Definition of Fourier Series MCQ Quiz in தமிழ் - Objective Question with Answer for Definition of Fourier Series - இலவச PDF ஐப் பதிவிறக்கவும்

If a function is even and real periodic then its Fourier series contains only cosine term:

f(x) = a₀ + a₁ cos ω₀ t + a₂ cos 2 ω₀ t + …

So, Statement P is correct.

If a function is odd and real periodic then its Fourier series contains only sine term.

f(x) = a₀ + b₁ sin ω₀ t + b₂ sin 2 ω₀ t + b₃ sin 3 ω₀ t + …

Statement S is correct.

Important Point:

• For a periodic function, Fourier series is given as
  $f\left(x\right)=a_0+{\sum }_{n=1}^{\mathrm{\infty }}(a_n\cos n{\omega }_{0}t+b_n\sin n{\omega }_{0}t)$
• A function is called even if f(x) = f(-x) and called odd if f(x) = -f(-x)

(-1)ⁿ can be written as (e^jπ)ⁿ

y(n) can now be written as:

${\left(-1\right)}^{n}x\left[n\right]={\left(e^{j\pi }\right)}^{n}x\left[n\right]$

This can also be written as:

$y\left[n\right]=e^{j\left(\frac{2\pi }{N}\right)\left(\frac{N}{2}\right)n}x\left[n\right]$

The Fourier series coefficient of y[n] will be:

$d_k=\frac{1}{N}\sum _{n=0}^{N-1}x\left[n\right]e^{j\left(\frac{2\pi }{N}\right)\frac{N}{2}n}{e}^{-j\left(\frac{2\pi }{N}\right)nk}$

$=\frac{1}{N}\sum _{n=0}^{N-1}x\left[n\right]e^{-j\left(\frac{2\pi }{N}\right)n\left(k-\frac{N}{2}\right)}$

$d_k=c_{k-\frac{N}{2}}$

Consider the following table for Relationship between time domain signal and trigonometric F.S coefficient

For an even symmetry, only cosine terms exist.
For an odd symmetry, only sine terms exist.

$f=\frac{1}{T}$

$T=\frac{1}{f}=\frac{1}{4}msec$

The given

Narrow bandpass filter

T = 0.25 msec

Now applying Fourier series formula on x(t), we get:

$a_k=\frac{1}{T}\underset{T}{\overset{}{\int }}x\left(t\right)e^{-jk{\omega }_{0}t}dt$ 

$a_k=\frac{1}{T}\underset{0}{\overset{T/2}{\int }}0.2e^{-jk{\omega }_{0}t}dt-\frac{1}{T}\underset{T/2}{\overset{T}{\int }}0.2e^{-jk{\omega }_{0}t}dt$ 

$a_k=\frac{0.2}{T}{\left[\frac{e^{-jk{\omega }_{0}t}}{-jk{\omega }_0}\right]}_0^{T/2}-\frac{0.2}{T}{\left[\frac{e^{-jk{\omega }_{0}t}}{-jk{\omega }_0}\right]}_{\frac{T}{2}}^T$ 

$a_k=\frac{0.2}{jk{\omega }_{0}T}\left[1-e^{-\frac{jk{\omega }_{0}T}{2}}\right]-\frac{0.2}{jk{\omega }_{0}T}\left[-e^{-jk{\omega }_{0}T}+e^{-jk{\omega }_{0}T/2}\right]$ 

ω₀T = 2π

$e^{-jk{\omega }_{0}T}=e^{-jk2\pi }={\left(1\right)}^k=1$ 

$e^{-\frac{jk{\omega }_{0}T}{2}}=e^{-jk\pi }={\left(-1\right)}^k$  

$a_k=\frac{0.2}{j2\pi k}\left[1-{\left(-1\right)}^k\right]-\frac{0.2}{j2\pi k}\left[{\left(-1\right)}^k-1\right]$ 

$a_k=\frac{2\times 0.2}{j2\pi k}\left[1-{\left(-1\right)}^k\right]$ 

$a_k=\frac{0.2}{j\pi k}\left[1-{\left(-1\right)}^k\right]$ 

$a_k=\{\begin{array}{c}\frac{0.4}{j\pi k};kodd\\ 0;keven\end{array}$ 

So, odd components are a₁, a₃, a₅, a₇, a₉…., and so on.

