Fourier Transform MCQ Quiz - Objective Question with Answer for Fourier Transform - Download Free PDF

Explanation:

\(\rm b_n=\frac{1}{l}\int_{-l}^lf(x)\cos\frac{(n\pi x)}{l}dx=\frac{1}{l}\int_{-l}^l(px^4+qx^5)\cos\left(\frac{n\pi x}{l}\right)dx\)​

\(=\rm \frac{1}{l}\int_{-l}^lpx^4\cos\left(\frac{n\pi x}{l}\right)dx+0\)  (∵2nd integral is an odd functions)

​Thus, bn depend on p​

Similarly, 

\(=\rm 0+\frac{1}{l}\int_{-l}^lqx^5\sin\left(\frac{n\pi x}{l}\right)dx\) (∵1st integral is an odd functions)

Thus, an depend on q.

Concept:

A Fourier series is an expansion of a periodic function f(x) in terms of an infinite sum of sines and cosines.

Explanation:

\(F(k) = \int_{ - \infty }^{+\infty}exp\, ({ikx})f(x) dx\)

∵ f(x) = exp (-x²)

\(F(k) = \int_{ - \infty }^{+\infty}exp (-x^2)exp\, ({ikx}) dx\)

\(= \int_{ - \infty }^{+\infty}exp (-x^2+{ikx}) dx\)

\(= \int_{ - \infty }^{+\infty}exp \left[-x^2+ikx-\frac{i^2k^2}{4}+\frac{i^2k^2}{4}\right] dx\)

= \(= \int_{ - \infty }^{+\infty}exp \left(\frac{i^2k^2}{4}\right) exp \left\{-\left(x-\frac{ik}{2}\right)^2 \right\}dx\)

= \(exp \left[i^2\frac{k^2}{4} \right] \int_{ - \infty }^{+\infty} exp \left\{-\left(x-\frac{ik}{2}\right)^2 \right\}dx\)

Let, x - \(\frac{ik}{2}\) = y ⇒ dy = dx

∴ \(F(k)=exp \left(\frac{-k^2}{4}\right)\int_{ - \infty }^{+\infty} exp\left\{-y^2\right\}dy\)

∵ \(\int_{ - \infty }^{+\infty} exp\left\{-x^2\right\}dx=\sqrt{\pi}\)

So, F(k) = \(\sqrt{\pi}\,\,\,exp \left(-\frac{k^2}{4}\right) \)

The correct answer is option (2).

Concept:

A Fourier series is an expansion of a periodic function into a sum of trigonometric functions. The Fourier series is an example of a trigonometric series, but not all Fourier series are trigonometric series

Calculation:

f(t) = e^-t -π

f(t) = ∑ c_ne^int

∑ |c_n|² =  \({1 \over 2 \pi} \int^{\pi}_{-\pi} e^{-2t} dt\)

= \({1 \over 2\pi}{e^{-2t} \over -2}|^{\pi}_{-\pi}\)

=  \({1 \over 2\pi}\) sinh (2π)

The correct answer is option (4).

The correct answer is option 1):(Circular convolution)

Concept:

• Circular convolution, also known as cyclic convolution, is a special case of periodic convolution, which is the convolution of two periodic functions that have the same period
• Convolution in the time domain is equal to multiplication in the frequency domain and vice versa.
• If X(ω) and Y(ω) are the Fourier transforms of x(t) and y(t)

then

x(t) * y(t) = X(ω) × Y(ω)

• If X(s) and Y(s) are the Laplace transforms of x(t) and y(t), t

hen x(t) * y(t) = X(s) × Y(s)

• Hence the multiplication of two discrete Fourier transforms (DFTs) is equal to the Circular convolution of two sequences in the time-domain.

Properties of ROC of Z-Transform:

•  The ROC of Z-transform X(z) is in the form of a circle centered on the origin.
   

