Question 1 |

The transfer function of a stable discrete-time LTI system is H(z)=\frac{K(z-\alpha )}{z+0.5}, where K
and \alpha are real numbers. The value of \alpha (rounded off to one decimal place) with |\alpha| \gt 1,
for which the magnitude response of the system is constant over all frequencies, is _____.

-1 | |

1 | |

2 | |

-2 |

Question 1 Explanation:

System is all-pass filter.

For digital all-pass filter, condition is

\text{zero}=\frac{1}{\text{pole}^{*}}\; \; ...(i)

By given transfer function,

\text{zero}=\alpha

\text{Pole}=-0.5

using condition(i),

\alpha =\frac{1}{-0.5}=-2

For digital all-pass filter, condition is

\text{zero}=\frac{1}{\text{pole}^{*}}\; \; ...(i)

By given transfer function,

\text{zero}=\alpha

\text{Pole}=-0.5

using condition(i),

\alpha =\frac{1}{-0.5}=-2

Question 2 |

Let H(z) be the z-transform of a real-valued discrete-time signal h[n]. If P(z)=H(z)H\left (\frac{1}{z} \right ) has a zero at z=\frac{1}{2}+\frac{1}{2}j, and P(z) has a total of four zeros, which one of the following plots represents all the zeros correctly?

A | |

B | |

C | |

D |

Question 2 Explanation:

P(Z)=H(Z)H\left ( \frac{1}{Z} \right )

(i) h(n) is real. Som p(n) will be also real

(ii) P(z)=P(z^{-1})

From (i) : if z_1 is a zero of P(z), then z_1^* will be also a zero of P(z).

From (ii): If z_1 is a zero of P(z), then \frac{1}{z_1} will be also a zero of P(z).

So, the 4 zeros are,

\begin{aligned} z_1&= \frac{1}{2}+\frac{1}{2}j\\ z_2&= z_1^*=\frac{1}{2}-\frac{1}{2}j\\ z_3&=\frac{1}{z_1}=\frac{1}{\frac{1}{2}-\frac{1}{2}j}=1-j \\ z_4&=\left ( \frac{1}{z_1} \right )^*=z_3^*=1+j \end{aligned}

(i) h(n) is real. Som p(n) will be also real

(ii) P(z)=P(z^{-1})

From (i) : if z_1 is a zero of P(z), then z_1^* will be also a zero of P(z).

From (ii): If z_1 is a zero of P(z), then \frac{1}{z_1} will be also a zero of P(z).

So, the 4 zeros are,

\begin{aligned} z_1&= \frac{1}{2}+\frac{1}{2}j\\ z_2&= z_1^*=\frac{1}{2}-\frac{1}{2}j\\ z_3&=\frac{1}{z_1}=\frac{1}{\frac{1}{2}-\frac{1}{2}j}=1-j \\ z_4&=\left ( \frac{1}{z_1} \right )^*=z_3^*=1+j \end{aligned}

Question 3 |

A discrete-time all-pass system has two of its poles at 0.25\angle 0^{\circ} and 2\angle 30^{\circ}. Which one of the following statements about the system is TRUE?

It has two more poles at 0.5\angle 0^{\circ} and 4\angle 0^{\circ} . | |

It is stable only when the impulse response is two-sided. | |

It has constant phase response over all frequencies. | |

It has constant phase response over the entire z-plane. |

Question 3 Explanation:

The AOC should include unit circle to make the system stable. From the given pole pattern it is clear that, to make the system stable, the ROC should be finite circular strip or finite annular strip and hence the impulse response of the system should be also two-sided.

Question 4 |

The ROC (region of convergence) of the z-transform of a discrete-time signal is represented by the shaded region in the z-plane. If the signal x[n]=(2.0)^{|n|}, -\infty \lt n \lt +\infty, then the ROC of its z-transform is represented by

A | |

B | |

C | |

D |

Question 4 Explanation:

\begin{array}{l} \begin{aligned} \text { Given, } & x[n]=(2.0)^{\ln 1},-\infty \lt n \lt +\infty \\ &=2^{n}(n]+\left(\frac{1}{2}\right)^{n} u(-n-1) \end{aligned} \\ \text{ROC}:|z| \gt 2 \text { and }|z| \lt \frac{1}{2}\\ \therefore \text{No Roc} \end{array}

Question 5 |

A discrete-time signal x[n]=\delta [n-3]+2\delta [n-5] has z-transform X(z). If X(z)=X(-z) is the z-transform of another signal y[n], then

y[n]=x[n] | |

y[n]=x[-n] | |

y[n]=-x[n] | |

y[n]=-x[-n] |

Question 5 Explanation:

Given,

\begin{aligned} x[n] &=\delta[n-3]+2 \delta[n-5] \\ X(z) &=z^{3}+2 z^{5} \\ X(-z) &=(-z)^{-3}+2(-z)^{-5} \\ Y(z) &=X(-z)=-z^{-3}-2 z^{-5} \\ &=-\left[z^{-3}+2 z^{-5}\right] \\ y[n] &=-x[n] \end{aligned}

\begin{aligned} x[n] &=\delta[n-3]+2 \delta[n-5] \\ X(z) &=z^{3}+2 z^{5} \\ X(-z) &=(-z)^{-3}+2(-z)^{-5} \\ Y(z) &=X(-z)=-z^{-3}-2 z^{-5} \\ &=-\left[z^{-3}+2 z^{-5}\right] \\ y[n] &=-x[n] \end{aligned}

