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

 Question 1
The period of the discrete-time signal $x[n]$ described by the equation below is $N=$ ___ (Round off to the nearest integer).

$x[n]=1+3 \sin \left(\frac{15 \pi}{8} n+\frac{3 \pi}{4}\right)-5 \sin \left(\frac{\pi}{3} n-\frac{\pi}{4}\right)$
 A 16 B 32 C 48 D 64
GATE EE 2023   Signals and Systems
Question 1 Explanation:
Given :
$x[n]=1+3 \sin \left(\frac{15 \pi}{8} n+\frac{3 \pi}{4}\right)-5 \sin \left(\frac{\pi}{3} n-\frac{\pi}{4}\right)$
\begin{aligned} \omega_{1}&=\frac{15 \pi}{8} \\ \therefore \quad \mathrm{N}_{1}&=\frac{2 \pi}{\omega_{1}} \mathrm{~K}-=\frac{2 \pi}{15 \pi} \times 8 \times \mathrm{K}=16 \\ \text { and } \quad \mathrm{N}_{2}&=\frac{2 \pi}{\omega_{2}} \mathrm{~K}=\frac{2 \pi}{\pi} \times 3 \times \mathrm{K}=6 \\ \therefore \quad \text { Period }&=\operatorname{LCM}\left(\mathrm{N}_{1}, \mathrm{~N}_{2}\right) \\ &=\operatorname{LCM}(16,6) \\ &=48 \end{aligned}
 Question 2
For the signals $x(t)$ and $y(t)$ shown in the figure, $z(t)=x(t) * y(t)$ is maximum at $t=T_{1}$. Then $T_{1}$ in seconds is (Round off to the nearest integer).

 A 2 B 3 C 4 D 5
GATE EE 2023   Signals and Systems
Question 2 Explanation:
Using convotution property
\begin{aligned} z(t) & =x(t) * y(t) \\ & =\int_{-\infty}^{\infty} \mathrm{y}(\tau) \mathrm{x}(\mathrm{t}-\tau) \mathrm{d} \tau \end{aligned}

For max. overlap

\begin{aligned} \therefore \quad -1+t&=3\\ \Rightarrow \quad t&=4 sec \\ or, \quad \quad T_1&=4 sec \end{aligned}

 Question 3
Two discrete-time linear time-invariant systems with impulse responses $h_{1}\left [ n \right ]=\delta \left [ n-1 \right ]+\delta \left [ n+1 \right ]$ and $h_{2}\left [ n \right ]=\delta \left [ n\right ]+\delta \left [ n-1 \right ]$ are connected in cascade, where $\delta \left [ n\right ]$ is the Kronecker delta. The impulse response of the cascaded system is
 A $\delta \left [ n-2\right ]+\delta \left [ n+1 \right ]$ B $\delta \left [ n-1\right ]\delta \left [ n\right ]+\delta \left [ n+1 \right ]\delta \left [ n-1 \right ]$ C $\delta \left [ n-2\right ]+\delta \left [ n-1\right ]+\delta \left [ n\right ]+\delta \left [ n+1 \right ]$ D $\delta \left [ n\right ]\delta \left [ n-1\right ]+\delta \left [ n-2\right ]\delta \left [ n+1 \right ]$
GATE EE 2021   Signals and Systems
Question 3 Explanation:
\begin{aligned} h(n) &=\text { Resultant impulse response } \\ &=h_{1}(n) * h_{2}(n) \end{aligned}
By applying z -transform
\begin{aligned} H(z) &=H_{1}(z) \cdot H_{2}(z) \\ &=\left(z+z^{-1}\right)\left(1+z^{-1}\right) \\ &=z+Z^{-1}+1+z^{2} \end{aligned}
By applying inverse ZT,
$h(n)=\delta(n+1)+\delta(n-1)+\delta(n)+\delta(n-2)$
 Question 4
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 4 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 5
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 5 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,
$h(t)=0,\;\;\; t \lt 0$
$h(t) \neq 0,\;\;\; t \gt 0$

There are 5 questions to complete.