Signals and System UNIT1 Lecture Material

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1. UNIT -1: SIGNALS AND SYSTEMS

2. 1.Signals, Systems & Communications - B.P. Lathi, 2013, BSP.
2.Signals and Systems - A.V. Oppenheim, A.S. Willsky and S.H.
Nawabi, 2 Ed.
3.Peyton Z. Peebles, “Probability, Random Variables & Random
Signal Principles”, 4th Edition, TMH, 2002.
PRAKASAM TUTORIALS:
https://www.youtube.com/playlist?list=PLNb3wUjRD8AkbD
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TEXT BOOKS

3. •Signals and Systems Subject is designed to cover analysis, types,
convolution, sampling and operations performed on signals. It also
describes various types of systems.
•Asignalcanbedefinedinoneofthefollowingways
➢Anything that conveys information can be termed as a signal.
➢A signal can also be defined as a single valued function of one or
more independent variables which has some information.
➢A signal may also be defined as any physical quantity that varies
with time or any other independent variable.

4. An electronic communications system usingelectronic signals

5. •Anything that carries information can be called a signal. Signals
constitute an important part of our daily life.
•Signal is a function of time. Generally signals are represented by
x(t), y(t), z(t), f(t), ……..
•A Signal is defined as a single- valued function of one or more
independent variables which contain some information.
What is a Signal?

6. •A signal may also be defined as any physical quantity that varies
with time, space or any other independent variable.
•A signal may be represented in time domain or frequency domain.
Some common examples of a signal are human speech, electric
current, electric voltage, etc.
•Examples:
➢Human speech is a familiar example of a signal.
➢Electric current and voltage are also examples of signals.
➢voice signal, video signal, signals on telephone wires, EEG,
ECG etc.

8. • Human Voice – best example
• Ear recognizes sounds 20KHz or less
• AM Radio – 535KHz to 1605KHz
• FM Radio – 88MHz to 108MHz

9. •A signal can be a function of one or more independent variables. A
signal can be a function of time, temperature, position, pressure,
distance etc.
•If a signal depends on only one independent variable, it is called a
one-dimensional signal, and if a signal depends on two independent
variables, it is called a two-dimensional signal.

10. •System is a device or combination of devices, which can operate on
signals and produces corresponding response.
•Input to a system is called as excitation and output from it is called
as response.
•Example: Communication System
What is a System?

11. CLASSIFICATION OF SIGNALS
•Basically, there are two types of signals:
1. Continuous-time signal
2. Discrete-time signal

12. Continuous-time signals
•Continuous time signals are defined for all values of time ‘t’ and it is
represented by x(t).
•Commonly found in the physical world
•ex. Human speech

13. •The temperature recorded over an interval of time. It is define at
every instant of time.
•Sine wave, cosine wave, triangular wave etc. similarly some
electrical signals derived from physical quantities like temperature,
pressure, sound etc. are also an examples of continuous signals.
x(t)
t

14. Discrete-time signals
•The discrete time signals are defined at discrete instant of time and is
represented by x(n), where n is index.
•Function of the sample value, n
•Write as x[n]
•Often called a sequence
•Commonly found in the digital world
•ex. wav file or mp3x[n]
n
12345678910

15. A CT signal is called a function.
A DT signal is called a sequence.

16. •Discrete time signal is obtained by sampling a continuous-time
signal at regular intervals.
•The discrete time signal can be denoted by x(nT) = x(t)/t=nT;
where T is called the sampling period and
n is integer ranging from -∞ to +∞.

17. ELEMENTARY SIGNALS AND SEQUENCES
•There are several elementary signals which play important role in
the study of signals and systems.
•These elementary signals used as basic building blocks for the
construction of more complex signals.
•These elementary signals are also called standard signals.

18. •The standard signals are:
1. Unit step function
2. Unit ramp function
3. Unit parabolic function
4. Unit impulse function
5. Sinusoidal function
6. Real exponential function
7. Complex exponential function, etc.

19. Unit Step Function (Heaviside Function)
•It have two different parameter such as CT unit step signal u(t) and
DT unit step signal u(n).
•Unit Step Signal: The unit step signal has amplitude of 1 for
positive value and amplitude of 0 for negative value of independent
variable.
•The mathematical representation of CT unit step signal u(t) is given
by, ut= ቊ
1 for t≥0
0 for t<0

20. •Unit Step Sequence: The mathematical representation of DT unit
step signal u(n) is given by,
u(n)=1;n≥0
=0;n<0

21. •The u(t) signal is shifted by ‘a’ times and it is represented by u(t-a).
•If ‘a’ is positive value, unit step signal is shifted right by a times and
if ‘a’ is negative value, unit step signal is shifted left by a times.
•The right shifted signal is represented by u(t-a) and left shifted signal
is represented by u(t+a).
•The delayed or right shifted unit step signal is
ut−??????= ቊ
1 for t≥a
0 for t<a
•The advanced unit step signal is ut+??????= ቊ
1 for t≥−a
0 for t<−a

22. •The delayed unit step signal is ut−??????= ቊ
1 for t≥a
0 for t<a
t
0
1
a
u(t-a)
Fig: Delayed or Right shifted unit step signal
u(t) = 1 for t ≥ 0
u(t-a) = 1 for t-a ≥ 0
==> t ≥ a

23. t
0
1
-a
u(t+a)
Fig: Advanced or Left shifted unit step signal
u(t) = 1 for t ≥ 0
u(t+a) = 1 for t+a ≥ 0
==> t ≥ -a
The advanced unit step signal is ut+??????= ቊ
1 for t≥−a
0 for t<−a

24. Example: Draw the graphical representation of u(t-1) and u(t+2)
signal.
t
0
1
1
u(t-1)
u(t-1) = 1 for t-1 ≥ 0
==> t ≥ 1
t
0
1
-2
u(t+2) = 1 for t+2 ≥ 0
==> t ≥ -2
u(t+2)

25. Example: Draw the graphical representation of u(t) - u(t-2) signal.
t
0
1
2
u(t-2)
u(t-2) = 1 for t-2 ≥ 0
==> t ≥ 2
t
0
1
u(t)
t
0
u(t)-u(t-2)
2
1
u(t) = 1 for t = 0 to ∞
u(t-2) = 1 for t = 2 to ∞
u(t)-u(t-2) = 1 for t = 0 to 2

28. •The mathematical representation of DT unit step signal u(n) is given
by,
•The graphical representation of Discrete Time Unit Step Function or
Unit Sequence Function is
u(n)=1;n≥0
=0;n<0
n
u(n)
0 1 2 3 4 . . .
. . . .
1
. . . -3 -2 -1
Fig: Unit Step Sequence

