Discrete Time Fourier Transform (DTFT) MCQ Quiz - Objective Question with Answer for Discrete Time Fourier Transform (DTFT) - Download Free PDF

Explanation:

Discrete-Time Fourier Transform (DTFT)

Definition: The discrete-time Fourier transform (DTFT) is a form of Fourier analysis that is applicable to a sequence of values, typically a function of time. The DTFT is used to analyze the frequency content of discrete signals. It transforms a sequence of discrete-time signals into a continuous function of frequency.

Mathematical Representation: The DTFT of a sequence \( x[n] \) is defined as:

\( X(\omega) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n} \)

where:

• \( x[n] \) is the discrete-time signal.
• \( \omega \) is the normalized angular frequency (in radians per sample).
• \( j \) is the imaginary unit.

Periodicity of the DTFT: One of the key properties of the DTFT is its periodicity. The DTFT is always periodic in \( \omega \) with a period of \( 2\pi \). This means that:

\( X(\omega + 2\pi) = X(\omega) \)

This periodicity arises due to the nature of discrete-time signals, which are inherently sampled. The Nyquist-Shannon sampling theorem indicates that the highest frequency that can be uniquely represented by a sampled signal is half the sampling rate, leading to a periodicity in the frequency domain.

Correct Option Analysis:

The correct option is:

Option 2: \( 2\pi \)

This option correctly identifies that the DTFT is periodic in \( \omega \) with a period of \( 2\pi \). This periodicity is a fundamental characteristic of the DTFT and is crucial for understanding the frequency representation of discrete-time signals.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: \( 4\pi \)

This option is incorrect because the DTFT is not periodic with a period of \( 4\pi \). The correct periodicity is \( 2\pi \), as discussed above.

Option 3: \( \pi \)

This option is incorrect because \( \pi \) is not the period of the DTFT. While \( \pi \) is half the period of \( 2\pi \), it does not represent the full periodicity of the DTFT.

Option 4: \( \pi/2 \)

This option is incorrect as well. \( \pi/2 \) is a quarter of the period \( 2\pi \) and does not represent the periodicity of the DTFT.

Conclusion:

Understanding the periodicity of the DTFT is essential for analyzing the frequency content of discrete-time signals. The DTFT is always periodic in \( \omega \) with a period of \( 2\pi \), which is a fundamental property derived from the nature of discrete-time signals. This periodicity allows for the analysis and representation of signals in the frequency domain, providing insights into their spectral characteristics.

The correct option is 3

Given:  𝑦[𝑛]  = x[n] ∗ h[n] 

We know that the convolution in one domain results in convolution in another domain.

Y(𝑒𝑗𝛺) =𝑋(𝑒𝑗𝛺) H(𝑒𝑗𝛺) 

= (1 − 𝑒−𝑗𝛺 + 2𝑒−3𝑗𝛺)(\(1-\frac{1}{2}e^{-j2\Omega}\))

=1 - 𝑒−𝑗𝛺 + 2.5𝑒−3𝑗𝛺 - 0.5𝑒−2𝑗𝛺 - 𝑒−5𝑗𝛺

Taking IDFT we have:

y[n] =\(\rm \delta[n]-\delta[n-1]-\frac{1}{2}\delta[n-2]+\frac{5}{2}\delta[n-3]-\delta[n-5]\)

Explanation:

Summation property of discrete-time Fourier transform:

The summation of the discrete time signal x[n] is given by:

\(\int_{−π}^{π} |X(e^{j\Omega})|^2d\Omega\) = [π - (-π )] × {1² + 2² + 3² + (-1)² + (-2)² + (-3)²}

\(\int_{−π}^{π} |X(e^{j\Omega})|^2d\Omega\) = 2π × 28

\(\int_{−π}^{π} |X(e^{j\Omega})|^2d\Omega=56\pi\)

 Given discrete time sequence x[n] = {1, 2, 3, 0, −3, −2, −1} , where x[0] is 1.