• The ROC of z-transform X(z) never contains any poles.
• X(n) is a right-sided signal and if the ROC contains r = r0, then all the circles for which r0 < r < ∞ will also be present in the ROC
• X(n) is a left-sided signal & if the circle (r = r0) is in the ROC then all values of r for which 0 < r < r0 will also be present in ROC.
• For a right-sided signal, the ROC is always outer to the outermost pole excluding infinity
• For a left-sided signal, the ROC is always left to the innermost non-zero pole
• For a two-sided signal, the ROC is between the two circles (poles) but does not contain any pole
• For finite duration absolutely summable signal, the ROC is the entire Z-plane except possibly for Z = 0 or Z = ∞.
• For discrete-time Dicea delta function δ(n), the ROC in the entire z plane.

Concept:

\(x\left( t \right)\mathop \leftrightarrow \limits^{LT} X\left( s \right)\)

\(\frac{{dx\left( t \right)}}{{dt}}\mathop \leftrightarrow \limits^{L.T} sX\left( s \right)\)

\(sX\left( s \right)\mathop \leftrightarrow \limits^{I.L.T} \frac{{dx\left( t \right)}}{{dt}}\)

\(L\left\{ {\frac{{x\left( t \right)}}{{t}}} \right\} = \frac{}{}\mathop \smallint \limits_s^0 \left\{ {x\left( t \right)} \right\}ds\\ \)

Calculation:

Given that Laplace transform \(f\left( t \right) = 2\sqrt {\frac{t}{\pi }}\) is \({s^{ - \frac{3}{2}}}\)

Given as \(g\left( t \right) = \frac{1}{{\sqrt {\pi t} }}\)

\(\begin{array}{l} \Rightarrow g\left( t \right) = \frac{{2\sqrt {\frac{t}{\pi }} }}{{2t}} = \frac{{f\left( t \right)}}{{2t}}\\ L\left\{ {g\left( t \right)} \right\} = L\left\{ {\frac{{f\left( t \right)}}{{2t}}} \right\} = \frac{1}{2}\mathop \smallint \limits_s^0 \left\{ {f\left( t \right)} \right\}ds\\ = \frac{1}{2}\mathop \smallint \limits_s^\infty {s^{ - \frac{3}{2}}}ds = \frac{1}{2}\left( {\frac{{{s^{\frac{{ - 3}}{2} + 1}}}}{{\frac{{ - 3}}{2} + 1}}} \right)_s^\infty \\ = \frac{1}{2}\left( { - 2} \right)\left[ {0 - {s^{ - \frac{1}{2}}}} \right] = {s^{ - \frac{1}{2}}} = \frac{1}{{\sqrt s }} \end{array}\)

The given fourier series is,

\(f\left( x \right) = \frac{\pi }{4} + \frac{2}{\pi }\left[ {\frac{{\cos x}}{1^2} + \frac{{\cos 3x}}{{{3^2}}} \ldots \ldots \ldots } \right] + \left[ {\frac{{\sin x}}{1} + \frac{{\sin 2x}}{2} + \frac{{\sin 3x}}{3} + \ldots \ldots \ldots } \right]\)

\(f\left( 0 \right) = \frac{\pi }{4} + \frac{2}{\pi }\left[ {\frac{{1}}{1^2} + \frac{{1}}{{{3^2}}} + \frac{{1}}{{{5^2}}} \ldots \ldots \ldots } \right]\)

The convergence of f(x) at x = 0 is valid if

\(f\left( 0 \right) = \frac{{f\left( {{0^ - }} \right) + f\left( {{0^ + }} \right)}}{2}\)

\(f\left( 0 \right) = \frac{{f\left( {{0^ - }} \right) + f\left( {{0^ + }} \right)}}{2}\)

where \(f\left( {{0^ - }} \right) = \mathop {\lim }\limits_{x \to 0\;} \left( {\pi - x} \right) = \pi\)