Question 6 |

Consider the sequence x[n]=a^{n}u[n]+b^{n}u[n] , where u[n] denotes the unit-step sequence and 0 \lt |a| \lt |b| \lt 1. The region of convergence (ROC) of the z-transform of x[n] is

|z|\gt|a| | |

|z|\gt|b| | |

|z|\lt|a| | |

|a|\lt|z|\lt|b| |

Question 6 Explanation:

\begin{array}{l} \text { Given, } x[n]=a^{n} u[n]+b^{n} u[n] \\ \text { Also given, } \quad \begin{aligned} 0 &<|a|<|b|<1 \\ & A O C=(|z|>|a|) \text { and }(|z|>|b|) \\ & R O C=|z|>|b| \end{aligned} \end{array}

Question 7 |

A realization of a stable discrete time system is shown in the figure. If the system is excited by a unit step sequence input x[n], the response y[n] is

4(-\frac{1}{3})^{n} u[n] - 5(-\frac{2}{3})^{n} u[n] | |

5(-\frac{2}{3})^{n} u[n] - 3(-\frac{1}{3})^{n} u[n] | |

5(\frac{1}{3})^{n} u[n] - 5(\frac{2}{3})^{n} u[n] | |

5(\frac{2}{3})^{n} u[n] - 5(\frac{1}{3})^{n} u[n] |

Question 7 Explanation:

\begin{aligned} \frac{Y(z)}{X(z)}&=\frac{\frac{5}{3}[1-z]}{z^{2}-z+\frac{2}{9}} \\ \Rightarrow \quad \eta (z)&=\frac{5}{3} \frac{(1-z)}{\left(z-\frac{1}{3}\right)\left(z-\frac{2}{3}\right)}\\ \eta(z) &=\frac{5 z}{z-\frac{1}{3}}-\frac{5 z}{z-\frac{2}{3}} \\ y(n) &=5\left(\frac{1}{3}\right)^{n} u(n)-5\left(\frac{2}{3}\right)^{n} u(n) \end{aligned}

Question 8 |

Suppose x[n] is an absolutely summable discrete-time signal. Its z-transform is a rational function with two poles and two zeroes. The poles are at z = \pm 2j. Which one of the following statements is TRUE for the signal x[n]?

It is a finite duration signal. | |

It is a causal signal | |

It is a non-causal signal | |

It is a periodic signal |

Question 8 Explanation:

Poles are at z=\pm 2

Since x[n] is absolutely summable hence, DTFT exists which implies ROC must include unit circle.

\Rightarrow x(t) is a non-causal signal.

Since x[n] is absolutely summable hence, DTFT exists which implies ROC must include unit circle.

\Rightarrow x(t) is a non-causal signal.

Question 9 |

Consider a four-point moving average filter defined by the equation y[n]=\sum_{i=0}^{3}\alpha _{i} x[n-i],

The condition on the filter coefficients that results in a null at zero frequency is

The condition on the filter coefficients that results in a null at zero frequency is

\alpha_{1}= \alpha_{2}=0; \alpha_{0} =-\alpha_{3} | |

\alpha_{1}= \alpha_{2}=1; \alpha_{0} =-\alpha_{3} | |

\alpha_{0}= \alpha_{3}=0; \alpha_{1} =\alpha_{2} | |

\alpha_{1}= \alpha_{2}=0; \alpha_{0} =\alpha_{3} |

Question 9 Explanation:

\begin{aligned} y(n)&=\sum_{i=0}^{3} \alpha_{i} x(n-i) \\ &=\alpha_{0}x[n]+\alpha_{1} x[n-1]+\alpha_{2} x[n-2]+\alpha_{3} x[n-3] \\ \Rightarrow Y(z)&=\alpha_{0} X(z)+\alpha_{1} X(z) z^{-1}+\alpha_{2} X(z) z^{-2} \\ & +\alpha_{3} X(z) z^{-3} \\ H(z)&=\alpha_{0}+\alpha_{1} z^{-1}+\alpha_{2} z^{-2}+\alpha_{3} z^{-3} \\ \text {Null at }&\text{zero frequency } \\ \Rightarrow \quad z&=\left.e^{j \omega}\right|_{\omega=0}\\ 0 &= \alpha_{0}+\alpha_{1}+\alpha_{2}+\alpha_{3} \end{aligned}

Option (A) satisfies the condition

Option (A) satisfies the condition

Question 10 |

Two causal discrete-time signals x[n]and y[n] are related as y[n]=\sum_{m=0}^{n}x[m]. If the z-transform of y[n] is \frac{2}{z(z-1)^{2}}, the value of x[2] is _______.

0 | |

1 | |

2 | |

3 |

Question 10 Explanation:

\begin{aligned} f(n)&=\sum_{m=0}^{n} x[m]=x[n] * u[n] \\ y(z)&=X(z) \cdot\left(\frac{z}{z-1}\right)\\ \text{Given} \quad Y(z)&=\frac{2}{z(z-1)^{2}}=X(z) \cdot \frac{z}{z-1} \\ \Rightarrow \quad A(z)&=\frac{2 \cdot z^{-2}}{z-1}=2 z^{-3}\left(\frac{z}{z-1}\right) \\ \Rightarrow \quad x[n]&=2 u[n-3] \\ \text{Therefore, } x[2]&=0 \end{aligned}

There are 10 questions to complete.