Region of Convergence of a Causal System MCQ Quiz - Objective Question with Answer for Region of Convergence of a Causal System - Download Free PDF

Concept:

• If a sequence is absolutely summable, then its DTFT will exist and the ROC will include the unit circle.
• If ROC includes the unit circle then the system will be a non-causal system, because causal systems are right-sided signals with ROC extending to the right side of z-plane.

Analysis:

• Given sequence is summable, so its ROC will include the unit circle, which implies that the DTFT exists.
• Poles location shown below:
  
  
   
• Here ROC includes the unit circle as can be easily seen from the above graph which means that the ROC is left sided.
• Hence it is a non-causal system

• So Option 3 is Correct.

Concept:

ROC is the region of the range of values for which the given summation converges:

$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left[n\right]z^{-n}$

Where z is a continuous complex variable defined in the polar form as:

z = rejω

This is explained with the help of the following:

Properties of Z-transform:

1) The ROC of x(z) consists of a ring in the z – plane centered about the origin.

2) ROC does not contain any pole.

3) If x(n) is a finite duration sequence, then the ROC is the entire z-plane except possibly z = 0 or z = ∞

4) If x(n) is a right sided sequence and if the circle |z| = a is in the ROC then all finite values of z for which |z| > a will also be in ROC

5) If x(n) is a left-sided sequence and if the circle |z| = a is the ROC then all finite values of z for which |z| < a will also be in the ROC.

6) If x(n) is two sided sequence and if the circle |z| = a is in the ROC, then the ROC will consist of a ring in the z-plane that includes the circle |z| = a

Explanation:

For F.T to exist the ROC must include the unit circle.

ROC is ½ < |z| < 2

Causal means that the ROC is facing away from the origin, i.e. ROC is to the right of the rightmost pole.

∴ Starting from outermost pole, the ROC is |z|> 5

Concept:

• A stable system requires the ROC of the Z transform to include the unit circle. 
• A causal system requires the ROC of the Z transform to be outside the outermost pole.
• A minimum phase requires all the poles and zeros to lie inside the unit circle. 

Analysis:

$H\left(z\right)=\frac{2-\frac{3}{4}{z}^{-1}}{\left(1-\frac{1}{2}{z}^{-1}\right)\left(1-\frac{1}{4}{z}^{-1}\right)}$

for |z| > ½ system is both stable and causal

for |z| < ¼ and ¼ < |z| < ½ system is neither stable nor causal.

The condition for the only causal system doesn’t exist.

Concept:

• If a sequence is absolutely summable, then its DTFT will exist and the ROC will include the unit circle.
• If ROC includes the unit circle then the system will be a non-causal system, because causal systems are right-sided signals with ROC extending to the right side of z-plane.

Analysis:

• Given sequence is summable, so its ROC will include the unit circle, which implies that the DTFT exists.
• Poles location shown below:
  
  
   
• Here ROC includes the unit circle as can be easily seen from the above graph which means that the ROC is left sided.
• Hence it is a non-causal system

• So Option 3 is Correct.

Concept:

ROC is the region of the range of values for which the given summation converges:

$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left[n\right]z^{-n}$

Where z is a continuous complex variable defined in the polar form as:

z = rejω

This is explained with the help of the following:

Properties of Z-transform:

1) The ROC of x(z) consists of a ring in the z – plane centered about the origin.

2) ROC does not contain any pole.

3) If x(n) is a finite duration sequence, then the ROC is the entire z-plane except possibly z = 0 or z = ∞

4) If x(n) is a right sided sequence and if the circle |z| = a is in the ROC then all finite values of z for which |z| > a will also be in ROC

5) If x(n) is a left-sided sequence and if the circle |z| = a is the ROC then all finite values of z for which |z| < a will also be in the ROC.

6) If x(n) is two sided sequence and if the circle |z| = a is in the ROC, then the ROC will consist of a ring in the z-plane that includes the circle |z| = a

Explanation:

For F.T to exist the ROC must include the unit circle.

ROC is ½ < |z| < 2

Causal means that the ROC is facing away from the origin, i.e. ROC is to the right of the rightmost pole.

∴ Starting from outermost pole, the ROC is |z|> 5

Concept:

• If a sequence is absolutely summable, then its DTFT will exist and the ROC will include the unit circle.
• If ROC includes the unit circle then the system will be a non-causal system, because causal systems are right-sided signals with ROC extending to the right side of z-plane.

Analysis:

• Given sequence is summable, so its ROC will include the unit circle, which implies that the DTFT exists.
• Poles location shown below:
  
  
   
• Here ROC includes the unit circle as can be easily seen from the above graph which means that the ROC is left sided.
• Hence it is a non-causal system

• So Option 3 is Correct.