\(f\left( 0 \right) = \frac{\pi}{2}\)

\(\frac{\pi }{2} = \frac{\pi }{4} + \frac{2 }{\pi} \left[ {\frac{1}{{{1^2}}} + \frac{1}{{{3^2}}} + \ldots } \right]\)

\( \frac{1}{1} + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \ldots =\frac{{{\pi^2}}}{8}\)

\(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{{{\left( {2n - 1} \right)}^2}}} = \frac{\pi^2 }{8}\)

\(f\left( t \right) = \begin{array}{*{20}{c}} {2;}&{0 < t < 1}\\ {0;}&{otherwise} \end{array}\)

Concept:

We know that

L{f(t)} = F(s) then f(t) = L^-1F(s)

Calculation:

\({\rm{L}}\left( {\sin {\rm{at}}} \right) = \frac{{\rm{a}}}{{{{\rm{s}}^2} + {{\rm{a}}^2}}}\)

Similarly,

\({\rm{L}}\left( {\sin {\rm{\omega t}}} \right) = \frac{{\rm{\omega }}}{{{{\rm{s}}^2} + {{\rm{\omega }}^2}}}\)

∴ \({L^{ - 1}}\left( {\frac{1}{{{s^2} + {\omega ^2}}}} \right) = \frac{{\sin\omega t}}{\omega } = f\left( t \right)\) 

The correct answer is option 1):(Circular convolution)

Concept:

• Circular convolution, also known as cyclic convolution, is a special case of periodic convolution, which is the convolution of two periodic functions that have the same period
• Convolution in the time domain is equal to multiplication in the frequency domain and vice versa.
• If X(ω) and Y(ω) are the Fourier transforms of x(t) and y(t)

then

x(t) * y(t) = X(ω) × Y(ω)

• If X(s) and Y(s) are the Laplace transforms of x(t) and y(t), t

hen x(t) * y(t) = X(s) × Y(s)

• Hence the multiplication of two discrete Fourier transforms (DFTs) is equal to the Circular convolution of two sequences in the time-domain.

Concept:

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

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx\)

where

\({a_o} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)dx;\;{a_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\cos nxdx;\;{b_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\sin nxdx\)

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

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

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

\(f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}\)

Calculation:

 f(x) = cos2(x)

\(f(x) = \frac{{1 + \cos 2x}}{2} = \frac{1}{2} + \frac{{\cos 2x}}{2}\)

On comparing with: \(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

a₀ = 1

a₁ = 0

a₂ = 1/2

Explanation:

\(X\left( \omega \right) = \frac{1}{{{{\left( {10 + j\omega } \right)}^2}}}\)

By applying inverse Fourier transform,

⇒ x(t) = t e^-10t u(t)

\(x\left( t \right) = {e^{ - at}}u\left( t \right) \leftrightarrow \frac{1}{{a + j\omega }}\)

By using frequency differentiation property,

\(t\;{x_1}\left( t \right) = t{e^{ - at}}u\left( t \right) \leftrightarrow j\frac{d}{{d\omega }} \times \left( \omega \right) = \frac{1}{{{{\left( {a + j\omega } \right)}^2}}}\)

\(\Rightarrow t{e^{ - 10t}}u\left( t \right) \leftrightarrow \frac{1}{{{{\left( {10 + j\omega } \right)}^2}}}\)

x(t) = te^-10t u(t)           

Taking logarithm (to base e ) on both sides,                              

\(\left| {lnx\left( t \right)} \right| = \left| {ln\left[ {t{e^{ - 10t}}u\left( t \right)} \right]} \right|\)

\(at\;t = 1, = \left| {ln\left[ {{e^{ - 10}}} \right]} \right| = \left| { - 10} \right| = 10\)

Fourier Transform:

It is used for frequency analysis of any Bounded Input and Bounded Output (BIBO) signal.

