Module-3

System Classification and Analysis

Objective: To understand the concept of systems, classification, signal transmission through

linear systems with bandwidth considerations.

Introduction:

A signal was defined as a mapping from a set of the independent variable (domain) to the

set of the dependent variable (co-domain). A system is also a mapping, but across signals, or

across mappings. That is, the domain set and the co-domain set for a system are both sets of

signals, and corresponding to each signal in the domain set, there exists a unique signal in the co-

domain set.

In signals and systems terminology, we say; corresponding to every possible input signal, a

system “produces” an output signal. In that sense, realize that a system, as a mapping is one step

hierarchically higher than a signal. While the correspondence for a signal is from one element of

one set to a unique element of another, the correspondence for a system is from one whole

mapping from a set of mappings to a unique mapping in another set of mappings.

Linear Shift-Invariant systems, called LSI systems for short, form a very important class

of practical systems, and hence are of interest to us. They are also referred to as Linear Time-

Invariant systems, in case the independent variable for the input and output signals is time.

Remember that linearity means that is y1(t) and y2(t) are responses of the system to signals x1(t)

and x2(t) respectively, then the response to ax1(t) + bx2(t) is ay1(t) + by2(t). Shift invariance

implies that the response of the system to x1(t - t0) is given by y1(t - t0) for all values of t and t0.

Linear systems are of interest to us for primarily two reasons: first, several real-life systems can

be well approximated by linear systems. Second, linear systems come with several properties

which make their analysis simple. Similarly, shift-invariant systems allow us to use simpler math

to analyze the system. As we proceed with our analysis, we will point out cases where some

results (which are rather intuitive) are valid for only LSI systems.

Description:

System: A system is defined as an entity that acts on an input signal and transforms it into an

output signal. A system may also be defined as a set of elements or functional blocks which are

connected together and produces an output in response to an input signal. There are various types

of systems: electrical systems, mechanical systems, biological systems, opto-electronic systems,

electromechanical systems and so on.

Classification of systems:

System is represented by a block diagram

x(t) y(t)

Input Output

The relation between the input x(t) and the output y(t) of a system has the form

y(t)= operation on x(t)

System Impulse

response h(t)

Mathematically, y(t)=T[x(t)]

Systems are broadly classified as

(a) Continuous – time systems

(b) Discrete –time systems

Continuous-time system

A continuous- time system is one wh8ich transforms continuous-time input signals into

continuous-time output signals.

y(t)=T[x(t)]

x(t) y(t)

Input Output

Amplifiers, filters, integrators and differentiators are the examples of continuous-time systems.

Discrete-time system

A discrete-time system is one which transforms discrete-time input signals into discrete-time

output signals.

y(n)=T[x(n)]

Microprocessors, semiconductor memories, shift registers, etc. are the examples of discrete-time

systems.

x(n) y(n)

Input Output

Both continuous-time and discrete-time systems may be classified as under

1. Lumped parameter and distributed parameter systems

2. Static (memory less) and dynamic (memory) systems

3. Causal and non-causal systems

4. Linear and non-linear systems

5. Time-invariant and time varying systems

6. Stable and unstable systems

7. Invertible and non-invertible systems

8. FIR and IIR systems

Lumped parameter and distribute parameter systems

Lumped parameter systems are the systems in which each component is lumped at one

point in space. These systems are described by ordinary differential equations. Distributed

parameter systems are the systems in which signals are functions of space as well as time.

These systems are described by partial differential equations.

Static and Dynamic systems

A system is said to be static or memoryless if the response is due to present input alone , i.e. for a

static or memoryless system, the output at any instant t(or n) depends only on the input applied at

that instant t (or n) but not on the past or future values of input.

For example, the system defined below are static or memoryless systems.

y(t)=x(t)

Continuous-time

system

Discrete-time system

y(t)=x2(t)

y(n)=x(n)

y(n)=2x2(n)

A System is said to be dynamic or memory system if the response depends upon past or future

inputs.

For example, the systems defined below are dynamic or memory systems.

y(t)=x(t-1)

y(t)=x(t)+x(t+2)

y(t)=𝑑²𝑥(𝑡)

𝑑𝑡² +x(t)

y(n)=x(2n)

y(n)=x(n)+x(n-2)

Any continuous-time system described by a differential equation or any discrete-time

system described by a difference equation is also a dynamic system.

A purely resistive electrical circuit is a static system, whereas an electric circuit having

inductors and/or capacitors is a dynamic system.

A summer or accumulator is an example of a discrete time system with memory. A delay

is also a discrete time system with memory.

Illustration

Example: Find whether the following systems are dynamic or not:

(a) y(t)=x(t-3) (b) y(t)=x(2t)

(c) y(t)= d²x(t)

dt²+2x(t) (d) y(n)=x(n+2)

Solution:

(a) y(t)=x(t-3)

The output depends on past value of input. Therefore, the system is dynamic.

(b) y(t)=x(2t)

The output depends on future value of input. Therefore, the system is dynamic.

(c) y(t)= 𝑑²𝑥(𝑡)

𝑑𝑡²+2x(t)

The system is described by a differential equation. Therefore, the system is dynamic.

(d) y(n)=x(n+2)

The output depends on future value of input. Therefore, the system is dynamic.

Causal and non-causal systems

A system is said to be causal if the output of the system at any time t depends only on the present

and past values of the input but not on future inputs. In other words, we can say that a system is

causal if the response or output does not begin before the input function is applied. A causal

system is non anticipatory. Causal systems are real time systems.

Examples for the causal systems are:

y(t)=x(t-2)+2x(t)

y(t)=tx(t)

y(n)=nx(n)

A system is said to be noncausal if the output of the system at any time t depends on future

inputs. They do not exists in real time. They are not physically realizable. They are anticipatory

systems. They produce an output even before the input is given.

Examples for the noncausal systems are:

y(t)=x(t+2)+2x(t)

y(t)=tx(t)+x2(t)

y(n)=nx(2n)+x(n)

A resistor is an example of a continuous time causal system. Image processing systems are

examples of noncausal systems.

Illustration

Example:

Check whether the following systems are causal or not

(a)y(t)=x2(t)+x(t-4) (b) y(t)=x(2-t)+x(t-4)

Solution

(a)y(t)=x2(t)+x(t-4)

For t=-2 y(-2)=x2(-2)+x(-6)

For t=0 y(0)=x2(0)+x(-4)

For t=2 y(2)=x2(2)+x(-2)

For all values of t, the output depends only on the present and past values of input. Therefore, the

system is causal.

(b) y(t)=x(2-t)+x(t-4)

For t=-1 y(-1)=x(3)+x(-5)

For t=0 y(0)=x(2)+x(-4)

For t=1 y(1)=x(1)+x(-3)

For some values of t, the output depends on the future input. Therefore, the system is noncausal.

Linear and nonlinear systems

A system which obeys the principle of superposition and principle of homogeneity is called a

linear system, and a system which does not obey the principle of superposition and principle of

homogeneity is called a non-linear system.

Homogeneity property means a system which produces an output y(t) for an input x(t)

must produce an output ay(t) for an input ax(t).

