The pole – zero diagram of a causal and stable discrete – time system is shown is shown in the figure. The zero at the origin has multiplicity 4. The impulse response of the system is h[n] . If h[0] = 1, we can conclude.

This question was previously asked in

GATE EC 2015 Official Paper: Shift 1

Option 1 : h[n] is real for all n

CT 1: Ratio and Proportion

3742

10 Questions
16 Marks
30 Mins

__ Concept__:

__Causality and Stability__:

For a system with rational transfer function H(z) to be causal, the ROC should lie outside the outermost pole, and for BIBO stability, the ROC should include the unit circle |z| = 1.

∴ An LTI discrete-time causal system with the rational system function H(z) is said to be stable if all the poles of H(z) lie inside the unit circle.

__ Calculation__:

The transfer function can be written as:

\({\rm{H}}\left( {\rm{z}} \right) = \frac{{{\rm{k}}{{\rm{z}}^4}}}{{\left( {{\rm{z}} - 0.5 - {\rm{j}}0.5} \right)\left( {{\rm{z}} - 0.5 + {\rm{j}}0.5} \right)\left( {{\rm{z}} + 0.5 + {\rm{j}}0.5} \right)\left( {{\rm{z}} + 0.5 - {\rm{j}}0.5} \right)}}\)

\({\rm{H}}\left( {\rm{z}} \right) = \frac{{{\rm{k}}{{\rm{z}}^4}}}{{\left( {{{\rm{z}}^2} - {\rm{z}} + \frac{1}{2}} \right)\left( {{{\rm{z}}^2} + {\rm{z}} + \frac{1}{2}} \right)}}\)

Now, since the system is causal and stable we have,

\({\rm{h}}\left[ 0 \right] = \mathop {\lim }\limits_{{\rm{z}} \to \infty } {\rm{H}}\left( {\rm{z}} \right) = {\rm{k}} = 1{\rm{\;}}\)

Thus, the transfer function will be:

\({\rm{H}}\left( {\rm{z}} \right) = \frac{{{{\rm{z}}^4}}}{{\left( {{{\rm{z}}^2} - {\rm{z}} + \frac{1}{2}} \right)\left( {{{\rm{z}}^2} + {\rm{z}} + \frac{1}{2}} \right)}}\)

H(z) = 1 - 0.25z^{-4}.......

h[n] = [1,0,0,0,-0.25,.....]

Now,

h[0] = 1

so,h[n] is real for all n

__ Note__:

__Causality__:

A linear time-invariant discrete-time system is said to be causal if the impulse response h[n] = 0, for n < 0 and it is therefore right-sided.

The ROC of such a system H(z) is the exterior of a circle. If H(z) is rational, then the system is said to be causal if:

1) The ROC is the exterior of a circle outside the outermost pole; and

2) The degree of the numerator polynomial of H(z) should be less than or equal to the degree of the denominator polynomial.

__Stability__:

A discrete-time LTI system is said to be BIBO stable if the impulse response h[n] is summable, i.e.

\(\mathop \sum \nolimits_{n = - \infty }^\infty \left| {h\left[ n \right]} \right| < \infty \)

z-transform of h[h] is given as:

\(H\left( z \right) = \mathop \sum \nolimits_{n = - \infty }^\infty h\left[ h \right]{z^{ - n}}\)

Let z = ejΩ (which describes a unit circle in the z-plane), then

\(\left| {H\left[ {{e^{j{\rm{\Omega }}}}} \right]} \right| = \left| {\mathop \sum \nolimits_{n = - \infty }^\infty h\left[ n \right]{e^{ - j{\rm{\Omega }}n}}} \right|\)

The above equality can be written as:

\( \le \;\mathop \sum \nolimits_{n = - \infty }^\infty \left| {h\left[ n \right]{e^{ - j{\rm{\Omega }}n}}} \right|\)

\(= \mathop \sum \nolimits_{n = - \infty }^\infty \left| {h\left[ n \right]} \right| < \infty \)

This is the condition for stability. Thus we can conclude that an LTI system is stable if the ROC of the system function H(z) contains the unit circle |z| = 1