Fourier Series – GATE EE Signals And SystemsPractice Questions & PYQs

Step 1: Find the period $T$ The signal $x(t)$ consists of cosine terms with frequencies $2\pi, 4\pi, 6\pi$ , which correspond to fundamental frequencies $1 , \text{Hz}, 2 , \text{Hz}, 3 , \text{Hz}$ . The period of the signal $T$ is the least common multiple (LCM) of the periods of these individual cosine terms. The period of $2\pi t$ (1 Hz) is $T_1 = 1 , \text{sec}$ . The period of $4\pi t$ (2 Hz) is $T_2 = 0.5 , \text{sec}$ . The period of $6\pi t$ (3 Hz) is $T_3 = \frac{1}{3} , \text{sec}$ . The LCM of $T_1, T_2, T_3$ is $T = 1 , \text{sec}$ . Step 2: Calculate $|x(t)|^2$ The signal $x(t)$ is real, so we can calculate $|x(t)|^2$ directly: $|x(t)|^2 = (1 + 2 \cos(2\pi t) + 2 \cos(4\pi t) + 2 \cos(6\pi t))^2$ Expanding the square: $|x(t)|^2 = 1 + 4 \cos(2\pi t) + 4 \cos(4\pi t) + 4 \cos(6\pi t) + 4 \cos^2(2\pi t) + 4 \cos(2\pi t) \cos(4\pi t) + \ldots$ Step 3: Integrate over one period We now need to compute the integral of $|x(t)|^2$ over one period $T = 1 , \text{sec}$ . Many of the cross-term integrals, like $\int_0^T \cos(2\pi t) \cos(4\pi t) dt$ , will be zero due to the orthogonality property of the cosine functions. For the terms that do not vanish: $\int_0^T 1 , dt = T = 1 , \text{sec}$ $\int_0^T \cos^2(2\pi t) , dt = \frac{1}{2} T = \frac{1}{2}$ (using the identity $\cos^2(x) = \frac{1 + \cos(2x)}{2}$ ) Similarly, $\int_0^T \cos^2(4\pi t) , dt = \frac{1}{2}$ and $\int_0^T \cos^2(6\pi t) , dt = \frac{1}{2}$ . Step 4: Sum up the integrals After evaluating the integrals and summing them up, the result of the integral is: $\int_0^T |x(t)|^2 , dt = 1 + 4 \times \frac{1}{2} + 4 \times \frac{1}{2} + 4 \times \frac{1}{2} = 1 + 2 + 2 + 2 = 7$ . Step 5: Compute the final result Now, we divide by the period [latex]T = 1] sec: $\frac{1}{T} \int_0^T |x(t)|^2 , dt = \frac{7}{1} = 7$ . Final Answer: The value of $\frac{1}{T} \int_0^T |x(t)|^2 , dt$ is closest to 7.

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.

Given that,

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.

Time period:

Since, x(t) is even-halt wave symmetric. So, its expansion will contain only odd harmonics of cos. Therefore, coefficient of fundamental harmonic is

Now, frequency components available in expansion are

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 $=a_1 \cos \omega _0 t=4000 \cos 2000 \pi t$

Given that, $f(-x)=-f(x)$ So, function is an odd function. So, the fourier series will have sine term only. So, $f(x)=\sum_{k=1}^{\infty }b_x \sin (kx)$

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.

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.

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

Currently no explanation available

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.

The given signal is odd as wel as having half wave symmetry. So, it has only sine terms with odd harmonics. So, for second harmonic term amplitude =0.

$y(t)=x(t-t_0)+x(t+x_0)$ Since, $x(t)$ is periodic with period T. Therefore, $x(t-t_0)$ and $x(t+t_0)$ will also be periodic with period T. $\therefore \; b_k=a_ke^{-jk\omega _0t_0}+a_ke^{jk\omega _0t_0}$ $a_k$ is Fourier series coefficient of signal $x(t)$ therefore,

According to defination of signal given in question the x(t) will be as So it is periodic with period, $T_0=\frac{7T}{4}-\left ( -\frac{T}{4} \right )=2T$ Therefore, fundamental angular frequency $\omega _0=\frac{2 \pi}{T_0}=\frac{2 \pi}{2T}=\frac{\pi}{T}$ Now, $C_1=$ (exponential series coefficient for k=1)

Since trigonometric fourier series has no sine terms and has only cosine terms therefore this will be an even signal i.e. it will satisfy. $x(t)=x(-t)$ or, we can write, $x(t-T)=x(-t+T)$ but signal is periodic with period T. therefore $x(t-T)=x(t)$ therefore, $x(t)=x(T-t)\;\;...(i)$ Now, since signal contains only odd harmonics i.e. no terms of frequency $\omega =\frac{2 \pi \times 2k}{T},\;\;\;k=1,2,3,4$ i.e. no even harmonics. This means signal contains half wave symmetry this implies that, $x(t)=-x\left ( t-\frac{T}{2} \right )\;\;...(ii)$ From eq. (i) and (ii), $x(t)=x(T-t)=-x\left ( t-\frac{T}{2} \right )$

$f(n)=\sin ^2 x$ for finding the fourier series expansion $f(n)=\sin ^2 x=\frac{1- \cos 2x}{2}=0.5-0.5 \cos 2x$

From the wave symmetry,

$\because$ f(t) is an even function with half waves symmetry, $\therefore$ dc term as well as sine terms=0 Only the cosine terms with odd harmonics will be present.