29. Example: Draw the graphical representation of u(n) - u(n-2) signal.
n
0 1 2 3 . .
. . . .
1
. . -2 -1
u(n)
n
0 1 2 3 4 . . .
. . . .
1
. . -2 -1
u(n-2)
n
0 1 2 3 . .
1
u(n) - u(n-2)
. . -2 -1
u(n-2) = 1 for n-2 ≥ 0
==> n ≥ 2

30. Unit Ramp Function
•Continuous time unit ramp signal is defined as
rt=t for t≥0or t u(t)
rt=0 for t<0
The graphical representation of
CT unit ramp signal is shown in
figure.
0
1
2
3
4
5
0 1 2 3 4 5
r(t)
Time
Fig: Unit Ramp Signal

31. •The unit ramp function is the integral of the unit step
function.
•In other words, unit step function can be obtained by differentiating
unit ramp signal (function).
•Area under unit ramp is unity.
r(t)=඲ut dt=න1.dt=t
&#3627408482;(&#3627408481;)=
)&#3627408465;&#3627408479;(&#3627408481;
&#3627408465;&#3627408481;
=
&#3627408465;
&#3627408465;&#3627408481;
(&#3627408481;)=1

32. Example: Draw the graphical representation of r(t), r(t-2),
r(t+2), -r(t), r(-t) and r(-t+2) signal.
t
r(t)
. . -1 0 1 2 3 4 . . .
3
2
1
r(t) graphical representation
t
-3 -2 -1 0 1 . . .
3
2
1
r(-t) graphical representation
r(-t)
t
-r(t)
. . -1 0 1 2 3 4 . . .
-1
-2
-3
-r(t) graphical representation
t
-r(-t)
. . . -2 -1 0 1 . .
-1
-2
-3
-r(-t) graphical representation

33. r(t-2) graphical representation
t
r(t-2)
. . -1 0 1 2 3 4 . . .
3
2
1
r(t+2) graphical representation
t
r(t+2)
. . -3 -2 -1 0 1 2 . . .
3
2
1
r(-t+2) graphical representation
t
-3 -2 -1 0 1 2 3 . . .
3
2
1
r(-t+2)
r(-t-2) graphical representation
t
-3 -2 -1 0 1 . . .
3
2
1
r(-t-2)

34. •The discrete time unit ramp signal or unit ramp
sequence is defined as
rn=n for n≥0or nu(n)
rn=0 for n<0
…... -2 -1 0 1 2 3 4 5 …..
n
r(n).
.
4
3
2
1
. . . .
Fig: Unit Ramp Sequence

35. Parabolic Signal
Parabolic signal can be defined as
pt=൝
t
2
2
for t≥0
0 for t<0
0
2
4
6
8
0 1 2 3 4 5
P(t)
Time
Fig: Unit Parabolic Signal

36. •The parabolic signal can obtained by integration of ramp signal
or double integration of unit step signal.
i.e.,ඵut dt= නrt dt=p(t)
ut=
d
2
p(t)
dt
2
and rt=
dp(t)
dt
ඵ1 dt= නt dt=
t
2
2
=p(t)

38. Unit Impulse Function
•Impulse function is denoted by δ(t). and it is defined as
δt= ቊ
1 for t=0
0 for t ≠0
t
δ(t)
1
. . . . -2 -1 0 1 2 3 . . . . .
Fig: Unit Impulse Signal

39. •So unit impulse function is the derivative of the unit step
function or unit step is the integral of the unit impulse function.
δt=
du(t)
dt
and න
−∞
∞
δt dt=ut()1
()0 0
tdt
t fort



−
=
=


40. •Discrete Time Unit Impulse Function or Unit Pulse Sequence.
δn= ቊ
1 for n=0
0 for n ≠0

41. 1)Shifting Property: (a)׬
−∞
∞
??????&#3627408533; ??????&#3627408533; &#3627408517;&#3627408533;=????????????
(b) ׬
−∞
∞
??????&#3627408533; ??????&#3627408533;−&#3627408533;
?????? &#3627408517;&#3627408533;=??????&#3627408533;
??????
Proof: (a) Let us consider the product of x(t) and δ(t). Which is x(t)
δ(t).
Let, the signal x(t) is continuous at t = 0, the value of x(t) at t = 0 is
x(0).
The impulse exists only at t = 0.
Representation of Impulse Function

42. Therefore, x(t) δ(t) = x(0) δ(t)
Integrating on both sides,
׬
−∞
∞
x(t) δ(t) &#3627408465;&#3627408481;= ׬
−∞
∞
x(0) δ(t) &#3627408465;&#3627408481;
න
−∞
∞
x(t) δ(t) &#3627408465;&#3627408481;=x(0)න
−∞
∞
δ(t) &#3627408465;&#3627408481;=x(0)∗1 = x(0)

We, Know that, ׬
−∞
∞
δ(t) &#3627408465;&#3627408481;=1
න
−∞
∞
x(t) δ(t) &#3627408465;&#3627408481;=x(0)

43. Proof: (b) Let us consider the product of x(t) and δ(t-t
0). Which is
x(t) δ(t-t
0).
Let, the signal x(t) is continuous at t = t
0 and the value of x(t) at t =
t
0 is x(t
0).
The impulse exists only at t = t
0.
Therefore, x(t) δ(t-t
0) = x(t
0) δ(t-t
0).
Integrating on both sides,
׬
−∞
∞
x(t) δ(t−t
0) &#3627408465;&#3627408481;= ׬
−∞
∞
x(t
0) δ(t−t
0) &#3627408465;&#3627408481;

44. න
−∞
∞
x(t) δ(t−t
0) &#3627408465;&#3627408481;=x(t
0)න
−∞
∞
δ(t−t
0) &#3627408465;&#3627408481;
න
−∞
∞
x(t) δ(t−t
0) &#3627408465;&#3627408481;=x(t
0)∗1 = x(t
0)
We, Know that, ׬
−∞
∞
δ(t−t
0) &#3627408465;&#3627408481;=1
න
−∞
∞
x(t) δ(t−t
0) &#3627408465;&#3627408481; = x(t
0)

45. 2) Scaling Property: (a) δat=
1
a
δt
(b) δa(t−&#3627408481;
0)=
1
a
δt−&#3627408481;
0
3) Replication Property: xt ⊗ δt=x(t)

46. Sinusoidal Signal
•Sinusoidal signal is in the form of
x(t) = A cos(wt ± ϕ) or A sin(wt ± ϕ) or A sin (ω
ot)
T
o = Time Period of the Sinusoidal Wave
ω
o = 2Пf
o = 2П/T
o

