Z-Transform

Question 1
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 1 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 2
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 2 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 3
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 3 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}
Question 4
The pole-zero plots of three discrete-time systems P, Q and R on the z-plane are shown below. Which one of the following is TRUE about the frequency selectivity of these systems?
A
All three are high-pass filters.
B
All three are band-pass filters.
C
All three are low-pass filters.
D
P is a low-pass filter, Q is a band-pass filter and R is a high-pass filter.
GATE EE 2017-SET-2   Signals and Systems
Question 4 Explanation: 
Since all 3 pole-zero plots have zero at z=1 and z=-1.
So, filter will not pass low frequency and high frequency components.
Therefore all are bandpass filter.
Note: in digital filter
For low frequency z=1
For high frequency z=-1
Question 5
Consider a causal and stable LTI system with rational transfer function H(z). Whose corresponding impulse response begins at n = 0. Furthermore, H(1)=\frac{5}{4}. The poles of H(z) are P_{k}=\frac{1}{\sqrt{2}}exp(j\frac{(2k-1)\pi }{4}) for k = 1,2,3,4. The zeros of H(z) are all at z = 0. Let g[n]=j^nh[n]. The value of g[8] equals ___________.
A
0.01
B
0.03
C
0.06
D
0.09
GATE EE 2017-SET-1   Signals and Systems
Question 5 Explanation: 
Pole locationof H(z) are given as,
\begin{aligned} z&=\frac{1}{\sqrt{2}}e^{j(2K-1)\frac{\pi}{4}};\;\;K=1,2,3,4\\ z_1&=\frac{1}{\sqrt{2}}e^{j\frac{\pi}{4}} =\frac{1}{2}(1+j)\\ z_2&=\frac{1}{\sqrt{2}}e^{j\frac{3\pi}{4}} =\frac{1}{2}(-1+j)\\ z_3&=\frac{1}{\sqrt{2}}e^{j\frac{5\pi}{4}} =\frac{1}{2}(-1-j)\\ z_4&=\frac{1}{\sqrt{2}}e^{j\frac{7\pi}{4}} =\frac{1}{2}(1-j)\\ \text{Now,}\\ H(z)&=\frac{K\cdot z^4}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)}...(i) \end{aligned}
[As h(n) is causal and it starts from n=0, so numerator will have same order as denominator is having]. by solving equation (i)
\begin{aligned} H(z)&=\frac{K\cdot z^4}{z^4+\frac{1}{4}}\;\;...(ii)\\ &\text{Given that,}\\ H(1)&=\frac{5}{4}\\ &\text{From equation (ii),}\\ H(1)&=\frac{K}{1+\frac{1}{4}}\Rightarrow \frac{5}{4}=\frac{K}{\frac{5}{4}}\\ \Rightarrow \; K&=\frac{25}{16}\\ H(z)&=\frac{\frac{25}{16}z^4}{z^4+\frac{1}{4}}\\ &\text{By using division rule,}\\ H(z)&=\frac{25}{16}\left [ 1+\frac{1}{4}z^{-4}+\frac{1}{16}z^{-8}+... \right ]\\ \text{Thus, }h(8)&=\frac{1}{16} \times \frac{25}{16}=0.097\\ &\text{it is given that,}\\ g(n)&=j^nh(n)\\ g(8)&=j^8h(8)=h(8)\\ \Rightarrow \; g(8)&=h(8)=0.097 \end{aligned}
Question 6
Let S=\sum_{n=0}^{\infty }n\alpha ^{n} \; where \; |\alpha | \lt 1. The value of \alpha in the range 0 \lt \alpha \lt 1, such that S= 2\alpha is _______.
A
0.1
B
0.9
C
0.6
D
0.3
GATE EE 2016-SET-1   Signals and Systems
Question 6 Explanation: 
The Z-transform of
\begin{aligned} \alpha ^n u(n)&\rightarrow \frac{1}{(1-\alpha Z^{-1})}\\ n\alpha ^n u(n)&\rightarrow \frac{\alpha Z^{-1}}{(1-\alpha Z^{-1})^2}\\ \frac{\alpha Z^{-1}}{(1-\alpha Z^{-1})^2}&=\sum_{n=0}^{\infty }n\alpha ^nZ^{-n} \end{aligned}
If we put Z=1 in above equation, we get,
\begin{aligned} \frac{\alpha }{(1-\alpha )^2}&=\sum_{n=0}^{\infty }n\alpha ^n\\ \text{Since, }\sum_{n=0}^{\infty }n\alpha ^n&=2\alpha =\frac{\alpha }{(1-\alpha )^2}\\ \text{So,}2&=\frac{1}{(1-\alpha )^2}\\ \Rightarrow \;\;\alpha &=0.29 \end{aligned}
Question 7
The z-Transform of a sequence x[n] is given as X(z)=2z+4-4/z+3/z^{2}. If y[n] is the first difference of x[n], then Y(z) is given by
A
2z+2-8/z+7/z^{2}-3/z^{3}
B
-2z+2-6/z-1/z^{2}-3/z^{3}
C
-2z+2-8/z-7/z^{2}-3/z^{3}
D
4z-2-8/z+7/z^{2}-3/z^{3}
GATE EE 2015-SET-2   Signals and Systems
Question 7 Explanation: 
y(n) is first difference of x(n)
\begin{aligned} \text{so, } y(n)&=x(n)-x(n-1) \\ \text{so, } Y(z)&=X(z)-z^{-1}X(z) \\ Y(z)&=(2z+4-4z^{-1}+3z^{-2})\\ &-(2+4z^{-1}-4z^{-2}+3z^{-3}) \\ Y(z) &=2z+2-8z^{-1}+7z^{-2}-3z^{-3} \end{aligned}
Question 8
Consider a discrete time signal given by
x[n]=(-0.25)^{n}u[n]+(0.5)^{n}u[-n-1]
The region of convergence of its Z-transform would be
A
the region inside the circle of radius 0.5 and centered at origin
B
the region outside the circle of radius 0.25 and centered at origin
C
the annular region between the two circles, both centered at origin and having radii 0.25 and 0.5
D
the entire Z plane.
