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
2+2e^{-\left ( j\frac{2\pi}{6}n \right )} \cos \left ( \frac{2\pi}{6}n \right ) | |
1+2e^{\left ( j\frac{2\pi}{6}n \right )} \cos \left ( \frac{2\pi}{6}n \right ) | |
1+2e^{\left ( j\frac{2\pi}{3}n \right )} \cos \left ( \frac{2\pi}{6}n \right ) | |
2+2e^{\left ( j\frac{2\pi}{6}n \right )} \cos \left ( \frac{2\pi}{6}n \right ) |
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.
\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 |
A periodic function f(t), with a period of 2\pi, is represented as its Fourier series,
f(t)=a_0+\sum_{n=1}^{\infty }a_n\cos nt+\sum_{n=1}^{\infty }a_n\sin nt.
If
f(t)=\left\{\begin{matrix} A \sin t, & 0\leq t\leq \pi\\ 0,& \pi\lt t\lt 2\pi \end{matrix}\right.
the Fourier series coefficients a_1 \; and \; b_1 of f(t) are
f(t)=a_0+\sum_{n=1}^{\infty }a_n\cos nt+\sum_{n=1}^{\infty }a_n\sin nt.
If
f(t)=\left\{\begin{matrix} A \sin t, & 0\leq t\leq \pi\\ 0,& \pi\lt t\lt 2\pi \end{matrix}\right.
the Fourier series coefficients a_1 \; and \; b_1 of f(t) are
a_1 =\frac{A}{\pi} ; \; b_1=0 | |
a_1 =\frac{A}{2} ; \; b_1=0 | |
a_1 =0 ; \; b_1=\frac{A}{\pi} | |
a_1 =0 ; \; b_1=\frac{A}{2} |
Question 2 Explanation:
\begin{aligned}
T_0 &=2\pi \Rightarrow \;\omega _0=\frac{2\pi}{T_0}=1 \\
\text{Now, }b_1 &=\frac{2}{T_0}\int_{0}^{T_0}f(t) \sin \omega _0 t \; dt \\
&= \frac{2}{2 \pi}\int_{0}^{2\pi}f(t) \sin t \; dt\;\;[\because \omega _0=1]\\
&= \frac{1}{\pi}\int_{0}^{\pi} A \sin t \cdot \sin t\; dt \\
&= \frac{A}{2 \pi}\int_{0}^{\pi}2 \sin ^2 t \; dt \\
&= \frac{A}{2 \pi}\int_{0}^{\pi}[1-\cos 2t] \; dt \\
&=\frac{A}{2 \pi} \left [ t-\frac{\sin 2t}{2} \right ]_0^{\pi} \\
&= \frac{A}{2 \pi} [(\pi-0)-(0-0)]=\frac{A}{2}\\
a_1&=\frac{2}{T_0}\int_{0}^{T_0}f(t) \cos \omega _0 t \; dt \\
&= \frac{2}{2 \pi}\int_{0}^{2\pi}f(t) \cos t \; dt \\
&= \frac{1}{\pi}\int_{0}^{\pi} A \sin t \cdot \cos t\; dt \\
&= \frac{A}{2 \pi}\int_{0}^{\pi} \sin 2t \; dt=0
\end{aligned}
Question 3 |
Let the signal x(t)=\sum_{k=-\infty }^{+\infty }(-1)^{k}\delta (t-\frac{k}{2000}) be passed through an LTI system with frequency response H(\omega) , as given in the figure below
The Fourier series representation of the output is given as
The Fourier series representation of the output is given as
4000+4000cos(2000\pi t) +4000cos(4000\pi t) | |
2000+2000cos(2000\pi t) +2000cos(4000\pi t) | |
4000cos(2000\pi t) | |
2000cos(2000\pi t) |
Question 3 Explanation:
\begin{aligned} x(t)&=\sum_{k=-\infty }^{\infty } (-1)^K \delta \left ( t-\frac{K}{2000} \right )\\ &= \sum_{k=-\infty }^{\infty } (-1)^K \delta (t-KT)\\ \text{where, } &=T=\frac{1}{2000} \end{aligned}
Time period:
\begin{aligned} T_0&=2T=\frac{1}{1000}\\ \omega _0&=\frac{2\pi}{T_0}=2000 \pi \; rad \end{aligned}
Since, x(t) is even-halt wave symmetric. So, its expansion will contain only odd harmonics of cos. Therefore, coefficient of fundamental harmonic is
\begin{aligned} a_1&=\frac{2}{T}\int_{-T_0/2}^{T_0/2}x(t) \cos \omega _0 t \; dt\\ &=\frac{4}{T_0}\int_{0}^{T_0/2}\delta (t) \cos \omega _0 t \; dt\\ &=\frac{4}{T_0}\int_{0}^{T_0/2}\delta (t) \cos 0 \; dt\\ &=\frac{4}{T_0}\left [\int_{0}^{T_0/2}\delta (t) \; dt \right ]\\ &=\frac{4}{T_0}=4000 \end{aligned}
Now, frequency components available in expansion are
\begin{aligned} \omega _0,3\omega _0,...\\ 2000\pi, 6000 \pi,... \end{aligned}
