Q21GATE 2010MCQ

Two discrete time system with impulse response $h_{1}[n]=\delta [n-1]$ and $h_{2}[n]=\delta [n-2]$ are connected in cascade. The overall impulse response of the cascaded system is

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📖 Explanation

\begin{aligned} h_{1}[n]&=\delta[n-1] \stackrel{Z}{\longleftrightarrow} H_{1}(z)=z^{1} \\ h_{2}[n]&=\delta[n-2] \stackrel{Z}{\longleftrightarrow} H_{2}(z)=z^{2} \end{aligned}

Overall impulse response in z-domain,

\begin{aligned} H(z) &=H_{1}(z) H_{2}(z) \\ &=z^{-1} z^{-2} \\ &=z^{-3} \end{aligned}

Overall impulse response in discrete-time domain. $h[n]=\delta(n-3)$

Q22GATE 2010MCQ

The transfer function of a discrete time LTI system is given by $H(z)=\frac{2-\frac{3}{4}z^{-1}}{1-\frac{3}{4}z^{-1}+\frac{1}{8}z^{-2}}$ Consider the following statements: S1: The system is stable and causal for ROC: |z| $\gt$ 1/2 S2: The system is stable but not causal for ROC:| z| $\lt$ 1/4 S3: The system is neither stable nor causal for ROC: 1/4 $\lt$ |z| $\lt$ 1/2 Which one of the following statements is valid ?

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📖 Explanation

A discrete-time LTI system is causal if and only if the ROC of its system function is the exterior of a circle, including infinity. A discrete-time LTI system is stable if and only if the ROC of its system function includes the unit circle, $|z|=1$

H(z)=\begin{array}{l}\left(1-\frac{1}{4} z^{-1}\right)+\left(1-\frac{1}{2} z^{-1}\right) \\ \left(1-\frac{1}{4} z^{-1}\right)\left(1-\frac{1}{2} z^{-1}\right)\end{array}

or $H(z)=\frac{1}{1-\frac{1}{4} z^{-1}}+\frac{1}{1-\frac{1}{2} z^{-1}}$ For $\mathrm{ROC}:|z| \gt \frac{1}{2},$ the system is stable and causal. For $\mathrm{ROC}:|z| \lt \frac{1}{4},$ ROC does not include unit circle. So, system is not stable. For $ROC: \frac{1}{4} \lt |z| \lt \frac{1}{2},$ ROC does not include unit circle. So, system is not stable. Also ROC is not the exterior of $|z|=\frac{1}{2}$ .So, it is not causal.

Q23GATE 2010MCQ

Consider the z-transform $x(z)=5z^{2}+4z^{-1}+3;0 \lt|z|\lt \infty$ . The inverse z-transform $x[n]$ is

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📖 Explanation

\begin{array}{c} \delta\left[n+n_{0}\right] \stackrel{Z}{\longleftrightarrow} z^{r 0} \\ X(z)=5 z^{2}+4 z^{1}+3 ; 0 \lt |z| \lt \infty \\ x(n)=5 \delta(n+2)+4 \delta(n-1)+3 \delta(n) \end{array}

Q24GATE 2009MCQ

The ROC of z -transform of the discrete time sequence $x(n)=(\frac{1}{3})^{n}u(n)-(\frac{1}{2})^{n}u(-n-1)$ is

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📖 Explanation

$x(n)=(1 / 3)^{n} u(n)-(1 / 2)^{n} u(-n-1)$ $(1 / 3)^{n} u(n)$ is right sided signal, so ROC will be $|z| \gt 1 / 3$ ;; ...(i) $-(1 / 2)^{n} u(-n-1)$ is left sided signal so ROC will be $|z| \lt 1 / 2 \; \; ...(ii)$ from (i) and (ii) we see that ROC of the function will be $1 / 3 \lt |z| \lt 1 / 2$

Q25GATE 2008MCQ

In the following network, the switch is closed at t= $0^{-}$ and the sampling starts from t=0. The sampling frequency is 10Hz. The expression and the region of convergence of the z-transform of the sampled signal are

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📖 Explanation

\begin{aligned} x(z) &=\sum_{n=0}^{\infty } 5 e^{-0.05 n} z^{-n} \\ &=5 \sum_{n=0}^{\infty }\left(e^{-0.05} z^{-1}\right)^{n} \\ \text { For }\left|e^{-0.05} z^{1}\right| &<1 \text { or }|z|>e^{-0.05} \\ x(z) &=\frac{5 z}{z-e^{-0.05}} \end{aligned}

Q26GATE 2007MCQ

The z-transform X[z] of a sequence x[n] is given by $X(z)=\frac{0.5}{1-2z^{-1}}$ . It is given that the region of convergence of X[z] includes the unit circle. The value of x[0] is

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📖 Explanation

$X(z)=\frac{0.5}{1-2 z^{-1}}$ $\because$ ROC includes unit circle $\Rightarrow$ Left handed system $\Rightarrow \quad x(n)=-(0.5)(2)^{n} u(-n-1)$ $\Rightarrow \quad x(0)=0$

Q27GATE 2006MCQ

A low-pass filter having a frequency response $H(j\omega )=A(\omega )e^{j\phi(\omega ) }$ does not produce any phase distortions if

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📖 Explanation

For distortionless transmission. $\frac{d \phi(\omega)}{d \omega}=\text { constant }$ Phase response should be linear $\phi(\omega)=k \omega$

