Question 1 |
The discrete-time Fourier transform of a signal x[n] is X(\Omega)=1(1+\cos \Omega) e^{-j \Omega}. Consider that x_{p}[n] is a periodic signal of period N=5 such that
\begin{aligned} x_{p}[n] & =x[n], \text { for } n=0,1,2 \\ & =0, \text { for } n=3,4 \end{aligned}
Note that x_{p}[n]=\sum_{k=0}^{N-1} a_{k} e^{j \frac{2 \pi}{N} k m}. The magnitude of the Fourier series coefficient a_{3} is ____ (Round off to 3 decimal places).
\begin{aligned} x_{p}[n] & =x[n], \text { for } n=0,1,2 \\ & =0, \text { for } n=3,4 \end{aligned}
Note that x_{p}[n]=\sum_{k=0}^{N-1} a_{k} e^{j \frac{2 \pi}{N} k m}. The magnitude of the Fourier series coefficient a_{3} is ____ (Round off to 3 decimal places).
0.038 | |
0.025 | |
0.068 | |
0.012 |
Question 1 Explanation:
Given : x_{p}(n) is a period signal of period N=5.
x_{p}(n)=\left\{\begin{array}{cc} x(n), & \text { for } n=0,1,2 \\ 0, & \text { for } n=3,4 \end{array}\right.
and \mathrm{x}(\mathrm{n}) \stackrel{\text { DTFT }}{\longrightarrow} \mathrm{X}(\Omega)=(1+\cos \Omega) \mathrm{e}^{-\mathrm{j} \Omega}
We have,
\begin{aligned} a_{K} & =\frac{X\left(\frac{2 \pi}{N} K\right)}{N} \\ & =\frac{\left[1+\cos \left(\frac{2 \pi}{N} K\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{N} R}}{N} \end{aligned}
For, \mathrm{N}=5 and \mathrm{K}=3
\begin{aligned} a_{3} & =\frac{\left[1+\cos \left(\frac{2 \pi \times 3}{5}\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{5} \times 3}}{5} \\ \left|a_{3}\right| & =\frac{1}{5}\left[1+\cos \frac{6 \pi}{5}\right] \\ & =0.0382 \end{aligned}
x_{p}(n)=\left\{\begin{array}{cc} x(n), & \text { for } n=0,1,2 \\ 0, & \text { for } n=3,4 \end{array}\right.
and \mathrm{x}(\mathrm{n}) \stackrel{\text { DTFT }}{\longrightarrow} \mathrm{X}(\Omega)=(1+\cos \Omega) \mathrm{e}^{-\mathrm{j} \Omega}
We have,
\begin{aligned} a_{K} & =\frac{X\left(\frac{2 \pi}{N} K\right)}{N} \\ & =\frac{\left[1+\cos \left(\frac{2 \pi}{N} K\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{N} R}}{N} \end{aligned}
For, \mathrm{N}=5 and \mathrm{K}=3
\begin{aligned} a_{3} & =\frac{\left[1+\cos \left(\frac{2 \pi \times 3}{5}\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{5} \times 3}}{5} \\ \left|a_{3}\right| & =\frac{1}{5}\left[1+\cos \frac{6 \pi}{5}\right] \\ & =0.0382 \end{aligned}
Question 2 |
The Fourier transform X(\omega) of the signal x(t) is given by
\begin{aligned} X(\omega) & =1, \text { for }|\omega| \lt W_{0} \\ & =0, \text { for }|\omega| \gt W_{0} \end{aligned}
Which one of the following statements is true?
\begin{aligned} X(\omega) & =1, \text { for }|\omega| \lt W_{0} \\ & =0, \text { for }|\omega| \gt W_{0} \end{aligned}
Which one of the following statements is true?
x(t) tends to be an impulse as \mathrm{W}_{0} \rightarrow \infty | |
x(0) decreases as W_{0} increases | |
At \mathrm{t}=\frac{\pi}{2 \mathrm{~W}_{0}}, \mathrm{x}(\mathrm{t})=-\frac{1}{\pi} | |
At \mathrm{t}=\frac{\pi}{2 \mathrm{~W}_{0}}, \mathrm{x}(\mathrm{t})=\frac{1}{\pi} |
Question 2 Explanation:
Given,
\quad X(\omega)=\left\{\begin{array}{l}1, \text { for }|\omega|<W_{0} \\ 0, \text { for }|\omega|>W_{0}\end{array}\right.
