Fourier Series MCQ Quiz in বাংলা - Objective Question with Answer for Fourier Series - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

The complex Exponential Fourier Series representation of a periodic signal x(t) with fundamental period To is given by,

\(x\left( t \right) = \mathop \sum \limits_{ - \infty }^{ + \infty } {C_k}{e^{jk{\omega _0}t}}\;\)

Where \(\;{\omega _0} = \frac{{2\pi }}{{{T_0}}}\) = fundamental frequency

Now, putting the value of k and expanding above series we get-

\(x\left( t \right) = \ldots + {C_{ - 2}}{e^{ - j2{\omega _0}t}} + {C_{ - 1}}{e^{ - j{\omega _0}t}} + {C_0} + {C_1}{e^{2{\omega _0}t}} + {C_2}{e^{j2{\omega _0}t}} + \ldots \)

Calculation:

Consider the signal, x(t) = A|sin(ω₁t)|

The period of the sinusoid is \({T_1} = \frac{{2\pi }}{{{\omega _1}}}\)

The period of the rectified sinusoid is one-half of this, \(T = \frac{{{T_1}}}{2} = \frac{\pi }{{{\omega _1}}}\)

Therefore, \({\omega _1} = \frac{{2\pi }}{{{T_1}}} = \frac{\pi }{T} = \frac{{{\omega _0}}}{2}\)

Thus, we can write \(x\left( t \right) = A\left| {\sin \left( {\frac{{{\omega _0}t}}{2}} \right)} \right|\)

The Fourier series coefficients are:

\({c_k} = \frac{1}{T}\mathop \smallint \limits_0^T x\left( t \right){e^{ - jk{\omega _0}t}}dt\)

\( = \frac{1}{T}\mathop \smallint \limits_0^T A\left| {\sin \left( {\frac{{{\omega _0}t}}{2}} \right)} \right|{e^{ - jk{\omega _0}t}}dt\)

From 0 to T, \(\left| {\sin \left( {\frac{{{\omega _0}t}}{2}} \right)} \right| = \sin \left( {\frac{{{\omega _0}t}}{2}} \right)\). Also, replace ω₀ by \(\frac{{2\pi }}{T}\)

\({c_k} = \frac{1}{T}\mathop \smallint \limits_0^T A\sin \left( {\frac{{\pi t}}{T}} \right){e^{ - \frac{{j2\pi kt}}{T}}}dt\)

\( = \frac{A}{T}\mathop \smallint \limits_0^T \left( {\frac{{{e^{\frac{{j\pi t}}{T}}} - {e^{ - \frac{{j\pi t}}{T}}}}}{{j2}}} \right){e^{ - \frac{{j2\pi kt}}{T}}}dt\)

\( = \frac{A}{{j2T}}\mathop \smallint \limits_0^T \left( {{e^{\frac{{j\pi t\left( {1 - 2k} \right)}}{T}}} - {e^{ - \frac{{j\pi t\left( {1 + 2k} \right)}}{T}}}} \right)dt\)

\( = \frac{A}{{j2T}}\left[ {\frac{{{e^{\frac{{j\pi t\left( {1 - 2k} \right)}}{T}}}}}{{j\pi \left( {1 - 2k} \right)/T}} - \frac{{{e^{ - \frac{{j\pi t\left( {1 + 2k} \right)}}{T}}}}}{{j\pi \left( {1 + 2k} \right)/T}}} \right]_0^T\)

\( = - \frac{A}{{2T}}\left[ {\frac{{{e^{j\pi \left( {1 - 2k} \right)}}}}{{\pi \left( {1 - 2k} \right)/T}} + \frac{{{e^{ - j\pi \left( {1 + 2k} \right)}}}}{{\pi \left( {1 + 2k} \right)/T}} - \frac{1}{{\pi \left( {1 - 2k} \right)/T}} - \frac{1}{{\pi \left( {1 + 2k} \right)/T}}} \right]\)

But \({e^{j\pi \left( {1 - 2k} \right)}} = {e^{j\pi }}{e^{ - j2\pi k}} = - 1\) and \({e^{ - j\pi \left( {1 + 2k} \right)}} = - 1\)

