Introduction of C.T. and D.T. Signals

Question 1
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   Signals and Systems
Question 1 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 2
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   Signals and Systems
Question 2 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 3
The signal energy of the continuous-time signal
x(t) =[(t -1) u(t -1)] -[(t -2)u(t -2)] -[(t -3)u(t -3)] +[(t -4)u(t -4)] is
A
11/3
B
7/3
C
1/3
D
5/3
GATE EE 2018   Signals and Systems
Question 3 Explanation: 
x(t)=r(t-1)-r(t-2)-r(t-3)+r(t-4)

\begin{aligned} Ex(t)&=\int_{-\infty }^{\infty }|x(t)|^2 dt \\ &=\int_{1}^{2}|(t-1)^2| dt +\int_{2}^{3}1^2 dt\\&+\int_{3}^{4}(-(t-4))^2dt\\ &= \frac{1}{3}+1+\frac{1}{3}=\frac{5}{3} \end{aligned}
Question 4
The mean square value of the given periodic waveform f(t) is_________
A
12
B
24
C
6
D
3
GATE EE 2017-SET-2   Signals and Systems
Question 4 Explanation: 
Mean square value = power of f(t)
Mean square value
\begin{aligned} &=\frac{1}{T_0}\int_{T_0} |f(t)|^2 dt\\ &= \frac{1}{4}[4^2 \times 1+2^2 \times 2]\\ &= \frac{16+8}{4}=6 \end{aligned}
Question 5
Consider the system with following input-output relation y[n]=(1+(-1)^{n})x[n]
where, x[n] is the input and y[n] is the output. The system is
A
invertible and time invariant
B
invertible and time varying
C
non-invertible and time invariant
D
non-invertible and time varying
GATE EE 2017-SET-1   Signals and Systems
Question 5 Explanation: 
Given relationship,
y(n)=[1+(-1)^n]x(n)
Time invariance test:

Since, y(n-1)\neq y'(n)
So, the system is time invariant.
Invertibility test:

Thus, we are getting many to one mapping between input and output. So, the system is non-invertible.
Question 6
The value of \int_{-\infty }^{+\infty }e^{-t}\delta (2t-2)dt, \; where \; \delta (t) is the Dirac delta function, is
A
\frac{1}{2e}
B
\frac{2}{e}
C
\frac{1}{e^{2}}
D
\frac{1}{2e^{2}}
GATE EE 2016-SET-1   Signals and Systems
Question 6 Explanation: 
To find the value of \int_{-\infty }^{\infty }e^{-t}\delta (2t-2)dt
Since, \delta (2t-2)=\frac{1}{2}\delta (1t-1) above integral can be written as
\int_{-\infty }^{\infty }e^{-t}\frac{1}{2}\delta (t-1)dt=\frac{1}{2}e^{-1}=\frac{1}{2e}
Question 7
The function shown in the figure can be represented as
A
u(t)-u(t-T)+\frac{(t-T)}{T}u(t-T) -\frac{(t-2T)}{T}u(t-2T)
B
u(t)+\frac{t}{T}u(t-T)-\frac{t}{T}u(t-2T)
C
u(t)-u(t-T)+\frac{(t-T)}{T}u(t-T) -\frac{(t-2T)}{T}u(t)
D
u(t)-u(t-T)+\frac{(t-T)}{T}u(t-T) -\frac{(t-2T)}{T}u(t-2T)
GATE EE 2014-SET-1   Signals and Systems
Question 7 Explanation: 


The given function can be realized as follow:

Combining the function (i0, (ii), (iii) and (iv), we get the given function

\begin{aligned} \text{Therefore, } f(t)&=u(t)-u(t-T)+\frac{(t-T)}{T}\\ &-u(t-T)-\frac{(t-2T)}{T}u(t-2T)u(t)\\&-u(t-T)+ \frac{1}{T}r(t-T)-\frac{1}{T}r(t-2T)\\ &OR \\ &u(t)-u(t-T)+\frac{1}{T}(t-T)u(t-T)\\&-\frac{1}{T}(t-2T)u(t-2T) \end{aligned}
Question 8
A zero mean random signal is uniformly distributed between limits -a and +a and its mean square value is equal to its variance. Then the r.m.s value of the signal is
A
\frac{a}{\sqrt{3}}
B
\frac{a}{\sqrt{2}}
C
a\sqrt{2}
D
a\sqrt{3}
GATE EE 2011   Signals and Systems
Question 9
Given a sequence x[n], to generate the sequence y[n]=x[3-4n], which one of the following procedures would be correct ?
A
First delay x[n] by 3 samples to generate z_{1}[n], then pick every 4^{th} sample of z_{1}[n] to generate z_{2}[n], and than finally time reverse z_{2}[n] to obtain y[n].
B
First advance x[n] by 3 samples to generate z_{1}[n], then pick every 4^{th} sample of z_{1}[n] to generate z_{2}[n], and then finally time reverse z_{2}[n] to obtain y[n]
C
First pick every fourth sample of x[n] to generate v_{1}[n], time-reverse v_{1}[n] to obtain v_{2}[n], and finally advance v_{2}[n] by 3 samples to obtain y[n]
D
First pick every fourth sample of x[n] to generate v_{1}[n], time-reverse v_{1}[n] to obtain v_{2}[n], and finally delay v_{2}[n] by 3 samples to obtain y[n]
GATE EE 2008   Signals and Systems
Question 9 Explanation: 
y[n]=x[3-4n]=x[-4n+3]
So to obtain y[n], we first advance x[n] by 3 unit.
i.e. z_1[n]=x[n+3]
Now we will take every fourth sample of z_1[n]
i.e. z_2[n]=z_1[4n]=x[4n+3]
Now reverse (time reverse) z_2[n] will give
y[n]=z_2[-n]=x[-4n+3]
Question 10
A continuous-time system is described by y(t)=e^{-|x(t)|}, where y(t) is the output and x(t) is the input. y(t) is bounded
A
only when x(t) is bounded
B
only when x(t) is non-negative
C
only for t\geq0 if x(t) is bounded for t \geq 0
D
even when x(t) is not bounded
GATE EE 2006   Signals and Systems
There are 10 questions to complete.
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