48. •A real exponential signal which is defined for every instant of time
is called continuous time real exponential signal.
•A continuous time real exponential signal is defined as x(t) = A e
αt
.
•Where, A and ?????? both are real.
•Here the parameter A is the amplitude of the exponential signal
measured at t = 0 and the parameter ?????? can be either positive or
negative.
Real Exponential Signal

49. •Depending upon the value of ??????, we obtain different exponential
signals as
➢When ?????? = 0, the exponential signal x(t) is a signal of constant
magnitude for all times.
If α = 0; x(t) = A e
0
= A
➢When ?????? > 0, i.e., ?????? is positive, then the exponential signal x(t) is
a growing exponential signal.
If, α > 0; i.e. +ve then x(t) = A e
αt
➢When ?????? < 0, i.e., ?????? is negative, then the signal x(t) is a decaying
exponential signal.
If, α< 0; i.e. -ve then x(t) = A e
-αt

51. •A discrete-time real exponential sequence is defined as
xn=a
n
for all n
•Depending upon the value of ‘a’ the discrete time real exponential
signal may be of following type
➢When a > 1, the exponential sequence x(n) grows
exponentially.
➢When 0 < a < 1, the exponential signal x(n) decays
exponentially.
➢When a < 0, the exponential sequence x(n) takes alternating
signs.

52. •The continuous time complex signal is defined as x(t) = Ae
st
Where, A is the amplitude of the signal & s is a complex variable.
•The complex variable s is defined as, s = σ + jω
•Therefore, the continuous time complex function can also be written
as
x(t) = A e
(σ+jω)t
= A e
σt
e
jωt
= Ae
σt
(cosωt + j sinωt)
Complex Exponential Signal

54. Signum Function
•Signum function is denoted as sgn(t). It is defined as


t
u(t)
1
0
-1
sgn&#3627408481;= ቐ
1 &#3627408467;&#3627408476;&#3627408479; &#3627408481;>0
0 &#3627408467;&#3627408476;&#3627408479; &#3627408481;=0
−1 &#3627408467;&#3627408476;&#3627408479; &#3627408481;<0

55. •The signum function can also be expressed in terms of the unit step
function as sgn(t) = 2u(t) – 1

56. •The signum function, is closely related to the unit-step function.
We know that,
sgnt= 2ut−1
ut= ቊ
1 for t≥0
0 for t<0
t
u(t)
1
0
u−t= ቊ
1 for t≤0
0 for t>0
t
-u(-t)
0
-1
t
u(-t)
0
1

57. •The signum function,
•Adding 1 both sides to above equation,
sgn&#3627408481;= ቐ
1 &#3627408467;&#3627408476;&#3627408479; &#3627408481;>0
0 &#3627408467;&#3627408476;&#3627408479; &#3627408481;=0
−1 &#3627408467;&#3627408476;&#3627408479; &#3627408481;<0
sgn&#3627408481;+1= ቐ
2 &#3627408467;&#3627408476;&#3627408479; &#3627408481;>0
1 &#3627408467;&#3627408476;&#3627408479; &#3627408481;=0
0 &#3627408467;&#3627408476;&#3627408479; &#3627408481;<0
sgn&#3627408481;+1=2u(t)
sgn&#3627408481;=2ut−1

58. •The mathematical formulation of the sinc function, also known as
the cardinal sine function. It is denoted as sinc(t).
•The sinc function is defined as,
Sinct=
sin(πt)
πt
for −∞<t<∞
Sinct=
sin(t)
t
for −∞<t<∞
Sinc Function

60. •To convert a signal from continuous time to discrete time, a process
called sampling is used. It is denoted as sa(t) and it is defined as
Sat=
sin(t)
t
for &#3627408481;=±??????,±2??????,±3??????,……….
Sampling Function

61. Rectangular Pulse Function

63. •Mathematically, the unit triangular pulse signal Δ(t/τ) is defined as,
Δ
t
τ
=
1−
2t
τ
fort<
τ
2
0 fort>
τ
2

&#3627408476;&#3627408479; Δt=ቐ
1−
t
a
fort≤a
0 fort>a
Triangular Pulse Function

64. CLASSIFICATION OF SIGNALS

65. •The signals are classified according to their characteristics. Some of
them are:
1.Continuous Time and Discrete Time Signals
2.Deterministic and Non-deterministic Signals
3.Even and Odd Signals
4.Periodic and Aperiodic Signals
5.Energy and Power Signals
6.Real and Imaginary Signals
7.Causal and Non-causal Signals
8.Analog and Digital Signals
CLASSIFICATION OF SIGNALS

66. •Dependinguponthenatureandcharacteristicsinthetimedomain,
thesignalsmaybeclassifiedintotwocategories
Thecontinuous-timesignalsare
alsocalledasanalogsignals.

67. •Continuous-Time Signal: A signal that is defined for every
instant of time is known as continuous time signal.
•Continuous time signals are continuous in amplitude and
continuous in time. It is denoted by x(t).
•Discrete-Time Signal: A signal that is defined for discrete
instants of time is known as discrete time signal.
•Discrete time signals are continuous in amplitude and discrete in
time. It is also obtained by sampling a continuous time signal. It is
denoted by x(n).
Continuous Time and Discrete Time Signals

69. •Analog signal: A signal that is defined for every instant of time is
known as analog signal.
•Analog signals are continuous in amplitude and continuous in time.
It is denoted by x(t). It is also called as Continuous time signal.
Example for Continuous time signal is shown in Fig.
•Digital signal: The signals that are discrete in time and quantized
in amplitude is called digital signal.
•An common example of a digital signal is a binary sequence, where
the values of the function can only be one or zero.
Analog Signal and Digital Signal

71. •Deterministic Signals: A signal is said to be deterministic if
there is no uncertainty with respect to its value at any instant of
time.
•A signal whose nature can be determined is known as deterministic
signal. They can be predicted.
•Random Signals: A signal is said to be non-deterministic if there is
uncertainty with respect to its value at some instant of time.
•Non-deterministic signals are random in nature hence they are called
random signals.
•A signal whose nature cannot be determined is known as
deterministic signal. They cannot be predicted.
Deterministic and Non-deterministic (Random) Signals

73. •Even signal & sequence: Those signals which are symmetrical
about the vertical axis or time origin are known as even signals. They
are also known as symmetrical signals.
•A signal is said to be even when it satisfies the condition x(t) = x(-t).
•Even signals can be easily marked as they are symmetric around the
vertical axis.
•A sequence x(n) is said to be even if it satisfies x(n) = x(−n) for all n.
•Example: t
2
, t
4
… cos(t) etc.
Even (Symmetric) and Odd (Anti-symmetric) Signal

