A continuous time periodic signal ${x}({t})$ is $x(t)=1+2 \cos 2 \pi t+2 \cos 4 \pi t+2 \cos 6 \pi t$ . If $T$ is the period of $x(t)$ , then $\frac{1}{T} \int_{0}^{T}| x(t)| ^{2} dt=$ _________ (round off to the nearest integer).

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📖 Explanation

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.

Q3GATE 2019MCQ

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

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📖 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}

Q5GATE 2017MCQ

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

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

Q13GATE 2009MCQ

The Fourier Series coefficients of a periodic signal x(t), expressed as $x(t)=\sum_{k=-\infty }^{\infty }a_{k}e^{j2 \pi kt/T}$ are given by $a_{-2}=2-j1$ , $a_{-1}=0.5+j0.2$ , $a_{0}=j2$ , $a_{1}=0.5-j0.2$ , $a_{2}=2+j1$ and $a_{k}=0$ for $|k| \gt 2$ Which of the following is true ?

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📖 Explanation

\begin{aligned} x(t)&=a_{-2}e^{-j2\omega _0t}+a_{-1}e^{-j\omega _0t}\\ &+a_{1}e^{j\omega _0t} +a_{2}e^{j2\omega _0t}\\ &=(2-j)e^{-j2\omega _0t}+(2+j)e^{j2\omega _0t}\\ &+(0.5+j0.2)e^{-j\omega _0t}+(0.5-j0.2)e^{j\omega _0t}+2j\\ &=2(e^{-j2\omega _0t}+e^{j2\omega _0t})+j(e^{j2\omega _0t}-e^{-j2\omega _0t})\\ &+0.5(e^{j\omega _0t}+e^{-j\omega _0t})-j0.2(e^{j\omega _0t}-e^{-j\omega _0t})+2j\\ &=4 \cos 2\omega _0t-2 \sin 2\omega _0t+ \cos \omega _0t \\ &+0.4 \sin \omega _0t+2j\\ &=\text{Real}[x(t)]+j\text{Img}[x(t)]\\ &\text{where,}\\ \text{Real}[x(t)]&=4 \cos 2\omega _0t-2 \sin 2\omega _0t+ \cos \omega _0t \\ &+0.4 \sin \omega _0t\\ &= \text{neither even nor odd signal}\\ \text{Img}[x(t)]&=2=\text{constant} \end{aligned}

Q15GATE 2007MCQ

A signal $x(t)$ is given by

x(t)=\left\{\begin{matrix} 1, & -T/4 \lt t \leq 3T/4\\ -1, & 3T/4 \lt t\leq 7T/4\\ -x(t-T) & \end{matrix}\right.

Which among the following gives the fundamental fourier term of x(t)?

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📖 Explanation

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)

\begin{aligned} C_1 &=\frac{1}{T_0}\int _{T_0}x(t)e^{-j\omega _0t}dt \\ C_1 &=\frac{1}{T_0}\left[ \int_{-T/4}^{3T/4} e^{-j\omega _0t}dt+\int_{3T/4}^{7T/4} -1e^{-j\omega _0t}dt \right]\$ \\ C_1 &=\frac{1}{j 2\pi}\$[2e^{j\pi/4}+2e^{j\pi/4}]\$ \\ C_1&=\frac{2}{j\pi}e^{j\pi/4}=\frac{2}{\pi}e^{-j\pi/4} =\frac{2}{\pi}\left ( \frac{1}{\sqrt{2}}-\frac{j}{\sqrt{2}} \right )\\ &\text{Comparing it with} \\ C_1 &=\frac{a_1}{2}-\frac{jb_1}{2} \\ a_1&=\frac{4}{\sqrt{2}\pi}, b_1=\frac{4}{\sqrt{2}\pi} \\ \therefore \; & \text{fundamental fourier term will be}\\ &= a_1 \cos \omega _0 t+b_1 \sin \omega _0 t \\ &= \frac{4}{\sqrt{2}\pi} \cos \left ( \frac{\pi}{T}t \right )+\frac{4}{\sqrt{2}\pi} \sin \left ( \frac{\pi}{T}t \right )\\ &=\frac{4}{\pi}\cos \left[\frac{\pi}{T}t-\frac{\pi}{4} \right] \end{aligned}

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