Since the filter passes frequencies only between 20 kHz and 40 kHz, the components lying between this band of frequency will be passed by the filter.

The fundamental frequency given is 4 kHz.

a₁ = 4 kHz

a₃ = 12 kHz

$\begin{array}{c}{a}_5=20kHz\\ a_7=28kHz\\ a_9=36kHz\end{array}\}$  

These 3 components of 20 kHz, 28 kHz, and 36 kHz will be passed and the rest of the components will be rejected by the filter.

a₁₁ → 44 kHz

a₁₃ → 52 kHz

       ⋮

And so on

∴ Option 2 is correct.

Given, $a_k=\{\begin{array}{cc}1;& k=0\\ -j{\left(\frac{1}{k}\right)}^2;& otherwise\end{array}$

For a real signal

$\begin{array}{l}{a}_k=a_{-k}^{\ast }\\ j{\left(\frac{1}{k}\right)}^2\ne j{\left(\frac{1}{-k}\right)}^2\end{array}$

The signal is not real

For even signal

a_k = a_-k

$j{\left(\frac{1}{k}\right)}^2=j{\left(\frac{1}{-k}\right)}^2$

The signal is even signal

Now, $\frac{d}{dt}x\left(t\right)\stackrel{F.S}{\leftrightarrow }{C}_k$

$\Rightarrow C_k=\left(jk\frac{2\pi }{T}\right).a_k$

$=\frac{2\pi }{T}\left(\frac{-1}{k}\right)$

C_k = -C_-k hence the signal is odd.

Only ii and iii are correct

Concept:

If a function f(x) is periodic with period T.

Then its Trigonometric Fourier series coefficients (a₀ , a_n, b_n )will be:

$a_0=\frac{1}{T}{\int }_T^{}f(x)dx$

$a_n=\frac{2}{T}{\int }_T^{}f(x)\cos (n{\omega }_{0}x)dx$

$b_n=\frac{2}{T}{\int }_T^{}f(x)\sin (n{\omega }_{0}x)dx$

Calculation:

Given;

f(x) = x²  in (0,2)

Period(T) = 2 

Calculation of a_n :

    $a_n=\frac{2}{T}{\int }_T^{}f(x)\cos (n{\omega }_{0}x)dx$

$\Rightarrow a_n=\frac{2}{2}{\int }_T^{}{x}^2\cos (n\pi x)dx[\because {\omega }_0=\frac{2\pi }{T}]$

Using integration by parts;

$\Rightarrow a_n=x^2{\int }_{}^{}\cos (n\pi x)dx-\int [\frac{\mathrm{d}(x^2)}{\mathrm{d}x}\int \cos (n\pi x)dx]dx$

$\Rightarrow a_n=\frac{x^2\sin (n\pi x)}{n\pi }-\frac{2}{n\pi }\int x\sin (n\pi x)dx$

$\Rightarrow a_n=\frac{x^2\sin (n\pi x)}{n\pi }-\frac{2}{n\pi }[x\int \sin (n\pi x)dx-\int [\frac{\mathrm{d}(x)}{\mathrm{d}x}\int \sin (n\pi x)dx]dx]$

$\Rightarrow a_n=\frac{x^2\sin (n\pi x)}{n\pi }-\frac{2}{n\pi }[\frac{-x(\cos n\pi x)}{n\pi }+\frac{\sin n\pi x}{(n\pi {)}^2}]$

$\Rightarrow a_n=\frac{x^2\sin (n\pi x)}{n\pi }+\frac{2x(\cos n\pi x)}{(n\pi {)}^2}-\frac{2\sin n\pi x}{(n\pi {)}^3}{|}_0^2$

$\therefore a_n=\frac{4}{(n\pi {)}^2}$

If a function is even and real periodic then its Fourier series contains only cosine term:

f(x) = a₀ + a₁ cos ω₀ t + a₂ cos 2 ω₀ t + …

So, Statement P is correct.