75. •Odd signal & sequence: The signals which are anti-symmetrical
about the vertical axis are said to be odd signals. The odd signals are
also known as anti-symmetrical signals.
•A signal is said to be odd when it satisfies the condition x(t) = -x(-t)
for all t.
•Odd signals can be easily marked as they are anti symmetric around
the horizontal axis (time origin).
•A sequence x(n) is said to be odd if it satisfies x(n) = −x(−n) for all n.
•Example: t, t
3
... And sin(t)

77. •Any signal x(t) can be expressed as
x(t) = x
e(t) + x
o(t) and x(−t) = x
e(t) − x
o(t)
•Adding & subtracting above two equations, we get, the even & odd part
of a function
x
et=
xt+x(−t)
2
& x
0t=
xt−x(−t)
2
•Any discrete time signal x(n) can be expressed as
x(n) = x
e(n) + x
o(n) and x(−n) = x
e(n) − x
o(n)
•Adding & subtracting above two equations, we get, the even & odd part
of a function
x
en=
xn+x(−n)
2
& x
0n=
xn−x(−n)
2

78. •Periodic signal & sequence: A signal is said to periodic if it repeats again
and again over a certain period of time.
•A CT signal is said to be periodic if it satisfies the condition
x(t) = x(t + T) for − ∞ < &#3627408481; < ∞.
•Where; T = fundamental time period = 2π/ω (or) 2π/ω
0
1/T = f = fundamental frequency.
•A discrete-time signal x(n) is said to be periodic if it satisfies the following
condition x(n) = x(n + N) for all integers ??????
•The fundamental time period (N) may be defined as N=
2π
ω
Periodic and Aperiodic Signals

81. •Aperiodic signal: A signal that does not repeat at a definite
interval of time is called aperiodic signal.
•The aperiodic signals are also called the non-periodic signals.
•For non-periodic signals x(t) ≠ x(t+Tₒ) (or) x(n) ≠ x(n + N). Where
N is the fundamental time period.
•Fundamental time period N=
2πm
ω
.
Where, ω is fundamental angular frequency in rad/sec, m is the
smallest positive integer that makes N as positive integer.

83. •Energy Signals: A signal is said to be an energy signal if and
only if its total energy is finite.
•The average power of an energy signal is zero, i.e., P = 0.
•Examples of energy signals are aperiodic signals.
•i.e., For energy signal, 0<E<∞ and P=0
Energy and Power Signal

84. •The total energy or normalized energy of a continuous time signal
is defined as,
E=lim
T→∞
න
−T/2
T/2
|xt|
2
dtOR E=lim
T→∞
න
−T
T
|xt|
2
dt Joules
E=න
−∞
∞
|xt|
2
dt Joules
•The total energy of the discrete time signal x(n) is defined as
E=෍
n=−∞
∞
|xn|
2

85. •Power signal: The signal which has finite average power
(0<P<∞) and infinite energy (E = ∞) is called power signal.
•The periodic signals like sinusoidal complex exponential signals
will have constant power and so periodic signals are power signals.
•Periodic signals are the examples of power signals.
•i.e., For power signal, 0<P<∞ and E=∞

86. •The average power or normalized power of a continuous time
signal x(t) is given by,
P=lim
T→∞
1
T
න
−T/2
T/2
|xt|
2
dtOR P=lim
T→∞
1
2T
න
−T
T
|xt|
2
dt Watts
•The average power of a discrete time signal x(n) is defined as
P=lim
N→∞
1
2N+1
෍
n=−N
N
|xn|
2

87. •Causal Signals: A continuous-time signal x(t) is said to be causal
signal if x(t) = 0 for t < 0.
•Similarly, a discrete time signal x(n) is said to be causal signal if
x(n) = 0 for n < 0.
•Thus, a causal signal does not exist for negative time. Example:
u(t) is a causal signal.
Causal and non-causal Signal

88. •Anti-Causal Signals: A continuous-time signal x(t) is said to be
anti-causal signal if x(t) = 0 for t > 0.
•Similarly, a discrete-time signal x(n) is said to be anti-causal signal
if x(n) = 0 for n > 0.
•Therefore, an anti-causal signal does not exist for positive time.
Example: u(-t) is an anti-causal signal.

89. •Non-Causal Signal: A signal which is not causal is called the
non-causal signal.
•Hence, by the definition, a signal that exists for positive as well as
negative time is neither causal nor anti-causal, it is non-causal
signal.
•The sine and cosine signals are examples of non-causal signal.

91. •In signal processing "real" denotes a signal component that is in
phase with (i.e. has a 0° phase shift from) some reference signal.
•A signal is said to be real when it satisfies the condition x(t) = x*(t).
•Imaginary denotes a signal component that is in quadrature with (i.e.
has a 90° phase shift from) the same reference signal.
•A signal is said to be odd when it satisfies the condition x(t) = -x*(t)
Real and Imaginary Signals

92. •In signals and systems, a system can be defined in a number of ways
as
➢A system is defined as a physical device that can produce an
output or response for the given input.
➢A system may also be defined as an entity which works on an
input signal and transforms it into an output signal.
➢A system can also be defined as a set of elements which are
connected together and generates an output signal corresponding
to an input signal.
What is a System ?

93. •The relationship between the input signal x(t) and the output signal
y(t) of a system is given as y(&#3627408481;) = ??????[&#3627408485;(&#3627408481;)]
•There are various types of systems such as mechanical system,
electrical system, electromechanical system, biological system, etc.
All the physical devices such as an electric motor, generator, filter,
turbine, etc. are also examples of systems.

94. •Depending upon the time domain, the systems may be classified
into two categories
1.Continuous-Time Systems
2.Discrete-Time Systems
CLASSIFICATION OF SYSTEMS

95. •Continuous-Time System: A system which transforms a
continuous-time input signal into a continuous time output signal is
called thecontinuous-time system.
•If x(t) and y(t) are the input and output signals of a continuous time
system respectively, then the relation between input and output
signals of the continuous time system is defined as y(&#3627408481;) = ??????[x(&#3627408481;)]
•Amplifiers, integrators, differentiators and filter circuits, etc. are
some examples of continuous time systems.

96. •Discrete-Time System: A system which processes the discrete-
time input signals and produces discrete time output signals is shown
as discrete-time system.
•If x(n) and y(n) are the input and output signals of a discrete time
system respectively, then the relation between input and output of the
discrete-time system is defined as y(??????) = ??????[x(??????)]
•Microprocessors, digital devices, semiconductor memories, shift
registers, etc. are some examples of discrete-time systems.