 To calculate the value of : \(X(e^{jπ})\)

The DTFT of a sequence x[n] is defined as:

X(e^jω) = \(\sum_{n = - \infty}^ \infty x[n] e^{-jω n}\)

The DTFT X(ejω) is a function of frequency

x[n] = {1, 2, 3, 0, -3, -2, -1}

L = 7, n = 0 to (7 - 1)

X(ejω) = \(\sum_{n = 0}^6 x[n] e^{-jω n}\)

X(ejω) = x[0] + x[1]e^-jω + x[2]e-j2 + x[3]e-j3 + x[4]e-j4 + x[5]e-j5 +x[6]e-j6

= 1 + 2e-jω + 3e-j2ω + 0 - 3e-j4ω - 2e-j5ω - 1e-j6ω 

= 1 - e-j6ω + 2[e-jω - e-j5ω ] + 3[e-j2ω - e-j4ω ]

X(e^jπ) = 1 - e^-j6π + 2[e-j^π - e-j5^π] + 3[e-j2^π - e-j4^π]  

e-j6^π = cos6π - isin6π = 1

e-j^π = cosπ - isinπ = -1

e-j5^π = cos5π - isin5π = -1

e-j2^π = cos2π - isin2π = 1

e-j4^π = cos4π - isin4π = 1

X(ej^π) = 1 - 1 + 2[-1 - (-1)] + 3 [1 - 1] 

= 0 + [-1 + 1] + 3[0]

=0

Concept:

The Fourier transform for a discrete time sequence y[n] is given by:

\(Y\left( {{e^{j\omega }}} \right) = \;\mathop \sum \limits_{n = - \infty }^\infty y\left( n \right){e^{ - j\omega n}}\)

Analysis:

\(x\left( n \right)={{\left( \dfrac{1}{2} \right)}^{n}}u\left( n \right)\)

\(y\left( n \right)={{x}^{2}}\left( n \right)={{\left( \dfrac{1}{2} \right)}^{2n}}{{u}^{2}}\left( n \right)\)

\(={{\left[ {{\left( \dfrac{1}{2} \right)}^{2}} \right]}^{n}}{{u}^{2}}\left( n \right)\)

\(={{\left( \dfrac{1}{4} \right)}^{n}}u\left( n \right)\)

The Fourier transform of the above signal will now be:

\(Y(e^{j\omega })= \mathop \sum \limits_{n\; = \;0}^\infty {\left( {\dfrac{1}{4}} \right)^n}{e^{ - j\omega n}}\)

\(Y\left( {{e^{j0}}} \right) = \mathop \sum \limits_{n = 0}^\infty {\left( {\dfrac{1}{4}} \right)^n} \)

\(= 1 + \dfrac{1}{4} + {\left( {\dfrac{1}{4}} \right)^2} + \ldots \; \)

\(= \dfrac{1}{{1 - \dfrac{1}{4}}} = \dfrac{4}{3}\)

The Discrete-Time Fourier transform of a signal of infinite duration x[n] is given by

\(X\left( \omega \right) = DTFT\left\{ {x\left[ n \right]} \right\} = \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{e^{ - j\omega n}}\)

Given signal is, x[n] = αⁿ u[n]

\(X\left( \omega \right) = \mathop \sum \limits_{n = - \infty }^\infty {\alpha ^n}u\left[ n \right]{e^{ - j\omega n}}\)

\( = \mathop \sum \limits_{n = 0}^\infty {\left( {\alpha {e^{ - j\omega }}} \right)^n}\)

\( = \frac{1}{{1 - \alpha {e^{ - j\omega }}}}\)

The extension of discrete Fourier series for discrete-time aperiodic signals gives discrete-time Fourier transform. The DTFT of a signal is periodic with a period of 2π. Therefore, all the information is contained in -π < ω < π. 

The discrete-time Fourier Transform of a sequence x(n) is given by:

\(X\left( ω \right) = \mathop \sum \limits_{ - ∞ }^∞ x\left( n \right){e^{ - jω n}}\)

Where ω is discrete-time frequency. Transform X(ω) is a frequency-domain description of x(n).

Thus x(n) and X(ω) make Fourier transform pairs.

x(n): Discrete Aperiodic

X(ω): Periodic Continuous

The inverse DTFT which maps the frequency domain X(ω) back into time is given as:

\(x\left( n \right)=\frac{1}{2π }\underset{-π }{\overset{π }{\mathop \int }}\,X\left( {{ω }} \right){{e}^{jω n}}dω\)

X(ω), the frequency domain representation is a continuous function of ω which can take any value over a continuous interval from -∞ to +∞. Also, X(ω) is a periodic function of ω with period 2π it follows:

X(ω + 2π) = X(ω)

Concept:

\(x\left[ n \right]*y\left[ n \right]\;\begin{array}{*{20}{c}} {\mathop \leftrightarrow \limits^{Fourier} }\\^{Transform} \end{array}X\left( {{e^{j\omega }}} \right)Y\left( {{e^{j\omega }}} \right)\)

