan
= Seem AO O tana
a
“Therefore, the non-tero Fourier series coefficients of za[n] in the range 0 < < 15 are
GT, ayy =p yet
Using the results derived in Section 35 the output ya] is given by
CRT Co etait)
win]
ESE
sen
(e) The signal zufn] may be written as
ai [() ane Bm un ator
where gin] = (3)" ufr] and rin) = = öfn— Ak]. Therefore, ys[n] may be obtained
by passing the signa ro] trough the filter with frequency response HH), and then
convolving the result with gn).
‘The signal rn] is periodic
period 4 and its Fourier series coeffients are
for all k (See Problem 3.14)
‘The output gfn] obtained by passing r{r] through the filter with frequency response
or) is
ant ero
IA) SIDA y SA + AD) ID)
=0
dl)
‘Therefore, the final output yaln] = a] + a
3.17. (a) Since complex exponentials are Eigen functions of LT systems, the input (1) =
has to produce an output ofthe form Ao, where A isa complex constant. But clearly,
in this case the output is not ofthis form. Therefore, system S) is definitely mot LTT.
(b) This aystem may be LTI because it satsfis the Eigen function property of LTT systems,
(©) In this caso, the output is of the form galt) = (1/2)0 + (1/2)e%. Clearly, the
‘output contains a complex exponential with frequency => which was not present in the
input 23(¢). We know that an LTI system can never produce e complex exponential of
frequency ~5 unless there was complex exponential of the same frequency at its input
Since this isnot the case in this problem, Sa is definitely not LT.
2
GT, ayy =p yet
Using the results derived in Section 35 the output ya] is given by
CRT Co etait)
win]
ESE
sen
(e) The signal zufn] may be written as
ai [() ane Bm un ator
where gin] = (3)" ufr] and rin) = = öfn— Ak]. Therefore, ys[n] may be obtained
by passing the signa ro] trough the filter with frequency response HH), and then
convolving the result with gn).
‘The signal rn] is periodic
period 4 and its Fourier series coeffients are
for all k (See Problem 3.14)
‘The output gfn] obtained by passing r{r] through the filter with frequency response
or) is
ant ero
IA) SIDA y SA + AD) ID)
=0
dl)
‘Therefore, the final output yaln] = a] + a
3.17. (a) Since complex exponentials are Eigen functions of LT systems, the input (1) =
has to produce an output ofthe form Ao, where A isa complex constant. But clearly,
in this case the output is not ofthis form. Therefore, system S) is definitely mot LTT.
(b) This aystem may be LTI because it satsfis the Eigen function property of LTT systems,
(©) In this caso, the output is of the form galt) = (1/2)0 + (1/2)e%. Clearly, the
‘output contains a complex exponential with frequency => which was not present in the
input 23(¢). We know that an LTI system can never produce e complex exponential of
frequency ~5 unless there was complex exponential of the same frequency at its input
Since this isnot the case in this problem, Sa is definitely not LT.
2
5.18. (a) By using an argument similar to the one used in part (a) ofthe previous problem, we
conclude that Si is defintely not LT
(6) The output inthis cas ee) = eX ae. Carly this iles the eigen
Function property of LTI systems. Therefore, Sa is definitely not ET.
= 2062/2 = 2080/00, This does not violate the eigen
coulé possibly be an LI system.
(6) The output in this case is af]
function property of LTT systems. Therefore, Ss
3.10. (9) Voltage across inductor = 24.
Current through resistor
Tapın current 2(t) = curren through resistor + current through inductor
‘Therefore,
Lay)
20 = = 2003
Substituting for R and L we obtain
4
EL 5 (0) = 20
outlined in Section 310.1, we know that the output of this system
‘cut, Substituting in the differential equation of
(b) Using the approach
‘will be HGu)eó when the input is
part (a),
jo jue" + Huet = dt
‘Therefore, 3
HG = 55
(6) The signal a) i parodie with period 2x. Since s() can be expres in the form
A
BE à
the nonsero Fourier series cooficients af 2) are
_
sais
‘Using the results derived in Section 3.8 (see eq.(3.124)), we have
ue) auge +
Re)
UNA
a)
ss
conclude that Si is defintely not LT
(6) The output inthis cas ee) = eX ae. Carly this iles the eigen
Function property of LTI systems. Therefore, Sa is definitely not ET.
= 2062/2 = 2080/00, This does not violate the eigen
coulé possibly be an LI system.
(6) The output in this case is af]
function property of LTT systems. Therefore, Ss
3.10. (9) Voltage across inductor = 24.
Current through resistor
Tapın current 2(t) = curren through resistor + current through inductor
‘Therefore,
Lay)
20 = = 2003
Substituting for R and L we obtain
4
EL 5 (0) = 20
outlined in Section 310.1, we know that the output of this system
‘cut, Substituting in the differential equation of
(b) Using the approach
‘will be HGu)eó when the input is
part (a),
jo jue" + Huet = dt
‘Therefore, 3
HG = 55
(6) The signal a) i parodie with period 2x. Since s() can be expres in the form
A
BE à
the nonsero Fourier series cooficients af 2) are
_
sais
‘Using the results derived in Section 3.8 (see eq.(3.124)), we have
ue) auge +
Re)
UNA
a)
ss
322. (a) ) T= 1,
(i) Here,
ln -2<t<nI
1 =I<t<l
3-1, 1<1<2
T=6,0=1/2, and
P ={% even
reinen, odd”
avn ze) ne KA
GT = 2, on = 1/2, a4 = }- (FO.
Note that op =D and ak esen
(Wi) T= 4, oy = 4/2, 09 =3/4 and
eI sine 2) + inte)
rr
From Example 3.5, we know that y(t) must be a periodic square wave which over one
period is
afk Me
wo e
Now, note that bo = 1/4. Let us define another signal x(£) = —1/4 whose only nonzero
FS coeficiont is co = 1/4, The signal p(t) = y(t) + (6) wil have FS coefficients
o k=0
Sell, otherwise.
inet, Therefore, the signal z(t) = plt+ 1) which is as shown
|
o
(i) Here,
ln -2<t<nI
1 =I<t<l
3-1, 1<1<2
T=6,0=1/2, and
P ={% even
reinen, odd”
avn ze) ne KA
GT = 2, on = 1/2, a4 = }- (FO.
Note that op =D and ak esen
(Wi) T= 4, oy = 4/2, 09 =3/4 and
eI sine 2) + inte)
rr
From Example 3.5, we know that y(t) must be a periodic square wave which over one
period is
afk Me
wo e
Now, note that bo = 1/4. Let us define another signal x(£) = —1/4 whose only nonzero
FS coeficiont is co = 1/4, The signal p(t) = y(t) + (6) wil have FS coefficients
o k=0
Sell, otherwise.
inet, Therefore, the signal z(t) = plt+ 1) which is as shown
|
o
‘Therefore, |
A ww.
=
2(0) =u) + rl
aa (oem ,
far fanaa
(0) The signa lt) = et shown in Figure 5924
Figure $8.24
‘the ES coefcionts by of lt) may be found as follows:
af
neifa-ifaro
(e) Note that
‘Therefore,
3.26. (a) The nonzero PS codes of 2() are as = RAT 12
{b) The vontero FS events of) ae b = Baa = ui
e
A ww.
=
2(0) =u) + rl
aa (oem ,
far fanaa
(0) The signa lt) = et shown in Figure 5924
Figure $8.24
‘the ES coefcionts by of lt) may be found as follows:
af
neifa-ifaro
(e) Note that
‘Therefore,
3.26. (a) The nonzero PS codes of 2() are as = RAT 12
{b) The vontero FS events of) ae b = Baa = ui
e
(©) Using the multiplication property, we know that
20) = alu) E à =
‘Therefore, í
axe = Laa) À
cr = an ee = Talk ~ 2] ~ ala
‘This implies that the nonzero Fourier series coeficiente of 2(0) are cz
(4) We have
1
a(t) = sin(4t) cost)
Fins.
‘Therefore, the nonzero Fourier serios codfiiente of x(t) are cz = e-+
3.26. (a) Uz(t) is real, then 2(¢) = 2*(0). This implies that for z{t) real ax =
ot true inthis case problem, 2() is not real
Since this is
(0) 1F2( is evn, then 2) = (4) and ay = a, Sine this rue for this cas, 2(0 is
© We have a.
10 EN ES = En
‘Therefor,
o, k=0
= {Samar Serre *
Since be is wot eve, of) is ot veo.
3:27. Using the Fou
ath) = 09 4 gern yeh 4er
LA 2 Rat y peine il HET gee AHO
24 doo(Aen/5) + 1/6] + 2coa(0mn/5)-+7/3)
24 dsl (nn /5) + 26/3} + 2sil(Ben/S) + 69/0]
y series synthesis ea. (3.88),
8.28. (a) N=7,
HT sin Snk/7)
Sl?)
= 6, ay over one period (0< k < 5) may be specified as: ao = 4/6,
1 asin) A
=;
(0)
20) = alu) E à =
‘Therefore, í
axe = Laa) À
cr = an ee = Talk ~ 2] ~ ala
‘This implies that the nonzero Fourier series coeficiente of 2(0) are cz
(4) We have
1
a(t) = sin(4t) cost)
Fins.
‘Therefore, the nonzero Fourier serios codfiiente of x(t) are cz = e-+
3.26. (a) Uz(t) is real, then 2(¢) = 2*(0). This implies that for z{t) real ax =
ot true inthis case problem, 2() is not real
Since this is
(0) 1F2( is evn, then 2) = (4) and ay = a, Sine this rue for this cas, 2(0 is
© We have a.
10 EN ES = En
‘Therefor,
o, k=0
= {Samar Serre *
Since be is wot eve, of) is ot veo.
