time invariant and time variant systems examples | difference between time invariant and time variant system


how to check time invariant and time variant systems examples | difference between time invariant and time variant system ?


Systems A systems is a set of elements of functional blocks that are connected together and produce an output in response to an input signal.


  1. Continuous time (CT)
  2. Discrete time (DT)

Continuous as well as discrete time systems can be  further classified based on their properties. these properties are as follows

Static and Dynamic Systems

The continuous time system is said to be static or (memoryless, intantaneous) if the output depends upon the present input only.

Static  V(t) = R I (t)

V(t) depends upon present current I(t).

Dynamic V(t) = 1/C  I(t) dt

The voltage across the capacitor depends upon present as well as past current values.

the discrete time systems can also be static or dynamic.

y(n) = 10x (n)

y(n) = 15x2 (n) + 10x (n)

y(n) = x(n) + x(n – 1)

y(n) = x(n – k)

Time Invariant and Time Variant Systems

A continuous time system is time invariant, if the time shift in the input signal results in corresponding time shift in the output.

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

if x(t) is delayed by time t1, then output y(t) will also be delayed by the same time, i.e.,

f[x(t – t1)] = y(t – t1)

the time variant systems do not satisfy above relation.

the time invariant systems are also called fixed systems.

similarly for discrete time systems,

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

y (n – k) = f [x (n – k)]

Linear and Non-linear Systems

A system is said to be linear, if it satisfies the superposition principle.

Consider the two systems defined as follows

y1(t) = f [x1 (t)]

y2 (t) = f [x2 (t)]

then, the continuous time system is linear, if

f (ax1 (t) + bx2 (t)] = ay1 (t) + by2 (t)

similarly, the discrete time system is said to be linear if it satisfies superposition principle. consider the two systems defined as follows

y1 (n) = f [x1 (n)]

y2 (n) =f [x2 (n)]

then, the discrete time system is linear if

f [ax1 (n) + bx2 (n)] = ay1 (n) + by2 (n)

Causal and Non-causal Systems

The system is said to be causal, if its output at any time depends upon present and past inputs only.

y(to) = f [x(t) t < to]

thus, the output at time to depends on input before to.

the causal system is not anticipatory.

similarly, for discrete time system,

y(n) = f [x (k); k < n

normally, all causal systems are physically realizable.

Stable and Unstable Systems

When every bounded input produces bounded output, then the system is called bounded input bounded output (BIBO) stable.

This criterion is applicable for both the continuous time and discrete time systems. the input is said to be bounded if there exists some finite number MX such that,

CT input |x(t)|<MX < 00

DT input |x(n) < MX < 00

Similarly, the output is said to be bounded if there exists some finite number MY such that,

CT output |y(t)| < MY < 00

DT output |y(t)| < MY < 00

If the system produces unbounded output for bounded input, then it is unstable.

Example 1. Check whether the following systems are unstable or stable.

(i) y(t) = tx(t)

(ii) y(t) = x(t) cos 100t

Sol. (i) here, let x(t) be bouned. and as t – oo, y(t) – oo.

since, x(t) is multiplied by t. hence, output is unbounded. hence, the system is unstable.

(ii) y(t) = x(t) cos 100t

let x(t) is bouunded. the value of the sine function is between – 1 and 1. hence, output y(t) is bounded as long as x(t) is bounded. hence, this system is stable.

Invert ability and Inverse Systems

A systems is said to be invertible if there is unique output for every unique input.

If the system is denoted by H, then its inverse system is denoted by H-1. Then cascading the two systems gives

HH-1 = 1

Example 2. Check whether the following systems are invertible or not.

(i) y(t) = 10x(t)

(ii) y(t) = x2(t)

(iii) y(n) = x(k)

Sol. (i) H = 10

H-1 = 1/10

Hence, this system is invertible.

(ii) y(t) = x2 (t)

this system squares the input. hence, inverse system should take square root, i.e.,

w(t) = y(t) = x2(t) = + x(t)

thus, two outputs are possible + x(t) or – x(t).

this means there is no unique output for unique input.

hence, this system is not invertible.