Let z(n) = x(n) * y(n)

Then Z(e^jω) = X(e^jω) Y(e^jω)

Also, using central ordinate theorem:

\(z\left[ 0 \right] = \frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } z\left( {{e^{j\omega }}} \right)d\omega \)

\( = \frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } X\left( {{e^{j\omega }}} \right)Y\left( {{e^{j\omega }}} \right)d\omega \)

Application:

x[n] = 2^n-1 u[-n + 2]

y[n] = 2^-n+2  u[n + 1]

Given:

\(\frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } X\left( {{e^{j\omega }}} \right)\;y\left( {{e^{ - j\omega }}} \right)d\omega \)

If y(n) = y(e^jω)

Then, using time reversal property:

y[-n] = y(e^-jω)

Hence z[n] = x[n] * y[-n]

z[n] = [2^n-1u[-n + 2]] * [2ⁿ⁺²u[-n + 1]]

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^\infty {2^{k - 1}}\;u\left[ { - k + 2} \right] \cdot {2^{n - k + 2}}\;u\left[ { - n + k + 1} \right]\)

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^2 {2^{k - 1}} \cdot {2^{n - k + 2}}\;u\left[ { - n + k + 1} \right]\)

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^2 {2^{k - 1 + n - k + 2}}\;u\left[ { - n + k + 1} \right]\)

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^2 {2^{n + 1}}\;u\left[ { - n + k + 1} \right]\;\)

\(\frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } X\left( {{e^{j\omega }}} \right)\;Y\left( {{e^{ - j\omega }}} \right) = z\left[ 0 \right]\)

Putting n = 0, we get:= 2

\(z\left[ 0 \right] = \mathop \sum \nolimits_{k = - \infty }^2 2 \cdot u\left[ {k + 1} \right]\)

\( = \mathop \sum \nolimits_{k = - 1}^2 2 \cdot \left( 1 \right)\)

= 2 × 4

= 8

The rectangular window is the most common windowing technique to design a finite impulse response (FIR) filter.

A rectangular window is defined as

\({w_R}\left( n \right) = \left\{ {\begin{array}{*{20}{c}} {1\;:0 \le n \le N - 1}\\ {0\;\;\;\;:otherwise} \end{array}} \right.\)

Where N = length of the FIR filter

The frequency response of w_R(n) is w_R(ω) which is calculated as:

w_R(ω) = DTFT of w_R(n), i.e.

\({w_R}\left( \omega \right) = \mathop \sum \limits_{n = - \infty }^\infty {w_R}\left( n \right){e^{ - j\omega n}}\)

ω ϵ [-π, π] rad/sec

For the given rectangular sequence, the Fourier transform will be a sinc sequence given by:

\({w_R}\left( \omega \right) \approx \frac{{\sin \left( {\frac{{\omega N}}{2}} \right)}}{{\sin \left( {\frac{\omega }{2}} \right)}}\)

The spectrum is as shown:

Main-lobe width = 2 × (zero-crossing points of sin c function), i.e.

\( = 2 \times \frac{{2\pi }}{N} = \frac{{4\pi }}{N}\)

Concept:

D.T.F.T

Discrete-time Fourier transform is defined by

\(H\left( {{{\rm{e}}^{{\rm{jω }}}}} \right) = \mathop \sum \limits_{n = - \infty }^\infty h\left[ n \right]{e^{ - jω n}}\)

Inverse function of x(n) is defined as:

h₁(n) = 1/h(n)

H₁(e^jω) = 1/H(e^jω)

H(e^jω) H₁(e^jω) = 1

Explanation:

Summation property of discrete-time Fourier transform:

The summation of the discrete time signal x[n] is given by:

\(\int_{−π}^{π} |X(e^{j\Omega})|^2d\Omega\) = [π - (-π )] × {1² + 2² + 3² + (-1)² + (-2)² + (-3)²}

\(\int_{−π}^{π} |X(e^{j\Omega})|^2d\Omega\) = 2π × 28

\(\int_{−π}^{π} |X(e^{j\Omega})|^2d\Omega=56\pi\)

Concept: 

Periodic Signal:

A signal x(t) is said to be periodic if for some positive constant T₀;

x(t) = x (t + T₀) for all t

The smallest value of T₀ that satisfies the periodicity condition of the above equation is called the fundamental period of x(t).