Superposition property means a system which produces an output y1(t) for an input x1(t)

and an output y2(t) for an input x2(t) must produce an output y1(t)+ y2(t) for an input x1(t)+ x2(t).

Combining them we can say that a system is linear if an arbitrary input x1(t) produces an

output y1(t) and an arbitrary input x2(t) produces an output y2(t), then the weighted sum of inputs

ax1(t)+ bx2(t), where a and b are constants, produces an output ay1(t)+ by2(t) which is the sum of

weighted outputs. That is

T[a x1(t)+ bx2(t)]= aT[x1(t)]+ bT[x2(t)]

For discrete time lineat system,

T[a x1(n)+ bx2(n)]= aT[x1(n)]+ bT[x2(n)]

Example of linear systems are filters, communication channels, etc.

In general, if the describing equation contains square or higher order terms of input

and/or output and/or product of input/output and its derivative or a constant, the system will

definitely be non-linear.

Illustration

Example:

Check whether the following systems are linear or not

(a) d²y(t)

dt²+2ty(t)=t2x(t) (b) 2

dy(t)

dt+5y(t)=x2(t)

Solution :

(a) d²y(t)

dt²+2ty(t)=t2x(t)

Let an input x1(t) produce an output y1(t).

Then d²y₁(t)

dt²+2ty₁(t)=t2x₁(t)

Let an input x₂(t) produce an output y₂(t).

Then d²y₂(t)

dt²+2ty₂(t)=t2x₂(t)

Linear combination of the above equations gives

ad²y₁(t)

dt²+a2ty₁(t)+𝑏

d²y₂(t)

dt²+b2ty₂(t)=at2x₁(t)+b t2x₂(t)

d²

dt²[ay₁(t)+by₂(t)]+2t[ay₁(t)+by₂(t)]=t2[ax₁(t)+bx₂(t)]

This shows that the weighted sum of inputs to the system produces an output which is

equal to the weighted sum of outputs to each of the individual inputs. Therefore the system is

linear.

(b) 2dy(t)

dt+5y(t)=x2(t)

If an input x₁(t) produces an output y₁(t), then

2dy₁(t)

dt+5y₁(t)=x₁2(t)

Similarly, if an input x₂(t) produces an output y₂(t), then

2dy₂(t)

dt+5y₂(t)=x₂2(t)

The Linear combination of the above equations can be written as

a2dy₁(t)

dt+a5y₁(t)+b 2

dy₂(t)

dt+b5y₂(t)=ax₁2(t)+b x₂2(t)

2d

dt[a y₁(t)+b y₂(t)]+5[a y₁(t)+b y₂(t)]= ax₁2(t)+b x₂2(t)

This is not a function of weighted sum of inputs. Hence superposition principle is not

satisfied. Therefore, the system is non-linear.

Time-invariant and Time-varying systems

Time-invariance is the property of system which makes the behaviour of the system

independent of time.

A system is said to be time-invariant if its input/output characteristics do not change with

time.

If x(t)→y(t)

Then x(t-T)→y(t-T)

A system not satisfying the above requirements is called a time-varying system.

The time-invariance property of the given continuous time system can be tested as

follows: let

x(t) be the input and x(t-T) be the input delayed by T units

y(t)= T[s(t)] be the outpu8t for an input x(t).

y(t,T)=T[x(t-T)]= 𝑦(𝑡)|x(t)=x(t-T) be the output for the delayed input x(t-T).

y(t-T)= 𝑦(𝑡)|t=t-T be the output delayed by T units.

If y(t,T)= y(t-T)

i.e. if the delayed output is equal to the output due to delayed input for all possible values of t,

then the system is time-invariant.

On the other hand, if y(t, T)≠y(t-T)

i.e. if the delayed output is not equal to the output due to delayed input, then the system is time-

variant.

Illustration

Example:

Determine whether the following systems are time-invariant or not:

(a) y(t)= t2x(t) (b)y(t)=е2x(t)

Solution:

(a) y(t)= t2x(t)

The output due to input delayed by T sec is:

y(t,T) = T[x(t-T)] =𝑦(𝑡)|x(t)=x(t-T) = t2x(t-T)

The output delayed by T sec

y(t-T) = 𝑦(𝑡)|t=t-T = (t-T)2x(t-T)

y(t, T) ≠ y(t-T)

i.e. the delayed output is not equal to the output due to delayed input.

Therefore, the system is time-variant.

(b)y(t)=е2x(t)

y(t) = T[x(t)] = е2x(t)

The output due to input delayed by T sec is

y(t, T) = T[x(t-T)] = 𝑦(𝑡)|x(t)=x(t-T) = е2x(t-T)

The output delayed by T sec is

y(t-T) = 𝑦(𝑡)|t=t-T = е2x(t-T)

y(t, T) = y(t-T)

i.e, the delayed output is equal to the output due to delayed input.

Therefore, the system is time-invariant.

Stable and Unstable systems

A bounded signal is signal whose magnitude is always a finite value (sinewave is a

bounded signal). A system is said to be bounded-input, bounded-output(BIBO) stable, if and

only if every bounded input produces a bounded output. The output of a stable system does not

diverge or does not grow unreasonably large.

Let the input signal x(t) be bounded

⎸x(t)⎹ ≤ Mx˂ ∞ for all t

Where Mx is a positive real number.

If ⎸y(t)⎹≤ My˂ ∞

i.e. if y(t) is also bounded, then the system is BIBO stable. Otherwise, the system is unstable.

That is, we say that a system is unstable even if one bounded input produces an unbounded

output.

The stability can be found from the impulse response of the system which is nothing but

the output of the system for a unit impulse input. If the impulse response is absolutely integrable

for a continuous-time system or absolutely summable for a discrete-time system, then the system

is stable.

Illustration

Example: Find whether the following systems are stable or not:

(a) y(t)= ex(t); ⎸x(t)⎹ ≤ 8 (b) y(t)= (t+5)u(t)

Solution:

(a) y(t)= ex(t); ⎸x(t)⎹ ≤ 8

Here the input is bounded, ⎸x(t)⎹ ≤ 8

Therefore for stability, the output must be bounded.

The output y(t) becomes

e-8 ≤ y(t) ≤ e8

Hence y(t) is also bounded. Therefore, the system is stable.

(b) y(t)= (t+5)u(t)

Therefore y(t)= t+5 for t≥0

So, as t→∞, y(t)→∞

Hence the output grows without any bound and hence the given system is unstable.

Invertible and non-invertible systems

If a system has a unique relationship between its input x(t) [or x(n)] and output y(t) [or

y(n)], the system is known as invertible. Therefore, for an invertible system if y(t) [or y(n)] is

known, x(t) [or x(n)] can be found out unambiguously and uniquely. On the other hand, if the

system does not have a unique relationship between its input and output, the system is said to be

non-invertible.

In other words a system is known as invertible only if an inverse system exists which

when cascaded with the original system produces an output equal to the input of the first system.