If a function is odd and real periodic then its Fourier series contains only sine term.

f(x) = a₀ + b₁ sin ω₀ t + b₂ sin 2 ω₀ t + b₃ sin 3 ω₀ t + …

Statement S is correct.

Important Point:

• For a periodic function, Fourier series is given as
  $f\left(x\right)=a_0+{\sum }_{n=1}^{\mathrm{\infty }}(a_n\cos n{\omega }_{0}t+b_n\sin n{\omega }_{0}t)$
• A function is called even if f(x) = f(-x) and called odd if f(x) = -f(-x)

Fourier series:

The Fourier series for the function f(x) in the interval α < x < α + 2π is given by

$f\left(x\right)=\frac{a_o}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos nx+\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin nx$

where

$a_o=\frac{1}{\pi }\underset{\alpha }{\overset{\alpha +2\pi }{\int }}f\left(x\right)dx;a_n=\frac{1}{\pi }\underset{\alpha }{\overset{\alpha +2\pi }{\int }}f\left(x\right)\cos nxdx;b_n=\frac{1}{\pi }\underset{\alpha }{\overset{\alpha +2\pi }{\int }}f\left(x\right)\sin nxdx$

An even function is any function f such that f(-x) = f(x)

Example: cos x, sec x, x2, x4, x6 …….., x-2, x-4 ……..

An odd function is any function f such that f(-x) = -f(x)

Example: sin x, tan x, cosec x, cot x, n, x3 ……., x-1, x-3 ……..

$\underset{-L}{\overset{L}{\int }}f\left(x\right)dx=\{\begin{array}{c}2\underset{0}{\overset{L}{\int }}f\left(x\right)dx,whenf\left(x\right)isanevenfunction\\ 0,whenf\left(x\right)isanoddfunction\end{array}$

When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.

$f\left(x\right)=\frac{a_o}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n\frac{\cos n\pi x}{L}$

$a_o=\frac{1}{L}\underset{-L}{\overset{L}{\int }}f\left(x\right)dx=\frac{2}{L}\underset{0}{\overset{L}{\int }}f\left(x\right)dx$

$a_n=\frac{1}{L}\underset{-L}{\overset{L}{\int }}f\left(x\right)\cos \frac{n\pi x}{L}dx=\frac{2}{L}\underset{0}{\overset{L}{\int }}f\left(x\right)\cos \frac{n\pi x}{L}dx$

When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.

$f\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin \frac{n\pi x}{L}$

$b_n=\frac{1}{L}\underset{-L}{\overset{L}{\int }}f\left(x\right)\sin \frac{n\pi x}{L}dx=\frac{2}{L}\underset{0}{\overset{L}{\int }}f\left(x\right)\sin \frac{n\pi x}{L}dx$

• Based on the symmetry of the periodic signal given we can conclude the Fourier series coefficients and the type of harmonics that are present in the signal given.
• With the Fourier series, the non-sinusoidal periodic waveform can be converted into the sinusoidal wave.

The below table shows the type of Fourier coefficient terms corresponding to the symmetry.

Concept:

The exponential representation of the Fourier series is given by:

$x\left(t\right)=\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}{C}_k{e}^{jk{\omega }_{0}t}$

Where, Ck in the Fourier coefficient given by:

$C_k=\frac{1}{T_0}\int x\left(t\right)e^{-jk{\omega }_{0}t}dt$

Given:

v(t) = 10 sin (2π100t)

ω₀ = 200π 

v(t) = (10 e^j200πt - 10 e^-j200πt )/2j

∴ The second harmonic = 0