As, LTI system given in the question will pass upto 5000\pi rad/sec frequency component of input.So, output will have only one component of sfrequency 2000\pi rad/sec
Thus, y(t)= expansion of output =a_1 \cos \omega _0 t=4000 \cos 2000 \pi t
Time period:
\begin{aligned} T_0&=2T=\frac{1}{1000}\\ \omega _0&=\frac{2\pi}{T_0}=2000 \pi \; rad \end{aligned}
Since, x(t) is even-halt wave symmetric. So, its expansion will contain only odd harmonics of cos. Therefore, coefficient of fundamental harmonic is
\begin{aligned} a_1&=\frac{2}{T}\int_{-T_0/2}^{T_0/2}x(t) \cos \omega _0 t \; dt\\ &=\frac{4}{T_0}\int_{0}^{T_0/2}\delta (t) \cos \omega _0 t \; dt\\ &=\frac{4}{T_0}\int_{0}^{T_0/2}\delta (t) \cos 0 \; dt\\ &=\frac{4}{T_0}\left [\int_{0}^{T_0/2}\delta (t) \; dt \right ]\\ &=\frac{4}{T_0}=4000 \end{aligned}
Now, frequency components available in expansion are
\begin{aligned} \omega _0,3\omega _0,...\\ 2000\pi, 6000 \pi,... \end{aligned}
As, LTI system given in the question will pass upto 5000\pi rad/sec frequency component of input.So, output will have only one component of sfrequency 2000\pi rad/sec
Thus, y(t)= expansion of output =a_1 \cos \omega _0 t=4000 \cos 2000 \pi t
Question 4 |
Consider g(t)=\left\{\begin{matrix} t-\left \lceil t \right \rceil, & t\geq 0 \\ t-\left \lceil t \right \rceil , & otherwise \end{matrix}\right. , \; \; where\; t \in \mathbb{R}
Here, \left \lfloor t \right \rfloor represents the largest integer less than or equal to t and \left \lceil t \right \rceil denotes the smallest integer greater than or equal to t. The coefficient of the second harmonic component of the Fourier series representing g(t) is _________.
Here, \left \lfloor t \right \rfloor represents the largest integer less than or equal to t and \left \lceil t \right \rceil denotes the smallest integer greater than or equal to t. The coefficient of the second harmonic component of the Fourier series representing g(t) is _________.
0 | |
1 | |
2 | |
3 |
Question 4 Explanation:
Given that, g(t)=\left\{\begin{matrix} t-\left \lfloor t \right \rfloor, & t \geq 0\\ t-\left \lceil t \right \rceil & \text{otherwise} \end{matrix}\right.
where,
\left \lfloor t \right \rfloor= greatest integer less than or equal to 't'.
\left \lceil t \right \rceil= smallest integer greater than or equal to 't'.
Now,
Since, g(t) is nonperiodic. So, there is no Fourier series expansion of this signal and hence no need to calculate harmonic here.
where,
\left \lfloor t \right \rfloor= greatest integer less than or equal to 't'.
\left \lceil t \right \rceil= smallest integer greater than or equal to 't'.
Now,
Since, g(t) is nonperiodic. So, there is no Fourier series expansion of this signal and hence no need to calculate harmonic here.
Question 5 |
Let f(x) be a real, periodic function satisfying f(-x)=-f(x). The general form of its Fourier series representation would be
f(x)=a_{0}+\sum_{k=1}^{\infty }a_{k}cos(kx) | |
f(x)=\sum_{k=1}^{\infty }b_{k}sin(kx) | |
f(x)=a_{0}+\sum_{k=1}^{\infty }a_{2k}cos(kx) | |
f(x)=\sum_{k=0}^{\infty }a_{2k+1}sin(2k+1)x |
Question 5 Explanation:
Given that,
f(-x)=-f(x)
So, function is an odd function.
So, the fourier series will have sine term only. So,
f(x)=\sum_{k=1}^{\infty }b_x \sin (kx)
f(-x)=-f(x)
So, function is an odd function.
So, the fourier series will have sine term only. So,
f(x)=\sum_{k=1}^{\infty }b_x \sin (kx)
There are 5 questions to complete.