X(\omega)=\operatorname{rect}\left(\frac{\omega}{2 W_{0}}\right)
We know,
\operatorname{Arect}\left(\frac{\mathrm{t}}{\tau}\right)=\operatorname{A\tau S}\left(\frac{\omega \tau}{2}\right)
By duality,
\operatorname{A} \tau \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right) \leftrightarrow 2 \pi \operatorname{Arect}\left(\frac{\omega}{\tau}\right)
Given,
2 \pi \mathrm{A}=1
\Rightarrow \quad A=\frac{1}{2 \pi}
\therefore \quad \mathrm{x}(\mathrm{t})=\frac{\tau}{2 \pi} \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right), where \tau=2 \mathrm{~W}_{0}
Thus,
\begin{aligned} x(t)&=\frac{W_0}{\pi} \mathrm{Sa}(W_0 t)\\ &=\frac{1}{\pi}\frac{\sin W_0 t}{t}\\ \therefore \quad \text{At } t&=\frac{\pi}{2W_0}, x(t)=\frac{1}{\pi t} \end{aligned}
From rectangular function, At W \Rightarrow \infty ,X(w)=1
Taking inverse fourier transform x(t)=\delta (t)
Option (A) will be correct.
\quad X(\omega)=\left\{\begin{array}{l}1, \text { for }|\omega|<W_{0} \\ 0, \text { for }|\omega|>W_{0}\end{array}\right.
X(\omega)=\operatorname{rect}\left(\frac{\omega}{2 W_{0}}\right)
We know,
\operatorname{Arect}\left(\frac{\mathrm{t}}{\tau}\right)=\operatorname{A\tau S}\left(\frac{\omega \tau}{2}\right)
By duality,
\operatorname{A} \tau \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right) \leftrightarrow 2 \pi \operatorname{Arect}\left(\frac{\omega}{\tau}\right)
Given,
2 \pi \mathrm{A}=1
\Rightarrow \quad A=\frac{1}{2 \pi}
\therefore \quad \mathrm{x}(\mathrm{t})=\frac{\tau}{2 \pi} \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right), where \tau=2 \mathrm{~W}_{0}
Thus,
\begin{aligned} x(t)&=\frac{W_0}{\pi} \mathrm{Sa}(W_0 t)\\ &=\frac{1}{\pi}\frac{\sin W_0 t}{t}\\ \therefore \quad \text{At } t&=\frac{\pi}{2W_0}, x(t)=\frac{1}{\pi t} \end{aligned}
From rectangular function, At W \Rightarrow \infty ,X(w)=1
Taking inverse fourier transform x(t)=\delta (t)
Option (A) will be correct.
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
h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos (40\pi t)
The output of the system is
2 \sin (10 \pi t)+5\cos (15 \pi t) | |
7 \sin (42 \pi t)+5\cos (15 \pi t) | |
7 \sin (42 \pi t)+4\cos (45 \pi t) | |
2 \sin (10 \pi t)+4\cos (45 \pi t) |
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
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 |
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 ___________.
1 | |
2 | |
3 | |
4 |
Question 4 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
\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 5 |
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
f\left ( 0 \right )\lt 1 | |
f\left ( 0 \right ) \gt 1 | |
f\left ( 0 \right )= 1 | |
f\left ( 0 \right )= 0 |
Question 5 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}
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}
There are 5 questions to complete.