Z-Transform

 Question 1
The Z-transform of a discrete signal $x[n]$ is

$X(z)=\frac{4z}{\left ( z-\frac{1}{5} \right ) \left ( z-\frac{2}{3} \right ) (z-3)}$ with ROC = R

Which one of the following statements is true?
 A Discrete-time Fourier transform of $x[n]$ converges if $R$ is $|z| \gt 3$ B Discrete-time Fourier transform of $x[n]$ converges if $R$ is $\frac{2}{3} \lt |z| \lt 3$ C Discrete-time Fourier transform of $x[n]$ converges if $R$ is such that $x[n]$ is a leftsided sequence D Discrete-time Fourier transform of $x[n]$ converges if $R$ is such that $x[n]$ is a rightsided sequence
GATE EE 2023   Signals and Systems
Question 1 Explanation:
$Z = 1$, unit circle include by ROC : $\frac{2}{3} \lt |z| \lt 3$.
Therefore DIFT of $x[n]$ conveys for this ROC.
 Question 2
The casual signal with z-transformer $Z^{2}\left ( Z-a \right )^{-2}$ is
($u[n]$ is the unit step signal)
 A $a^{2n}u\left [ n \right ]$ B $\left ( n+1 \right )a^{n}u\left [ n \right ]$ C $n^{-1}a^{n}u\left [ n \right ]$ D $n^{2}a^{n}u\left [ n \right ]$
GATE EE 2021   Signals and Systems
Question 2 Explanation:
As we know,
\begin{aligned} n \cdot a^{n} u(n) & \rightleftharpoons \frac{a z}{(z-a)^{2}} \\ \frac{1}{a} \cdot n \cdot a^{n} u(n) & \rightleftharpoons \frac{z}{(z-a)^{2}} \\ f(n) &=n \cdot a^{n-1} u(n) \rightleftharpoons \frac{z}{(z-a)^{2}}=F(z) \end{aligned}
Time-shifting property,
\begin{aligned} f(n) &\rightleftharpoons F(z) \\ f(n+1) &\rightleftharpoons z \cdot F(z) & \\ x(n)&=(n+1) a^{n} u(n+1) \rightleftharpoons \frac{z^{2}}{(z-a)^{2}}=X(z) & \\ \text{Thus},\qquad \qquad x(n)&=(n+1) a^{n} u(n+1)=(n+1) a^{n} \cdot u(n) \\ &[\because(n+1) u(n+1)=(n+1) u(n) \end{aligned}

 Question 3
The causal realization of a system transfer function H(s) having poles at (2,-1), (-2,1) and zeroes at (2,1), (-2,-1) will be
 A stable, real, allpass B unstable, complex, allpass C unstable, real, highpass D stable, complex, lowpass
GATE EE 2020   Signals and Systems
Question 3 Explanation:
Since pole zero plot of given transfer function

Since, given pole zero is symmetrical about origin hence it is a all pass system.

Since, one pole on the RHS thus unstable.

Since, pole doesnot have complex conjugate poles and zeros present thus system is not real means system is complex.
 Question 4
Consider a signal $x[n]=\left ( \frac{1}{2} \right )^n \; 1[n]$, where 1[n]=0 if $n \lt 0$, and 1[n]=1 if $n \geq 0.$ The z-transform of $x[n-k], k \gt 0$ is $\frac{z^{-k}}{1-\frac{1}{2}z^{-1}}$ with region of convergence being
 A $|z| \lt 2$ B $|z| \gt 2$ C $|z| \lt 1/2$ D $|z| \gt 1/2$
GATE EE 2020   Signals and Systems
Question 4 Explanation:
\begin{aligned}x(n)&=\left (\frac{1}{2} \right )^{n} u(n) , \; \; \; \text{ROC of }x(n):\left | z \right | \gt \frac{1}{2} \\ x(n-k)\rightleftharpoons X(z)&=\frac{z^{-k}}{1-\frac{1}{2}z^{-1}} , \; \; \; \text{ROC of }x(n-k): \left | z \right | \gt \frac{1}{2}\\ \text{For } x(n-k) \; \; \; &\text{ROC will be } \left | z \right |\gt \frac{1}{2}\end{aligned}.
 Question 5
A cascade system having the impulse responses
$h_1(n)=\{\overset{1}{\uparrow},-1\}$ and
$h_2(n)=\{\overset{1}{\uparrow},1\}$
is shown in the figure below, where symbol $\uparrow$ denotes the time origin.

The input sequence x(n) for which the cascade system produces an output sequence $y(n)=\{\overset{1}{\uparrow},2,1,-1,-2,-1\}$ is
 A $x(n)=\{\overset{1}{\uparrow},2,1,1\}$ B $x(n)=\{\overset{1}{\uparrow},1,2,2\}$ C $x(n)=\{\overset{1}{\uparrow},1,1,1\}$ D $x(n)=\{\overset{1}{\uparrow},2,2,1\}$
GATE EE 2017-SET-2   Signals and Systems
Question 5 Explanation:

Now,
\begin{aligned} h(n)&=\text{overall system impulse response}\\ &=h_1(n)*h_2(n)\\ &=\{1,-1\}*\{1,1\}\\ &=\{ 1,0,-1\}\underset{*}{\rightleftharpoons }H(z)=1-z^2\\ &\text{As we know,}\\ H(z)&=\frac{Y(z)}{X(z)}\Rightarrow X(z)=\frac{Y(z)}{H(z)}\\ &=\frac{1+2z^{-1}+z^{-2}-z^{-3}-2z^{-4}-z^{-5}}{1-z^{-2}}\\ &=1+2z^{-1}+2z^{-2}+z^{-3}\\ \Rightarrow \; x(n)&=\{1,2,2,1\} \end{aligned}

There are 5 questions to complete.