97. •Systems are classified into the following categories:
1.Linear and Non-linear Systems
2.Time Variant and Time Invariant Systems
3.Linear Time variant and linear Time invariant systems
4.Static and Dynamic Systems
5.Causal and Non-causal Systems
6.Invertible and Non-Invertible Systems
7.Stable and Unstable Systems

98. •A system is said to be linear when it satisfies superposition and
homogenate principles.
•Consider two systems with inputs as x
1(t), x
2(t), and outputs as
y
1(t), y
2(t) respectively.
•Then, according to the superposition and homogenate principles,
T[a
1x
1(t)+a
2x
2(t)]=a
1T[x
1(t)]+a
2T[x
2(t)]
T[a
1x
1(t)+a
2x
2(t)]=a
1y
1(t)+a
2y
2(t)
Linear And Non-linear Systems

99. •A system is said to be time variant if its input and output
characteristics vary with time. Otherwise, the system is considered as
time invariant.
•The condition for time invariant system is: yt,t
0=y(t−t
0)
•The condition for time variant system is: yt,t
0≠y(t−t
0)
•Where, yt,t
0=T[xt−t
0] which is input change
y(t−t
0) which is output change
Time Variant and Time Invariant Systems

100. •If a system is both linear and time variant, then it is called linear
time variant (LTV) system.
Example:
??????
2
??????(&#3627408481;)
??????&#3627408481;
2
+4&#3627408481;
????????????(&#3627408481;)
??????&#3627408481;
+5&#3627408486;&#3627408481;=&#3627408485;&#3627408481;.
•If a system is both linear and time Invariant then that system is
called linear time invariant (LTI) system.
Example: 2
??????
2
??????(&#3627408481;)
??????&#3627408481;
2
+4
????????????(&#3627408481;)
??????&#3627408481;
+5&#3627408486;&#3627408481;=5&#3627408485;&#3627408481;.
linear Time variant (LTV) and linear Time Invariant
(LTI) Systems

101. •Static system is memory-less whereas dynamic system is a memory
system.
•Static system: A system is said to be memoryless or static if the
response of the system is due to present input alone.
Example: &#3627408486;(&#3627408481;) = 2&#3627408485;(&#3627408481;)
•Dynamic system: A system is said to be memory or dynamic if the
response of the system depends on factors other than present input
also.
Example: &#3627408486;(&#3627408481;) = 2&#3627408485;(&#3627408481;) + &#3627408485;(−&#3627408481;)
Static and Dynamic Systems

102. •A system is said to be causal if its output depends upon present and
past inputs, and does not depend upon future input.
Example:y(n)=2x(t)+3x(t-3)
•For non causal system, the output depends upon future inputs also.
Example:y(n)=2x(t)+3x(t-3)+6x(t+3)
Causal and Non-Causal Systems

103. •The system is said to be stable only when the output is bounded for
bounded input. For a bounded input, if the output is unbounded in
the system then it is said to be unstable.
•Example:y(t)=x
2
(t).Let the input is u(t) (unit step bounded
input) then the output y(t) = u
2
(t) = u(t) = bounded output. Hence,
the system is stable.
Stable and Unstable Systems

104. Basic Operations ON Signals

105. Basic Operations ON Signals
•Different operations are required for different purposes of
arbitrary signals.
•The operations on signals can be
1.Time Shifting
2.Time Reversal
3.Time Scaling
4.Amplitude Scaling
5.Signal Multiplier
6.Signal Addition

106. Time shifting
•Time shifting (also called translation) maps the input signal x(t) to
the output signal y(t) as given by, y(t) = x(t−T); where T is a real
number.
•Such a transformation shifts the signal (to the left or right) along the
time axis.
•If T is positive, the shifting delays the signal as shown in figure 1(b)
and if T is negative, the shifting advances the signal as shown in
figure 1(c).

107. •If T>0 ,y(t) is shifted to the right by |T|, relative to x(t) (i.e., delayed
in time).
•If T<0, y(t) is shifted to the left by |T|, relative to x(t) (i.e., advanced
in time).

108. The time shifting operation of a discrete time signal x(n) is represented as &#3627408486;(??????) = &#3627408485;(?????? − ??????
0)

110. Sketch Y(t) = U(t – 2)

112. Solution: Given signal is, x(&#3627408481;) = 3??????(&#3627408481; − 3)
Example: Sketch the following signals. x(&#3627408481;) = 3??????(&#3627408481; − 3) and
x(&#3627408481;) = Π(&#3627408481; + 2)

113. •Given signal is, x(&#3627408481;) = Π(&#3627408481; + 2)

114. Time reversal
•Time reversal (also known as reflection) maps the input signal x(t) to
the output signal y(t) as given by, y(t) = x(−t).
•Geometrically, the output signal y(t) is a reflection of the input signal
x(t) about the (vertical) line t = 0.

117. •For a discrete time sequence x(n), the time reversal is given by,
&#3627408486;(??????) = &#3627408485;(−??????)
•An arbitrary discrete-time signal x(n) and its time reversal x(-n) are
shown in Figure.

118. •Solution: Given signal is, &#3627408485;(&#3627408481;) = 3&#3627408482;(−&#3627408481;)
Example: Sketch the following signals: &#3627408485;(&#3627408481;) = 3&#3627408482;(−&#3627408481;)
and &#3627408485;(&#3627408481;) = 2??????(−&#3627408481;).

119. •Given signal is, &#3627408485;(&#3627408481;) = 2??????(−&#3627408481;)

120. Determine Time Reversal of Unit Step Signal, -u(t) &
-u(-t)

121. Amplitude Scaling
•Amplitude scaling maps the input signal x(t) to the output signal
y(t) as given by y(t) = a x(t), where a is a real number.
x(t) a y(t) = a x(t)
•Geometrically, the output signal y(t) is expanded/compressed in
amplitude and/or reflected about the horizontal axis.

122. The amplitude scaling of a continuous time signal x(t) is defined as,
&#3627408486;(&#3627408481;) = ?????? &#3627408485;(&#3627408481;)

123. The amplitude scaling of a discrete time sequence x(n) is defined as,
&#3627408486;(??????) = ?????? &#3627408485;(??????)

124. Amplitude Shifting
•Amplitude shifting maps the input signal x(t) to the output signal y(t)
as given by, y(t) = x(t) + b, where b is a real number.
•Geometrically, amplitude shifting adds a vertical displacement to
x(t).

125. TIME SCALING
•The process of multiplying a constant to the time axis of a signal is
known as time scaling of the signal.
•The time scaling of signal may be time compression or time
expansion depending upon the value of the constant or scaling factor.

