Signals and Systems

Question 1
Which of the following options is true for a linear time-invariant discrete time system that obeys the difference equation:

y[n]-ay[n-1]=b_0x[n]-b_1x[n-1]
A
y[n] is unaffected by the values of x[n - k]; k \gt 2
B
The system is necessarily causal.
C
The system impulse response is non-zero at infinitely many instants.
D
When x[n] = 0, n \lt 0, the function y[n]; n \gt 0 is solely determined by the function x[n].
GATE EE 2020      Linear Time Invariant Systems
Question 1 Explanation: 
\begin{aligned} y(n)-ay(n-1)&=b_{0}x(n)-b_{1}x(n-2) \\ &\text{By applying ZT,} \\ Y(z)-az^{-1}Y(z)&=b_{0}X(z)-b_{1}z^{-1}X(z)\\ \Rightarrow \, \, H(z)&=\frac{Y(z)}{X(z)}=\frac{b_{0}-b_{1}z^{-1}}{1-az^{-1}} \end{aligned}
By taking right-sided inverse ZT,
h(n)=b_{0}a^{n}u(n)-b_{1}a^{n-1}u(n-1)
By taking left-sided inverse ZT,
h(n)=-b_{0}a^{n}u(-n-1)+b_{1}a^{n-1}u(-n)
Thus system is not necessarily causal.
The impulse response is non-zero at infinitely many instants.
Question 2
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      Z-Transform
Question 2 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 3
Suppose for input x(t) a linear time-invariant system with impulse response h(t) produces output y(t), so that x(t)*h(t)=y(t). Further, if |x(t)|*|h(t)|=z(t), which of the following statements is true?
A
For all t \in (-\infty ,\infty ), z(t)\leq y(t)
B
For some but not all t \in (-\infty ,\infty ), z(t)\leq y(t)
C
For all t \in (-\infty ,\infty ), z(t)\geq y(t)
D
For some but not all t \in (-\infty ,\infty ), z(t)\geq y(t)
GATE EE 2020      Introduction of C.T. and D.T. Signals
Question 3 Explanation: 
\begin{aligned} \text{Since, } y(t) &= x(t) + h(t)\\ \text{and }z(t)&=\left | x(t) \right |\times \left | h(t) \right | \end{aligned}
Case-1:


case 2:

\text{then, }y(t)=z(t)


