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)

Question 6 |

The signum function is given by

sgn(x)=\left\{\begin{matrix} \frac{x}{|x|}; &x\neq 0 \\ 0; & x=0 \end{matrix}\right.

The Fourier series expansion of sgn(cos(t)) has

sgn(x)=\left\{\begin{matrix} \frac{x}{|x|}; &x\neq 0 \\ 0; & x=0 \end{matrix}\right.

The Fourier series expansion of sgn(cos(t)) has

only sine terms with all harmonics | |

only cosine terms with all harmonics. | |

only sine terms with even numbered harmonics. | |

only cosine terms with odd numbered harmonics. |

Question 6 Explanation:

So, cos(t) is

So, sgn(cos t) is a rectangular signal which is even and has half wave symmetry.

So, Fourier series will have only cosine terms with add harmonics only.

Question 7 |

Let g : [0,\infty )\rightarrow [0,\infty ) be a function defined by g(x)=x-[x], where [x] represents the integer part of x . (That is, it is the largest integer which is less than or equal
to x ). The value of the constant term in the Fourier series expansion of g(x) is____.

0 | |

0.5 | |

0.75 | |

1 |

Question 7 Explanation:

Given function g(t)=x-[x]

where, [x] is a integer part of x

Then function g(x) will be

The value of the constant term (or) dc term in the Fourier series expansion of g(x) is

\begin{aligned} a_0=\frac{1}{T} \int_{0}^{T}f(x)dx&=\frac{\text{Area in one period}}{\text{one period}}\\ &= \frac{1}{2} \times 1 \times 1=0.5 \end{aligned}

where, [x] is a integer part of x

Then function g(x) will be

The value of the constant term (or) dc term in the Fourier series expansion of g(x) is

\begin{aligned} a_0=\frac{1}{T} \int_{0}^{T}f(x)dx&=\frac{\text{Area in one period}}{\text{one period}}\\ &= \frac{1}{2} \times 1 \times 1=0.5 \end{aligned}

Question 8 |

For a periodic square wave, which one of the following statements is TRUE ?

The Fourier series coefficients do not exist | |

The Fourier series coefficients exist but the reconstruction converges at no point | |

The Fourier series coefficients exist but the reconstruction converges at most point | |

The Fourier series coefficients exist and the reconstruction converges at every
point. |

Question 8 Explanation:

Reconstruction of signal by its Fourier series coefficient is not possible at those points where signal is discontinuous.

In the above figure, at integer multiples of 'T/2', signal recovery is not possible by using its coefficient.

Therefore, reconstruction of x(t) by using its coefficient is possible at most of the points except those instants where x(t) is discontinous.

Question 9 |

For a periodic signal

v(t)=30 sin100t + 10 cos 300t + 6 sin(500t +\pi/4),

the fundamental frequency in rad/s

v(t)=30 sin100t + 10 cos 300t + 6 sin(500t +\pi/4),

the fundamental frequency in rad/s

100 | |

300 | |

500 | |

1500 |

Question 10 |

The fourier series expansion f(t)=a_{0}+\sum_{n=1}^{\infty }a_{n}cosn\omega t+b_{n}sin n\omega t of the periodic signal shown below will contain the following nonzero terms

a_{0} \; and \; b_{n},n=1,3,5,...\infty | |

a_{0} \; and \; a_{n},n=1,2,3,...\infty | |

a_{0},a_{n} \; and \; b_{n},n=1,3,5,...\infty | |

a_{0} \; and \; a_{n},n=1,3,5,...\infty |

Question 10 Explanation:

Let, x(t)= Even and Hws

Fourier series expansion of x(t) contains cos terms with odd harmonics.

NOw, f(t)=1+x(t)

Fourier series of f(t) contains dc and cos terms with odd harmonics.

Fourier series expansion of x(t) contains cos terms with odd harmonics.

NOw, f(t)=1+x(t)

Fourier series of f(t) contains dc and cos terms with odd harmonics.

There are 10 questions to complete.