Fourier Transform for any function x(t) is given by

\(X\left( w \right) = \mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( t \right){e^{ - jwt}}dt\)

Inverse Fourier Transform for any function X(w) is given by

\(x\left( t \right) = \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \infty }^{ + \infty } X\left( w \right){e^{jwt}}dw\)

Calculation:

δ (w) is impulse function that exists only at t = 0

so its inverse fourier transform will be

\(x\left( t \right) = \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \infty }^{ + \infty } \delta \left( w \right){e^{jwt}}dw\)

\(x\left( t \right) = \frac{1}{{2\pi }}{e^{jw0}}dw = \frac{1}{{2\pi }}\)

Important Points

The delta function is a normalized impulse, that is, sample number zero has a value of one, while all other samples have a value of zero.

For this reason, the delta function is frequently called the unit impulse. 

Fourier Transform of δ (t) is 1.

Properties of ROC of Z-Transform:

•  The ROC of Z-transform X(z) is in the form of a circle centered on the origin.
   

• The ROC of z-transform X(z) never contains any poles.
• X(n) is a right-sided signal and if the ROC contains r = r0, then all the circles for which r0 < r < ∞ will also be present in the ROC
• X(n) is a left-sided signal & if the circle (r = r0) is in the ROC then all values of r for which 0 < r < r0 will also be present in ROC.
• For a right-sided signal, the ROC is always outer to the outermost pole excluding infinity
• For a left-sided signal, the ROC is always left to the innermost non-zero pole
• For a two-sided signal, the ROC is between the two circles (poles) but does not contain any pole
• For finite duration absolutely summable signal, the ROC is the entire Z-plane except possibly for Z = 0 or Z = ∞.
• For discrete-time Dicea delta function δ(n), the ROC in the entire z plane.

Fourier series of f(x) of period 2T is given as:

\(f_T(x)=a_0+\sum_{n=1}^{∞} (a_n cos ~ω_n x+b_n sin~ ω_n x)\)

Putting the values of a₀, a_n and b_n we get,

\(f_T(x)=\frac{1}{2T}\int_{-T}^{T} f_T(v) dv + \frac{1}{T}\sum_{n=1}^{∞} [cos ~ω_nx\int_{-T}^{T} f_T(v) cos ~(ω_nv )dv+sin ~ω_nx\int_{-T}^{T} f_T(v) sin~(ω_nv) dv]\)

Δ ω = ωn+1 - ωn

= \(\frac{2\pi (n+1)}{2T}-\frac{2\pi n}{2T}\)

= \(\frac{\pi}{T}\)

\(\frac{1}{T}=\frac{\Delta \omega}{\pi}\)

∴ \(f_T(x)=\frac{1}{2T}\int_{-T}^{T} f_T(v) dv + \frac{1}{\pi}\sum_{n=1}^{∞} [cos ~(ω_nx) ~\Delta{\omega}\int_{-T}^{T} f_T(v) cos ~(ω_nv )dv+sin ~(ω_nx) \Delta\omega\int_{-T}^{T} f_T(v) sin~(ω_nv) dv]\)

When T → ∞ 

\(f(x)=\)

 \(\frac{1}{\pi}\int_{0}^{∞} [cos ~(ωx) \int_{-\infty}^{\infty} f(v) cos ~(ωv )dv]~d{\omega} +\frac{1}{\pi}[sin ~(ωx)\int_{-\infty}^{\infty} f(v) sin~(ωv) dv]~d{\omega}\)    ---(1)

In the question f(x) is given as:

 \(f(x)=\int_0^{\infty} [A(ω) cos~ω x + B(ω) sin~ω x]dω \)  ------(2)

By comparing (1) and (2), we get

\(A(\omega)= \frac{1}{\pi} \int_{-\infty}^{\infty} f(v) cos ~(ωv )dv \)

\(B(\omega)=\frac{1}{\pi}\int_{-\infty}^{\infty} f(v) sin~(ωv) dv\)