Concept:

• A stable system requires the ROC of the Z transform to include the unit circle. 
• A causal system requires the ROC of the Z transform to be outside the outermost pole.
• A minimum phase requires all the poles and zeros to lie inside the unit circle. 

Analysis:

$H\left(z\right)=\frac{2-\frac{3}{4}{z}^{-1}}{\left(1-\frac{1}{2}{z}^{-1}\right)\left(1-\frac{1}{4}{z}^{-1}\right)}$

for |z| > ½ system is both stable and causal

for |z| < ¼ and ¼ < |z| < ½ system is neither stable nor causal.

The condition for the only causal system doesn’t exist.

z-transform of a discrete all pass system is given as

$H\left(z\right)=\frac{z^{-1}-z_0^{\ast }}{1-z_0{z}^{-1}}$

It has a pole at z₀ and a zero at $\frac{1}{z_0^{\ast }}$

Now, the given system has a pole at

$z=2\mathrm{\angle }{30}^{\circ }=\frac{2\left(\sqrt{3}+J\right)}{2}=\sqrt{3}+J$

System is stable if |z|<1 and for this it is anti-causal.

$X\left(z\right)=\frac{1}{\left(1+z^{-1}\right){\left(1-z^{-1}\right)}^2}$

$=\frac{z^3}{\left(z+1\right){\left(z-1\right)}^2}$

$\Rightarrow \frac{X\left(z\right)}{z}=\frac{z^2}{\left(z+1\right){\left(z-1\right)}^2}$

$=\frac{1}{4\left(z+1\right)}+\frac{1}{2{\left(z-1\right)}^2}+\frac{3}{4}\frac{1}{\left(z-1\right)}$

For causal system, ROC is right sided sequence

$\Rightarrow \frac{X\left(z\right)}{z}=\frac{z}{4\left(z+1\right)}+\frac{z}{2{\left(z-1\right)}^2}+\frac{3z}{4\left(z-1\right)}$

By applying inverse z transform

$\Rightarrow x\left(n\right)=\frac{1}{4}{\left(-1\right)}^{n}u\left(n\right)+\frac{1}{2}nu\left(n\right)+\frac{3}{4}u\left(n\right)$

The equivalent system can be represented as

The transfer function from the block diagram can be written as

$\$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}=\left(1+\frac{9}{8}{z}^{-1}\right)\left(\frac{1}{1+\frac{1}{3}{z}^{-1}-\frac{2}{9}{z}^{-2}}\right)$

$=\frac{1+\frac{9}{8}{z}^{-1}}{\left(1+\frac{2}{3}{z}^1\right)\left(1-\frac{1}{3}{z}^{-1}\right)}$

H(z) has poles at $z=\frac{1}{3}andz=\frac{-2}{3}$ for system to be causal ROC must lie outside the outermost pole.

$\left|z\right|>\frac{2}{3}$

Since ROC contains unit circle the system is stable.

The above block diagram can be redrawn as

We have one intermediate sequence marked as $\mathrm{w}[\mathrm{n}]$

Now, $\frac{\mathrm{W}\left(\mathrm{z}\right)}{\mathrm{X}\left(\mathrm{z}\right)}=\frac{1}{1+\frac{\mathrm{k}{\mathrm{z}}^{-1}}{5}}$

and $\frac{\mathrm{Y}\left(\mathrm{z}\right)}{\mathrm{W}\left(\mathrm{z}\right)}=1-\frac{\mathrm{k}{\mathrm{z}}^{-1}}{4}$

$\begin{array}{l}\Rightarrow \frac{Y\left(z\right)}{X\left(z\right)}=\frac{Y\left(z\right)}{W\left(z\right)}.\frac{W\left(z\right)}{X\left(z\right)}\\ \Rightarrow \frac{\mathrm{Y}\left(\mathrm{z}\right)}{\mathrm{X}\left(\mathrm{z}\right)}=\frac{1-\frac{\mathrm{k}{\mathrm{z}}^{-1}}{4}}{1+\frac{\mathrm{k}{\mathrm{z}}^{-1}}{5}}\end{array}$

Now for a system to be causal, power of numerator should not be greater than the power of denominator. Thus the system is already causal if the ROC is outside the outmost pole.

$\Rightarrow \left|\mathrm{z}\right|>\frac{\left|\mathrm{k}\right|}{5}$

And for system to be stable $\frac{\left|\mathrm{k}\right|}{5}<1$

$\Rightarrow \left|\mathrm{k}\right|<5$

$\Rightarrow \mathrm{k}\in \left(-5,5\right)$

Thus, the maximum value of $\mathrm{k}$ for which system is stable is $4$.

For a system to be causal the power of numerator cannot be greater than the power of denominator only option a satisfies this condition