3:27. Using the Fou
ath) = 09 4 gern yeh 4er
LA 2 Rat y peine il HET gee AHO
24 doo(Aen/5) + 1/6] + 2coa(0mn/5)-+7/3)
24 dsl (nn /5) + 26/3} + 2sil(Ben/S) + 69/0]
y series synthesis ea. (3.88),
8.28. (a) N=7,
HT sin Snk/7)
Sl?)
= 6, ay over one period (0< k < 5) may be specified as: ao = 4/6,
1 asin) A
=;
(0)
3.98. We will Ast evaluate the frequency response ofthe sytem. Consideran input (1) of e
sre rit Prom the discussion in Section 3.9.2 we know thatthe response to this input wil
peut) = HGule™". Therefore, substituing thse in the given differential equation, we Bet
Huet +a! = et,
Therefore, :
HG) = Er
Prom eq, (8.124), we know that
ul) = Y alan
‘when the input is z(0. (4) has the Fourier series coefcients ay and fundamental frequency
se Therefore, the Fourier series coefficients of u(t) are ak)
(6) Here, op = Dr and the nonzero FS coeficont of 2) ae a = a
the nonzero FS coefcints of y(t) are
1/2. Therefore,
1
amd = a
(0) Mere, oy = 2x and the nonzero ES coeficiente of 20) ae os = ala = 1/25 md
(FIR, Therefor, the nonzero FS coefficients of y(t) are
1 5 i
DENE Er er CIO AS
a en
b= este) = gen IO = Fa Fy
8.34. The frequency response ofthe system is given by
co A
s hat =
nu = [e dez E
(a) Here, 7 = 1 and uy =2r and ay = 1 for al The ES coeticints ol the ott ase
1 1
dy = ont Ghee) = pj TE
(b) Here, T = 2 and uy = # and
0 kom
m= La, Rodd
“Therefore, the FS counts ofthe output are
o even
amo | Oa + shes kodd
101
sre rit Prom the discussion in Section 3.9.2 we know thatthe response to this input wil
peut) = HGule™". Therefore, substituing thse in the given differential equation, we Bet
Huet +a! = et,
Therefore, :
HG) = Er
Prom eq, (8.124), we know that
ul) = Y alan
‘when the input is z(0. (4) has the Fourier series coefcients ay and fundamental frequency
se Therefore, the Fourier series coefficients of u(t) are ak)
(6) Here, op = Dr and the nonzero FS coeficont of 2) ae a = a
the nonzero FS coefcints of y(t) are
1/2. Therefore,
1
amd = a
(0) Mere, oy = 2x and the nonzero ES coeficiente of 20) ae os = ala = 1/25 md
(FIR, Therefor, the nonzero FS coefficients of y(t) are
1 5 i
DENE Er er CIO AS
a en
b= este) = gen IO = Fa Fy
8.34. The frequency response ofthe system is given by
co A
s hat =
nu = [e dez E
(a) Here, 7 = 1 and uy =2r and ay = 1 for al The ES coeticints ol the ott ase
1 1
dy = ont Ghee) = pj TE
(b) Here, T = 2 and uy = # and
0 kom
m= La, Rodd
“Therefore, the FS counts ofthe output are
o even
amo | Oa + shes kodd
101
Ya k=0
=10 koumk£0.
: sD kodd
‘Therefore, the FS coeficiente ofthe output ae
va, ko
nan [on Lon k#0
99 [re + oe] boda
3-35. We know that the Fourier series coefficient of y(t) are dy = H(jkwy)a,, where wy ls the
fundamen rquency of 0) and a ae the BS colle of).
If y(t) is identical to x(t), then ax for all k. Noting that H(jw) = 0 for kw] > 250,
we know that H(jkun) = 0 for |k] > 18 (because wy = 14). Therefore, a; must be zero for
tal 2 18.
3.30. We wll rst evaluate the frequency response ofthe sytem. Consider an input sf) ofthe
farm oS. Pom the iu in Sn 98 we ku tat the reponse t thi ape wll
te vn] = HG, Therefore, subettuing these nthe given aint equation me eet
Home) = oh
1
Mn
7
‘Therefore,
HG) = mee
rom og. (8.151), ve know that
dl Y either
von th input fl, zn as be Four series oeficints as and fundamental frequency
2/1. Therefore, the Fourier erie coeficionte ol yn] are axif(o#"),
(a) Here, N m 4 andthe nomero PS oefiens of af] are ay = a, = 1/2). Therefore,
the nonzero ES coeficiente of yp ae
= aol 1 a = ae) = EL u,
N Se N O
(0) Here, N = 8 and the nonzero FS coficents of zn] ae as
&.2=1. Theo, the nosso PS recents of yl) ae
1
E
—
AMA
a = 1/2 and az =
o
ners
ba
teh 1
= ger
1
bn = ae) = Tey
ba nH (HM) me
102
=10 koumk£0.
: sD kodd
‘Therefore, the FS coeficiente ofthe output ae
va, ko
nan [on Lon k#0
99 [re + oe] boda
3-35. We know that the Fourier series coefficient of y(t) are dy = H(jkwy)a,, where wy ls the
fundamen rquency of 0) and a ae the BS colle of).
If y(t) is identical to x(t), then ax for all k. Noting that H(jw) = 0 for kw] > 250,
we know that H(jkun) = 0 for |k] > 18 (because wy = 14). Therefore, a; must be zero for
tal 2 18.
3.30. We wll rst evaluate the frequency response ofthe sytem. Consider an input sf) ofthe
farm oS. Pom the iu in Sn 98 we ku tat the reponse t thi ape wll
te vn] = HG, Therefore, subettuing these nthe given aint equation me eet
Home) = oh
1
Mn
7
‘Therefore,
HG) = mee
rom og. (8.151), ve know that
dl Y either
von th input fl, zn as be Four series oeficints as and fundamental frequency
2/1. Therefore, the Fourier erie coeficionte ol yn] are axif(o#"),
(a) Here, N m 4 andthe nomero PS oefiens of af] are ay = a, = 1/2). Therefore,
the nonzero ES coeficiente of yp ae
= aol 1 a = ae) = EL u,
N Se N O
(0) Here, N = 8 and the nonzero FS coficents of zn] ae as
&.2=1. Theo, the nosso PS recents of yl) ae
1
E
—
AMA
a = 1/2 and az =
o
ners
ba
teh 1
= ger
1
bn = ae) = Tey
ba nH (HM) me
102
19.37, The frequency response of the system may be easily shown to be
1 1
He) = I -
N) = fen Toa
(6) The Fourier series cosficiente of af) are
1
=p hralk
4, Therefore, the Fourier series coeficiente of yin} are
Ho
(6) In this case, the Fourier eros cociente of af] are
ay = gi +2cos(br/9), forall
‘Also, N = 6. Therefore, the Fourier series coefficients of yn] are
Dem Te
;
Au opera
en tn ay ni
vom
a ee
Bor air], N
and wo
/2. The ES coeficiente of the input zfn] are
1
agai, foralln.
Therefore, the ES coefficients of the output are
dem) =
Au oh eit
3.89. Let the FS coefcienta of the input be ax. The FS coefients of the output are of the form
alto),
were wy = 27/8. Note that in the range 0 < k <2, Ho) = 0 for k = 1,2. Therefore,
‘nly by has a nonzero value among by in the range OS À 52.
3.40. Let the Fourier serie coefficients of 2(() be ar.
103
1 1
He) = I -
N) = fen Toa
(6) The Fourier series cosficiente of af) are
1
=p hralk
4, Therefore, the Fourier series coeficiente of yin} are
Ho
(6) In this case, the Fourier eros cociente of af] are
ay = gi +2cos(br/9), forall
‘Also, N = 6. Therefore, the Fourier series coefficients of yn] are
Dem Te
;
Au opera
en tn ay ni
vom
a ee
Bor air], N
and wo
/2. The ES coeficiente of the input zfn] are
1
agai, foralln.
Therefore, the ES coefficients of the output are
dem) =
Au oh eit
3.89. Let the FS coefcienta of the input be ax. The FS coefients of the output are of the form
alto),
were wy = 27/8. Note that in the range 0 < k <2, Ho) = 0 for k = 1,2. Therefore,
‘nly by has a nonzero value among by in the range OS À 52.
3.40. Let the Fourier serie coefficients of 2(() be ar.
103
(8) zit - to) is also periodic with period 7. ‘The Fourier seres coeficiente by of z(E — to)
A
n= fume
open MENT ge
SAM [serio
BT gy
Similarly, the Fourier serios coefficients of 2(¢ fa) are
e = MONIT Hag,
‘Finally, the Fourier series coeficiente of x(t ~ 1) + 2( +14) are
Org, 4 ¿HNOS TY, = 2cos(kbrto/ Tt
dy =e ce
(0) Note eat Ev(2(0) = 1 + (0/2 The ES cots of (4) are
n= Escocia
= Efacenorrrar
‘hereto he FS coins of Es) ae
MENTE
2 7
(©) Note that Refz(0)) = f2(@) + 2° (0/2. The FS coefficients of (E) are
a. [; (he HOM
Conjugating both sides, we get
4
Therefore, the FS coeficiente of Re(z(0) are
EST
7
ten
?
= J AD e ar
(a) The Fourier series synthesis equation gives
Fane
x
14
A
n= fume
open MENT ge
SAM [serio
BT gy
Similarly, the Fourier serios coefficients of 2(¢ fa) are
e = MONIT Hag,
‘Finally, the Fourier series coeficiente of x(t ~ 1) + 2( +14) are
Org, 4 ¿HNOS TY, = 2cos(kbrto/ Tt
dy =e ce
(0) Note eat Ev(2(0) = 1 + (0/2 The ES cots of (4) are
n= Escocia
= Efacenorrrar
‘hereto he FS coins of Es) ae
MENTE
2 7
(©) Note that Refz(0)) = f2(@) + 2° (0/2. The FS coefficients of (E) are
a. [; (he HOM
Conjugating both sides, we get
4
Therefore, the FS coeficiente of Re(z(0) are
EST
7
ten
?