•Time scaling maps the input signal x(t) to the output signal y(t) as
given by y(t) = x(at), Where a is a non zero real number.

126. •If a>1, y(t) is compressed along the horizontal axis by a factor of a,
relative to x(t).
•If a<1, y(t) is expanded (i.e., stretched) along the horizontal axis by
a factor of 1/a, relative to x(t).
•If a > 1, then the signal is compressed in time by a factor α and the
time scaling of the signal is called the time compression. Whereas,
if a < 1, then the signal is expanded in time by the factor α and the
time scaling is said to be time expansion.

130. •The sum of two CT signals x1(t) and x2(t) can be obtained by
adding their values at every instant of time.
•Similarly, the subtraction of two CT signals x1(t) and x2(t) can be
obtained by subtracting their values at every instant of time.
Signal Addition Operation

133. •The multiplication of two CT signals can be obtained by
multiplying their values at every instant of time. The multiplication
of two CT signals is shown in figure.
Signal Multiplication Operation

136. Signal Analysis EXAMPLES

137. Problem 1: Find the fundamental period T of the
following continuous-time signals.
(i) xt=je
j5t
(v) xt=sin50πt
(ii) xt=20cos10πt+
π
6
(iii) xt=sin10πt ut
(iv) xt=4cos(5πt) (vi) xt=e
−t

138. Solution: (i) Given &#3627408485;&#3627408481;=&#3627408471;&#3627408466;
??????5&#3627408481;
We know that, complex exponential signal is &#3627408485;&#3627408481;=&#3627408466;
????????????0&#3627408481;
The fundamental period, ??????=
2??????
??????
0
From given signal,??????
0=5
∴??????=
2??????
5
=0.4??????

139. (ii) &#3627408485;&#3627408481;=sin50??????&#3627408481;
The sinusoidal signal is &#3627408485;&#3627408481;=sin50??????&#3627408481;
From given signal, ??????
0=50??????
∴??????=
2??????
50??????
=
1
25
&#3627408480;&#3627408466;&#3627408464;.

140. (iii) &#3627408485;&#3627408481;=20cos10??????&#3627408481;+
??????
6
From given signal , ??????
0=10?????? and ∅=
??????
6
Fundamental time period ??????=
2??????
??????
0
=
2??????
10??????
=
1
5
sec.
(iv) Not a periodic signal.
(v) Periodic signal with period
2
5
(vi) Not a periodic signal.

141. Problem 2: Find whether the following signals
are periodic or not.
(i) xt=2 cos10t+1−sin4t−1
(ii)cos60πt+sin50πt
(iii) 2ut+2sin2t
(iv) 3cos4t+2sin2πt
(v) ut−
1
2
(vi) sin
2
(t)

142. Solution: (i) given &#3627408485;&#3627408481;=2 cos10&#3627408481;+1−sin4&#3627408481;−1
&#3627408485;&#3627408481;=&#3627408485;
1&#3627408481;+&#3627408485;
2(&#3627408481;)
Time period of 2cos10&#3627408481;+1 is T
1 =
2??????
10
=
??????
5
&#3627408480;&#3627408466;&#3627408464;.
Time period ofsin4&#3627408481;−1 is T
2 =
2??????
4
=
??????
2
&#3627408480;&#3627408466;&#3627408464;.
The ratio
??????1
??????
2
is a rational number or ratio of two integers.

143. Therefore, the sum of two signals [i.e., &#3627408485;
1&#3627408481; &&#3627408485;
2(&#3627408481;)] are periodic
and the period is given by T = 2T
2 = 5T
1
T
1 = ??????∗
1
5
and T
2 = ??????∗
1
2
The least common multiple of T
1 and T
2 is ??????
∴ The time period, T = ?????? sec.
T
1
T
2
=
π
5
π
2
=
2
5

144. &#3627408470;&#3627408470;&#3627408470; &#3627408468;&#3627408470;&#3627408483;&#3627408466;&#3627408475; &#3627408485;&#3627408481;=2&#3627408482;&#3627408481;+2sin2&#3627408481;
&#3627408485;&#3627408481;=&#3627408485;
1&#3627408481;+&#3627408485;
2(&#3627408481;)
Where &#3627408485;
1&#3627408481;=2&#3627408482;&#3627408481;=2 &#3627408467;&#3627408476;&#3627408479; &#3627408481;≥0.
&#3627408485;
1&#3627408481; is sketched as shown in fig.
∴&#3627408485;
1&#3627408481; is aperiodic signal.
The time period of &#3627408485;
2(&#3627408481;) is T
2=
2??????
2
=??????
∴ the signal &#3627408485;&#3627408481; is aperiodic signal.

145. (v) Given, &#3627408485;&#3627408481;=&#3627408482;&#3627408481;−
1
2
=
2&#3627408482;&#3627408481;−1
2
The signal &#3627408482;&#3627408481;−
1
2
is not a periodic

146. Problem 3: Determine the power and R.M.S value of
the signal ??????&#3627408533;=??????&#3627408516;??????&#3627408532;(??????
??????&#3627408533;+??????)
Solution: Given, &#3627408485;&#3627408481;=??????cos(??????
0&#3627408481;+??????)
The power, p=lim
??????→∞
1
2??????
׬
−??????
??????
&#3627408485;(&#3627408481;)
2
&#3627408465;&#3627408481;
??????=lim
??????→∞
1
2??????
න
−??????
??????
??????cos??????
0&#3627408481;+??????
2
&#3627408465;&#3627408481;

147. ??????=lim
??????→∞
1
2??????
න
−??????
??????
??????
2
cos
2
??????
0&#3627408481;+??????&#3627408465;&#3627408481;
cos
2
??????=
1+cos2??????
2
??????=lim
??????→∞
??????
2
2??????
න
−??????
??????
1+cos2??????
0&#3627408481;+??????
2
&#3627408465;&#3627408481;

148. ??????=lim
??????→∞
??????
2
4??????
න
−??????
??????
1.&#3627408465;&#3627408481;+න
−??????
??????
cos2??????
0&#3627408481;+??????&#3627408465;&#3627408481;
??????=lim
??????→∞
??????
2
4??????
&#3627408481;
−??????
??????
+
sin2??????
0&#3627408481;+2??????
2??????
0
−??????
??????
??????=lim
??????→∞
??????
2
4??????
&#3627408481;
−??????
??????
+
sin2??????
0&#3627408481;+2??????
2??????
0
−??????
??????

149. P=lim
??????→∞
??????
2
4??????
2??????+
1
2??????
0
sin2??????
0.
2??????
??????
0
+2??????+sin2??????
0.
2??????
??????
0
−2??????
??????
=lim
??????→∞
??????
2
4??????
2??????+lim
??????→∞
??????
2
4??????
.
1
2??????
0
sin4??????+2??????+sin4??????−2??????