Q28GATE 2006MCQ

If the region of convergence of $x_{1}[n]+x_{2}[n]$ is $\frac{1}{3} \lt |z| \lt \frac{2}{3}$ then the region of convergence of $x_{1}[n]- x_{2}[n]$ includes

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Q29GATE 2005MCQ

The region of convergence of z-transform of the sequence $(\frac{5}{6})^{n}u(n)-(\frac{6}{5})^{n}u(-n-1)$ must be

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📖 Explanation

\begin{array}{l} z \text {-transform of }\left(\frac{5}{6}\right)^{n} u(n)-\left(\frac{6}{5}\right)^{n} u(-n-1)\\ =\frac{1}{1-\frac{5}{6} z^{-1}}+\left[1-\frac{1}{1-\left(\frac{6}{5}\right)^{-1} z}\right]\$ \\ \begin{array}{cc} (\mathrm{ROC}) &\quad(\mathrm{ROC})\\ =|z| \gt \frac{5}{6}&\quad |z| \lt \frac{6}{5} \\ \end{array}\\ \frac{5}{6}\lt |z| \lt \frac{6}{5} \end{array}

Q30GATE 2004MCQ

The z-transform of a system is $H(z)=\frac{z}{z-0.2}$ . If the ROC is $z \lt 0.2$ , then the impulse response of the system is

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📖 Explanation

\begin{aligned} H(z)&=\frac{z}{z-0.2}=\frac{z}{z\left(1-0.2 z^{-1}\right)} \\ H(z)&=\frac{1}{1-0.2 z^{-1}} \end{aligned}

ROC is $|z|\lt 0.2$ Comparing with, $-a^{n} u(-n-1) \rightarrow \frac{1}{1-a z^{-1}},|z|\lt a$ We get, $h(n)=-(0.2)^{n} u(-n-1)$

Q31GATE 2004MCQ

A causal LTI system is described by the difference equation 2y[n]=ay[n-2]-2x[n]+bx[n-1] The system is stable only if

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📖 Explanation

$2 y[n]=\alpha y[n-2]-2 x[n]+\beta x[n-1]$ Taking z-transform

\begin{array}{c} 2 Y(z)=\alpha Y(z) z^{2}-2 X(z)+\beta X(z) z^{-1} \\ Y z)\left[2-\alpha z^{2}\right]\$=X(z)\left[\beta z^{-1}-2\right]\$ \\ \frac{Y(z)}{X(z)}=\left(\begin{array}{c} \beta z^{-1}-2 \\ 2-\alpha z^{-2} \end{array}\right) \end{array}

For system to be stable, $\beta$ can be of any value.

\begin{aligned} 2-\alpha z^{-2} &=0 \\ 2 z^{2}-\alpha &=0 \\ z &=\sqrt{\frac{\alpha}{2}} \lt 1 \\ |\alpha| & \lt 2 \end{aligned}

For system to be stable all poles should be inside unity circle.

Q32GATE 2003MCQ

A sequence x(n) with the z-transform $X(z) = z^{4}+z^{2}-2z+2-3z^{-4}$ is applied as an input to a linear, time-invariant system with the impulse response $h(n) =2 \delta (n - 3)$ where

\delta (n)=\left\{\begin{matrix} 1, &n=0 \\ 0, &otherwise \end{matrix}\right.

The output at n = 4 is

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📖 Explanation

\begin{aligned} Y(z) &=H(z) X(z) \\ H(z) &=2 z^{-3} \\ \therefore Y(z) &=2 z^{3}\left(z^{4}+z^{2}-2 z+2-3 z^{-4}\right) \\ &=2\left(z+z^{-1}-2 z^{2}+2 z^{3}-3 z^{7}\right) \end{aligned}

Taking inverse of z -transform

\begin{aligned} y(n)=& 2[\delta(n+1)+\delta(n-1)-2 \delta(n-2)\\ &+2 \delta(n-3)-3 \delta(n-7)]\\ \text{At }n&=4, y(4)=0 \end{aligned}

Q33GATE 2002MCQ

If the impulse response of a discrete-time system is $h[n]=-5^{n}u[-n-1]$ , then the system function H(z) is equal to

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📖 Explanation

\begin{aligned} h(n) &= -5^{n}u[- n-1] \\ H(z) &=\sum_{n=-\infty }^{\infty } h(n) z^{-n} \\ H(z) &=\sum_{n=-\infty }^{-1}-5^{n} z^{-n}, \text { Let } n=-k \\ &=-\sum_{n=\infty }^{1}\left(5 z^{-1}\right)^{-k} \\ &=1-\sum_{K=0}^{\infty }\left(5^{-1} z\right)^{k} \\ &=1-\frac{1}{1-5^{-1} z}, \; \left|5^{-1} z\right| \lt 1 , \; |z| \lt 5 \mid \\ &=1-\frac{5}{5-z} \\ &=-\frac{z}{5-z}=\frac{z}{z-5} \end{aligned}

The system is stable because ROC includes unit circle.

Q34GATE 2001MCQ

The region of convergence of the z-transform of a unit step function is

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📖 Explanation

\begin{aligned} h(n)&=u(n) \\ H(z)&=\sum_{n=0}^{\infty } 1 . z^{-n} \end{aligned}

For the convergence of H(z) $\sum_{n=0}^{n}\left(z^{-1}\right)^{n}\lt \infty$ \therefore ROC is the range of values of z for which $\left|z^{-1}\right| \lt 1\text{ or } |z|\gt |1|$