= J AD e ar
(a) The Fourier series synthesis equation gives
Fane
x
14
Differentiating both sides wrt ¢ twice, we get
Patt)
ae
$ -rha tent
By inspection, we know that te Fourier eis coefficients of it) /d are ka.
(0) The period of (St) i a third of the period of z(t). Therefore, the signal af - 1)
is porodi with period 7/9. The Fourier seres coficnts of x31) ae sill ax. Using
them of por (we Btw ta he rie ee cose a (4 D
, we require that 2(t) = 2(-1). Also, note that since ay = 0442, we require
20) = a(t Xe.
‘This in turn implies that x(t) may have nonzero values only for t= 0, 41.5, 43,245,
, we may conclude that z(t) = 6(@) for -05 S 4 £ 05. Also, since,
fine pe
[ (ajdt = 2, we may conclude that 2(@) = 24(¢ - 3/2) in the range 05 < # < 3/2
‘Therefore, z{) may be written as
=
Eumafu-u-m
ES
3.42. (a) From Problem 340 (and Table 3.1), we know that FS coefficients of 2°(t) are al.
Now, we know that is z(t) is real, then zit) = 2*(2)- Therefore, a: = 0! y. Note that
this implies ag = aj. Therefore, ag must be seal
(b) From Problem 3.40 (and Table 3.1), we know that FS evelicients of 2(-1) are a+. Tf
24) is even, then 2(¢) = 2-0). This implies that
ao. (65420)
‘This implics that the PS coefficients are even. From the previous part, we know that
if x(t) is real, then
Per (2-2)
Using eqs. (S3.42-1) and (53.422), we know that ay = af. Therefore, ax is real for all
Ik. Hence, we may conclude that a is real and even.
(©) From Problem 3.40 (and Table 3.1), we know that ES coefficients of 2(—t) are a. If
20 is odd, then z(t) = ~2(-t)- This implies that
as (65.42-3)
105
Patt)
ae
$ -rha tent
By inspection, we know that te Fourier eis coefficients of it) /d are ka.
(0) The period of (St) i a third of the period of z(t). Therefore, the signal af - 1)
is porodi with period 7/9. The Fourier seres coficnts of x31) ae sill ax. Using
them of por (we Btw ta he rie ee cose a (4 D
, we require that 2(t) = 2(-1). Also, note that since ay = 0442, we require
20) = a(t Xe.
‘This in turn implies that x(t) may have nonzero values only for t= 0, 41.5, 43,245,
, we may conclude that z(t) = 6(@) for -05 S 4 £ 05. Also, since,
fine pe
[ (ajdt = 2, we may conclude that 2(@) = 24(¢ - 3/2) in the range 05 < # < 3/2
‘Therefore, z{) may be written as
=
Eumafu-u-m
ES
3.42. (a) From Problem 340 (and Table 3.1), we know that FS coefficients of 2°(t) are al.
Now, we know that is z(t) is real, then zit) = 2*(2)- Therefore, a: = 0! y. Note that
this implies ag = aj. Therefore, ag must be seal
(b) From Problem 3.40 (and Table 3.1), we know that FS evelicients of 2(-1) are a+. Tf
24) is even, then 2(¢) = 2-0). This implies that
ao. (65420)
‘This implics that the PS coefficients are even. From the previous part, we know that
if x(t) is real, then
Per (2-2)
Using eqs. (S3.42-1) and (53.422), we know that ay = af. Therefore, ax is real for all
Ik. Hence, we may conclude that a is real and even.
(©) From Problem 3.40 (and Table 3.1), we know that ES coefficients of 2(—t) are a. If
20 is odd, then z(t) = ~2(-t)- This implies that
as (65.42-3)
105
Figure $3.43,
(@) (Ian oF a-ı is nonzero, then
zit) = ent PT 4.
and
a(t to) = an FI) +
‘The smallest value of [fal (other than [f= 0 for which ete = 1 is the funda
mental period. Only then is
alte to) = amet 4 = a(t),
‘Therefore, lo has to be the fundamental period.
(2) The period of z(1) is the least common multiple of the periods of e/M26/7% and
TT, The period of 2/70 ig T/k and the period of 72/7" and T/L, Since
and 1 have no common factors, the leat common multiple of T/k and 7/1 is T.
8.44, ‘The only unknown FS coofcients are 07, a, a2, and a-2. Since z(t) is real, a; = al and
2 = ag. Since a i real, a = 0.1. Now, 2( is of the form
zit) = Ay conlunt) + Az cos(Zunt +0),
where wp = 21/6. From this we got
{t= 3) = Aycoolunt — Bun) + Arcosläunt + 8 Gun).
Now if we need x(() = ~2(¢~3), then Say and Guy should both be odd multiples of.
Clearly, this is impossible. Therefore, az = 0-1 = 0 and
2) = Ar cos(unt).
Nom, using Parseval's relation on Clue 5, we get
$ la? = la? +la-ı
1
2
‘Therefore, lay) = 1/2. Since ay is positive, we have ay = a-ı = 1/2. Therefore, =(t) =
cos(rt/3)-
107
(@) (Ian oF a-ı is nonzero, then
zit) = ent PT 4.
and
a(t to) = an FI) +
‘The smallest value of [fal (other than [f= 0 for which ete = 1 is the funda
mental period. Only then is
alte to) = amet 4 = a(t),
‘Therefore, lo has to be the fundamental period.
(2) The period of z(1) is the least common multiple of the periods of e/M26/7% and
TT, The period of 2/70 ig T/k and the period of 72/7" and T/L, Since
and 1 have no common factors, the leat common multiple of T/k and 7/1 is T.
8.44, ‘The only unknown FS coofcients are 07, a, a2, and a-2. Since z(t) is real, a; = al and
2 = ag. Since a i real, a = 0.1. Now, 2( is of the form
zit) = Ay conlunt) + Az cos(Zunt +0),
where wp = 21/6. From this we got
{t= 3) = Aycoolunt — Bun) + Arcosläunt + 8 Gun).
Now if we need x(() = ~2(¢~3), then Say and Guy should both be odd multiples of.
Clearly, this is impossible. Therefore, az = 0-1 = 0 and
2) = Ar cos(unt).
Nom, using Parseval's relation on Clue 5, we get
$ la? = la? +la-ı
1
2
‘Therefore, lay) = 1/2. Since ay is positive, we have ay = a-ı = 1/2. Therefore, =(t) =
cos(rt/3)-
107
3.47. Considering 2(¢) to be periodic with period 1, the nonzero PS coefficients of 2(0) are ay =
/2. IE we now consider 2(£) to be periodic with period 3, then the the nonzero FS
coefficients of z(¢) are by = b-g = 1/2.
348. (a) The FS coefficient of fn = na ae
(b) Using the results of part (a), the FS couficients of fa] — sin — 1} are given by
ag [675888 og
(©) Using the results of part (a), the FS eoeicients of fr) — 2{n — N/2] are given by
(A) Note that 2[x]+2[n-+N/2 has a period of N/2. The ES cooficients of x(n] +z[n—N/2)
are given by
for 0 < k < (N/2-1).
(0) The FS cooicients of {=n} are
CEE PH = a
(£) With N even the PS coefficients of (-1)"a{r) aro
EN peered "ann
(g) With N odd, the period of (~1)"z{n] is 2N. Therefore, the FS coccients are
E
ka
FREE à Yap APM HM
Note that for k odd in an integer and k — N is an even integer. Also, fork even,
RN is an odd integer and e-9-") = —1, ‘Therefore,
ir
wie
no
/2. IE we now consider 2(£) to be periodic with period 3, then the the nonzero FS
coefficients of z(¢) are by = b-g = 1/2.
348. (a) The FS coefficient of fn = na ae
(b) Using the results of part (a), the FS couficients of fa] — sin — 1} are given by
ag [675888 og
(©) Using the results of part (a), the FS eoeicients of fr) — 2{n — N/2] are given by
(A) Note that 2[x]+2[n-+N/2 has a period of N/2. The ES cooficients of x(n] +z[n—N/2)
are given by
for 0 < k < (N/2-1).
(0) The FS cooicients of {=n} are
CEE PH = a
(£) With N even the PS coefficients of (-1)"a{r) aro
EN peered "ann
(g) With N odd, the period of (~1)"z{n] is 2N. Therefore, the FS coccients are
E
ka
FREE à Yap APM HM
Note that for k odd in an integer and k — N is an even integer. Also, fork even,
RN is an odd integer and e-9-") = —1, ‘Therefore,
ir
wie
no
(0) Here,
sin = Sto] + CA)" eh)
or N gen,
ac lr
For odd,
age se
aa a
3.49. (a) The ES coeficiente are given by
yo
a woe sige
en Ener
= FE ae + E PT =0
En Ener
= 0 “fork even
(6) By adopting an approach similar to part (a), we may show that
ES E
a = À Len see ane
= 0 Inkemrez
(©) IE NM is an integer, we may generale the approach of part (a) to how that
A [Enemies eo].
ner = N/A and = km. Bram the above equation, it is cur that
ka et.
8.50. From Table 32, we know that if
a) Fon,
m
sin = Sto] + CA)" eh)
or N gen,
ac lr
For odd,
age se
aa a
3.49. (a) The ES coeficiente are given by
yo
a woe sige
en Ener
= FE ae + E PT =0
En Ener
= 0 “fork even
(6) By adopting an approach similar to part (a), we may show that
ES E
a = À Len see ane
= 0 Inkemrez
(©) IE NM is an integer, we may generale the approach of part (a) to how that
A [Enemies eo].
ner = N/A and = km. Bram the above equation, it is cur that
ka et.
8.50. From Table 32, we know that if
a) Fon,
m
me) ww
D nn 2
a Y TAN u
ry \/ ar + "| Tr art
a
Figure 83.02
(e) The de component of the input is 0. The de component of the output is 2/2.