For y(t)= 3x(t) [ or y(n)= 3x(n)], the system is said to be invertible, whereas for y(t)=

2x2(t) [ or y(n)= 2x2(n)], the system is said to be non-invertible. Mathematically, a system is to

be invertible if

x(t) = T-1 {T[x(t)]}

or x(n) = T-1 {T[x(n)]}

The block diagram representation of both an invertible and non-invertible system is

shown in figure

x(t) T-1{T[x(t)]} = x(t) x(t) T-1{ T[x(t)]}≠ x(t)

x(n) T-1{T[x(n)]} = x(n) x(n) T-1{T[x(n)]}≠x(n)

Invertible system non-invertible system

T T-1 T-1

T

T T-1

FIR and IIR systems

Linear time invariant systems can be classified according to the type of impulse response.

If the impulse response sequence is of finite duration, the system is called a finite impulse

response (FIR) system and if the impulse response sequence is of infinite duration, the system is

called an infinite impulse response (IIR) system.

An example of FIR system is described by

-2 n=2,4

h(n) = 2 n=1,3

0 otherwise

An example of IIR system is described by

h(n) =2n u(n)

Properties of linear time invariant systems

An extremely important representation of continuous- time LTI systems is in terms of its

unit impulse response. It takes the form of convolution integral. The importance of impulse

response is the characteristics of an LTI system are completely determined by its impulse

response. The unit impulse response of a nonlinear system does not completely characterize the

behavior of the system.

LTI systems have a number of properties not possessed by other systems beginning with

convolution integral. Some of the most basic and important properties of continuous-time LTI

systems are—

1. The commutative property

A basic property of convolution in continuous-time is that it is a commutative operation.

x(t)*h(t)= h(t)*x(t)= ∫ 𝑥(𝜏)ℎ(𝑡 − 𝜏)𝑑𝜏∞

−∞ = ∫ ℎ(𝜏)𝑥(𝑡 − 𝜏)𝑑𝜏

∞

−∞

i.e. The output of an LTI system with input x(t) and unit impulse response h(t) is identical to the

output of an LTI system with input h(t) and impulse response x(t).

2. The distributive property

Another basic property of convolution is the distributive property. Specifically convolution

distributes over addition so that

x(t)*[h1(t)+h2(t)] = x(t)*h1(t)+ x(t)*h2(t)

The distributive property has a useful interpretation in terms of system interconnection.

Two systems with impulse response h1(t) and h2(t) connected in parallel can be replaced by a

single system with impulse response h1(t)+h2(t). The distributive property of convolution can be

used to break a complicated convolution into several simpler ones. Figure illustrates the

distributive property.

x(t) y(t) = x(t) y(t)

3. The associative property

Another important and useful property of convolution is that it is associative. That is

x(t)*[h1(t)*h2(t)] = [x(t)*h1(t)]*h2(t)

i.e. it is immaterial in which order the signals are convolved.

According to the associative property, the series interconnection of two systems is equivalent to

a single system. Figure illustrates the associative property.

x(t) y(t) = x(t) y(t)

System 1 System 2

The impulse response of the cascade of two LTI systems is the convolution of their individual

impulse responses.

4. Systems with and without memory

A system is said to be static or memoryless if its output at any time depends only on the value of

the input at that time. A continuous-time LTI system is memoryless if h(t)=0 for t≠0 and such a

memoryless LTI system has the form.

y(t) = kx(t)

for some constant k and has the impulse response

h(t)=kδ(t)

therefore, if h(t)≠0 for t≠0

the continuous-time system has memory. A memory system is also called a dynamic system. If

k=1, h(t)=δ(t), the system becomes an identity system.

5. Causality

A causal system is non anticipatory and does not produce an output before an input is applied. Its

output depends only on the present and past values of input but not on future inputs.

Therefore for a causal LTI system we have h(t)=0 for t˂0.

The output of a causal LTI system for a noncausal input is

y(t)= ∫ ℎ(𝜏)𝑥(𝑡 − 𝜏)𝑑𝜏∞

0 = ∫ 𝑥(𝜏)ℎ(𝑡 − 𝜏)𝑑𝜏

𝑡

−∞

The output of a causal LTI system for a causal signal is

y(t)= ∫ ℎ(𝜏)𝑥(𝑡 − 𝜏)𝑑𝜏𝑡

0 = ∫ 𝑥(𝜏)ℎ(𝑡 − 𝜏)𝑑𝜏

𝑡

0

A noncausal system is anticipatory and h(t)≠0 for t˂0

6. Stability

A system is stable if every bounded input produces a bounded output. The BIBO stability of an

LTI system can be easily determined from its impulse response. For a continuous-time LTI

system to be BIBO stable, its impulse response h(t) must be absolutely intergrable.

∫ ℎ(𝜏)𝑑𝜏∞

−∞˂∞

h1(t)

+h2(

t)

h1(t)

+h2(

t)

+ h1(t)+h2(t)

H1(t) H2(t) h(t)= h1(t)*h2(t)

7. Invertibility

A continuous-time LTI system with impulse response h(t) is said to be invertible, if an inverse

system with impulse response h1(t) which when connected in series with the original system

produces an output equal to the input of the first system

i.e. h(t)*h1(t) = δ(t)

Figure illustrates Invertibility of system

x(t) y(t) x(t) ≡ x(t) x(t)

8. The unit step response

The unit step response s(t) of an LTI system is the output of the system for a unit step input u(t).

The unit step response can be obtained by convolving the input unit step u(t) with the impulse

response h(t) of the system.

Therefore s(t)=h(t)*u(t)=u(t)*h(t)

s(t)=∫ ℎ(𝜏)𝑑𝜏𝑡

−∞

i.e. The unit step response of a continuous-time LTI system is the running integral of its impulse

response.

h(t)=ds(t)/dt

The unit impulse response is the first derivative of the unit step response.

Therefore the unit step response can also be used to characterize an LTI system because a

unit impulse response can be characterized from the unit step response.

Transfer Function of an LTI System:

The transfer function of a continuous-time LTI system may be defined using

Fourier transform or Laplace transform. The transfer function is defined only under zero initial

conditions.

A continuous time system is shown in fig:

The i/p signal The o/p signal

X(t) y(t)

The i/p spectrum The o/p spectrum

X(s) or X(ω) Y(s) or Y(ω)

Fig: A system

The transfer function of a LTI system H(ω) is defined as the ratio of the

Fourier transform of the output signal to the Fourier Transform of the input signal when the

initial conditions are zero.

𝐻(𝜔) = 𝑌(𝜔)/𝑋(𝜔) H(ω) is a complex quantity having magnitude and phase.

𝐻(𝜔) = |𝐻(𝜔)|𝑒𝑗𝜃(𝜔) The transfer function in frequency domain H(ω) is also called frequency response of the system.

The frequency response is amplitude response plus phase response.

|𝐻(𝜔)|= Amplitude response of the system.

h(t) h1(t) Identity system δ(t)

The system h(t)or H(s)or H(ω)

θ(ω)=⌊𝐻(𝜔)= Phase response of the system.

We can say that H(ω) is a frequency domain representation of a system.

Since Y(ω)=H(ω)X(ω)

|𝑌(𝜔)| = |𝐻(𝜔)||𝑋(𝜔)| ⌊𝑌(𝜔) = ⌊𝐻(𝜔) + ⌊𝑋(𝜔)

H(ω) has conjugate symmetry property.