150. ??????=lim
??????→∞
??????
2
2
+lim
??????→∞
??????
2
4??????
+
1
2??????
0
sin2??????−sin2??????
??????=
??????
2
2
+0=
??????
2
2
R.M.S value =&#3627408477;&#3627408476;&#3627408484;&#3627408466;&#3627408479;=
??????
2
2
=
??????
2

151. Problem 4: Determine the power and R.M.S value of
the following signals: (i) ) &#3627408485;&#3627408481;=5cos50&#3627408481;+
??????
3
(ii) &#3627408485;&#3627408481;=10sin50&#3627408481;+
??????
4
+16) cos100&#3627408481;+
??????
3

152. Problem 5: Determine whether a unit step signal u(t) is
energy or power signal.
Solution: We know that, &#3627408482;&#3627408481;= ቊ
1;&#3627408481;≥0
0;&#3627408481;<0
Energy, ??????= ׬
−∞
∞
|&#3627408485;&#3627408481;|
2
&#3627408465;&#3627408481; &#3627408476;&#3627408479;׬
−∞
∞
&#3627408485;
2
(&#3627408481;) &#3627408465;&#3627408481;
??????= න
−∞
∞
|&#3627408482;&#3627408481;|
2
&#3627408465;&#3627408481;=න
0
∞
|1|
2
&#3627408465;&#3627408481;=න
0
∞
1 &#3627408465;&#3627408481;

??????=[&#3627408481;]
0
∞
=∞

153. Power, P= lim
T→∞
1
2T
׬
−T
T
|xt|
2
dt
P=lim
T→∞
1
2T
න
−T
T
|ut|
2
dt=lim
T→∞
1
2T
න
0
T
|1|
2
dt

P=lim
T→∞
1
2T
න
0
T
1 dt=lim
T→∞
1
2T
[t]
0
T
=lim
T→∞
1
2T
[T−0]
P=lim
T→∞
1
2T
T=lim
T→∞
1
2
=
1
2
The Energy of the signal is infinity and average power is zero.
Therefore, the u(t) is a power signal.

154. Assignment Question:
Problem 6: Find the energy and
power of a signal shown in figure.

155. Problem 7: sketch the following signals.
(i) u(-t+1) (ii) -2u(t-1) (iii) 3r(t-1) (iv) -2r(t)
(v) r(-t+2) (vi) π(t+3)
Solution: Given that, x(t)=u(-t+1)
The signal x(t) can be obtained by folding the unit step signal u(t)
about t=0 and shifting the signal u(-t) right by one unit of time.

156. Fig(a): unit step signal
Fig (b) : folded unit step signal
Folded signal about t=0 is shown in
fig (b)

157. The signal u(-t) is shifted right by one sample, which is shown in
fig.(c).
Fig (c)

158. (ii)Given x(t)=-2u(t-1).
Draw the unit step signal.
➢The signal u(t) is shifted right by one sample.

160. (iii) Given x(t)=3r(t-1)
The signal r(t) is a unit ramp signal with slope equal to 1 and the
signal Ar(t) is a ramp signal with slope equal to A.
First sketch 3r(t) and then and then shift it by one unit sample.

162. (iv) Given x(t)=-2r(t)
The signal x(t)is a ramp signal with slope -2.

163. (v) Given r(t)=r(-t+2)
➢First draw the unit sample signal.

164. ➢Fold the signal r(t) about t=0.

165. (vi) Given x(t)=π(t+3)
Π(t) is a rectangular pulse function, which is shown in fig.
Π(t)=1 for&#3627408481;≤
1
2

166. ➢The signal x(t) can be obtained by shifting π(t) left by 3 units of
times as shown in fig.

167. Problem 8: For the signal x(t) shown in fig. find
the signals (i) x(t-2) (ii) x(2t+3) (iii) x(
&#3627409361;
&#3627409360;
t) (iv)
x(-t+1)

168. Problem 9: Determine if the following signals are Energy
signals, Power signals, or neither, and evaluate E and P for
each signal.

169. So, the signal a(t) is definitely an Power signal.

171. So, the signal b(t) is definitely an energy signal.

172. ANALOGY BETWEEN SIGNALS
AND VECTORS

173. •Signals & Vectors: There is a perfect analogy between signals
and vectors.
•A vector can be represented as a sum of its components in various
ways.
•A signal can also be represented as a sum of its components in
various ways.
•Vector: A vector contains magnitude and direction. The name of
the vector is denoted by bold face (A, B, C, ….) type and their
magnitude is denoted by light face type.
ANALOGY BETWEEN SIGNALS AND VECTORS

174. •Example: ഥV is a vector with magnitude V. Consider two vectors V
1
and V
2 as shown in the following diagram.
•Let the component of ഥV
1 along with ഥV
2 is given by C
12
ഥV
2.
•The component of a vector ഥV
1 along with the vector ഥV
2 can obtained
by taking a perpendicular from the end of ഥV
1 to the vector ഥV
2 as
shown in diagram.

175. •The vector ഥV
1 can be expressed in terms of vector ഥV
2,
ഥV
1=C
12
ഥV
2+ഥV
e
Where ഥV
e is the error vector, C
12 is coefficient.
•But this is not the only way of expressing vector ഥV
1 in terms of ഥV
2.
The alternate possibilities are:
ഥV
1=C
1
ഥV
2+ഥV
e1 and ഥV
1=C
2
ഥV
2+ഥV
e2

176. •The error signal is minimum for large component value. If C
12=0,
then two signals are said to be orthogonal.
•The Dot Product of Two Vectors, ഥA.ഥB=AB cosθ
•Similarly, ഥV
1.ഥV
2=V
1V
2cosθ
Where, θ = Angle between V
1 and V
2
•The dot product obeys commutative law, ഥV
1.ഥV
2=ഥV
2.ഥV
1

177. •The components of V
1 along V
2 is V
2cos(θ)
V
2cosθ=
ഥV
1.ഥV
2
V
1
and V
1cosθ=
ഥV
1.ഥV
2
V
2
•From the diagram, components of V
1 along V
2 is C
12V
2
==>V
1cosθ=
ഥV
1.ഥV
2
V
2
=C
12V
2
==>C
12=
ഥV
1.ഥV
2
V
2
2
•If the two vectors ഥV
1 & ഥV
2 are orthogonal, then the dot product of
ഥV
1 & ഥV
2 is zero. i.e.,ഥV
1.ഥV
2=0
•So, C
12=0

