LTI -LaplacesignalsandsystemEEEBEmaterial.pdf

Laplace transform

1. CONTINUOUS
TIME
LTI SYSTEMS
Dr.R.Helen

2. Laplace Tranfromation
•we saw that for a continuous-time LTI system with
impulse response h(t),the output y(t) of the system
to the complex exponential input of the form e
st
is

6. a +3)

144513 "(s+

X(s)=

N

Re(s) > -1

E. ku

N

11. Transfer function

A. The System Function:

In Sec. 2.2 we showed that the output y(t) of a continuous-time LTI system equals the
convolution of the input x(t) with the impulse response A(t); that is,

y(t) =x(1)* A(t) (3.35)
Applying the convolution property (3.23), we obtain
Y(s) = X(s)H(s) (3.36)

where Y(s), X(s), and H(s) are the Laplace transforms of y(t), x(t), and A(t), respec-
tively. Equation (3.36) can be expressed as

고 (3.37)

50) (の =xO # hit)

Xs) Vis)=X(s)H(s)
Hs)

Fig. 3-7 Impulse response and system function.

14. Transfrom circuits

1. Signal Sources:
u(t) V(s) (1) —1I(s)
where v(t) and ¿(1) are the voltage and current source signals, respectively.
2. Resistance R:
v(t) = Ri(t) <> V(s) = RI(s)

3. Inductance L:
di(s)
u) = LE Vs) =sLI(s) — Li”) (3.50)
The second model of the inductance L in Fig. 3-10 is obtained by rewriting Eq. (3.50) as
‘i 1 1 ロー
ie) >15) = VS) + 2007) (3.51)
4. Capacitance C:
| の (のり
ir) = CS Hs) =sCV(s) - Co(0”) (3.52)

The second model of the capacitance C in Fig. 3-10 is obtained by rewriting Eq. (3.52) as

1 1
VL) = HS) + 0") (3.53)

18. eu + e*u( 一 7)

SS
E

19. Find inverse LT

X(t)=?
X(t)=?

20. Find Inverse LT


C1=1,c2=1

21. C1=1,c2=(-1/2)(1+j), C3= =(-1/2)(1-
j)

22. The output y(t) of a continuous-time LTI system is found to be 2e~“u(t) when the
input x(t) is u(t),

(a) Find the impulse response A(t) of the system.

(b) Find the output y(#) when the input x(1) is e 'u(t).
ADS ulby
gb se à eZ (も)

Take LT
Del(t) LT =1
(925 Ys Reldro 1 leo
2
Nod: 43 Pes 7-2
mod: 7 の = as
X(S Srs
> 2Stb—b
S+3
= 2.0S#D -6

S43