3.09. The average energy per period is
fora Die = De
We want N such that
‘This implies that
Solving,
log. AB? +09)
Tige
and
EN (N-Dx
BN cw DE
3.64. (a) Due to linearity, we have
1) = Dedalo
(b) Let
a) a a(t) walt).
Mo, ee
2) = an) +) +)
Then,
ml) = Parto + ba" (0] + Ha (0 + be (01
OO)
12
D nn 2
a Y TAN u
ry \/ ar + "| Tr art
a
Figure 83.02
(e) The de component of the input is 0. The de component of the output is 2/2.
3.09. The average energy per period is
fora Die = De
We want N such that
‘This implies that
Solving,
log. AB? +09)
Tige
and
EN (N-Dx
BN cw DE
3.64. (a) Due to linearity, we have
1) = Dedalo
(b) Let
a) a a(t) walt).
Mo, ee
2) = an) +) +)
Then,
ml) = Parto + ba" (0] + Ha (0 + be (01
OO)
12
Valen 3.22
TZ = we
E
ap (4)
ed (Ck)
(ok)
. yu 4a, periodic with TEA
- gie gt) +p)
- 1-25(t) (1464!)
. gle t = =0 Tab \
per mai TE
quo AS) = des day Lae thr ait 6)
(050) time. Shun property Y 22 ew _ in k=0
-(- Ke
oe OK = + dr (Using Unennty. property, Table, 31)
260 K=O
Ln po
Qe = Dz bes o ko
3 lo y KT .
ptes Lp"
Cinkegrabion, property Table 3.1)
TZ = we
E
ap (4)
ed (Ck)
(ok)
. yu 4a, periodic with TEA
- gie gt) +p)
- 1-25(t) (1464!)
. gle t = =0 Tab \
per mai TE
quo AS) = des day Lae thr ait 6)
(050) time. Shun property Y 22 ew _ in k=0
-(- Ke
oe OK = + dr (Using Unennty. property, Table, 31)
260 K=O
Ln po
Qe = Dz bes o ko
3 lo y KT .
ptes Lp"
Cinkegrabion, property Table 3.1)
xt
(0%) T26 => we =
a, DS
aw a
LN — >t M2 413 =
6 z
g - gt
fs m gu 9216)
u = | co | (aa
&
zu + ri
qe wave giten in tebe 42
sur 16 te periodic S
4 a chifled by b=%
hw Tat & b=-2 (eis)
with Ti =
2 Sing for y 2H) y wi
. A he ong ved in abe à 42, à
- bbe ME & dk: Es eE
na AU nares 1
2j sin Me) RC») sin (HH)
- Using indegfaon- Pe aa 55
> = Les sin UE) sin (ET)
(0%) T26 => we =
a, DS
aw a
LN — >t M2 413 =
6 z
g - gt
fs m gu 9216)
u = | co | (aa
&
zu + ri
qe wave giten in tebe 42
sur 16 te periodic S
4 a chifled by b=%
hw Tat & b=-2 (eis)
with Ti =
2 Sing for y 2H) y wi
. A he ong ved in abe à 42, à
- bbe ME & dk: Es eE
na AU nares 1
2j sin Me) RC») sin (HH)
- Using indegfaon- Pe aa 55
> = Les sin UE) sin (ET)
1-5 20
> 0-2
c
© Won ly S24
3
- yle\
dey € 4 yl get)
| (x) (&) (dx)
pi nl Ion
+ Using, formula, for fooner coeflicints of pevodic Sqm
tabe 42 & Shiffing popety in ble 3.1
wave given in
y jr
= ne) & dez2sin TE > 3
kr kar
( Unsanty property)
be = Ce + de
de > Oe = be (Integrabion property.)
GK JA
>
a ar
> a [a (A eS 2500") 7]
2 ar ar
js “UY
a Caan ce? A
> 0-2
c
© Won ly S24
3
- yle\
dey € 4 yl get)
| (x) (&) (dx)
pi nl Ion
+ Using, formula, for fooner coeflicints of pevodic Sqm
tabe 42 & Shiffing popety in ble 3.1
wave given in
y jr
= ne) & dez2sin TE > 3
kr kar
( Unsanty property)
be = Ce + de
de > Oe = be (Integrabion property.)
GK JA
>
a ar
> a [a (A eS 2500") 7]
2 ar ar
js “UY
a Caan ce? A
T2 => wesfr
@ + fc s10-2860]
unit
xD = Gl +SCE-1) (ES)
| Using fourier coeffivents for SL) given in fable UT
y sitting pop, we ge:
La ae ely En
On = che u ae = 4 ED
xt)
(o - ea | xy hue equal y)
= A of pe tel
> ae be (pak) = ca sin AE) si (ET)
@ + fc s10-2860]
unit
xD = Gl +SCE-1) (ES)
| Using fourier coeffivents for SL) given in fable UT
y sitting pop, we ge:
La ae ely En
On = che u ae = 4 ED
xt)
(o - ea | xy hue equal y)
= A of pe tel
> ae be (pak) = ca sin AE) si (ET)
A xt
” : (0) Ted = wo = AP
Os = dyed
ns
Lo (09
a un
+ L $ L
Gen “4
lo) & (a) Gan be oblivud using fourier cefficunt for
pericdic square wave given in fale 42 Y Shitting
prets in tebe 21.
be = Sn (2) ee
tr
B Ca sin) e IS
kr
Cle = brete Cine propedly)
= sa (De JS de AS
KIT AT
” : (0) Ted = wo = AP
Os = dyed
ns
Lo (09
a un
+ L $ L
Gen “4
lo) & (a) Gan be oblivud using fourier cefficunt for
pericdic square wave given in fale 42 Y Shitting
prets in tebe 21.
be = Sn (2) ee
tr
B Ca sin) e IS
kr
Cle = brete Cine propedly)
= sa (De JS de AS
KIT AT
SO
Chapter 4 Answers
Ad. (0) Let s(t) = cult 1), Then the Fourier transform Xu) of) à
xq) = [eran
Ferena
= ens)
sta,
[XG] sas shown in Figure St.
(0) Lat 2) = AA) Then the Fourier transforms Ko) of (1) is
xu) = [tea
[erverar [arena
= eM (2-4 ju) + HH io
= ar)
x(a) ie as shown in Figure SH.
beet ' sige
2
Q "a a ù
Figure $4.1 w
ta. Win nät 4-2 Tan ee m 60 40
ao = [nenne
= Sienne
wis a sketch in Figure $42.
(6) The signal za) = u(—2 0) + u(t—2) ina shown in the figure below. Clery,
4 3
Laa 0 alt=} = 42-8042)
2
Chapter 4 Answers
Ad. (0) Let s(t) = cult 1), Then the Fourier transform Xu) of) à
xq) = [eran
Ferena
= ens)
sta,
[XG] sas shown in Figure St.
(0) Lat 2) = AA) Then the Fourier transforms Ko) of (1) is
xu) = [tea
[erverar [arena
= eM (2-4 ju) + HH io
= ar)
x(a) ie as shown in Figure SH.
beet ' sige
2
Q "a a ù
Figure $4.1 w
ta. Win nät 4-2 Tan ee m 60 40
ao = [nenne
= Sienne
wis a sketch in Figure $42.
(6) The signal za) = u(—2 0) + u(t—2) ina shown in the figure below. Clery,
4 3
Laa 0 alt=} = 42-8042)
2
Therefore,
in = [Twe-n-amera
a an
I a tél a Fg S42
{oe Dell
a A
igure $42
43. (a) The signal 2(t) = sin(2r + x/4) is periodic with a fondamental period of 7
This trancates to a fundamental frequency of uy = 2x. The nonzero Fourier series
cocficints of this signal may be found by writing fin the form
20 = 2 (otter em)
1
Lori
po
Beraten
E
“Therefore, the onzero Four sere coflets of (2) are
aL, a Leti
un poe, a e
Flom Serio 4.2, me know that for period signal, Fourier transform consiste of
à tain of mpc occuring ot hay Prtbermor, the area under each pls a 2
{ines the Borer serien cofiient ay. Therefore, for (0) the orrnponding Fourer
Atom X Gu) i ven by
Xiu) = Prado an) à rôle +)
= (re Ho ~ Be) = (rte élu + 2e)
(©) The signa 23) = 1006+ 5/8) is period with fundamenta) period of = 1/3.
‘hig tantes toa fundamental fequeney of uy = Gr. The nono Fourier seres
caen of tis sal may be found by wet a the orm
= 142 (eerie sorte
a = 14 (0! E )
1 1
- en y Leite
= 14 belting L
10
in = [Twe-n-amera
a an
I a tél a Fg S42
{oe Dell
a A
igure $42
43. (a) The signal 2(t) = sin(2r + x/4) is periodic with a fondamental period of 7
This trancates to a fundamental frequency of uy = 2x. The nonzero Fourier series
cocficints of this signal may be found by writing fin the form
20 = 2 (otter em)
1
Lori
po
Beraten
E
“Therefore, the onzero Four sere coflets of (2) are
aL, a Leti
un poe, a e
Flom Serio 4.2, me know that for period signal, Fourier transform consiste of
à tain of mpc occuring ot hay Prtbermor, the area under each pls a 2
{ines the Borer serien cofiient ay. Therefore, for (0) the orrnponding Fourer
Atom X Gu) i ven by
Xiu) = Prado an) à rôle +)
= (re Ho ~ Be) = (rte élu + 2e)
(©) The signa 23) = 1006+ 5/8) is period with fundamenta) period of = 1/3.
