# Signals and Systems

 Question 1
The discrete time Fourier series representation of a signal $x[n]$ with period $N$ is written as $x[n]=\sum_{k=0}^{N-1}a_ke^{j(2kn\pi/N)}$. A discrete time periodic signal with period $N=3$, has the non-zero Fourier series coefficients: $a_{-3}=2$ and $a_4=1$. The signal is
 A $2+2e^{-\left ( j\frac{2\pi}{6}n \right )} \cos \left ( \frac{2\pi}{6}n \right )$ B $1+2e^{\left ( j\frac{2\pi}{6}n \right )} \cos \left ( \frac{2\pi}{6}n \right )$ C $1+2e^{\left ( j\frac{2\pi}{3}n \right )} \cos \left ( \frac{2\pi}{6}n \right )$ D $2+2e^{\left ( j\frac{2\pi}{6}n \right )} \cos \left ( \frac{2\pi}{6}n \right )$
GATE EE 2022      Fourier Series
Question 1 Explanation:
Given: $x(n)=\sum_{k=0}^{N-1}a_ke^{j\frac{2 \pi kn}{N}}$
$\therefore \; x(n)=a_0+a_1e^{\frac{j2\pi n}{3}}+0+... \;\;\;...(1)$
We have,
$a_k=a_{k+N}$
$a_0=a_3$
$a_1=a_4=1$
$a_{-3}=a_0=2$
Put n = 0 in eq. (1)
$x(0)=a_0+a_1=2+1=3$
Put n = 1 in eq. (1)
$x(1)=a_0+a_1e^{j 2 \pi /3}=2+1\left ( \cos \frac{2\pi}{3} +j \sin \frac{2\pi}{3}\right )=2+(-0.5+j0.866)=1.5+j0.866$
These two conditions satisfy by the option (B). Hence, option (B) will be correct.
 Question 2
Consider the system as shown below where $y(t)=x(e^t)$. The system is
 A linear and causal. B linear and non-causal. C non-linear and causal D non-linear and non-causal
GATE EE 2022      Linear Time Invariant Systems
Question 2 Explanation:
We know, a linear system follows the law of superposition.
It is a combination of two laws: Both results are same, hence, it follows law of additivity.
(ii) Law of Homogeneity: Here also both results are same, hence it follows law of Homogeneity.
Therefore, System is linear.
We know, a causal system is independent of future values of input at each & every instant of time them system will be causal.
Given : $y(t)=x(e^t)$
Put $t=0$
$y(0)=x(e^0)=x(1)$
Because its depends on future value.
Therefore, system is non-causal.
 Question 3
Let an input $x(t)=2 \sin (10 \pi t)+5 \cos (15 \pi t)+7 \sin (42 \pi t)+4 \cos (45 \pi t)$ is passed through an LTI system having an impulse response,
$h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos (40\pi t)$
The output of the system is
 A $2 \sin (10 \pi t)+5\cos (15 \pi t)$ B $7 \sin (42 \pi t)+5\cos (15 \pi t)$ C $7 \sin (42 \pi t)+4\cos (45 \pi t)$ D $2 \sin (10 \pi t)+4\cos (45 \pi t)$
GATE EE 2022      Fourier Transform
Question 3 Explanation:
Given: $h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos 40 \pi t$
Fourier transform of signal $\frac{\sin (10 \pi t)}{\pi t}$ is given by Now, impulse response
$h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right )\cdot \left ( \frac{e^{+j40\pi t}+e^{-j40\pi t}}{2} \right )$
Using property, $e^{-j\omega _0t}x(t)\rightleftharpoons X(\omega +\omega _0)$
Therefore, Fourier transform of impulse response Cut-off frequencies,
$\omega_L=30 \pi rad/sec$
$\omega_H=50 \pi rad/sec$
Thus, output of the system $=7 \sin 42 \pi t+4 \cos 45 \pi t$
 Question 4
Let a causal LTI system be governed by the following differential equation $y(t)+\frac{1}{4}\frac{dy}{dt}=2x(t)$, where $x(t)$ and $x(t)$ are the input and output respectively. Its impulse response is
 A $2e^{-\frac{1}{4}t}u(t)$ B $2e^{-{4}t}u(t)$ C $8e^{-\frac{1}{4}t}u(t)$ D $8e^{-{4}t}u(t)$
GATE EE 2022      Linear Time Invariant Systems
Question 4 Explanation:
Given:
$y(t)+\frac{1}{4}\frac{dy}{dt}=2x(t)$
Taking Laplace transform,
$Y(s)+\frac{1}{4}(sY(s))=2X(s)$
Now, $H(s)=\frac{Y(s)}{X(s)}=\frac{2}{\frac{s}{4}+1}=\frac{8}{s+4}$
Taking inverse Laplace, transform,
$h(t)=8e^{-4t}u(t)$
 Question 5
Consider a continuous-time signal $x(t)$ defined by $x(t)=0$ for $\left | t \right |> 1$, and $x\left ( t \right )=1-\left | t \right | for \left | t \right |\leq 1$. Let the Fourier transform of $x(t)$ be defined as $X\left ( \omega \right )=\int\limits_{-\infty }^{\infty }\:x\left ( t \right )e^{-j\omega t}\:dt$. The maximum magnitude of $X\left ( \omega \right )$ is ___________.