H(-ω)=H*(ω)

i.e. |𝐻(−𝜔)| = |𝐻(𝜔)| and ⌊𝐻(ω)=-⌊𝐻(𝜔)

The impulse response h(t) of a system is the inverse Fourier transform of its transfer function

H(ω).

H(ω)=F[h(t)]

h(t)=𝐹−1 [H(ω)]

The transfer function of a system in s-domain (Laplace domain) is defined as the ratio of the

Laplace transform of the output of the system to the Laplace transform of the input of the system

when the initial conditions are neglected.

𝐻(𝑠) = 𝑌(𝑠)/𝑋(𝑠) H(s)=L[h(t)]

The impulse response is nothing but the inverse Laplace transform of the transfer function H(s)

h(t)=𝐿−1 [H(s)]

Once the transfer function is s-domain H(s)is known, the transfer function in frequency domain

H(ω) can be found by replacing s in H(s) by jω.

H(ω)=𝐻(𝑠)|s=jω

Illustration

Example: Let the system function of an LTI system be 1/(jω+2). What is the output of the

system for an input (0.8)𝑡u(t)?

Solution: Given that the transfer function of the system

H(ω)=1/jω+2

The input X(t)= (0.8)𝑡u(t)

The impulse response h(t)= 𝐹−1 [H(ω)]= 𝐹−1 (1/jω+2)=𝑒−2𝑡u(t)

We know that the output y(t) is the convolution of input x(t) and impulse response h(t).

Y(t)=x(t)*h(t)=∫ ℎ(𝜏)𝑥(𝑡 − 𝜏) 𝑑𝜏∞

−∞

=∫ 𝑒−2𝜏∞

−∞𝑢(𝜏)(0.8)𝑡−𝜏u(t-𝛕) d𝛕

=∫ 𝑒−2𝜏∞

0(0.8)𝑡 (0.8)−𝜏 𝑑𝝉

=(0.8)𝑡 ∫ (0.8𝑒2∞

0)-𝛕 d𝛕

Let (0.8e2)-1=a

Y (t) = (0.8)t∫ 𝑎∞

0𝛕 d𝛕

= (0.8)t[a𝛕 /log a ]₀ ∞ (𝑠𝑖𝑛𝑐𝑒 ∫ 𝑎x dx = ax/ log a)

= (0.8)t [(0.8e2)-𝛕 /log [(0.8e2)-1]∞0 = (0.8)t

log[(0.8e2)−1][(0.8e2)-∞ -1]

= (0.8)𝑡

−log (0.8𝑒2)[0-1]

=(0.8)𝑡

log (0.8𝑒2) =

(0.8)𝑡

log (0.8+2)

Output of the system y (t) = (0.8)𝑡

log (0.8+2)

Filter Characteristics of linear systems:

For a given system an input signal x(t) gives rise to a response signal y(t).

The system processes the signal x(t) in a way that is characteristic of the system. The spectral

density function of the input signal x(t) is given by X(s) or X(ω), and the spectral density

function of the response signal y(t) is given by Y(s) or Y(ω). Y(s)=H(s)X(s) or Y(ω)=H(ω)X(ω)

where H(s) or H(ω) is the transfer function or system function of the system.

The system modifies the spectral density function of the input. The system

acts as a kind of filter for various frequency components. Some frequency components are

boosted in strength, i.e. they are amplified. some frequency components are weakened in

strength, i.e. they are attenuated and some may remain unaffected. Similarly, each frequency

component suffers a different amount of phase shift in the process of transmission. The system,

therefore, modifies the spectral density function of the input according to its filter characteristics.

The modification is carried out according to the transfer function H(s) or H(ω), which represents

the response of the system to various frequency components. H(ω) acts as a weighting function

or spectral shaping function to the different frequency components in the input signal. An LTI

system, acts as a filter. A filter is a basically a frequency selective network.

(i) Some LTI systems allow the transmission of only low frequency components and

stop all high frequency components. They are called low – pass filters (LPFs).

(ii) Some LTI systems allow the transmission of only high frequency components and

stop all low frequency components. They are called high – pass filters (HPFs).

(iii) Some LTI systems allow the transmission of only a particular band of frequencies and

stop all other frequency components. They are called band pass filters (BPFs).

(iv) Some LTI systems reject the transmission of only a particular band of frequencies and

allow all other frequency components. They are called band-rejection filters (BRFs).

The band of frequency that is allowed by the filter is called pass-band.

The band of frequency that is severely attenuated and not allowed to pass through the filter is

called stop-band or rejection-band. An LTI system may be characterized by its pass-band, stop-

band and half power band width.

Distortion less Transmission through a system:

The change of shape of the signal when it is transmitted through a system is called

distortion. Transmission of a signal through a system is said to be distortion less if the output is

an exact replica of the input signal. This replica may have different magnitude and also it may

have different time delay. A constant change in magnitude and a constant time delay are not

considered as distortion. Only the shape of the signal is important. Mathematically, We can say

that a signal x(t) is transmitted without distortion if the output

Y(t) = kx(t-td)

Where k is a constant representing the change in amplitude and td is delay time.

Taking Fourier transform on both sides of the equation for y(t) and using the shifting property,

we have

Y(ω)= k e-jωtd

Therefore inverse Fourier transform, the corresponding impulse response must be

h(t)=kδ(t-td)

it is evident that the magnitude of the transfer function

|𝐻(𝜔)|=k

And that it is constant for all values of ω.

The phase shift

θ (ω) = ⌊𝐻(𝜔)= -ωtd

and it varies linearly with frequency

θ (ω) = n𝜋-ωtd ( n integral )

So for distortion less transmission of a signal through a system, the magnitude

|𝐻(𝜔)|should be a constant, i.e. all the frequency components of the input signal must undergo

the same amount of amplification or attenuation, i.e. the system bandwidth is infinite and the

phase spectrum should be proportional to frequency as shown in figure below:

But, in practice, no system can have infinite bandwidth and hence distortion less conditions are

never met exactly.

Linear phase systems:

For distortion less transmission, there should not be any phase distortion. No phase

distortion means the phase should be linear. Therefore, for distortion less transmission, the

system must be of linear phase type. For linear phase systems, the impulse response is

symmetrical about t=td. This can be proved as follows:

For linear phase system,

H(ω)=⎸H(ω)⎹ e-jωtd

The impulse response of such a system is obtained by finding the inverse Fourier transform.

h (t)= F-1[H(ω)]

=F-1[⎹H(ω)⎸e-jωtd]

=1

2𝜋 ∫ {|𝐻(𝜔)|𝑒−𝑗𝜔𝑡𝑑∞

−∞ }ejωt dω

= 1

2𝜋 ∫ |𝐻(𝜔)|𝑒𝑗𝜔(𝑡−𝑡𝑑)𝑑𝜔

∞

−∞

= 1

2𝜋 [ ∫ |𝐻(𝜔)|𝑒𝑗𝜔(𝑡−𝑡𝑑)𝑑𝜔

0

−∞ + ∫ |𝐻(𝜔)|𝑒𝑗𝜔(𝑡−𝑡𝑑)𝑑𝜔

∞

0

]

