Question 1 |
Let x_{1}(t)=u(t+1.5)-u(t-1.5) and x_{2}(t) is shown in the figure below. For y(t)=x_{1}(t) * x_{2}(t), the \int_{-\infty}^{\infty} y(t) d t is ___. (rounded off to the nearest integer).
12 | |
14 | |
15 | |
18 |
Question 1 Explanation:
x_{1}(t)=u(t+1.5)-u(t-1.5)
]\Rightarrow x_1(t)=rect\left ( \frac{t}{3} \right )
] x_1(t)=rect\left ( \frac{t}{3} \right ) \overset{FT}{\leftrightarrow} 3Sa(1.5\omega )
Now, x_{2}(t)=\delta(t+3)+\operatorname{rect}\left(\frac{t}{2}\right)+2 \delta(t-2)
Taking Fourier transform
\begin{aligned} X_{2}(\omega) & =e^{3 j \omega}+2 S a(\omega)+2 e^{-2 j \omega} \\ y(t) & =x_{1}(t) * x_{2}(t) \\ Y(\omega) & =X_{1}(\omega) \cdot X_{2}(\omega) \\ Y(\omega) & =\int_{-\infty}^{\infty} y(t) \cdot e^{-j \omega t} \cdot d t \end{aligned}
We know
\therefore \quad \int_{-\infty}^{\infty} y(t)=Y(0)
\therefore \quad Y(0)=X_{1}(0) \cdot X_{2}(0) =3[1+2+2]=15
]\Rightarrow x_1(t)=rect\left ( \frac{t}{3} \right )
] x_1(t)=rect\left ( \frac{t}{3} \right ) \overset{FT}{\leftrightarrow} 3Sa(1.5\omega )
Now, x_{2}(t)=\delta(t+3)+\operatorname{rect}\left(\frac{t}{2}\right)+2 \delta(t-2)
Taking Fourier transform
\begin{aligned} X_{2}(\omega) & =e^{3 j \omega}+2 S a(\omega)+2 e^{-2 j \omega} \\ y(t) & =x_{1}(t) * x_{2}(t) \\ Y(\omega) & =X_{1}(\omega) \cdot X_{2}(\omega) \\ Y(\omega) & =\int_{-\infty}^{\infty} y(t) \cdot e^{-j \omega t} \cdot d t \end{aligned}
We know
\therefore \quad \int_{-\infty}^{\infty} y(t)=Y(0)
\therefore \quad Y(0)=X_{1}(0) \cdot X_{2}(0) =3[1+2+2]=15
Question 2 |
The Fourier transform X(\omega) of x(t)=e^{-t^{2}} is
\sqrt{\pi} e^{\frac{\omega^{2}}{2}} | |
\frac{e^{-\frac{\omega^{2}}{4}}}{2 \sqrt{\pi}} | |
\sqrt{\pi} e^{-\frac{\omega^{2}}{4}} | |
\sqrt{\pi} e^{-\frac{\omega^{2}}{2}} |
Question 2 Explanation:
We know; \quad e^{-a t^{2}} ; a>0 \stackrel{F T}{\rightleftharpoons} \sqrt{\frac{\pi}{a}} \cdot e^{-\omega^{2} / 4 a}
Here; a=1
\therefore \quad X(\omega)=\sqrt{\pi} \cdot e^{-\omega^{2} / 4}
Here; a=1
\therefore \quad X(\omega)=\sqrt{\pi} \cdot e^{-\omega^{2} / 4}
Question 3 |
The exponential Fourier series representation of a continuous-time periodic signal x(t) is defined as
x\left ( t \right )=\sum_{k=-\infty }^{\infty }a_{k}e^{jk\omega _{0}t}
where \omega _0 is the fundamental angular frequency of x(t) and the coefficients of the series are a_{k}. The following information is given about x(t) and a_{k}.