\({c_k} = - \frac{A}{{2T}}\left[ {\frac{{ - 1}}{{\pi \left( {1 - 2k} \right)/T}} + \frac{{ - 1}}{{\pi \left( {1 + 2k} \right)/T}} - \frac{1}{{\pi \left( {1 - 2k} \right)/T}} - \frac{1}{{\pi \left( {1 + 2k} \right)/T}}} \right]\)

\( = - \frac{A}{2}\left[ { - \frac{2}{{\pi \left( {1 - 2k} \right)}} - \frac{2}{{\pi \left( {1 + 2k} \right)}}} \right]\)

\( = \frac{A}{\pi }\left[ {\frac{1}{{1 - 2k}} + \frac{1}{{1 + 2k}}} \right]\)

\( = \frac{{2A}}{\pi }\left( {\frac{1}{{1 - 4{k^2}}}} \right)\)
The given function is x(t) = |A sin πt|

\({T_1} = \frac{{2\pi }}{\pi } = 2\)

\({T_0} = \frac{{{T_1}}}{2} = \frac{2}{2} = 1\)

Exponential Fourier series, \(x\left( t \right) = \mathop \sum \limits_{ - \infty }^{ + \infty } {C_k}{e^{jk{\omega _0}t}}\)

Concept:

The signal x(t) can be found using:

\(x\left( t \right)=\underset{=-\infty }{\overset{+\infty }{\mathop \sum }}\,{{c}_{n}}{{e}^{jn{{\omega }_{o}}t}}\)

Application

\({{C}_{n}}=\left| {{C}_{n}} \right|{{e}^{j\angle n}}\)

\({{C}_{-2}}=2{{e}^{j\left( -\frac{\pi }{6} \right)}}=2{{e}^{-j\frac{\pi }{6}}}\)

\({{C}_{-1}}=1{{e}^{j\left( \frac{\pi }{3} \right)}}={{e}^{j\frac{\pi }{3}}}\)

\({{C}_{1}}=1{{e}^{-j\left( \frac{\pi }{3} \right)}}={{e}^{-j\frac{\pi }{3}}}\)

\({{C}_{2}}=2{{e}^{+j\left( \frac{\pi }{6} \right)}}=2{{e}^{j\frac{\pi }{6}}}\)

\(x\left( t \right)={{C}_{-2}}{{e}^{-j2\pi t}}+{{C}_{-1}}{{e}^{-j\pi t}}+{{C}_{1}}{{e}^{j\pi t}}+{{C}_{2}}{{e}^{j2\pi t}}\)

\(x\left( t \right)=~2{{e}^{-j\left( 2\pi t+\frac{\pi }{6} \right)}}+{{e}^{-j\left( \pi t-\frac{\pi }{3} \right)}}+{{e}^{+j\left( \pi t-\frac{\pi }{3} \right)}}+2{{e}^{j\left( \frac{\pi }{6}+2\pi t \right)}}~\)

\(2\pi t+\frac{\pi }{6}=\theta \)

\(\left( \pi t-\frac{\pi }{3} \right)=ϕ\)

\(x\left( t \right)=2\left( {{e}^{-j\theta }}+{{e}^{j\theta }} \right)+1\left( {{e}^{-jϕ }}+{{e}^{jϕ }} \right)\)

x (t) = 4 cosθ + 2 cos ϕ 

\(x\left( t \right)=4\cos \left( 2\pi t+\frac{\pi }{6} \right)+2\cos \left( \pi t\frac{-\pi }{3} \right)~\)

Fourier Series:

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

Fourier Series is used for the orthogonality relationships of the sine and cosine functions.

For functions that are not periodic, the Fourier series is replaced by the Fourier transform.

For periodic signals, this representation becomes the discrete-time Fourier series, and for aperiodic signals, it becomes the discrete-time Fourier transform.

Continuous Fourier Transform:

The Fourier transform pair in the most general form for a continuous and aperiodic time signal x(t).

Discrete-time Fourier series:

The Fourier transform pair in the most general form for a discrete and periodic time signal x(t).