= 1

2𝜋 [ ∫ |𝐻(𝜔)|𝑒−𝑗𝜔(𝑡−𝑡𝑑)𝑑𝜔

∞

0 +

∫ |𝐻(𝜔)|𝑒𝑗𝜔(𝑡−𝑡𝑑)𝑑𝜔∞

0 ]

=1

2𝜋 [ ∫ |𝐻(𝜔)|[𝑒𝑗𝜔(𝑡−𝑡𝑑) + 𝑒−𝑗𝜔(𝑡−𝑡𝑑)]𝑑𝜔

∞

0]

=1

𝜋 ∫ |𝐻(𝜔)|𝑐𝑜𝑠𝜔(𝑡 − 𝑡𝑑)𝑑𝜔

∞

0

h (t d+ t)= 1

𝜋 ∫ |𝐻(𝜔)|𝑐𝑜𝑠𝜔𝑡𝑑𝜔

∞

0

h (td - t)= 1

𝜋 ∫ |𝐻(𝜔)|𝑐𝑜𝑠𝜔𝑡𝑑𝜔

∞

0

h (t d+ t)= h (td - t)

This shows that for a linear phase system, the impulse response h(t) is symmetrical about td , and

it is non-coaxial. The maximum value of h(t) is at t=td and is given by

hmax = h(td)= 1

𝜋 ∫ |𝐻(𝜔)|𝑑𝜔

∞

0

Signal Bandwidth:

The spectral components of a signal extend from -∞ 𝑡𝑜∞. Any practical signal has

finite energy. As a result, the spectral components approach zero as ω tends to ∞. Therefore, we

neglect the spectral components which have negligible energy and select only a band of

frequency components which have most of the signal energy is known as the bandwidth of the

signal. Normally, the band is selected such that it contains 95% of total energy depending on the

precision.

Illustration

Example: There are several possible ways of estimating an essential bandwidth of non-band

limited signal. For a low pass signal , for example , the essential band width may be chosen as a

frequency where the amplitude spectrum of the signal decays to K% of its peak value. The

choice of K depends on the nature of application. Choosing K=5, determine the essential

bandwidth of g(t)= e-atu(t).

Solution:

Given g(t)=e-atu(t)

G(ω)=F[e-atu(t)]=1

𝑎+𝑗𝜔

|𝐺(𝜔)|=1

√𝑎2+𝜔2

The peak value of|𝐺(𝜔)| is obtained at ω=0.

⎹G(ω)⎸peak = 1

𝑎

Given K=5%=5

100

Let the new frequency be ωc

5

100𝑎=

1

√𝑎2+𝜔𝑐2

1

2𝑎=

1

√𝑎2+𝜔𝑐2

(20a)2=400a2=a2+ω2c

ω2c=399a2

fc2=

399a2

(2𝜋)2 =10.10a2

fc = 3.179a Hz

Essential bandwidth = 3.179a Hz

System Bandwidth:

For distortion less transmission, we need a system with infinite bandwidth. Due to

physical limitations, it is impossible to construct a system with infinite bandwidth. Actually a

satisfactory distortion less transmission can be achieved by a system with finite, but fairly large

band widths , if the magnitude ⎹H(ω)⎸is constant over this band.

The constancy of the magnitude⎹ H(ω)⎸ in a system is usually

specified by its bandwidth. The bandwidth of a system is defined as the range of frequencies

over which the magnitude ⎹ H(ω)⎸ remain within 1/√2 times (within 3 dB) of its value at mid

band . The bandwidth of a system ⎹ H(ω)⎸plot is shown in below figure is(ω2-ω1) where ω2 is

called the upper cut off frequency or upper 3 dB frequency or upper half power frequency and ω1

is called lower cut off frequency or lower 3dB frequency or lower half frequency.

The band limited signals can be transmitted without distortion , if the system bandwidth is at

least equal to the signal bandwidth.

Worked out Problems:

Example 1: check whether the following system is:

1. Static or dynamic

2. Linear or non-linear

3. Causal or non-causal

4. Time-invariant or time-variant

(a) d³y(t)

dt³+2

d²y(t)

dt²+4

dy(t)

dt+3y²(t)=x(t+1)

Solution:

(a) d³y(t)

dt³+2

d²y(t)

dt²+4

dy(t)

dt+3y²(t)=x(t+1)

1. The system is described by a differential equation. Hence the system is dynamic.

2. There is a square term of output. So the system is non-linear. This can be proved.

Let an input x₁(t) produce an output y₁(t). so the differential equation becomes

d³y₁(t)

dt³+2

d²y₁(t)

dt²+4

dy₁(t)

dt+3y₁²(t)=x₁(t+1)

Let an input x₂(t) produce an output y₂(t). so the differential equation becomes

d³y₂(t)

dt³+2

d²y₂(t)

dt²+4

dy₂(t)

dt+3y₂²(t)=x₂(t+1)

The linear combination of the above equations becomes

ad³y₁(t)

dt³+a2

d²y₁(t)

dt²+a4

dy₁(t)

dt+3ay₁²(t)+b

d³y₂(t)

dt³+b2

d²y₂(t)

dt²+b4

dy₂(t)

dt+b3y₂²(t)

= ax₁(t+1)+bx₂(t+1)

i.e. d³

dt³[ay₁(t)+by₂(t)]+2

d²

dt²[ay₁(t)+by₂(t)]+ 4

d

dt[ay₁(t)+by₂(t)]+3[ay₁²(t)+by₂²(t)]

= ax₁(t+1)+bx₂(t+1) Since one term in LHS is not a weighted sum of outputs, the superposition principle is not valid. Hence the system is non-linear.

3. The output depends on future values of input. Hence the system is non-causal.

4. All the coefficients of the differential equation are constant. Hence the system is time-

invariant.

So the given system is dynamic, non-linear, non-causal and time- invariant.

Example 2 Comment about the linearity, stability, time-invariance and causality for the

following filter: y(n)=2x(n+1)+[x(n-1)]2

Solution: y(t)=at²x(t)+btx(t-4)

1. there is a square term of delayed input in [i.e., x(n-1)2] the difference equation. So the

system is nonlinear.

2. The output depends on the future value of input [i.e., 2x(n+1)]. So the system is non-

causal.

3. For x(n)=δ(n), y(n)=h(n)

Therefore h(n)= 2δ(n+1)+{δ(n-1)}2

h(0)= 2δ(1)+{δ(-1)}2=0+0=0

h(1)= 2δ(2)+{δ(0)}2=0+1=1

h(-1)= 2δ(0)+{δ(-2)}2=2+0=2

h(-2)= 2δ(-1)+{δ(-3)}2=0+0=0

h(n)=0, for any other n

∑ ⎹ℎ(𝑛)⎸∞𝑛=−∞ =0+1+2+0+0+….=3 < ∞

Impulse response is absolutely summable. So the system is stable. Also we can say that

since the output depends only on the delayed and advanced inputs, if the input is

bounded, the output is bounded. So the system is BIBO stable.

4. The output due to delayed input is given by

y(n,k)= 2x(n+1-k)+{x(n-1-k)}2

the delayed output is

y(n-k)= 2x(n-1+k)+{x(n-k-1)}2

y(n,k)= y(n-k)

Therefore, the system is time-invariant.