‘hig tantes toa fundamental fequeney of uy = Gr. The nono Fourier seres
caen of tis sal may be found by wet a the orm
= 142 (eerie sorte
a = 14 (0! E )
1 1
- en y Leite
= 14 belting L
10
Zn >
‘Therefore, the nonzero Fourie series colin of z(t) ae
1 1
jo gym Lee ht
pod any
rom Section 42, we know that fr periodi signals, the Fourier transform consist of
Rai of impulses occurring at ku, Furthermore, the area under each impulse is 27
Ginn the Fourier seres coeicint a. Therefore, fr (0), the corresponding Fourier
transform Kali) i given by
Kae) = Bro) + Zend un) +20. o + uv)
= tas) + ro Gx) + re o + 67)
44, (0) The inverse Fourier transform is
md = We) fee) rate An) + a nee
Ra + nett 4e
= 1 + (4/2) A = Lem)
(0) Te tve Feuer eo is
alt) = an Ko
| - (ara [nes + up
(rn Ue
(as (et)
445, From the give informan,
20) = ann [x00
| = ann [Funes
= (0120) [aerate
= zu et 2)
| “The signal 2() i zero when 3(¢ 5/2) is nomerointager multi of x. This gives
be 8
for ke, and k #0.
1m
‘Therefore, the nonzero Fourie series colin of z(t) ae
1 1
jo gym Lee ht
pod any
rom Section 42, we know that fr periodi signals, the Fourier transform consist of
Rai of impulses occurring at ku, Furthermore, the area under each impulse is 27
Ginn the Fourier seres coeicint a. Therefore, fr (0), the corresponding Fourier
transform Kali) i given by
Kae) = Bro) + Zend un) +20. o + uv)
= tas) + ro Gx) + re o + 67)
44, (0) The inverse Fourier transform is
md = We) fee) rate An) + a nee
Ra + nett 4e
= 1 + (4/2) A = Lem)
(0) Te tve Feuer eo is
alt) = an Ko
| - (ara [nes + up
(rn Ue
(as (et)
445, From the give informan,
20) = ann [x00
| = ann [Funes
= (0120) [aerate
= zu et 2)
| “The signal 2() i zero when 3(¢ 5/2) is nomerointager multi of x. This gives
be 8
for ke, and k #0.
1m
a
“Therefore the desired resul is
FE {On port otto) = PS
4.10, (0) We kaow fom Table 42 that
SE nectanglr fonction Yu) es Figure Su]
Three
(=) ¿la (1/25) Rectangulas funcion Y Gu) + Rectangular function Y Gs)
“This sa triangular function Yi (jo) as shown in the Figure $4.10,
7) m4) m
1 a
= err ja az 0
ert
a "
Bien 54:10
ring Tbe 4, ve may vite
(BY E xû sb)
“Thi nas chown in the figure above. XGu) may be expressed mathematically as
ie, 2£u<0
E À js, Sur
D
(b) Using Parsera' elation,
the
=
m
E
“Therefore the desired resul is
FE {On port otto) = PS
4.10, (0) We kaow fom Table 42 that
SE nectanglr fonction Yu) es Figure Su]
Three
(=) ¿la (1/25) Rectangulas funcion Y Gu) + Rectangular function Y Gs)
“This sa triangular function Yi (jo) as shown in the Figure $4.10,
7) m4) m
1 a
= err ja az 0
ert
a "
Bien 54:10
ring Tbe 4, ve may vite
(BY E xû sb)
“Thi nas chown in the figure above. XGu) may be expressed mathematically as
ie, 2£u<0
E À js, Sur
D
(b) Using Parsera' elation,
the
=
m
E
4.1, We know that
200 5 2x0), nay € 5468)
Time,
Gu) = FT{E(36) + M30) = 5XGZ)HUZ)
Now note that
Y (ju) = FT) + ME) = X ju) HG)
From this, we may write
vop=* 63) (5)
‘Using this in og. (+), we have ;
Ge) = rus)
and a
90 = Ius,
‘Therefore, As and B= 3
4.12, (a) From Psamplo 42 we know that
2
mt
‘Using the differentiation in frequency property we bare
el sie
ci
(0) Th duty property ttes that i
10 5 Gt)
se
Gt) gl
Ct
0 may we duality o wie
gime
Multiplying both sides by j, we obtain
a a
aoe inne
16
200 5 2x0), nay € 5468)
Time,
Gu) = FT{E(36) + M30) = 5XGZ)HUZ)
Now note that
Y (ju) = FT) + ME) = X ju) HG)
From this, we may write
vop=* 63) (5)
‘Using this in og. (+), we have ;
Ge) = rus)
and a
90 = Ius,
‘Therefore, As and B= 3
4.12, (a) From Psamplo 42 we know that
2
mt
‘Using the differentiation in frequency property we bare
el sie
ci
(0) Th duty property ttes that i
10 5 Gt)
se
Gt) gl
Ct
0 may we duality o wie
gime
Multiplying both sides by j, we obtain
a a
aoe inne
16
MM
413. (a) Taking the inverse Fourier transform of Xu), we obtain
Loy Leu
mot
Pt
‘Po signal z(t) is therefore constant gummed with two complex esponental whose
‘omental frequencies are 2r/5 rad/sec and 2 rad/sec. These teo complex expo
ink are not harmoniealy related. That is, the fundamental frequence of thse
Sole cmploe exponentials con never be integral multiples a common fundamen
{at rouen, Therefore, the signal is not periodic.
(6) Consider the signal 4 = 2) + AO, From the convolution property, we know tat
Yo) = XGu)H (jo). Aso, from M1), we know that
Hijo) = 282.
., where sa nonzero integer. Therefore,
(Ws
The funcion 1 (ju) is ero wen.
Ys) = Xul)
“This gives
Ly tye
MESA
“Therefor, y(t i a complex exponential summed with a constant. We know shat a
aces exponen in perdi. Adding a constant to a complex exponential doesnot
Sr pedi, Therefore, (0) wil be a signa with a fundamental frequency of
EN
(6) From the results of part (a) and (), we se Eat the answer is yes.
4.34. Taking the Fourier transform ofboth sides ofthe equation
FL HR = AT MUA,
ve obtain
2
laa)
“Taking the inverse Fourie trazo ofthe above equation
20 = Atul) = Aue)
ring Pasova's relation, we bave
[ora - nf CO
Using te fact ta [Go = 20, we have
[[rura=ı
| ví
413. (a) Taking the inverse Fourier transform of Xu), we obtain
Loy Leu
mot
Pt
‘Po signal z(t) is therefore constant gummed with two complex esponental whose
‘omental frequencies are 2r/5 rad/sec and 2 rad/sec. These teo complex expo
ink are not harmoniealy related. That is, the fundamental frequence of thse
Sole cmploe exponentials con never be integral multiples a common fundamen
{at rouen, Therefore, the signal is not periodic.
(6) Consider the signal 4 = 2) + AO, From the convolution property, we know tat
Yo) = XGu)H (jo). Aso, from M1), we know that
Hijo) = 282.
., where sa nonzero integer. Therefore,
(Ws
The funcion 1 (ju) is ero wen.
Ys) = Xul)
“This gives
Ly tye
MESA
“Therefor, y(t i a complex exponential summed with a constant. We know shat a
aces exponen in perdi. Adding a constant to a complex exponential doesnot
Sr pedi, Therefore, (0) wil be a signa with a fundamental frequency of
EN
(6) From the results of part (a) and (), we se Eat the answer is yes.
4.34. Taking the Fourier transform ofboth sides ofthe equation
FL HR = AT MUA,
ve obtain
2
laa)
“Taking the inverse Fourie trazo ofthe above equation
20 = Atul) = Aue)
ring Pasova's relation, we bave
[ora - nf CO
Using te fact ta [Go = 20, we have
[[rura=ı
| ví
416.
416.
‘Substituting the previously obtained expression for 2{) in the above equation, wo have
[wer pes awenuon
1
[werswer-wenan
nm
Sauve
We choos A to be intend of -VT because me Luo that x8 ion negativo
Sine a) is ol
stato) = ECO ES, Rx Go)
We are given that
IAT(RAXGO)) = Me™.
These,
20 + 0-0
3
Eolo tue
We also know that 2(@) = 0 fr # € O. This implis that (6) is zero for ¢ > 0. We may
conclude that
= Ale forte
‘Therefore,
O)
(a) We may writ
ay = Euro
EEC
een Esso
(b) Sine o(t) x an impulso train, its Fourier transform G(ja) is also an impulse tain,
From Table 42,
416.
‘Substituting the previously obtained expression for 2{) in the above equation, wo have
[wer pes awenuon
1
[werswer-wenan
nm
Sauve
We choos A to be intend of -VT because me Luo that x8 ion negativo
Sine a) is ol
stato) = ECO ES, Rx Go)
We are given that
IAT(RAXGO)) = Me™.