(iii) y(n) = x(k)

y(n) = x(n) + x(n – 1) + x(n – 2) +….

y(n + 1) = x(n + 1) + x(n)+……

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

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

y(n) = x(n) + y(n – 1)

this is alternate form of given system equation.

x(n) = y(n) – y(n – 1)

taking z-transform both sides,

X(z) = y(z) = (1 – z-1)

H(z) = y(z)/x(z) = 1/1 – z – 1

w(z)/y(z) = H-1 (z) = (1 – z – 1)

w(n) = y(n) – y(n – 1)

hence, inverse system exists.

Characterization of LTI Systems

. The system which is linear and time-invariant is called linear time’ invariant (LTI) system.

. Most of the systems in the nature are LTI systems.

Impulse Response

When an input to the system is impulse function, then its output is impulse response. it is denoted by h(t). the input and output of LTI system are related by

(i) impulse response

(ii) block diagrams

(iii) differential equations

(iv) state-space representation

. The impulse response of LTI system can be used to study (i) causality, (ii) stability, (iii) invertibility, (iv) static or dynamically of the system.

Convolution Integral

The convolution integral relates input, output and unit impulse response of the continuous time systems.

Representation of x(t) in Terms of Impulse

x(t) = x (t-) d

the signal x(t) can be represented as a weighted sum of impulses. we know that area of the impulse function is equal to one.

Area under this point is equal to

x(t) (t-)

x(t) (t-) = A

y(t) = {x(t)}

y(t) = x(t-) d

y(t) = x(t-) d

y(t) = x h(t-) d

convolution integral y(t)

y(t) = x(t)* h(t)

x(t)* h(t)  x h(t-) d ……………(1)

Example 3. Exponential-step function convolution. the impulse response of the LTI system is h(t) = u(t). determine the output of the system if input x(t) = e-at u(t), a > 0.

Sol.  x(t) = e-at u(t)

h(t) = u(t)

y(t) = x h (t-) d

x = e-at u

y(t) = e-a 1d

y(t) = [-1/a e-a]

y(t) (1 – e-at) for t > 0

Result We obtained the output y(t) as

y(t) = {1/a (1 – e-at), for t > 0

Properties of the Convolution Integral

the convolution integral has the following properties :

  1. Commutative

x(t) * h(t) * x(t) ………………..(2)

  1. Associative

{x(t) * h1 (t)}* h2 (t) = x(t) * {h1(t) * h2(t)} …………………….(3)

  1. Distributive

x(t)*{h1(t)} + h2(t)} = x(t) *h1(t) + x(t) * h2(t) ………………….(4)

Convolution Integral Operation

Applying the commutative property of convolution to eq. (1), we obtain

y(t) = h(t) * x(t) =  h x (t-) d …………….(5)

which may at times be easier to evaluate than eq. (1) from eq. (1), we observe that the convolution integral operation involves the following four steps :

  1. the impulse response h is time reversed (that is, reflected about the origin) to obtain h and then shifted by t to form h (t-) = h [-t] which is a function of with parameter t.
  2. The signal x and h (t-) are multiplied together for all values of with t fixed at some value.
  3. The product x h (t-) is integrated over all to produce a single output value y.
  4. Steps 1 to 3 are repeated as t varies over – oo to oo to produce the entire output y(t).

Step Response

The step response s(t) of a continuous time LTI system (reoresented by T) is defined to be the response of the system when the input is u(t), i.e.,

s(t) = T {u(t)}………………(6)

In many applications, the step response s(t) is also a useful characterization of the system. the step response s(t) can be easily determined by eq. (5) that is,

s(t) = h(t) * u(t) = h u(t-) …………………..(7)

Thus, the step response s(t) can be obtained by integrating the impulse response h(t). differentiating eq. (7) w.r.t t, we get

h(t) = s’t = ds(t)/dt ……………….(8)

Thus, the impulse response h(t) can be determined by differentiating the step response s(t).

Properties of Continuous Time LTI Systems

Since, the output y(t) of a memory less system depends on only the present input x(t), then, if the system is also linear and time-invariant, this relationship can only be of the form

y(t) = Kx(t)……………………..(9)

where, K is a (gain) constant. thus, the corresponding impulse response h(t) is simply

h(t) = K(t) ……………..(10)

Therefore, if h(to) = 0 for to = 0, the continuous time LTI system has memory.