Also, we can say that since the system is described by constant coefficient difference equation,

the system is time-invariant. So the given system is non-linear, stable, time-invariant and non-

causal.

Example 3 State whether the following system is linear, causal, time-invariant and stable:

y(n)+y(n-1)=x(n)+x(n-2)

Solution:

y(n)+y(n-1)=x(n)+x(n-2)

1. Let an input x₁(n) produce an output y₁(n) and an input x₂(n) produce an output y₂(n).

Therefore, the weighted sum of outputs is:

ay₁(n)+by₂(n) = -[ay₁(n-1)+by₂(n-1)]+[ax₁(n)+bx₂(n)]+[ax₁(n-2)+bx₂(n-2)]

The output due to weighted sum of inputs is:

y₃(n)= -{ay₁(n-1)+by₂(n-1)}+{ax₁(n)+bx₂(n)}+{ax₁(n-2)+bx₂(n-2)}

y₃(n)= ay₁(n)+by₂(n)

so the system is linear

2. The output depends only on the present and past inputs and past outputs. So the system is

causal.

3. All the coefficients of the differential equation are constants. So the system is time-

invariant.

4. For x(n)=δ(n), y(n)=h(n)

h(n)= -h(n-1)+δ(n)+δ(n-2)

h(0)= -h(-1)+δ(0)+δ(-2) = 1

h(1)= -h(0)+δ(1)+δ(-1) = -1

h(2)= -h(1)+δ(2)+δ(0) = 1+0+1 = 2

h(3)= -h(2)+δ(3)+δ(1) = -2+0+0 = -2

∑ ⎹ℎ(𝑛)⎸∞𝑛=−∞ =1+1+2+2+…. = ∞

i.e., the impulse response is not absolutely summable. So the system is unstable.

Therefore, the given system is non-linear, causal, time-variant and unstable.

Example 4 Determine whether the following system is linear, stable, causal and time-invariant

using appropriate tests:

y(n)= nx(n)+x(n+2)+y(n-2)

Solution:

y(n)= nx(n)+x(n+2)+y(n-2)

1. Let an input x1(n) produce an output y₁(n) and an input x₂(n) produce an output y₂(n).

Then the weighted sum of outputs is:

ay₁(n)+by₂(n)= n[ax₁(n)+bx₂(n)]+[ax₁(n+2)+bx₂(n+2)]+[ay₁(n-2)+by₂(n-2)]

The output due to weighted sum of inputs is:

y₃(n)= ay₁(n)+by₂(n)= n[ax₁(n)+bx₂(n)]+[ax₁(n+2)+bx₂(n+2)]+[ay₁(n-2)+by₂(n-2)]

so the system is linear.

2. For x(n)=δ(n), y(n)=h(n)

h(n)= nδ(n)+δ(n+2)+h(n-2)

h(-2)= -2δ(-2)+δ(0)+h(-4)=1

h(0)= 0δ(0)+δ(2)+h(-2)= 0+0+1= 1

h(1)= 1δ(1)+δ(3)+h(-1)=0

h(2)= 2δ(2)+δ(4)+h(0)= 1

∑ ⎹ℎ(𝑛)⎸∞𝑛=−∞ = 1+0+1+0+….. = ∞

So the system is unstable.

3. y(2)= 2x(2)_x(4)+y(0)

The output depends on future inputs. So the system is non-causal.

4. The coefficient of x(n) is a function of time. So it is a time-varying system.

Therefore, the given system is linear, unstable, noncausal and time-varying.

Example 5: check whether the following systems are:

1. Static or dynamic

2. Linear or non-linear

3. Causal or non-causal

4. Time-invariant or time-variant

(a)d²y(t)

dt²+2y(t)

dy(t)

dt+3ty(t)=x(t)

Solution: d²y(t)

dt²+2y(t)

dy(t)

dt+3ty(t)=x(t)

1. The system is described by a differential equation. Hence the system is dynamic.

2. There is a term with product of output and its derivative. Hence the system is non-

linear. This can be proved.

Let an input x₁(t) produce an output y₁(t)

Then d²y₁(t)

dt²+2y(t)

dy₁(t)

dt+3ty₁(t)=x₁(t)

Let an input x₂(t) produce an output y₂(t).

Then d²y₂(t)

dt²+2y(t)

dy₂(t)

dt+3ty₂(t)=x₂(t)

The linear combination of the above equations gives

a d²y₁(t)

dt²+a2y(t)

dy₁(t)

dt+a3ty₁(t) + b

d²y₂(t)

dt²+b2y(t)

dy₂(t)

dt+b3ty₂(t) = ax₁(t)+bx₂(t)

i.e. d²

dt²[ay₁(t)+by₂(t)] +2[ay₁(t)

dy₁(t)

dt+by₂(t)

dy₂(t)

dt]+3t[ay₁(t)+by₂(t)]

= [ax₁(t)+bx₂(t)] Since one term in LHS is not a weighted sum of outputs, the superposition principle is

not valid. Hence the system is causal.

3. The output depends on present input only. Hence the system is causal.

4. All the coefficients of the differential equation are not constants. One coefficient is a

function of time. Hence the system is time-variant.

So the given system is dynamic, non-linear, causal and time-variant.

Example 6: check whether the following systems are:

1. Static or dynamic

2. Linear or non-linear

3. Causal or non-causal

4. Time-invariant or time-variant

(a) y(t)=at²x(t)+btx(t-4)

Solution: y(t)=at²x(t)+btx(t-4)

1. The output depends on past inputs. So it requires memory. Hence it is a dynamic system

2. y(t)=at²x(t)+btx(t-4)

For an input x₁(t), y₁(t)=at²x₁(t)+btx₁(t-4)

For an input x₂(t), y₂(t)=at²x₂(t)+btx₂(t-4)

The weighted sum of outputs is :

py₁(t)+qy₂(t) = pat²x₁(t)+pbtx₁(t-4)+ qat²x₁(t)+qbtx₁(t-4)

= at²[px₁(t)+qx₂(t)]+bt[px₁(t-4)+qx₂(t-4)]

The output due to weifhted sum of inputs is:

y₃(t)= T[px₁(t)+qx₂(t)] =at²[px₁(t)+qx₂(t)]+bt[px₁(t-4)+qx₂(t-4)]

y₃(t)= ay₁(t)+by₂(t)

superposition principle is satisfied. Hence the system is linear.

3. The output depends only on the present and past inputs. It does not depend onfuture

inputs. Hence the system is causal.

4. y(t)=T[x(t)] =at²x(t)+btx(t-4)

The output due to input delayed by T sec is:

y(t,T) = T[x(t-T)] = y(t)⎸x(t)=x(t-T)= at2x(t-T)+btx(t-4-T)

the output delayed by T sec is:

y(t-T)=y(t)⎸t=t-T =a(t-T)2x(t-T)+b(t-T)x(t-T-4)

y(t,T)≠y(t-T)

hence the system is time-variant.

So the given is dynamic , linear, causal and time-variant.

Example:7

Find the transfer function of the Lattice network shown in fig

Solution:

The given circuit diagram is shown in figure (a). The potential across points A

and B is similar to the potential across points C and D. Also the potential across point A and D is

similar to potential across the points B and C. Therefore , The given circuit diagram can be

rearranged as shown in figure (b).