These,
20 + 0-0
3
Eolo tue
We also know that 2(@) = 0 fr # € O. This implis that (6) is zero for ¢ > 0. We may
conclude that
= Ale forte
‘Therefore,
O)
(a) We may writ
ay = Euro
EEC
een Esso
(b) Sine o(t) x an impulso train, its Fourier transform G(ja) is also an impulse tain,
From Table 42,
es ere
rs :
ave er {ét} cu]
IE we denote FT ($3!) by AG) then
us = uma ote Ea
= ¿aa
periodic
‘Using Table 42, we obtain
an(s ME
“Therefore, we may spesly (ja) over one period as
4 Wet
won [e Fes
Be X Ca). The function ¿XGu) will easly be rel and od
statement i falo,
swe may write
a(t) +21) El Xan) = Xi Xu
ET
Late Jo (EY
so
We ses that G(ju) i periodic with a period of . Using the multiplication property
Gu) may thus be viewed a a rpletion of AG) every 8 alee. Tit i obviously
4417, (a) Fro Tao 4, ve now tha a ea and od signal signal if) has a purely imaginary
From poner transform X (ju). Let us now consider the purely imaginary and
SA signal je(t). Using inne, we obtain the Fourier transform of this signal to
I. "Therefore, the given
(6) And Power transform corseponds o an od senal, wile an even Fourier transform
onda to an even signal The convoltion of an even Fourier transfer with an
ono may bo viewed in the Gime domain as a multiplication of am even and
| Ut anal Such a moltiphation wi always result in an od tie signal. The Fourier
dare ofthis ad signal will always De odd, Therefore, the given statement ls true,
418. Using Tube 42, we ses that the rectangular pubs (0) shown in Figure 54.18 has a Fourier
o Jaja) = sinfä)/o. Using the convolution property ofthe Fourier transoran
“the signal zl) e shown in Figure S38. Using te lis property we alo note that
rs :
ave er {ét} cu]
IE we denote FT ($3!) by AG) then
us = uma ote Ea
= ¿aa
periodic
‘Using Table 42, we obtain
an(s ME
“Therefore, we may spesly (ja) over one period as
4 Wet
won [e Fes
Be X Ca). The function ¿XGu) will easly be rel and od
statement i falo,
swe may write
a(t) +21) El Xan) = Xi Xu
ET
Late Jo (EY
so
We ses that G(ju) i periodic with a period of . Using the multiplication property
Gu) may thus be viewed a a rpletion of AG) every 8 alee. Tit i obviously
4417, (a) Fro Tao 4, ve now tha a ea and od signal signal if) has a purely imaginary
From poner transform X (ju). Let us now consider the purely imaginary and
SA signal je(t). Using inne, we obtain the Fourier transform of this signal to
I. "Therefore, the given
(6) And Power transform corseponds o an od senal, wile an even Fourier transform
onda to an even signal The convoltion of an even Fourier transfer with an
ono may bo viewed in the Gime domain as a multiplication of am even and
| Ut anal Such a moltiphation wi always result in an od tie signal. The Fourier
dare ofthis ad signal will always De odd, Therefore, the given statement ls true,
418. Using Tube 42, we ses that the rectangular pubs (0) shown in Figure 54.18 has a Fourier
o Jaja) = sinfä)/o. Using the convolution property ofthe Fourier transoran
“the signal zl) e shown in Figure S38. Using te lis property we alo note that
and
1
7
ding the two above equations, me obtain
1
IR
aa le
“ey
ana)?
alt +1)+ junte 1) £4 co) (EEE
‘The signal A(t) i as shown in Figure S418. We note that h() has the given Fourier
transform HG),
Puto
x0
Ne
Figure $4.18
Mathematically AE) may be expresed as
von} att
o
4.29, We know that
Sur,
Wea
ETES
= seis?
otherwise
1
7
ding the two above equations, me obtain
1
IR
aa le
“ey
ana)?
alt +1)+ junte 1) £4 co) (EEE
‘The signal A(t) i as shown in Figure S418. We note that h() has the given Fourier
transform HG),
Puto
x0
Ne
Figure $4.18
Mathematically AE) may be expresed as
von} att
o
4.29, We know that
Sur,
Wea
ETES
= seis?
otherwise
qm TT
Since, HG) =1/0 + ju), we have
XG)
a=
44.20, From the answer to Problem 3:20, we know that the frequency response of the circuit Is
node rar
rei thi up ito pat factions, we may veto
a a
alate réel
ld in Table 42, wo obtain the Fourier ransom
He)
Using the Fourier transform pais provi
of Hu) tobe Ñ 4 E
De q Den acido e,
Sinplifyng,
4.21, (a) The given signal is
1
eh)
7 de.
“Therefore,
7 oa
Maria Maar
(0) The given signal is
ade + inkl)
40
We have
1/25
e
Ft sin(atul®) Dx) =
HE
Ao,
Mai
LR AR
anto = Main 2u(-8) = =) ED Kae) = Klo) = 3
Si
u
Go) = Kibo Ke) = TT TRIER
1m
Since, HG) =1/0 + ju), we have
XG)
a=
44.20, From the answer to Problem 3:20, we know that the frequency response of the circuit Is
node rar
rei thi up ito pat factions, we may veto
a a
alate réel
ld in Table 42, wo obtain the Fourier ransom
He)
Using the Fourier transform pais provi
of Hu) tobe Ñ 4 E
De q Den acido e,
Sinplifyng,
4.21, (a) The given signal is
1
eh)
7 de.
“Therefore,
7 oa
Maria Maar
(0) The given signal is
ade + inkl)
40
We have
1/25
e
Ft sin(atul®) Dx) =
HE
Ao,
Mai
LR AR
anto = Main 2u(-8) = =) ED Kae) = Klo) = 3
Si
u
Go) = Kibo Ke) = TT TRIER
1m
(6) Using the Fourier transform analysis equation (4.9) we have
Dring, sine ino
XGu)
| am un |
l zit) = (1/26 Hude) — (1/2 Me Ml) |
“Therefore,
15 un.
KU) Tee BER
4) We have
o { Lo bier
0, bom
A
. Mieze
D 4, xatju) = ee
20 = 210) E% Xe) = F(X) Kol)
| These,
E e, Men
5 E <-+
| as (as: r<w<ie |
ñ o, there
ll (6) Using the Four traer nas o. (4) vo bain
i
200 fee |
N wu Ea
|! a= DM,
cn
f 20 = (0 +0) |
if ‘Therefore, |
FS tte bee
XG) Xi
Dring, sine ino
XGu)
| am un |
l zit) = (1/26 Hude) — (1/2 Me Ml) |
“Therefore,
15 un.
KU) Tee BER
4) We have
o { Lo bier
0, bom
A
. Mieze
D 4, xatju) = ee
20 = 210) E% Xe) = F(X) Kol)
| These,
E e, Men
5 E <-+
| as (as: r<w<ie |
ñ o, there
ll (6) Using the Four traer nas o. (4) vo bain
i
200 fee |
N wu Ea
|! a= DM,
cn
f 20 = (0 +0) |
if ‘Therefore, |
FS tte bee
XG) Xi
OM
(6) Using the Fourier transform analysis eq. (49) we obtain
1 „ch meh
Us) +S
5
G) 24 is periodie with period 2. Therefore,
XGu)=" Y Kükn)dle kn),
where (ju) is the Fourier transform of one period of le). That is
u 0 ME
CHARME
(6) The Fourier transform syathosis eq. (48) may be written as
GUA A a
a=
From the given figure we have
= [oes tan
tba to
(a) (0) = Hit + Joso)
(6) Using the Fours trnsform ato equation (48),
SH , int = int
tz
428. For the given signal zo), we use the Fourier transform analysis u. (4.8) to erate the
‘corresponding Pourer transform 2
Solio) =
(We know that
EEE
Using the Iinearity and time reversal properties of the Fourier transform we have
Pape
Abu) =
ob) + Kol fi) =
16
(6) Using the Fourier transform analysis eq. (49) we obtain
1 „ch meh
Us) +S
5
G) 24 is periodie with period 2. Therefore,
XGu)=" Y Kükn)dle kn),
where (ju) is the Fourier transform of one period of le). That is
u 0 ME
CHARME
(6) The Fourier transform syathosis eq. (48) may be written as
GUA A a
a=
From the given figure we have
= [oes tan
tba to
(a) (0) = Hit + Joso)
(6) Using the Fours trnsform ato equation (48),
SH , int = int
tz
428. For the given signal zo), we use the Fourier transform analysis u. (4.8) to erate the
‘corresponding Pourer transform 2
Solio) =
(We know that
EEE
Using the Iinearity and time reversal properties of the Fourier transform we have
Pape
Abu) =
ob) + Kol fi) =
16
(i) We know that
salt) = zal) = ol)
‘sing the nest and time reversal propertios of the Fourier transform we have
ab) = Kol) - Kal-je) =
ad = + der)
| (ui) We ata
| ‘sing the ty ad im tin propano th Fro transform we ve
Loto)
Ha = Xoju) + Nol fu) Te
i (iv) We know that
i al) = tal)
Using the differentiation in frequency property
| Haba) = GL Xe)
i ‘Therefore,
Hl Latte m= jure er
i
H
Fe Te
| 424. (a) (i) For Re(X(ju)) to be 0, ie signal 2() must be sea and odd. ‘Therefore, signals
| in Ares (8) and () have this property
i (i) For Zm{X(ju)} to be D, the signal 2(¢) must be real and even. Therefore, signal
| in figures () and (1) have this property.
(Gi) For ther to exist a real a such that cs (ju) is real, we require that 2(¢-+a) be
‘real and even signal. Therefore, signals in gues (9) (), (), and (F) have this
property:
| (Gs) For this condition o be rue, 2(0) =
‘Therefore, signals in figure (a), (0), (0s
| (8), ad (D bave his property. -
| 0) For thisoniton to be ru the derivative of() has to beer at t= 0. Therefore,
y signals in figures (), (), (0), and (0) have this property |
(i) Por this t be tre, the signa 2{) has to be periodic. Only the signal in figure (a) |
bas this property. |
(6) Fora signal to sata only properties (), (iv), and (x), it must be real and od and
x)=0, 2(0)=0.
‘The signal shown below i an example ofthat.
salt) = zal) = ol)
‘sing the nest and time reversal propertios of the Fourier transform we have
ab) = Kol) - Kal-je) =
ad = + der)
| (ui) We ata
| ‘sing the ty ad im tin propano th Fro transform we ve
Loto)
Ha = Xoju) + Nol fu) Te
i (iv) We know that
i al) = tal)
Using the differentiation in frequency property
| Haba) = GL Xe)
i ‘Therefore,
Hl Latte m= jure er
i
H
Fe Te
| 424. (a) (i) For Re(X(ju)) to be 0, ie signal 2() must be sea and odd. ‘Therefore, signals
| in Ares (8) and () have this property
i (i) For Zm{X(ju)} to be D, the signal 2(¢) must be real and even. Therefore, signal
| in figures () and (1) have this property.
