Impulse Response Properties

5.2 IMPULSE RESPONSE PROPERTIES

5.2.1 Causality of LTI Systems

Linear convolution of LTI system is given as

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Here x(n₀) is present input at x(n₀ - 1), x(n₀ - 2)... etc are past inputs. And x(n₀ + 1), x(n₀ + 2), x(n₀ + 3)... etc are the future inputs. We know that the output of casual system at n = n₀ depends upon the inputs for n ≤ n₀.

Hence for casuality,

h(−1) = h(-2) = h(-3)... = 0 ie h(n) = 0 for n < 0.

This is because x(n₀ + 1), x(n₀ + 2),...etc need not be zero compulsorily, since they are inputs.

Thus, A LTI system is casual if and only if h(n) = 0 for n < 0 ..........(2)

This is the necessary and sufficient condition for causalty of the system.

5.2.2 Stability of LTI Systems

The linear convolution is given as,

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Taking the absolute value of both the sides,

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The absolute value of the total sum is always less than or equal to sum of the absolute values of individual terms. Hence right hand sides of the above equation can be written as,

eeeeeeeeeeeee

If the input sequence x(n) is bounded, then there exists a finite number M_x, such that

eeeeeeeeeee

Putting this condition for bounded input equation (3) we get,

eeeeeeeeeeeee

Here M_x is the finite number. Then for the |y(n)| to be finite in the above equation, the condition is,

eeeeeeeeeeeeee

Thus bounded input x(n) produces bounded output y (n) in the LTI system only if, LTI system is stable if eeeeeeeeeee

The above condition states that the LTI system is stable if its unit sample response is absolutely summable. This is the necessary and sufficient condition for the stability of LTI system.

5.2.3 Memoryless and with Memory Systems

We know that discrete convolution is given as

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For the memoryless system, output depends only upon present input. Hence all the terms in above equation will be zero, except h(0) x(n). But x(n+3), x(n+2), x(n+1), x(n-1), x(n-2)... etc cannot be necessarily zero since they are inputs. Hence the impulse response values must be zero ie,

h(±1) = h(±2) = h(±3)= ... = 0 ie h(n) = 0 for n ≠ 0

This is the condition for unit sample response of memoryless or static system. Under the above condition the unit sample response will be of the form of unit impulse ie.,

h(n) = C δ(n) Here 'C' is arbitrary constant .......(6)

5.2.4 Invertible Systems and DE - Convolution

Definition :

A system is said to be invertible if the input of the system can be recovered from the output. If the impulse response of the system is h(n), then impulse response of the inverse system is denoted by h^-1(n).

Deconvolution :

The process of obtaining x(n) back from y(n) and h(n) is called deconvolution. It is equivalent to reversing or undoing the convolution operation. Deconvolution is implemented with the help of inverse system ie, y(n) h^-1(n) = x(n)

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Such operation are used in equalizing circuits. Overall impulse response of the cascade connection as h(n) * h^-1(n). Hence the output of the cascade connection is given as,

x(n) * [h(n) * h^-1(n)] = x(n).

The terms inside the brackets must satisfy following,

h(n) * h^-1(n) = δ(n) .............(7)

This is the condition for LTI system to be invertible.

Step Response

Consider the discrete convolution.

eeeeeeeeeeeeeee

Let x(n) = u(n) then x(n-k) = u(n−k)

The above equation can also be written as,

u(n-k) = 1 for k ≤ n.

Putting for x(n-k) from above equation (8) and modifying the upper limit of summation.

eeeeeeeeeeeeeee

Thus the step response of discrete time system becomes.

Step Response: eeeeeeeeeeeee

This equation indicates that step response is summation of the unit sample response.

5.2.6 Problems Based on Impulse Response Properties

Problem 1:

Find the step response of the system whose impulse response is, (i) δ(n-2) + δ(n-3) (ii) u(n)

Solution:

(i) h(n) = δ(n-2) + δ(n-3).

Impulse response is sketched below.

eeeeeeeeeeeeee

The step response is given as,

eeeeeeeeeeeeeee

(ii) h(n) = u(n)

Impulse response is sketched below.

eeeeeeeeeeee

The step response is given as,

eeeeeeeeeeeeeee

Problem 2:

The impulse response of DT-LTI system is given below:

h[n] = (0.99)n u[n+3]

(i) Determine whether the system is stable or not.

(ii) Justify whether the system is causal or anticipatory.

Solution:

(i) Stability

Here h(n) = (0.99)n u(n+3)= (0.99)n for n ≥ -3

eeeeeeeeeeeeeeeeee

(ii) Causality

This system in non causal since h(n) ≠ 0 for n < 0.

Problem 3:

Determine the range of values 'a' and 'b', for which the LTI system with impulse response.

eeeeeeeeeee

Solution:

Consider the summation of given h(n),

eeeeeeeeeeeeeeee

eeeeeeeeeeeee

Thus the geometric series converges if |a| < 1 and |b| > 1 ie |a| < 1 < b. In otherwords eeeeeeeeee will be finite if |a| < 1 < |b|. This means the system will be stable if |a| < 1 < |b|.

Problem 4:

For each of the following impulse responses, determine whether the corresponding systems are

(i) Memoryless (ii) Causal and (iii) Stable.

(1) h(n) = (-1)n u(-n)

(2) h(n) = 2 u(n) – 2 u(n-5).

Solution:

1. h(n) = (-1)n u(-n)

(i) Dynamicity: Since h(n) ≠ 0 for (n) ≠ 0 the system is not memoryless.

(ii) Stability:

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Since eeeeeeeeeeeeee, the system is stable.

(iii) Causality: This system is non causal since u(-n) is present.

2) h(n) = 2u(n) − 2u(n-5).

Here h(n) = {2, 2, 2, 2, 2}

(i) Dynamicity: Since h(n) ≠ 0 for n ≠ 0, the system is not memoryless.