Fourier Transform

Question 1
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 1 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 2
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 2 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 3
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 3 Explanation: 
Maximum possible frequency of x(t)(2000 \pi t)=f_1+f_2=5+1=6kHz
Question 4
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 4 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 5
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 5 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 6
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 6 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 7
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 7 Explanation: 
\Rightarrow \;\; Symmetrically located pole and zero.
\Rightarrow \;\; All pass filter.
\Rightarrow \;\; Constant magnitude
(-\infty \leq \omega \leq \infty )
Question 8
A signal is represented by
x(t)=\left\{\begin{matrix} 1 &|t| \lt 1 \\ 0& |t|\gt 1 \end{matrix}\right.
The Fourier transform of the convolved signal y(t)= x(2t)* x(t/2) is
A
\frac{4}{\omega ^{2}}sin(\frac{\omega }{2})sin(2 \omega )
B
\frac{4}{\omega ^{2}}sin(\frac{\omega }{2})
C
\frac{4}{\omega ^{2}}sin(2 \omega )
D
\frac{4}{\omega ^{2}}sin^{2} \omega
GATE EE 2014-SET-3   Signals and Systems
Question 8 Explanation: 
Given signal can be drawn as

Therefore,
\begin{aligned} &x(t)\leftrightarrow X(\omega )=2Sa(\omega ) \\ &\text{Now, } x(t)\leftrightarrow X(\omega )\\ &\text{then by time scaling,} \\ &x(at)\leftrightarrow \frac{1}{|a|}X(\omega /a) \\ &\therefore \; x(2t)\leftrightarrow Sa \left ( \frac{\omega }{2} \right )\;\;...(i) \\ &x\left ( \frac{t}{2} \right ) \leftrightarrow 4Sa(2\omega )\;\;...(ii)\\ &\text{Now, }y(t)=x(2t) \times x(t/2) \end{aligned}
Convolution in time domain multiplication in frequency domain
\begin{aligned} Y(\omega )&=4Sa\left ( \frac{\omega }{2}\right ) Sa(2\omega ) \\ Y(\omega )&=\frac{4\sin \left ( \frac{\omega }{2}\right ) }{\left ( \frac{\omega }{2}\right ) } \frac{\sin (2\omega )}{2\omega }\\ Y(\omega )&=\frac{4}{\omega ^2}\sin \left ( \frac{\omega }{2}\right ) \sin (2\omega ) \end{aligned}
Question 9
A function f(t) is shown in the figure.

The Fourier transform F(\omega) of f(t) is
A
real and even function of w
B
real and odd function of w
C
imaginary and odd function of w
D
imaginary and even function of w
GATE EE 2014-SET-3   Signals and Systems
Question 9 Explanation: 
Fiven signal f(t) is an odd signal. Hence, F(\omega ) is imaginary and odd function of \omega .
Question 10
Let f(t) be a continuous time signal and let F(\omega) be its Fourier Transform defined by

F(\omega)=\int_{-\infty }^{\infty }f(t)e^{-j \omega t}dt

Define g(t) by
g(t)=\int_{-\infty }^{\infty }F(u)e^{-jut}du

What is the relationship between f(t) and g(t) ?
A
g(t) would always be proportional to f (t)
B
g(t) would be proportional to f(t) if f(t) is an even function
C
g(t) would be proportional to f(t) only if f(t) is a sinusoidal function
D
g(t) would never be proportional to f(t)
GATE EE 2014-SET-1   Signals and Systems
Question 10 Explanation: 
Given that,
\begin{aligned} f(t)&\rightleftharpoons F(\omega )\\ g(t)&=\int_{-\infty }^{\infty }F(u)e^{-jut}du\\ \text{Now, }f(t)&=\frac{1}{2\pi}\int_{-\infty }^{\infty }F(\omega)e^{j \omega t}d\omega \\ \Rightarrow \; 2 \pi f(t)&=\int_{-\infty }^{\infty }F(\omega)e^{j \omega t}d\omega \\ \omega &\rightarrow u,\\ 2 \pi f(t)&=\int_{-\infty }^{\infty }F(u)e^{j u t}du \\ t &\rightarrow -t,\\ 2 \pi f(-t)&=\int_{-\infty }^{\infty }F(u)e^{-j u t}du \\ &=g(t)\\ \Rightarrow \; g(t)&=2 \pi f(-t)\\ &= 2\pi f(t) \end{aligned}
[If f(t) is even then f(t)=f(-t)
\Rightarrow \;\; g(t) \propto f(t) \text{ if }f(t) \text{ is even.}
There are 10 questions to complete.
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