(Gi) For ther to exist a real a such that cs (ju) is real, we require that 2(¢-+a) be
‘real and even signal. Therefore, signals in gues (9) (), (), and (F) have this
property:
| (Gs) For this condition o be rue, 2(0) =
‘Therefore, signals in figure (a), (0), (0s
| (8), ad (D bave his property. -
| 0) For thisoniton to be ru the derivative of() has to beer at t= 0. Therefore,
y signals in figures (), (), (0), and (0) have this property |
(i) Por this t be tre, the signa 2{) has to be periodic. Only the signal in figure (a) |
bas this property. |
(6) Fora signal to sata only properties (), (iv), and (x), it must be real and od and
x)=0, 2(0)=0.
‘The signal shown below i an example ofthat.
KE 1 nm eal and even signal, Therefore, Y (jo) l o real and
=’. Alb, since Yu) = XU), we know that
425. (a) Note that yl
‘even, This implies that <¥ Gs)
por
om 7
x¢0) = [rabat =?
(0 Wei -
[xm arte
(ot G0 ue Cod
Lo -3<i<-t
co me
“Then the given integral is
[run us Ae «sho =
(om - -
[rrootae=tef pra =2e
(tine nen RK eo) wi AA
RUT See en
1 pa
a
A
Figure $4.25,
4.26. (a) 6) We have
1
vn xe = [etl [ets
lus vos tn
dejo da Br
15
=’. Alb, since Yu) = XU), we know that
425. (a) Note that yl
‘even, This implies that <¥ Gs)
por
om 7
x¢0) = [rabat =?
(0 Wei -
[xm arte
(ot G0 ue Cod
Lo -3<i<-t
co me
“Then the given integral is
[run us Ae «sho =
(om - -
[rrootae=tef pra =2e
(tine nen RK eo) wi AA
RUT See en
1 pa
a
A
Figure $4.25,
4.26. (a) 6) We have
1
vn xe = [etl [ets
lus vos tn
dejo da Br
15
‘Taking the inverse Fourier transform we obtain
wo + Let)
“=
(i) We have
ros = num [al a
. am, 04) 0/0, WN
dijo Res Erden
“Taking the inverse Fourier transform we obtain
1
wo = tera eto Jen uo,
(4) We have
YG) = Xola)
Taking the inverse Fourier transform, we obtain
w= je
(b) By direct convolution of (4) with AE) we obtain
o, tea
POI reess
Taking the Fourier transform of y(t),
[pro] an
427. (a) The Fourier transform XGio) la
20 = [Linea fera
ia
tha
eyes
16
wo + Let)
“=
(i) We have
ros = num [al a
. am, 04) 0/0, WN
dijo Res Erden
“Taking the inverse Fourier transform we obtain
1
wo = tera eto Jen uo,
(4) We have
YG) = Xola)
Taking the inverse Fourier transform, we obtain
w= je
(b) By direct convolution of (4) with AE) we obtain
o, tea
POI reess
Taking the Fourier transform of y(t),
[pro] an
427. (a) The Fourier transform XGio) la
20 = [Linea fera
ia
tha
eyes
16
(6) The Fourier series coefcents os are
ae Ase
fora Love
(doy nr
mp meni) en we
ara
PATES
where 7
428. (a) From Table 42 we ow that
= D ment El, Pa) 2 É exile ka
roa: this,
La HG} = E 4X Glo ~ hon)
Ye)
(6) The spectra ae sete in Figure $4.28,
420. () We have
dm Lu A der = Xe te.
property we know that
rom the time shit
sl) =)
(i) Weave
so = KGa PROM = XG
rom the time siting property we know that
a) = 249).
(Gi) We have
alia) = XGA = XG.
From the conjugation and time reversal properties we know that
=
Since 2(0 is red, ul) = (1
ur
|
ae Ase
fora Love
(doy nr
mp meni) en we
ara
PATES
where 7
428. (a) From Table 42 we ow that
= D ment El, Pa) 2 É exile ka
roa: this,
La HG} = E 4X Glo ~ hon)
Ye)
(6) The spectra ae sete in Figure $4.28,
420. () We have
dm Lu A der = Xe te.
property we know that
rom the time shit
sl) =)
(i) Weave
so = KGa PROM = XG
rom the time siting property we know that
a) = 249).
(Gi) We have
alia) = XGA = XG.
From the conjugation and time reversal properties we know that
=
Since 2(0 is red, ul) = (1
ur
|
run ©
| 2200) jan e
ii 60 ps o Messe)
! SANA a
Figure 542877
rear = jo
From the conjugation, time several, and une shifting properties, we now that
240 = ta)
Since z(t) is real, zul) = 24 4
430. (8) We know that
ao) = cost & WG) =
lo +5(0-+8)
and 1
909 = 200 cost EB Glin) = Fe (Xin) » WUD)
1 ‘Therefore,
| Sw)
1 1
Axt) + XG + D)
i us
| 2200) jan e
ii 60 ps o Messe)
! SANA a
Figure 542877
rear = jo
From the conjugation, time several, and une shifting properties, we now that
240 = ta)
Since z(t) is real, zul) = 24 4
430. (8) We know that
ao) = cost & WG) =
lo +5(0-+8)
and 1
909 = 200 cost EB Glin) = Fe (Xin) » WUD)
1 ‘Therefore,
| Sw)
1 1
Axt) + XG + D)
i us
u ere
Since Glu) is as shown in Figure 4.30, is clear From the above equation that Xu)
is as shown in the Figure $4.30.
ie REG
1 TO
‘Therefore,
(0) Xin) e as shown in Figur $4.30.
as. (0) Wehe
at = coat EB XG) = lo +1) D
() Wehe
wu =o = E +
‘Therefore,
Vu) = XGu)th Ge) = Flle +1) élu — 1)
Tang the inverse Fourier transform, we obtain
v(t) = sin
(8) Weave _ .
daft) 280) +) EB Ho) = 2+ 7
‘Therefore,
| Y (ju) = X Ga) Go) = Tole +1) o 1)
| ‘Pking the inverse Fourie transform, we obtain
109 = sin.
Since Glu) is as shown in Figure 4.30, is clear From the above equation that Xu)
is as shown in the Figure $4.30.
ie REG
1 TO
‘Therefore,
(0) Xin) e as shown in Figur $4.30.
as. (0) Wehe
at = coat EB XG) = lo +1) D
() Wehe
wu =o = E +
‘Therefore,
Vu) = XGu)th Ge) = Flle +1) élu — 1)
Tang the inverse Fourier transform, we obtain
v(t) = sin
(8) Weave _ .
daft) 280) +) EB Ho) = 2+ 7
‘Therefore,
| Y (ju) = X Ga) Go) = Tole +1) o 1)
| ‘Pking the inverse Fourie transform, we obtain
109 = sin.
(Gi) We have
s(t) = 2e tu) € a) =
2.
Tr
‘Therefor,
Fo) = Xu) th i) = H+ 1) = Ble — D)
‘Taking the inverse Fourier transform, wo obtain
u(t) = ain)
(6) An LTT system with
Auf)
Lim +00)
wll have the same response to 2) = ons). We can fad other such Impulse responses
‘y suitably sealing and linearly combining ha), h(t), and f(t).
4.82, Note that AO = fe), where
go E
“The Fourier transform Hg) of h(t) sas shown in Figure $4.32.
From the above Sigur iti lear that f(t i the impulse response of an ideal lowpass
filter whose passband sin tbe range Jl <4 Therefore, h(i the Impulse response of an
‘dea lowpas filter sited by one tothe right. Using the sift propery,
a Me
(0) We have
Xj ju) metio — 6) + nef +6).
D tt
Yi) = Xiu) Hin) = 0 + mld) = 0.
‘This rales equivalent to saping that X (jo) is ero in the passband of A (ju).
(6) We have .
Xabi) [Bere Tak) + »)]
‘Therefore,
Valin) = Xe) = F [0/28 9) — Blo 4 A
Mis implis that
lt) = Zune 0)
‘We may have obtained the sre ret by noting that only the sinsoid with frequency
in Xola) ls in e passband of Hu).
10
s(t) = 2e tu) € a) =
2.
Tr
‘Therefor,
Fo) = Xu) th i) = H+ 1) = Ble — D)
‘Taking the inverse Fourier transform, wo obtain
u(t) = ain)
(6) An LTT system with
Auf)
Lim +00)
wll have the same response to 2) = ons). We can fad other such Impulse responses
‘y suitably sealing and linearly combining ha), h(t), and f(t).
4.82, Note that AO = fe), where
go E
“The Fourier transform Hg) of h(t) sas shown in Figure $4.32.
From the above Sigur iti lear that f(t i the impulse response of an ideal lowpass
filter whose passband sin tbe range Jl <4 Therefore, h(i the Impulse response of an
‘dea lowpas filter sited by one tothe right. Using the sift propery,
a Me
(0) We have
Xj ju) metio — 6) + nef +6).
D tt
Yi) = Xiu) Hin) = 0 + mld) = 0.
‘This rales equivalent to saping that X (jo) is ero in the passband of A (ju).
(6) We have .
Xabi) [Bere Tak) + »)]
‘Therefore,
Valin) = Xe) = F [0/28 9) — Blo 4 A
Mis implis that
lt) = Zune 0)
‘We may have obtained the sre ret by noting that only the sinsoid with frequency
in Xola) ls in e passband of Hu).
10
ee rrr
(e) Ve ae
e las
ae) {Shee
Yale) = X GG = Rule”.
‘This imp Eat
sie)
mt = ste is
We may have obtained the same result by noting that (ju) Ties entirely in the
passband of Hu)
(2) XG) i a shown in Figure 54:32.
o
f
Io
Eo
ker
hs —,
Figure 54.32
“Therefore,
Valin) = KG) © Klee.
sin(ate DIV
A
Wie may have obtained the same result ‘By noting that Xela) es entirely in the
passband of Gu).
4488. (a) Tung the Fourier transform ofboth sos ofthe given dtforential equation,
Ya) 2 _
Me) = Ga) © =F Bo TS
“This implies that
=
wit)
ve obtain
Using partial facion expansion, we obtain
Em
jara jara
HG)
1
(e) Ve ae
e las
ae) {Shee
Yale) = X GG = Rule”.