 A 1 B 2 C 3 D 4
GATE EE 2021      Fourier Transform
Question 5 Explanation:
Fourier transform, $F(\omega)=A \tau S a^{2}\left(\frac{\omega \tau}{2}\right)$
\begin{aligned} \text { As } A=1, \tau=1 \\ H \omega) &=S a^{2}\left(\frac{\omega}{2}\right) \\ \left.F(\omega)\right|_{\text {peak }} &=F(0)=S a^{2}(0)=1 \end{aligned}
$\because$ Peak value of sampling function occurs at $\omega=0$
Peak value $=1$ Question 6
Let $f(t)$ be an even function, i.e. $f(-t)=f(t)$ for all t. Let the Fourier transform of $f(t)$ be defined as $F\left ( \omega \right )=\int\limits_{-\infty }^{\infty }\:f\left ( t \right )e^{-j\omega t}dt$. Suppose $\dfrac{dF\left ( \omega \right )}{d\omega }=-\omega F\left ( \omega \right )$ for all $\omega$, and $F(0)=1$. Then
 A $f\left ( 0 \right )\lt 1$ B $f\left ( 0 \right ) \gt 1$ C $f\left ( 0 \right )= 1$ D $f\left ( 0 \right )= 0$
GATE EE 2021      Fourier Transform
Question 6 Explanation:
$f(t) \rightleftharpoons F(\omega)=\int_{-\infty}^{\infty} f(t) e^{-j \omega t} d t$
The following informations are given about $f(t) \rightleftharpoons F(\omega).$
$\begin{array}{l} (i) f(t)=f(-t)\\ (ii) \left.F(\omega)\right|_{\omega=0}=1\\ (iii) \frac{d F(\omega)}{d \omega}=-\omega F(\omega)\\ \text{From }(iii), \frac{d F(\omega)}{d \omega}+\omega F(\omega)=0 \end{array}$
By solving the above linear differential equations, (by mathematics)
\begin{aligned} \ln F(\omega) &=-\frac{\omega^{2}}{2} \\ \Rightarrow \qquad\qquad F(\omega) &=K \cdot e^{-\omega^{2} / 2} \\ \text { Put } \omega=0, \qquad\qquad F(0) &=K \\ \Rightarrow \qquad\qquad 1 &=K \text { (from info. } \\ \text { From (iv), } \qquad\qquad F(\omega) &=e^{-\omega^{2} / 2} \\ \text{As we know}, e^{-a t^{2}}, a \gt 0 &\rightleftharpoons \sqrt{\frac{\pi}{a}} e^{-\omega^{2} / 4 a}\\ \text{At }a=\frac{1}{2}, \qquad\qquad\quad e^{-t^{2} / 2} &\rightleftharpoons \sqrt{\frac{\pi}{1 / 2}} \cdot e^{-\omega^{2} / 2}\\ f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \rightleftharpoons e^{-\omega^{2} / 2}=F(\omega)\\ \text { Thus, }\qquad\qquad f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \\ \text { At } t=0, \qquad\qquad f(0)&=\frac{1}{\sqrt{2 \pi}} \lt 1 \end{aligned}
 Question 7
The casual signal with z-transformer $Z^{2}\left ( Z-a \right )^{-2}$ is
($u[n]$ is the unit step signal)
 A $a^{2n}u\left [ n \right ]$ B $\left ( n+1 \right )a^{n}u\left [ n \right ]$ C $n^{-1}a^{n}u\left [ n \right ]$ D $n^{2}a^{n}u\left [ n \right ]$
GATE EE 2021      Z-Transform
Question 7 Explanation:
As we know,
\begin{aligned} n \cdot a^{n} u(n) & \rightleftharpoons \frac{a z}{(z-a)^{2}} \\ \frac{1}{a} \cdot n \cdot a^{n} u(n) & \rightleftharpoons \frac{z}{(z-a)^{2}} \\ f(n) &=n \cdot a^{n-1} u(n) \rightleftharpoons \frac{z}{(z-a)^{2}}=F(z) \end{aligned}
Time-shifting property,
\begin{aligned} f(n) &\rightleftharpoons F(z) \\ f(n+1) &\rightleftharpoons z \cdot F(z) & \\ x(n)&=(n+1) a^{n} u(n+1) \rightleftharpoons \frac{z^{2}}{(z-a)^{2}}=X(z) & \\ \text{Thus},\qquad \qquad x(n)&=(n+1) a^{n} u(n+1)=(n+1) a^{n} \cdot u(n) \\ &[\because(n+1) u(n+1)=(n+1) u(n) \end{aligned}
 Question 8
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      Introduction of C.T. and D.T. Signals
Question 8 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 9
If the input x(t) and output y(t) of a system are related as $y\left ( t \right )=\text{max}\left ( 0,x\left ( t \right ) \right )$, then the system is
 A linear and time-variant B linear and time-invariant C non-linear and time-variant D non-linear and time-invariant
GATE EE 2021      Linear Time Invariant Systems
Question 9 Explanation:
\begin{aligned} y(t) &=\max (0, x(t)) \\ &=\left\{\begin{array}{cl} 0, & x(t) \lt 0 \\ x(t), & x(t)\gt 0 \end{array}\right. \end{aligned} Linearity check:
at input $x_{1}(t)=-2$, output $y_{1}(t)=0$
at input $x_{2}(t)=1$, output $y_{2}(t)=1$ $\therefore$ system is non-linear because it violates law of additivity.
Check for time-invariance :
Delayed $\mathrm{O} / \mathrm{P}:$
$y\left(t-t_{0}\right)=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right) \gt 0 \\ 0 & x\left(t-t_{0}\right)\lt 0 \end{array}\right.$
$\mathrm{O} / \mathrm{P}$ of system when input is $x\left(t-t_{0}\right)=f(t)$
$y_{1}(t)=\left\{\begin{array}{cl} f(t), & f(t) \gt 0 \\ 0, & f(t)\lt 0 \end{array}=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right)\gt 0 \\ 0, & x\left(t-t_{0}\right)\lt 0 \end{array}\right.\right.$
Therefore, system is time-invariant.
 Question 10
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 10 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.

There are 10 questions to complete.