\text{Thus, } z(t)\geq y(t), \text{ for all 't'}
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      Z-Transform
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
Which of the following statements is true about the two sided Laplace transform?
A
It exists for every signal that may or may not have a Fourier transform.
B
It has no poles for any bounded signal that is non-zero only inside a finite time interval.
C
The number of finite poles and finite zeroes must be equal.
D
If a signal can be expressed as a weighted sum of shifted one sided exponentials, then its Laplace Transform will have no poles.
GATE EE 2020      Laplace Transform
Question 5 Explanation: 
It has no poles for any bounded signal that is nonzero in a finite time interval. This is true as we know for finite amplitude finite width signal ROC is entire s plane and ROC never includes any pole.
It implies for such signals there is no poles. Hence the correct answer is option (B).
Question 6
A periodic function f(t), with a period of 2\pi, is represented as its Fourier series,
f(t)=a_0+\sum_{n=1}^{\infty }a_n\cos nt+\sum_{n=1}^{\infty }a_n\sin nt.
If
f(t)=\left\{\begin{matrix} A \sin t, & 0\leq t\leq \pi\\ 0,& \pi\lt t\lt 2\pi \end{matrix}\right.
the Fourier series coefficients a_1 \; and \; b_1 of f(t) are
A
a_1 =\frac{A}{\pi} ; \; b_1=0
B
a_1 =\frac{A}{2} ; \; b_1=0
C
a_1 =0 ; \; b_1=\frac{A}{\pi}
D
a_1 =0 ; \; b_1=\frac{A}{2}
GATE EE 2019      Fourier Series
Question 6 Explanation: 
\begin{aligned} T_0 &=2\pi \Rightarrow \;\omega _0=\frac{2\pi}{T_0}=1 \\ \text{Now, }b_1 &=\frac{2}{T_0}\int_{0}^{T_0}f(t) \sin \omega _0 t \; dt \\ &= \frac{2}{2 \pi}\int_{0}^{2\pi}f(t) \sin t \; dt\;\;[\because \omega _0=1]\\ &= \frac{1}{\pi}\int_{0}^{\pi} A \sin t \cdot \sin t\; dt \\ &= \frac{A}{2 \pi}\int_{0}^{\pi}2 \sin ^2 t \; dt \\ &= \frac{A}{2 \pi}\int_{0}^{\pi}[1-\cos 2t] \; dt \\ &=\frac{A}{2 \pi} \left [ t-\frac{\sin 2t}{2} \right ]_0^{\pi} \\ &= \frac{A}{2 \pi} [(\pi-0)-(0-0)]=\frac{A}{2}\\ a_1&=\frac{2}{T_0}\int_{0}^{T_0}f(t) \cos \omega _0 t \; dt \\ &= \frac{2}{2 \pi}\int_{0}^{2\pi}f(t) \cos t \; dt \\ &= \frac{1}{\pi}\int_{0}^{\pi} A \sin t \cdot \cos t\; dt \\ &= \frac{A}{2 \pi}\int_{0}^{\pi} \sin 2t \; dt=0 \end{aligned}
Question 7
The output response of a system is denoted as y(t), and its Laplace transform is given by
Y(s)=\frac{10}{s(s^2+s+100\sqrt{2})}
The steady state value of y(t) is
A
\frac{1}{10\sqrt{2}}
B
10\sqrt{2}
C
\frac{1}{100\sqrt{2}}
D
100\sqrt{2}
GATE EE 2019      Laplace Transform
Question 7 Explanation: 
Steady state value of y(t)
\begin{aligned} &=\lim_{s \to 0}sY(s)\\ &=\lim_{s \to 0}\frac{10s}{s(s^2+s+100\sqrt{2})}\\ &=\frac{10}{100\sqrt{2}}=\frac{1}{10\sqrt{2}} \end{aligned}
Question 8
The symbols, a and T, represent positive quantities, and u(t) is the unit step function. Which one of the following impulse responses is NOT the output of a causal linear time-invariant system?
A
e^{+at}u(t)
B
e^{-a(t+T)}u(t)
C
1+e^{-at}u(t)
D
e^{-a(t-T)}u(t)
GATE EE 2019      Introduction of C.T. and D.T. signals
Question 8 Explanation: 
a and T represents positive quantities.
u(t) is unit step function.
h(t)=1+e^{-at}u(t), is non-causal
Here '1' is a constant and two sided so the system will be non-causal, because for causal system,
h(t)=0,\;\;\; t \lt 0
h(t) \neq 0,\;\;\; t \gt 0
Question 9
A system transfer function is H(s)=\frac{a_1s^2+b_1s+c_1}{a_2s^2+b_2s+c_2}. If a_1=b_1=0, and all other coefficients are positive, the transfer function represents a
A
low pass filter
B
high pass filter
C
band pass filter
D
notch filter
GATE EE 2019      Laplace Transform
Question 9 Explanation: 
\begin{aligned} H(s)&=\frac{c_1}{a_2s^2+b_2s+c_2}\\ & as \;\; a_1=b_1=0\\ &=\frac{c_1}{(1+s\tau _1)(1+s\tau _2)}\\ \text{Put }s&=0,\; H(0)=\frac{c_1}{c_2}\\ \text{Put }s&=\infty ,\; H(\infty )=0 \end{aligned}

which represents second order low pass filter.
Question 10
The inverse Laplace transform of
H(s)=\frac{s+3}{s^2+2s+1} \; for \; t\geq 0 is
A
3te^{-t}+e^{-t}
B
3e^{-t}
C
2te^{-t}+e^{-t}
D
4te^{-t}+e^{-t}
GATE EE 2019      Laplace Transform
Question 10 Explanation: 
\begin{aligned} L^{-1}\left ( \frac{s+3}{s^2+2s+1} \right )&=L^{-1}\left ( \frac{s+1+2}{(s+1)^2} \right )\\ &=L^{-1}\left ( \frac{1}{s+1}+\frac{2}{(s+1)^2} \right )\\ &=e^{-t}+2te^{-t} \end{aligned}


There are 10 questions to complete.
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