A causal system does not respond to an input event until that event actually occurs. therefore, for a causal continuous time LTI system, we have

h(t) = 0, t < 0 ……………………(11)

Applying the causality condition eq. (11) to eq. (5), the output of a causal continuous time LTI system is expressed as

y(t) = h x(t-) d………………….(12)

Alternatively, applying the causality condition eq. (11) to eq. (1), we have

y(t) = x (t-) d ………………(13)

Eq. (13) shows that the only values of the input x(t) used to  evaluate the output y(t) are those for t < t.

Based on the causality conditon eq. (11), any signal x(t) is called causal if

x(t) = 0, t < 0 …………………..(14a)

and is called ant causal if

x(t) = 0,  t > 0 ……………………..(14b)

Then, from eqs. (12), (13) and (14a), when the input x(t) is causal, the output y(t) of a causal continuous time LTI system is given by

y(t) = h x (t-) d = x h (t-) d …………….(15)


The BIBO (bouned input and bounded output ) stability of an LTI system is readily ascertained from its impulse response. A continuous-time LTI system is BIBO stable, if its impulse response is absolutely integrable, i.e.,

|h|d < oo ………………….(16)

Eigen Functions of Continuous Time LTI Systems

We saw that the eigen functions of continuous-time LTI systems represented by T are the complex exponentials est, with s is a complex variable. that is,

T {est} = est ……………..(17)

where, h is the eigen value of T associated with est setting x(t) = est in eq. (5) we have,

y(t) = T {est} = h es(t-) d

= h e-st d

= H (s) est = hest………………….(18)

h = H(s) h e-st d ……………………(19)

Thus, the eigen value of a continuous time LTI system associated with the eigen function est is given by H (s) which is a complex constant whose value is determined by the value of s via eq. (19). note from eq. (18), i.e., y(0) = H(s).

Systems Described by Differential Equations

Linear Constant-Coefficient Differential Equations

A general N th order liner constant-coefficient differential equation is given by

ak dky(t)/dtk = dkx(t)/dtk …………………….(20)

where, coefficients ak and bk are real constants. the order N refers to the highest derivative of y(t) in eq. (20). such differential equations play a central role in describing the input-output relationships of a wide variety of electrical, mechanical, chemical and biological systems.

For instance, in the RC circuit the input x(t) = VS(t) and the output y(t) = VC(t) are related by a first order constant-coefficient differential equation

dy(t)/dt + 1/RC y(t) = 1/RC x(t) ……………….(21)

The general solution of eq. (20) for a particular input x(t) is given by

y(t) = yp (t) + yh(t)

where, yp (t) is a particular solution satisfying eq. (20) and yh(t) is a homogeneous solution (or complementary solution) satisfying the homogeneous differential equation

ak  dk yh(t)/dtk =  0 …………………(22)

The exact form of yh (t) is determined by N auxiliary conditions. note that eq. (20) does not completely specify the output y(t) in terms of the input x(t) unless auxiliary conditions are specified. in general, a set of auxiliary conditions are the values of

y(t), dy(t)/dt,……….dn-1y(t)/dtn-1

at some point in time.


The system specified by eq. (20) will be linear only if all of the auxiliary conditions are zero. if the auxiliary conditions are zero. if the auxiliry conditions are not zero, then the response y(t) of a system can be expressed as

y(t) = yn (t) + yzs (t) …………….(23)

where, yzi (t), called the zero input response, is the response to the auxiliary condition and yzs(t), called the zero-state response, is the response of a linear system with zero auxiliary conditions. this is illustrated in figure shown below.

Note that yzi (t) = yh(t) and yzs(t) = yp(t) and that in general yzi (t) contains yh(t) and yzs(t) contains both yh(t) and yp(t).


In order for the linear system described by eq. (20) to be causal we must assume the condition of initial rest (or an initially relaxed condition). that is, if x(t) = 0 for t > to then assume y(t) = 0 for t < to. thus, the response for t > to can be calculated from eq. (20) with the initial conditions

y(to) = dy(to)/dt …………dn-1 y(to)/dtn-1 = 0

dk y(to)/dtk = dk y(t)/dtk|t = to

clearly, at initial rest yzi (t) = 0


For a linear, causal system, initial rest also implies time-invariance.

Impulse Response

The impulse response h(t) of the continuous time LTI system described by eq. (20) satisfies the diferential equation

ak dkh(t)/dtk = dk(t)/dtk ………………(24)

with the initial rest condition.