‘This imp Eat
sie)
mt = ste is
We may have obtained the same result by noting that (ju) Ties entirely in the
passband of Hu)
(2) XG) i a shown in Figure 54:32.
o
f
Io
Eo
ker
hs —,
Figure 54.32
“Therefore,
Valin) = KG) © Klee.
sin(ate DIV
A
Wie may have obtained the same result ‘By noting that Xela) es entirely in the
passband of Gu).
4488. (a) Tung the Fourier transform ofboth sos ofthe given dtforential equation,
Ya) 2 _
Me) = Ga) © =F Bo TS
“This implies that
=
wit)
ve obtain
Using partial facion expansion, we obtain
Em
jara jara
HG)
1
Taking the inverse Fourier transform,
A = eus) eu,
(b) For the given signal 2(), we have
1
260 = oF
‘Therefore,
N Yo) = KUH) = y
CAGA E
Using partial fraction expansion, wo obtain
A
»r
‘Taking the inverse Fourier transform,
Lau = Heerttu(g à eV
ut) = Lett) = Herat) + Pea)
(6) Taking the Fourier transform of both sides f he given differential uation, we obtain
= Ye) 20
E © = + iio tT
|
| sing partial econ expansion, we obtain
|
|
TE LERNTE
HG) = 2+
“Taking the ve Fourier tracto,
At) = 260) = VA + Be Mu) = VL JN),
| (9) We have
u Yo +4
KG © Pure
Crow-multpying and taking bo inverse Fourie tanfor, wo obtain
do
LU 52019 cg a
I En oye) E + a0,
(0) We hae Doa
E E
“Taking o Ines Fourier tner we ain,
A =
A = eus) eu,
(b) For the given signal 2(), we have
1
260 = oF
‘Therefore,
N Yo) = KUH) = y
CAGA E
Using partial fraction expansion, wo obtain
A
»r
‘Taking the inverse Fourier transform,
Lau = Heerttu(g à eV
ut) = Lett) = Herat) + Pea)
(6) Taking the Fourier transform of both sides f he given differential uation, we obtain
= Ye) 20
E © = + iio tT
|
| sing partial econ expansion, we obtain
|
|
TE LERNTE
HG) = 2+
“Taking the ve Fourier tracto,
At) = 260) = VA + Be Mu) = VL JN),
| (9) We have
u Yo +4
KG © Pure
Crow-multpying and taking bo inverse Fourie tanfor, wo obtain
do
LU 52019 cg a
I En oye) E + a0,
(0) We hae Doa
E E
“Taking o Ines Fourier tner we ain,
A =
(0 We have
1
4 rr |
herir,
| 1
Ys) = KG) Ge) = a
(is) = XGA GE) = ETI
Finding the pst faction expansion of Y ja) an taking the inverse Fuer ene
om, ;
male fet,
35. (a) From the given information, |
|
|
|
Also,
Also,
> Maza
(b) fa =1, wo have
HG = 1 jo) = tan tu
Therefore,
DURE
=
4:36. (a) The frequency response is
| y HUE) = 640)
MGV KG) © Wr GOI
(b) Finding the partial faction expansion of answer in part (a) and taking its inverse
Fourier transform, we obtain u
“real
(6) We bare
Otte.
rie
rosal and taking the averse Fourier tano, we bin
Ey D nt) ;
153 1
A UUUTUTUTTT{UU!
1
4 rr |
herir,
| 1
Ys) = KG) Ge) = a
(is) = XGA GE) = ETI
Finding the pst faction expansion of Y ja) an taking the inverse Fuer ene
om, ;
male fet,
35. (a) From the given information, |
|
|
|
Also,
Also,
> Maza
(b) fa =1, wo have
HG = 1 jo) = tan tu
Therefore,
DURE
=
4:36. (a) The frequency response is
| y HUE) = 640)
MGV KG) © Wr GOI
(b) Finding the partial faction expansion of answer in part (a) and taking its inverse
Fourier transform, we obtain u
“real
(6) We bare
Otte.
rie
rosal and taking the averse Fourier tano, we bin
Ey D nt) ;
153 1
A UUUTUTUTTT{UU!
as. (a) Noe
a(t) = a(t) + zu),
he
el a
We ina o) a
Abo)
Using the convolution property we have
XG) = Xo)X Gi)
(0) The signal (2 is as shown in Figure S437
A
A fs e
ES
wo
| y e
| ES Figure $4.37 ur
(6) One possible oie of of) is as shown in Figure 4.37.
i (a) Note that
(ju) = XGa)
"This may also be weiten as
KGa) = E Y xGn mito KD)
Cleary, isis posible only if
Gljnk/2) = XGnk/2
E sate Ap) 060,
+
Gurke).
a(t) = a(t) + zu),
he
el a
We ina o) a
Abo)
Using the convolution property we have
XG) = Xo)X Gi)
(0) The signal (2 is as shown in Figure S437
A
A fs e
ES
wo
| y e
| ES Figure $4.37 ur
(6) One possible oie of of) is as shown in Figure 4.37.
i (a) Note that
(ju) = XGa)
"This may also be weiten as
KGa) = E Y xGn mito KD)
Cleary, isis posible only if
Gljnk/2) = XGnk/2
E sate Ap) 060,
+
Gurke).
u: A
a.
4.48.
446.
Therefore, an LM system with Impulse response hf) = 46(@) may be und to obtain al)
tom st),
(4) Tag the Fourier transform of Both sides ofthe given differential equation, we bave
Vs) + jo] = XGullZG) ~ 1
Since, Zu) = ph +, we obtain from the above equation
(D) Finding the partial fraction expansion of Mu) and then taking e inverse Fourier
transform we obtain
ao)
to + eal
tut) + Luto.
We have
HO ==) M0) + ¥Gu) = KUH).
rom Parseval’s relation the total energy in pl) is
5 = [worn roots
IS
[tao ra
i Lega
= Lixtsalta + LG rs
For real z(t), |X(-ju0)l* = LX Gon)? Therefore,
ela
Let g(t) be the response of Hu) tos) coswnt. Lat g(t) bo the response of HU) to
2 einust. Then, wth reference to Figure 4 10,
aft) = atte = at) cnet + (Dino
and
st) =) + al
Also,
10) = a(t) = eos ain + SO
‘Therefore,
RASO) = su count + lO si,
"Tiss exactly what Figure P446 implements
19
a.
4.48.
446.
Therefore, an LM system with Impulse response hf) = 46(@) may be und to obtain al)
tom st),
(4) Tag the Fourier transform of Both sides ofthe given differential equation, we bave
Vs) + jo] = XGullZG) ~ 1
Since, Zu) = ph +, we obtain from the above equation
(D) Finding the partial fraction expansion of Mu) and then taking e inverse Fourier
transform we obtain
ao)
to + eal
tut) + Luto.
We have
HO ==) M0) + ¥Gu) = KUH).
rom Parseval’s relation the total energy in pl) is
5 = [worn roots
IS
[tao ra
i Lega
= Lixtsalta + LG rs
For real z(t), |X(-ju0)l* = LX Gon)? Therefore,
ela
Let g(t) be the response of Hu) tos) coswnt. Lat g(t) bo the response of HU) to
2 einust. Then, wth reference to Figure 4 10,
aft) = atte = at) cnet + (Dino
and
st) =) + al
Also,
10) = a(t) = eos ain + SO
‘Therefore,
RASO) = su count + lO si,
"Tiss exactly what Figure P446 implements
19
sar
(0) We have
n= MOEA
Since hf) is causal, the non-sero portions of At) and AL) overlap only a £ = 0
‘Therefore,
o 1<0
xo -{ Mo 10 cuero
Dhl, 1>0
Also, from Table 4.1 we have
nd) Reto).
Given Re{HGa), we can obtain fe). From Ae) we can recover h(t) (and conse-
quently (ja) by using ea. (ATI). Therefore, HG) is completely specified by
Re(UGu)
wi
qa Lett
He
Roll) = cos
then,
ni ne hun,
Therefore rom en, ($4.71),
HO = 56-1)
(©) We hare
nay = AO ACO
Since Al) is causal, he non-zero portions of ME) and (6) overlap only at t= 0 and
a) vil be aro at £= 0; Therefore,
o t<o
Ae) = 4 unknown, 420 (ara
MO
‘Also, from Table 4. wo have
10 Tm (HG).
Given Zm{HGu), we can obtain Al). From hal), we can recover h(t) except for
10 by wing oy, (S471). If here are no singularities in A) at £=0, then His)
can be reoweted from A(t) even if (0) is unknown. Therefore Hu) is completly
‘specie by Zo (2) inthis ease
(0) We have
n= MOEA
Since hf) is causal, the non-sero portions of At) and AL) overlap only a £ = 0
‘Therefore,
o 1<0
xo -{ Mo 10 cuero
Dhl, 1>0
Also, from Table 4.1 we have
nd) Reto).
Given Re{HGa), we can obtain fe). From Ae) we can recover h(t) (and conse-
quently (ja) by using ea. (ATI). Therefore, HG) is completely specified by
Re(UGu)
wi
qa Lett
He
Roll) = cos
then,
ni ne hun,
Therefore rom en, ($4.71),
HO = 56-1)
(©) We hare
nay = AO ACO
Since Al) is causal, he non-zero portions of ME) and (6) overlap only at t= 0 and
a) vil be aro at £= 0; Therefore,
o t<o
Ae) = 4 unknown, 420 (ara
MO
‘Also, from Table 4. wo have
10 Tm (HG).
Given Zm{HGu), we can obtain Al). From hal), we can recover h(t) except for
10 by wing oy, (S471). If here are no singularities in A) at £=0, then His)
can be reoweted from A(t) even if (0) is unknown. Therefore Hu) is completly
‘specie by Zo (2) inthis ease
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