GATE EE 2015-SET-1   Signals and Systems
Question 8 Explanation: 
x[n]=(0.25)^nu(n)+(0.5)^n u(-n-1)
Signal x[n] is sum of two signals, one is right sided [(-0.25)^n u(n)] and other is left sided [(0.5)^n u(-n-1)]. The right sided signal will have pole at location with magnitude 0.25. So, ROC is |z| \gt 0.25. The left sided signal will have pole at location with magnitude 0.5. So, ROC is |z| \lt 0.5. So, ROC of X(z) (Z-transform of x(n) will be ) 0.25 \lt |z| \lt 0.5.
Question 9
An input signal x(t)=2+5sin(100\pit) is sampled with a sampling frequency of 400 Hz and applied to the system whose transfer function is represented by
\frac{Y(z)}{X(z)}=\frac{1}{N}(\frac{1-z^{-N}}{1-z^{-1}})
where, N represents the number of samples per cycle. The output y(n) of the system under steady state is
A
0
B
1
C
2
D
5
GATE EE 2014-SET-2   Signals and Systems
Question 9 Explanation: 


\begin{aligned} f_s&= 400Hz\\ x(t)&=2+5 \sin (100 \pi t) \\ t\rightarrow xT_s &=\frac{n}{f_s}=\frac{n}{400} \\ x(n)&=2+5 \sin \left ( 100 \pi\frac{n}{400} \right ) \\ &= 2+5 \sin \frac{\pi}{4}n\\ \text{Here, }\omega _0 &=\frac{\pi}{4} \end{aligned}
Therefore, N= time-period of x(n)=\frac{2 \pi }{\omega _0}=\frac{2 \pi}{\pi/4}=0
\begin{aligned} \text{Now, }H(z)&=\frac{1}{N}\left [ \frac{1-z^{-N}}{1-z^{-1}} \right ] \\ \text{Applying, }z &=e^{j\omega } \\ H(e^{j\omega }) &= \frac{1}{N}\left [ \frac{1-e^{-j\omega N}}{1-e^{-j\omega }} \right ] \\ \text{For, } &\text{sinusoidal part of }x(n): \\ H(e^{j\omega _0 }) &= \frac{1}{N}\left [ \frac{1-e^{-j\omega _0 N}}{1-e^{-j\omega _0 }} \right ] \\ [\text{Note: }] \omega _0 N&=\frac{\pi}{4} \times 8 =2 \pi] \\ &= \frac{1}{N}\left [ \frac{1-e^{-j2 \pi N}}{1-e^{-j \pi/4}} \right ]=0 \end{aligned}
Therefore, for sinusoidal part of input, system output is zero. For dc part of input,
\begin{aligned} y(n)&=\text{output}\\ &=H(e^{j0}) \times \text{input dc value}\\ [\because \; &H(e^{j\omega })_{\omega =0}=1]\\ &=1 \times 2=2 \end{aligned}
Thus, steady state output = 2
Question 10
A discrete system is represented by the difference equation
\begin{bmatrix} X_{1}(k+1)\\ X_{2}(k+1) \end{bmatrix}=\begin{bmatrix} a & a-1\\ a+1& a \end{bmatrix}\begin{bmatrix} X_{1}(k)\\ X_{2}(k) \end{bmatrix}
It has initial conditions X_1(0)=1;X_2(0)=0. The pole locations of the system for a = 1, are
A
1\pmj0
B
-1 \pm j0
C
\pm 1+j0
D
0 \pm j1
GATE EE 2014-SET-2   Signals and Systems
Question 10 Explanation: 
Given that,
\begin{bmatrix} X_1(K+1)\\X_2(K+1) \end{bmatrix}=\begin{bmatrix} a & a-1\\ a+1 & 1 \end{bmatrix}\begin{bmatrix} X_1(K)\\X_2(K) \end{bmatrix}
With initial conditions,
X_1(0)=1,\;\;X_2(0)=0
For a=1, we can write,
\begin{bmatrix} X_1(K+1)\\X_2(K+1) \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 2 & 1 \end{bmatrix}\begin{bmatrix} X_1(K)\\X_2(K+) \end{bmatrix}
\Rightarrow \; X_1(K+1)=X_1(K) \;\;...(i)
\Rightarrow \; X_2(K+1)=2X_1(K)+X_2(k) \;\;...(ii)
Applying z-transform on equation (i),
\begin{aligned} zX_1(z)-zX_1(0)&=X_1(z) \\ X_1(z)[z-1]&=zX_1(0) \\ X_1(z) =\frac{z}{z-1}\;\;&[Given, \; X_1(0)=1]...(iii) \\ \text{from equation (ii),}\\ zX_2(z)-zX_2(0)&=2X_1(z)+X_2(z)\\ X_2(z)[z-1]&=2X_1(z)+zX_2(0)\\ [Given,\;\; X_2(0)&=0]\\ \frac{X_2(z)}{X_1(z)}&=\frac{2}{z-1} \end{aligned}
Thus, transfer function, H(z)=\frac{2}{z-1}
Therefore, pole location is z=1.
There are 10 questions to complete.
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