I. x(t) is real and even, having a fundamental period of 6
II. The average value of x(t) is 2
III. a_{k}=\left\{\begin{matrix} k, & 1\leq k\leq 3\\ 0,& k> 3 \end{matrix}\right.
The average power of the signal x(t) (rounded off to one decimal place) is ____________
x\left ( t \right )=\sum_{k=-\infty }^{\infty }a_{k}e^{jk\omega _{0}t}
where \omega _0 is the fundamental angular frequency of x(t) and the coefficients of the series are a_{k}. The following information is given about x(t) and a_{k}.
I. x(t) is real and even, having a fundamental period of 6
II. The average value of x(t) is 2
III. a_{k}=\left\{\begin{matrix} k, & 1\leq k\leq 3\\ 0,& k> 3 \end{matrix}\right.
The average power of the signal x(t) (rounded off to one decimal place) is ____________
14 | |
45 | |
63 | |
32 |
Question 3 Explanation:
1. x(t) is real and even so a_{k} is also real and even a_{k}=a_{-k}
2. Average of x(t) is 2 i.e., a_{0}=2.
\begin{array}{llll} 3. \;x(t) \rightarrow a_{k}=k &1 \leq k \leq 3 \quad& a_{1}=1 \quad& a_{-1}=1 \\ \quad \quad \quad \quad \;\;\;\;\;\;\;\;0 & k>3 & a_{2}=2 & a_{-2}=2 \\ && a_{3}=3 & a_{-3}=3 \end{array}
4. \quad T_{0}=6
Parsval's Power Theorem
\begin{aligned} \frac{1}{T} \int_{0}^{T}\left|x(t)^{2}\right| d t &=\sum_{n=-\infty}^{+\infty}\left|a_{k}\right|^{2} \\ P(t) &=\sum_{n=-\infty}^{+\infty}\left|a_{k}\right|^{2}=\left|a_{-3}\right|^{2}+\left|a_{-2}\right|^{2}+\left|a_{-1}\right|^{2}+\left|a_{0}\right|^{2}+\left|a_{1}\right|^{2}+\left|a_{2}\right|^{2}+\left|a_{3}\right|^{2} \\ &=2\left|a_{1}\right|^{2}+2\left|a_{2}\right|^{2}+2\left|a_{3}\right|^{2}+\left|a_{0}\right|^{2} \\ &=2 \times 1^{2}+2 \times(2)^{2}+2(3)^{2}+(2)^{2} \\ &=2+8+18+4 \\ P_{x(t)} &=32 \end{aligned}
2. Average of x(t) is 2 i.e., a_{0}=2.
\begin{array}{llll} 3. \;x(t) \rightarrow a_{k}=k &1 \leq k \leq 3 \quad& a_{1}=1 \quad& a_{-1}=1 \\ \quad \quad \quad \quad \;\;\;\;\;\;\;\;0 & k>3 & a_{2}=2 & a_{-2}=2 \\ && a_{3}=3 & a_{-3}=3 \end{array}
4. \quad T_{0}=6
Parsval's Power Theorem
\begin{aligned} \frac{1}{T} \int_{0}^{T}\left|x(t)^{2}\right| d t &=\sum_{n=-\infty}^{+\infty}\left|a_{k}\right|^{2} \\ P(t) &=\sum_{n=-\infty}^{+\infty}\left|a_{k}\right|^{2}=\left|a_{-3}\right|^{2}+\left|a_{-2}\right|^{2}+\left|a_{-1}\right|^{2}+\left|a_{0}\right|^{2}+\left|a_{1}\right|^{2}+\left|a_{2}\right|^{2}+\left|a_{3}\right|^{2} \\ &=2\left|a_{1}\right|^{2}+2\left|a_{2}\right|^{2}+2\left|a_{3}\right|^{2}+\left|a_{0}\right|^{2} \\ &=2 \times 1^{2}+2 \times(2)^{2}+2(3)^{2}+(2)^{2} \\ &=2+8+18+4 \\ P_{x(t)} &=32 \end{aligned}
Question 4 |
X(\omega ) is the Fourier transform of x(t) shown below. The value of \int_{-\infty }^{\infty }|X(\omega )|^2 d\omega (rounded
off to two decimal places) is ______.