Concept:

The Fourier transform of a signal in time domain is given as:

\(X\left( \omega \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - j\omega t}}\)

For f(t) = e-t u(t)

\( F\left( \omega \right) = \frac{1}{{1 + j\omega }}\)

Calculation:

\({e^{ - 3t}}u\left( t \right) \leftrightarrow \frac{1}{{3 + j\omega }}\)

\({e^{ - 3\left( {t - 2} \right)}}u\left( {t - 2} \right) \leftrightarrow \frac{{{e^{ - 2j\omega }}}}{{3 + j\omega }}\)

\({e^{ - 3t}}u\left( {t - 2} \right) \leftrightarrow \frac{{{e^{ - 6}} \cdot {e^{ - 2j\omega }}}}{{3 + j\omega }} = \frac{{{e^{ - 2\left( {3 + j\omega } \right)}}}}{{3 + j\omega }}\)

A = 2, B = 3

Concept:

Discrete-time Fourier series is given by:

\(\rm x[n] = \sum_{k = 0}^{N - 1} a_k e^{j(2kn\pi/N)}\)

where N = time period
In discrete-time Fourier series, coefficients are periodic with the same time period of input signal x[n].

Therefore, a_k = a_k+N

m = any positive integer

Calculation:

Given, N = 3 
a_-3 = 2 and a₄ = 1
a_k = a_k+N
a_-3 = a_-3+3
a_-3 = a₀ = 2
a₁ = a₁₊₃
a₁ = a₄ = 1

\(\rm x[n] = a_0 e^{j(2(0)n\pi/3)} + a_1 e^{j(2(1)n\pi/3)}\)

\(\rm x[n] = a_0 e^{j0} + a_1 e^{j(2n\pi/3)} \)

\(\rm x[n] = 2 + 1 e^{j(2n\pi/3)} \)

\(\rm x[n] = 1 +1 + 1 e^{j(2n\pi/3)} \)

\(\rm x[n] = 1 + e^{j(2n\pi/6)} e^{j(-2n\pi/6)} + 1 e^{j(2n\pi/3)} \)

\(\rm x[n] = 1 + e^{j(2n\pi/6)} \times (e^{j(2n\pi/6)}+e^{-j(2n\pi/6)})\)

\(\rm x[n] = 1 + e^{j(2n\pi/6)} \times 2{e^{j(2n\pi/6)}+e^{-j(2n\pi/6)}\over2}\)

∵ \(e^{j(2n\pi/6)}+e^{-j(2n\pi/6)}\over2\) = \(cos \left( \frac{2 \pi}{6} n\right)\)

\(\rm x[n] =\rm 1 + 2e^{ \left( j \frac{2\pi}{6} n \right) }\cos \left( \frac{2 \pi}{6} n\right)\)

Parseval's Theorem:

For continuous-time, periodic signal, the energy is given by:

\(\frac{1}{T}\int\limits_T {{{\left| {x(t)} \right|}^2}dt = {{\sum\limits_{k = - \infty }^{ + \infty } {\left| {{a_k}} \right|} }^2}} \)

Where ak is the Fourier series coefficient of x(t), and T is the period of the signal.

For average power in one period of the periodic signal x(t), we write:

\(\frac{1}{T}\int\limits_T {{{\left| {{a_k}{e^{jk{\omega _o}t}}} \right|}^2}dt = \frac{1}{T}{{\int\limits_T {\left| {{a_k}} \right|} }^2}} dt = {\left| {{a_k}} \right|^2}\)

∴\({\left| {{a_k}} \right|^2}\) is the average power in the kth harmonic of x(t).

∴ Parseval's relation states that the total average power in a periodic signal equals the sum of the average powers in all of its harmonic components.