178. •Signal (Component of a Signal): The concept of orthogonality can
be applied to signals. Let us consider two signals f
1(t) and f
2(t).
•Similar to vectors, we can approximate f
1(t) in terms of f
2(t) as
f
1t=C
12f
2t+f
et for t
1<t<t
2
==>f
et=f
1t−C
12f
2t
Where, f
e(t) is error function.
•One possible way of minimizing the error is integrating over the
interval t
1 to t
2.
i.e.,
1
t
2−t
1
න
t
1
t2
f
et dt

179. ==>
1
t
2−t
1
න
t1
t
2
f
1t−C
12f
2t dt
•However, this step also does not reduce the error to appreciable
extent.
•This can be corrected by taking the square of error function and it is
denoted by ϵ.
??????=
1
t
2−t
1
න
t1
t
2
f
e
2
(t) dt==>??????=
1
t
2−t
1
න
t1
t
2
f
1t−C
12f
2t
2
dt

180. •Where ε is the mean square value of error signal. The value of C
12
which minimizes the error, we need to calculate
dϵ
dC12
=0
==>
d
dC
12
1
t
2−t
1
න
t1
t
2
f
1t−C
12f
2t
2
dt=0
==>
d
dC
12
1
t
2−t
1
න
t
1
t
2
f
1
2
t+C
12
2
f
2
2
t−2C
12f
1tf
2tdt=0
==>
1
t
2−t
1
න
t1
t
2
d
dC
12
f
1
2
t+C
12
2
f
2
2
t−2C
12f
1tf
2tdt=0

181. =>
1
t
2−t
1
න
t1
t2
d
dC
12
f
1
2
t+
d
dC
12
C
12
2
f
2
2
t−
d
dC
12
2C
12f
1tf
2tdt=0
==>
1
t
2−t
1
න
t
1
t
2
0+f
2
2
t 2C
12−2f
1tf
2tdt=0
==>න
t1
t
2
f
2
2
t 2C
12−2f
1tf
2tdt=0
==>2C
12න
t1
t
2
f
2
2
tdt−2න
t1
t
2
f
1tf
2tdt=0

182. ==>2C
12න
t
1
t2
f
2
2
tdt=2න
t
1
t2
f
1tf
2tdt
==>C
12=
׬
t1
t2
f
1tf
2tdt
׬
t
1
t2
f
2
2
tdt
•If C
12 component is zero, then two signals are said to be orthogonal.
•Put C
12 = 0 to get condition for orthogonality.
0=
׬
t1
t
2
f
1tf
2tdt
׬
t1
t2
f
2
2
tdt
==>න
t1
t
2
f
1tf
2tdt=0

183. •The average of square of error function f
e(t) is called as mean
square error. It is denoted by ε.
??????=
1
t
2−t
1
න
t
1
t2
f
et
2
dt
??????=
1
t
2−t
1
න
t
1
t2
ft−෍
??????=1
??????
C
rg
rt
2
dt
MEAN SQUARE ERROR [MSE]

184. ??????=
1
t
2−t
1
න
t
1
t2
f
2
tdt+෍
??????=1
??????
C
r
2
න
t
1
t2
g
r
2
tdt−2෍
??????=1
??????
C
rන
t
1
t2
ftg
rtdt−
→(1)
•But, C
j=
׬
t1
t2
ftg
jtdt
׬
t1
t2
g
j
2
tdt
==>C
j=
1
K
j
׬
t1
t
2
ftg
jtdt
(or) C
r=
׬
t1
t2
ftgrtdt
׬
t1
t2
g
r
2
tdt
==>C
r=
1
K
r
׬
t1
t2
ftg
rtdt
==>න
t1
t2
ftg
rtdt=C
rන
t1
t2
g
r
2
tdt=C
rK
r−−−→(2)

185. •Substituting eq(2) in eq(1), we get,
??????
=
1
t
2−t
1
ቐ
ቑ
න
t
1
t2
f
2
tdt+෍
??????=1
??????
C
r
2
න
t
1
t2
g
r
2
tdt
−2෍
??????=1
??????
C
rන
t1
t
2
ftg
rtdt
=⇒??????=
1
t
2−t
1
න
t
1
t
2
f
2
tdt+෍
??????=1
??????
C
r
2
K
r−2෍
??????=1
??????
C
r(C
rK
r)

186. =⇒??????=
1
t
2−t
1
න
t
1
t2
f
2
tdt+෍
??????=1
??????
C
r
2
K
r−2෍
??????=1
??????
C
r
2
K
r
=⇒??????=
1
t
2−t
1
න
t1
t
2
f
2
tdt−෍
??????=1
??????
C
r
2
K
r
??????=
1
t
2−t
1
න
t1
t2
f
2
tdt−C
1
2
K
1+C
2
2
K
2+⋯……+C
n
2
K
n
•The above equation is used to evaluate the mean square error.

187. •The MSE is given by,
??????=
1
t
2−t
1
න
t
1
t
2
f
2
tdt−෍
??????=1
??????
C
r
2
K
r
•Let us consider a set of n mutually orthogonal functions g
1(t), g
2(t),
…..., g
n(t) over the interval t
1 to t
2.
•This is called as closed and complete set when there exist no
function f(t) satisfying the condition ׬
t1
t2
ftg
rtdt=0
Closed and Complete Set of Orthogonal Functions

188. •If this function is satisfying the equation ׬
t
1
t2
ftg
rtdt=0 for r =
1, 2, 3, ……, then f(t) is said to be orthogonal to each and every
function of orthogonal set. This set is incomplete without f(t). It
becomes closed and complete set when f(t) is included.
•f(t) can be approximated with this orthogonal set by adding the
components along mutually orthogonal signals
i.e. ft=C
1g
1t+C
2g
2t+C
3g
3t+⋯……+C
ng
nt+f
et
•If the infinite series C
1g
1t+C
2g
2t+C
3g
3t+⋯……+
C
ng
nt converges to f(t) then mean square error is zero.

189. •If f
1(t) and f
2(t) are two complex functions, then f
1(t) can be
expressed in terms of f
2(t) as f
1t=C
12f
2t with negligible error
•Where,
C
12=
׬
t1
t
2
f
1tf
2
∗
tdt
׬
t
1
t2
|f
2t|
2
dt
•Where, f
2
∗
t is complex conjugate of f
2(t)
Orthogonality in Complex Functions

190. •If f
1(t) and f
2(t) are orthogonal then C
12 = 0
0=
׬
t1
t
2
f
1tf
2
∗
tdt
׬
t
1
t2
|f
2t|
2
dt
==>න
t
1
t
2
f
1tf
2
∗
tdt=0
•The above equation represents orthogonality condition in complex
functions.