58.61 | |
42.45 | |
26.12 | |
78.92 |
Question 4 Explanation:
\begin{aligned} \int_{-\infty }^{\infty }\left | X(\omega ) \right |^{2}d\omega &=2\pi \int_{-\infty }^{\infty }\left | x(t) \right |^{2}dt \\ &=2\pi \int_{-\infty }^{\infty }\left | y(t) \right |^{2}\\ &=2\times 2\pi \int_{-2}^{0}\left | y(t) \right |^{2}dt\\ &=2\times 2\pi\left [ \int_{-2}^{-1}(t+2)^{2}dt+\int_{-1}^{0}(2t+3)^{2}dt \right ]\\&=4\pi \left [\left \{ \frac{(t+2)^{3}}{3} \right \} _{-1}^{-2}+\left \{ \frac{(2t+3)^3}{3\times 2} \right \}_{0}^{-1} \right ]\\&=4\pi \left [ \frac{1-0}{3}+\frac{3^{3}-1}{6} \right ]\\&=4\pi \left [ \frac{1}{3}+\frac{26}{6} \right ]\\&= 4\pi \times \left [ \frac{1}{3}+\frac{26}{6} \right ] \\ &=4\pi \times \left [ \frac{1}{3}+\frac{13}{3} \right ]\\&=4\pi \times \frac{14}{3} \\&=\frac{56\pi }{3}=58.61\end{aligned}
Question 5 |
The input 4sinc(2t) is fed to a Hilbert transformer to obtain y(t), as shown in the figure
below:
Here sinc(x)=\frac{sin(\pi x)}{\pi x}. The value (accurate to two decimal places) of \int_{-\infty }^{\infty } |y(t)|^{2}dt is _______.
Here sinc(x)=\frac{sin(\pi x)}{\pi x}. The value (accurate to two decimal places) of \int_{-\infty }^{\infty } |y(t)|^{2}dt is _______.
7 | |
8 | |
9 | |
10 |
Question 5 Explanation:
Hilbert transform does not alter the amplitude spectrum of the signal.
\begin{aligned} \text{So, }\quad \int_{-\infty}^{\infty}|y(t)|^{2} d t&=\int_{-\infty}^{\infty}|x(t)|^{2} d t=\int_{-\infty}^{\infty}|X(f)|^{2} d f \\ x(t)&=4 \sin \mathrm{c}(2 t) \\ \text{sinc}(t) &\stackrel{CTFT}{\longleftrightarrow} \text{rect}(f) \\ 4 \text{sinc}(2 t) &\stackrel{CTFT}{\longleftrightarrow}+\frac{4}{2} \text{rect}\left(\frac{f}{2}\right)=2 \text{rect}\left(\frac{f}{2}\right) \end{aligned}\\ \begin{aligned} \int_{-\infty}^{\infty}|X(f)|^{2} d f &=2 \times(2)^{2}=8 \\ \text { So. } \int_{-\infty}^{\infty}|y(t)|^{2} d t &=8 \end{aligned}
\begin{aligned} \text{So, }\quad \int_{-\infty}^{\infty}|y(t)|^{2} d t&=\int_{-\infty}^{\infty}|x(t)|^{2} d t=\int_{-\infty}^{\infty}|X(f)|^{2} d f \\ x(t)&=4 \sin \mathrm{c}(2 t) \\ \text{sinc}(t) &\stackrel{CTFT}{\longleftrightarrow} \text{rect}(f) \\ 4 \text{sinc}(2 t) &\stackrel{CTFT}{\longleftrightarrow}+\frac{4}{2} \text{rect}\left(\frac{f}{2}\right)=2 \text{rect}\left(\frac{f}{2}\right) \end{aligned}\\ \begin{aligned} \int_{-\infty}^{\infty}|X(f)|^{2} d f &=2 \times(2)^{2}=8 \\ \text { So. } \int_{-\infty}^{\infty}|y(t)|^{2} d t &=8 \end{aligned}
There are 5 questions to complete.