Let i1(t) and i2(t) be the loop currents in figure (b) . The loop equation around loop 1 is:

𝑣i(t)=(1)𝑑𝑖1(𝑡)

𝑑𝑡 +

1

(1)∫[i1(t) − i2(t)]dt +(1)

𝑑𝑖1(𝑡)

𝑑𝑡

Taking Laplace transform on both sides, and neglecting the initial conditions, we have

Vi(s)=sI1(s) + 1

𝑠[I1(s)-I2(s)] + sI1(s)

= [2s+1

𝑠] I1(s)-

1

𝑠I2(s)

The loop equation around loop 2 is:

(1)i2(t)+ 1

(1)∫[i1(t) − i2(t)]dt =0

Taking Laplace transform on both sides, we have

I2(s) + 1

𝑠[I2(s)-I1(s)]=0

[1+1

𝑠] I2(s)-

1

𝑠I1(s)=0

I1(s) = (s+1)I2(s)

The output voltage o(t) is given by

𝑣o(t)=i2(t)(1)

Taking Laplace transform on both sides, we get

Vo(s) I2(s)

Substituting this value of I2(s) in the equation for I1(s), we get

I1(s) = (s+1)Vo(s)

Vi(s)= [2s+1

𝑠] [(s+1)Vo(s) ]-

1

𝑠Vo(s)

=[(2s+1

𝑠)(s+1)]-

1

𝑠]Vo(s)

=Vo(s)(2s2+2s+1+1

𝑠 -

1

𝑠)

= Vo(s)(2s2+2s+1)

Vo(s)

Vi(s) =

1

2s2+2s+1

So the transfer function of the system

H(s) = Vo(s)

Vi(s) =

1

2s2+2s+1

Example 8:

Consider the filter circuit show in bellow figure:

(a) Write the input/output relationship.

(b) Obtain its impulse response.

(c) Find the step response.

Solution:

The Laplace transformed network for the circuit of above figure is shown in

figure(a). Its equivalent obtained by combining the parallel R and C is shown in figure (b).

From basic circuit theory, the input/output relationship for the circuit of figure(b)is:

Y(s)

X(s)=

R/(1+RCs)

R+[R

1+RCs]=

R

2R+R2Cs=

1

2+RCs

The transfer function of the system

H(s) = Y(s)

X(s) =

1

2+RCs =

1

RC[

1

s+(2

RC)]

The impulse response of the system

h(t) = L-1[H(s)] = L-1[1

𝑅𝐶

1

𝑠+(2

𝑅𝐶)] =

1

𝑅𝐶𝑒−(

2

𝑅𝐶)𝑡𝑢(t)

For a step input

x(t) = 𝑢(t)

X(s) = 1

𝑠

The output for a unit step input is :

Y(s) = X(s)( 1

2+𝑅𝐶𝑠) = (

1

𝑠)(

1

2+𝑅𝐶𝑠) =

1

𝑅𝐶[

1

𝑠[𝑠+(2

𝑅𝐶)]

]

Taking partial fractions, we have

Y(s) = 1

𝑅𝐶{

1

𝑠[𝑠+(2

𝑅𝐶)]

} = 1

2 [

1

𝑠 -

1

𝑠+(2

𝑅𝐶)]

Taking inverse Laplace transform, we have

(t) = L-1[Y(s)] = 1

2 L-1 [

1

𝑠 -

1

𝑠+(2

𝑅𝐶)]

= 1

2 [(t)- 𝑒−(

2

𝑅𝐶)𝑡𝑢(t)]

So the step response of the system is:

y(t) =1

2[1- 𝑒−(

2

𝑅𝐶)𝑡

]𝑢(t)

Example 9:

Find the current i(t) in a series RLC circuit as shown in figure, when a voltage of

100 volts is switched on across the terminals aa’ at t=0.

Solution:

The given RLC circuit of the above figure is shown in bellow figure for t>0,i.e.

after the switch is closed.

Since the given switch is closed at t=0,for t<0, it was open and so the initial

current through the inductor and the initial voltage across the capacitor are zero, i.e. the initial

conditions are zero.

The loop equation around the single loop in the circuit is:

100 = 2i(t) = (1)𝑑𝑖(𝑡)

𝑑𝑡 +

1

0.2∫ 𝑖(𝑡)𝑑𝑡

Taking Laplace transform on both sides, and neglecting initial conditions, we get

100

s = 2I(s) + sI(s) +

1

0.2

I(s)

s

=I(s)(2+s+5

s)

=I(s)(s2+2s+5

s )

I(s)[ 𝑠2 + 2𝑠 + 5]=100

I(s) = 100

s2+2s+5 =

100

(s+1)2+22

Taking inverse Laplace transform on both sides, we get

i(t) = L-1[100

(s+1)2+22] = 50 L-1[

2

(s+1)2+22]

= 50 e-t sin 2t (t)

So the current in the RLC circuit for t.0 is i(t)= 50 e-t sin 2t 𝑢(t).

Example 10: check whether the following system is:

i) Static or dynamic

ii) Linear or non-linear

iii) Causal or non-causal

iv) Time-invariant or time-variant

y(t) =e𝑣en{x(t)}

Solution:

y(t) =e𝑣en{x(t)}

= ½ [x(t)+x(-t)]

i) For positive values of t, the output depends on past values of input and for negative

values of t, the output depends on future values of input. Hence the system is dynamic.

ii) y(t)= T[x(t)] = ½ [x(t)+x(-t)]

For an input x₁(t), y₁(t)= = ½ [x₁(t)+x₁(-t)]

For an input x₁(t), y₂(t)= = ½ [x₂(t)+x₂(-t)]

The weighted sum of outputs is :

ay₁(t)+by₂(t) =a½ [x₁(t)+x₁(-t)]+b½ [x₂(t)+x₂(-t)]

=1/2{[ax₁(t)+bx₂(t)]+[ax₁(-t)+bx₂(-t)]}

The output due to weighted sum of inputs is:

y₃(t)= T[ax₁(t)+bx₂(t)] =1/2{[ax₁(t)+bx₂(t)]+[ax₁(-t)+bx₂(-t)]}

y₃(t) = ay₁(t)+by₂(t)

The weighted sum of outputs is equal to the output due to weighted sum of inputs. Hence

superposition principle is valid and the system is linear.

iii) y(-2) = ½[x(-2)+x(2)]

i.e. For negative values of t, the output depends on future values of input. Hence the

system is non-causal

iv) y(t) = ½ [x(t)+x(-t)]

The output due to input delayed by T units is:

y(t,T) = T[x(t-T)] = y(t)⎸x(t)=x(t-T)= ½[x(t-T)+x(-t-T)]

The output delayed by T units is

y(t-T) =y(t)⎸t=t-T = ½ [x(t-T)+x(-t-T)]

y(t-T)≠y(t-T)

So the system is time-variant.

So the given system is dynamic, linear, non-causal and time-variant.

