# 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$\geq$0 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.