Fourier Series

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.

\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}

\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

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.

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.

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}

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.

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.