Assignment:

1. check whether the following systems are causal or not:

(a) y(t)= x2(t)+x(t-3) (b) y(t)=∫ 𝑥(𝜏)𝑑𝜏𝑡

−∞

(c)y(t) =x(t+2) (d) y(n)= x(n)+1/[2x(n-2)]

(e)y(n)=x(-2n)

2. check whether the following systems are linear or not

(a) y(n)=Ax(n)+B (b) dy(t)

dt +y2(t)=3x(t)

(c)y(n)= n²x(2n)

3. Determine whether the following systems are time-invariant or not

(a) y(t)= x(t)+tx(t-1) (b) y(t)= x(t)cos20t

(c)y(n)=x²(n-2) (d)y(n)=sin[x(n)]

4. Determine whether the following systems are stable or not

(a)y(t)=r(t)+2 (b) y(t)=5e-2tu(t)

(c)y(n)=8x(n-4) (d)h(n)=2-nu(n)

5. check whether the following systems are

1. static or dynamic

2. linear or nonlinear

3. causal or non-causal

4. time-invariant or time variant

(a)y(t)=atx(t)+bt2x(t-2) (b) d³y(t)

dt³+5

d²y(t)

dt²+6t

dy(t)

dt+2y(t)=x2(t)

(c)y(n)=⎹x(n)⎸ (d)y(t)=od[x(t)]

6. Let the system function of an LTI system be 1/(jω+3) . What is the output of the system

y(t) for an input (0.5)t u(t)?

7. Consider a stable LTI system that is characterized by the differential equation

d2y(t)

dt2 + 6

dy(t)

dt + 5y(t) =

dx(t)

dt + 3x(t)

Find its response for input x(t) = e-3t 𝑢(t).

8. A system produces an output of (t) = e-3t(t) for an input of x(t) = e-5t𝑢(t). Determine the

impulse response and frequency response of the system.

9. For a system excited by x(t) = e-t𝑢(t), the impulse response is

h(t) = e-3t𝑢(t) + et𝑢(-t)

Find the output for the system

10. Consider a stable LTI system characterized by the differential equation . d2y(t)

dt2 +

2dy(t)

dt + y(t) = x(t). Find its impulse response.

Simulation:

%Verification of Linearity and Time-Invariant Properties of a given Continuous / Discrete

System

% Program to Verify Linearity

clc

clear all

close all

x1 = input (' type the samples of x1 '); % Input sequence 1

x2 = input (' type the samples of x2 '); % Input sequence 2

if (length(x1) ~= length(x2))

disp(' ERROR : Lengths of x1 & x2 are different ')

return;

end;

h = input (' type the samples of impulse response ');

N = length(x1) + length(h) -1;

disp('length of the output signal will be ');

disp(N);

a1 = input (' The scale factor a1 is '); % Scaling factor 1

a2 = input (' The scale factor a2 is '); % Scaling factor 2

x = a1 * x1 + a2 * x2;

yo1 = conv(x,h);

% Response due to combined input

y1 = conv(x1,h);

y1s = a1 * y1;

y2 = conv(x2,h);

y2s = a2 * y2;

yo2 = y1s + y2s; % Response due to individual inputs and

%combining outputs

disp ('Input signal x1 is '); disp(x1);

disp ('Input signal x2 is '); disp(x2);

disp ('Output Sequence yo1 is ');

disp(yo1);

disp ('Output Sequence yo2 is ');

disp(yo2);

if ( yo1 == yo2 ) % Verifying Linearity Property

disp(' yo1 = yo2. Hence the LTI system is LINEAR ')

end;

INPUT:

type the samples of x1 [1 2 3 4]

type the samples of x2 [1 1 1 1]

type the samples of impulse response [1 4]

length of the output signal will be

5

The scale factor a1 is 2

The scale factor a2 is 3

OUTPUT:

Input signal x1 is

1 2 3 4

Input signal x2 is

1 1 1 1

Output Sequence yo1 is

5 27 37 47 44

Output Sequence yo2 is

5 27 37 47 44

yo1 = yo2. Hence the LTI system is LINEAR

%Program for Time-Invariance Verification

x = input( ' Type the samples of signal x(n) ' );% Input

%Sequence x(n)

h = input( ' Type the samples of signal h(n) ' );% Input

%Sequence h(n)

y = conv(x,h);

% Performing convolution

disp( ' Enter a POSITIVE number for delay ' );

d = input( ' Desired delay of the signal is ' );

xd = [zeros(1,d), x];

nxd = 0 : length(xd)-1;

yd = conv(xd,h);

nyd = 0:length(yd)-1;

disp(' Original Input Signal x(n) is ');

disp(x);

disp(' Original Output Signal y(n) is ');

disp(y);

disp(' Delayed Input Signal xd(n) is ');

disp(xd);

disp(' output yd(n) obtained for Delayed input Signal xd(n)is');

disp(yd);

xp = [x , zeros(1,d)];

figure

subplot(2,1,1);

stem(nxd,xp); grid;

xlabel( ' Time Index n ' );ylabel( ' x(n) ' );

title( ' Original Input Signal x(n) ' );

subplot(2,1,2);

stem(nxd,xd); grid;

xlabel( ' Time Index n ' );ylabel( ' xd(n) ' );

title( ' Delayed Input Signal xd(n) ' );

yp = [y zeros(1,d)];

figure

subplot(2,1,1);

stem(nyd,yp); grid;

xlabel( ' Time Index n ' );ylabel( ' y(n) ' );

title( ' Original Output Signal y(n) ' );

subplot(2,1,2);

stem(nyd,yd); grid;

xlabel( ' Time Index n ' );ylabel( ' yd(n) ' );

title( ' Delayed Output Signal yd(n) ' );

yo1=[zeros(1,d),y] % delayed original output

yo2=yd %output obtained for delayed input

if ( yo1 == yo2 ) % Verifying Time invariance

disp(' yo1 = yo2. Hence the system is TIME INVARIANT ')

end;

INPUT:

Type the samples of signal x(n) [1 2 3 4 5]

Type the samples of signal h(n) [1 4 7]

Enter a POSITIVE number for delay

Desired delay of the signal is 2

OUTPUT:

Original Input Signal x(n) is

1 2 3 4 5

Original Output Signal y(n) is

1 6 18 30 42 48 35

Delayed Input Signal xd(n) is

0 0 1 2 3 4 5

output yd(n) obtained for Delayed input Signal xd(n) is

0 0 1 6 18 30 42 48 35

yo1 =

0 0 1 6 18 30 42 48 35

yo2 =

0 0 1 6 18 30 42 48 35

yo1 = yo2. Hence the system is TIME INVARIANT

References:

[1] Alan V.Oppenheim, Alan S.Willsky and S.Hamind Nawab, “Signals & Systems”, Second

edition, Pearson Education, 8th Indian Reprint, 2005.

[2] M.J.Roberts, “Signals and Systems, Analysis using Transform methods and MATLAB”,

Second edition, McGraw-Hill Education,2011

[3] John R Buck, Michael M Daniel and Andrew C.Singer, “Computer explorations in Signals

and Systems using MATLAB”,Prentice Hall Signal Processing Series

[4] P Ramakrishna rao, “Signals and Systems”, Tata McGraw-Hill, 2008

[5] Tarun Kumar Rawat, “Signals and Systems”, Oxford University Press,2011

[6] A.Anand Kumar, “Signals and Systems” , PHI Learning Private Limited ,2011