Fourier Transform

Question 1
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   Signals and Systems
Question 1 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 2
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   Signals and Systems
Question 2 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 3
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   Signals and Systems
Question 3 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 4
The Fourier transform of a continuous-time signal x(t) is given by
X(\omega )=\frac{1}{(10+j\omega )^{2}},-\infty \lt \omega \lt \infty,
where j=\sqrt{-1} \; and \; \omega denotes frequency. Then the value of |lnx(t)| at t=1 is ___________ (up to 1 decimal place). ( ln denotes the logarithm to base e )
A
10.0
B
7.5
C
11.8
D
2.8
GATE EE 2018   Signals and Systems
Question 4 Explanation: 
x(t)\rightleftharpoons X(j\omega )=\frac{1}{(10+j\omega )^2}
By taking inverse Fourier transform,
\begin{aligned} x(t)&=te^{-10t}u(t) \\ \text{Now, } x(t)_{t=1}&= 1 \times e^{-10} \times 1=e^{-10}\\ \text{Thus, }|\ln (x(t))| &= |\ln (e^{-10})|=|-10|=10 \end{aligned}
Question 5
The value of the integral 2\int_{-\infty }^{\infty }(\frac{sin 2\pi t}{\pi t})dt is equal to
A
0
B
0.5
C
1
D
2
GATE EE 2016-SET-2   Signals and Systems
Question 5 Explanation: 
The Fourier transform of
\begin{aligned} &\frac{2 \sin (t\tau /2)}{t}\rightarrow 2 \pi \text{rect}\left ( \frac{\omega }{\tau } \right ) \\ &\frac{\sin (2 \pi t)}{\pi t}\rightarrow \text{rect} \left ( \frac{\omega }{4 \pi} \right )\\ &\text{So, }\int_{-\infty }^{\infty } \frac{\sin (2 \pi t)}{\pi t} e^{-j\omega t}dt=\text{rect}\left ( \frac{\omega }{4 \pi} \right )\\ &\text{Putting }\omega =0 \text{ in the above equation} \\ & \int_{-\infty }^{\infty } \frac{\sin (2 \pi t)}{\pi t} dt=1\\ &2\int_{-\infty }^{\infty } \frac{\sin (2 \pi t)}{\pi t} dt=2 \end{aligned}
Question 6
Suppose the maximum frequency in a band-limited signal x(t) is 5 kHz. Then, the maximum frequency in x(t)\cos (2000\pi t), in kHz, is ________.
A
5
B
6
C
7
D
8
GATE EE 2016-SET-2   Signals and Systems
Question 6 Explanation: 
Maximum possible frequency of x(t)(2000 \pi t)=f_1+f_2=5+1=6kHz
Question 7
Suppose x_{1}(t) \; and \; x_{2}(t) have the Fourier transforms as shown below.

Which one of the following statements is TRUE?
A
x_{1}(t) \; and \; x_{2}(t) are complex and x_{1}(t) x_{2}(t) is also complex with nonzero imaginary part
B
x_{1}(t) \; and \; x_{2}(t) are real and x_{1}(t) x_{2}(t) is also real
C
x_{1}(t) \; and \; x_{2}(t) are complex but x_{1}(t) x_{2}(t) is real
D
x_{1}(t) \; and \; x_{2}(t) are imaginary but x_{1}(t) x_{2}(t) is real
GATE EE 2016-SET-1   Signals and Systems
Question 7 Explanation: 
By observing X_1(j \omega ) and X_2(j \omega ), we can say that they are not conjugate symmetric. Since, the fourier transform is not conjugate symmetric the signal will not be real. So, x_1(t), x_2(t) are not real. Now the fourier transform of x_1(t) \cdot x_2(t) will be \frac{1}{2\pi}X_1(j \omega )*X_2(j \omega ) and by looking at X_1(j \omega ) and X_2(j \omega ), we can say that X_1(j \omega ) *X_2(j \omega ) will be conjugate symmetric and thus x_1(t) \cdot x_2(t) will be real.
By observing X_1(j \omega ) and X_2(j \omega ), we can say,
X_2( \omega )=X_1(- \omega )\overset{IFT}{\rightarrow}x_2(t)=x_1(-t)\;\;...(i)
Now, X_1(j \omega ) is real. Therefore, x_1(t) will be conjugate symmetric.
\begin{aligned} \text{i.e., } x_1(t)&=x_1^*(-t) \\ \downarrow t&= -t\\ x_1(-t)&= x_1^*(t)\\ \text{Now, }x_1(t)\cdot x_2(t) &= x_1(t)\cdot x_1(-t) \;\;(from \;(i))\\ &= x_1(t)\cdot x_1^*(t)\\ &= |x_1(t)|^2=\text{real function} \end{aligned}
Question 8
Consider a signal defined by
x(t)=\left\{\begin{matrix} e^{j10t} & for |t|\leq 1\\ 0 & for |t|\gt 1 \end{matrix}\right.
Its Fourier Transform is
A
\frac{2sin(\omega -10)}{\omega -10}
B
2e^{j10}\frac{sin(\omega -10)}{\omega -10}
C
\frac{2sin \omega }{\omega -10}
D
e^{j10\omega }\frac{2sin \omega }{\omega}
GATE EE 2015-SET-2   Signals and Systems
Question 8 Explanation: 
\begin{aligned} \text{Since, }x(t) &=\left\{\begin{matrix} e^{j10t} &\text{for }|t|\leq 1 \\ 0& \text{for }|t|\gt 1 \end{matrix}\right. \\ \text{as, } X(\omega )&=\int_{-\infty }^{\infty } x(t)e^{-j\omega t}dt\\ \text{so, } X(\omega )&=\int_{-1}^{1 } e^{j10t}e^{-j\omega t}dt\\ &=\int_{-1}^{1 } e^{jt(10-\omega )}dt\\ &= \frac{1}{j(10-\omega )}[e^{j(10-\omega )}- e^{-j(10-\omega )}]\\ X(\omega ) &=\frac{2 \sin (\omega -10)}{(\omega -10)} \end{aligned}
Question 9
A differentiable non constant even function x(t) has a derivative y(t), and their respective Fourier Transforms are X(\omega) and Y(\omega). Which of the following statments is TRUE ?
A
X(\omega) and Y(\omega) are both real
B
X(\omega) is real and Y(\omega) is imaginary
C
X(\omega) and Y(\omega) are both imaginary
D
X(\omega) is imaginary and Y(\omega) is real
GATE EE 2014-SET-3   Signals and Systems
Question 9 Explanation: 
For real even function x(t), the Fourier transform X(\omega ) is always real even. y(t) is a derivative of x(t) which is a real odd function because derivative of even function is an odd function and hence, Fourier transform Y(\omega ) is imaginary odd.
Question 10
A continuous-time LTI system with system function H(\omega) has the following polezero plot. For this system, which of the alternatives is TRUE ?
A
|H(0)|\gt |H(\omega )|; |\omega |\gt 0
B
|H(\omega )| has multiple maxima, at \omega_1 \; and \; \omega _2
C
|H(0)|\lt |H(\omega )|; |\omega |\gt 0
D
|H(\omega )|=constant; -\infty \lt \omega \lt \infty
GATE EE 2014-SET-3   Signals and Systems
Question 10 Explanation: 
\Rightarrow \;\; Symmetrically located pole and zero.
\Rightarrow \;\; All pass filter.
\Rightarrow \;\; Constant magnitude
(-\infty \leq \omega \leq \infty )
There are 10 questions to complete.