T = 6, \({\omega _0} = \frac{{2\pi }}{6},\; = \frac{\pi }{3}\)

\({\rm{x}}\left( {\rm{t}} \right){\rm{\;}} = \mathop \sum \limits_{n = \; - \infty }^e {C_n}{e^{jn{\omega _0}t}}\)

\({C_n} = \frac{1}{6}\mathop \smallint \nolimits_{ - 1}^5 x\left( t \right){e^{ - jn{\omega _0}t}}dt\)

\(= \frac{1}{6}\mathop \smallint \nolimits_{ - 1}^0 - 2\;{e^{ - jn{\omega _0}t\;\;}}dt + \frac{1}{6}\mathop \smallint \nolimits_2^3 2\;{e^{ - jn{\omega _0}t}}dt\)

\( = \frac{{ - 1}}{3}\;\left[ {\frac{{ - 1}}{{jn{\omega _0}\;}}{e^{ - jn{\omega _0}}}t} \right]_{ - 1}^0 + \frac{1}{3}\left[ {\frac{{ - 1}}{{jn{\omega _0}}}\;{e^{ - jn{\omega _0}t}}} \right]_2^3\)

\(= \frac{1}{{jn\pi }}\left( {1 - {e^{jn\frac{\pi }{3}}}} \right) - \frac{1}{{jn\pi }}\left( {{e^{ - jn\pi \;}} - {e^{ - jn\frac{{2\pi }}{3}}}} \right)\)

\(= \frac{1}{{jn\pi }}[\;{e^{\frac{{jn\pi }}{6}}}\left( {{e^{ - jn\frac{\pi }{6}}} - {e^{jn\frac{\pi }{6}}}} \right) - {e^{ - jn\frac{{5\pi }}{6}\;}}\left( {{e^{ - jn\frac{\pi }{6}}} - {e^{ - jn\frac{\pi }{6})}}} \right]\)

\(= \frac{2}{{n\pi }}\left[ { - {e^{jn\frac{\pi }{6}}}\sin \left( {\frac{{n\pi }}{6}} \right) + {e^{ - jn5\frac{\pi }{6}}}\sin \left( {n\frac{\pi }{6}} \right)} \right]\)

\(= \frac{2}{{n\pi }}\sin \left( {\frac{{n\pi }}{6}} \right)\left[ {{e^{ - jn\frac{{3\pi }}{6}}} + {e^{ - j\;3n\frac{\pi }{6}}}} \right]{e^{ - j\frac{{2\pi }}{6}n}}\)

\(= \frac{{ - 4j}}{{n\pi }}\sin \left( {\frac{{n\pi }}{6}} \right)\sin \left( {\frac{{n\pi }}{2}} \right){e^{ - jn\frac{\pi }{3}}}\)

• Fourier Series:

• A Fourier series is a way of representing a periodic function as a (possibly infinite) sum of sine and cosine functions.
• For functions that are not periodic, the Fourier series is replaced by the Fourier transform.
• For periodic signals this representation becomes the discrete-time Fourier series, and for aperiodic signals it becomes the discrete-time Fourier transform.
   

• Continuous Fourier Transform:

  The Fourier transform pair in the most general form for a continuous and aperiodic time signal x(t).

  Discrete-time Fourier series:

  The Fourier transform pair in the most general form for a discrete and periodic time signal x(t).

Effect of symmetry on Fourier series coefficients:

A periodic function with half-wave symmetry contains only odd harmonics.

Concept:

Convolution is a mathematical operation on two functions that produces a third function that expresses how the shape of one is modified by the other.

The term convolution refers to both the result function. It is a mathematical way of combining two signals to form a third signal.

It is the single most important technique in Digital Signal Processing. Tabular Method is one of the methods to solve it.

x1 (n)  =  [a₁, a₂, a₃] 

x2 (n)  =  [b₁, b₂, b₃]

Calculation:

x1 (n)  =  [1, 2, 3] 

x2 (n)  =  [3, 2, 1] 

Solving by the help of Tabular Method

y(n) = x1 (n) ∗ x2 (n)

y(n) = x1 (n) ∗ x2 (n) =  [3 , 2 + 6 , 1 + 4 + 9 , 2 + 6 , 3] =  [3,8,14,8,3]

Important Points

Length Property:

The length of the y(n) signal can be calculated as:

Length of x1 (n)  +  Length of x2 (n)  -  1

3 + 3 - 1 = 5