Periodic signal repeat with some fixed time period T  and continue with -∞  to +∞

The sum of two periodic functions is often periodic, but not always.

Suppose that x(t) is periodic with fundamental period P, and g(t) is periodic with fundamental period Q. Then x(t) + g(t) will be periodic with a period R, if P and Q both have a common multiple R.

Example: sin x + sin πx

sin x is periodic with a period 2π 

sin πx is periodic with a period 2

But since 2π and 2 have no common multiple, the sum of the two signals will not be periodic.

Given:

x(t) = \(sin (\sqrt{2 \pi t})\)

• For negative values of ‘t, sine-function will apply to imaginary values.
• For positive values of ‘t’, sine-function will apply on real-values.
• So, there will not be repetition in the waveform.

So We can draw waveform only in the RHS but we cannot draw waveform in the LHS.
So, sin( 2πt ) is non-periodic.

Important Points

• Any analog sinusoid or harmonic signal is always periodic. As the sinusoids are periodic and harmonic are expressed in terms of sinusoids only.
• Additional phase or Amplitude will not affect the periodicity of a signal.
• The Sum of two periodic discrete-time signals is always periodic.
• All periodic signals are power signals but vice versa may not be true.
• The sequence obtained by uniform sampling of a continuous-time signal may not be periodic.

The Discrete-Time Fourier transform of a signal of infinite duration x[n] is given by:

\(X\left( \omega \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{e^{ - j\omega n}}\)

For a signal x(n) = aⁿ u(n), the DTFT will be:

\(X\left( \omega \right) = \mathop \sum \limits_{n = - \infty }^\infty a^nu\left[ n \right]{e^{ - j\omega n}}\)

Since u[n] is zero for n < 0 and a constant 1 for n > 0, the above summation becomes:

\(X\left( \omega \right) = \mathop \sum \limits_{n = 0 }^\infty a^n{e^{ - j\omega n}}\)

\(X\left( \omega \right) = \mathop \sum \limits_{n = 0 }^\infty (a{e^{ - j\omega n})}\)   ---(1)

The geometric series states that:

\(\mathop \sum \limits_{n = 0 }^\infty r^n=\frac{1}{1-r}\), for 0 < r < 1

Equation (1) can now be written as:

\(X\left( \omega \right) =\frac{1}{1-ae^{-j\omega}}\)    ---(2)

Application:

Given x[n] = 0.5n u[n], i.e. a = 0.5

The DTFT of the above given sequence using Equation (2) will be:

\(X\left( \omega \right) =\frac{1}{1-0.5e^{-j\omega}}\)

Concept:

The Fourier transform of aⁿu(n) is given as:

\({a^n}u\left( n \right)\mathop \leftrightarrow \limits^{F.T.} \frac{1}{{1 - a{e^{ - j\omega }}}}\)

Also, the time-shifting property of Fourier transforms states that:

If x(n) ↔ X(jω)

\(x\left( {n - {n_0}} \right) \leftrightarrow {e^{ - j{n_0}\omega }} \cdot X\left( {j\omega } \right)\)

Application:

Let \(f\left( n \right) = {\left( {\frac{2}{3}} \right)^n}u\left( n \right)\)

The Fourier Transform of f(n) will be:

\(F\left( {j\omega } \right) = \frac{1}{{1 - \left( {\frac{2}{3}} \right){e^{ - j\omega }}}}\)

Replacing ‘ω’ with 2nf, we get:

\(F\left( f \right) = \frac{1}{{1 - \left( {\frac{2}{3}} \right){e^{ - j2\pi f}}}}\)

Applying Time shifting property, we can write:

\({\left( {\frac{2}{3}} \right)^{n + 3}}u\left( {n + 3} \right) \leftrightarrow \frac{{{e^{j \cdot 3\omega }}}}{{1 - \left( {\frac{2}{3}} \right){e^{ - j2\pi f}}}}\)

The above expression can be written as:

\({\left( {\frac{2}{3}} \right)^3} \cdot {\left( {\frac{2}{3}} \right)^n}u\left( {n + 3} \right) \leftrightarrow \frac{{{e^{j \cdot 6\pi f}}}}{{1 - \left( {\frac{2}{3}} \right){e^{ - j2\pi f}}}}\)

Multiplying both sides with \({\left( {\frac{3}{2}} \right)^3}\), we get:

\({\left( {\frac{2}{3}} \right)^n}u\left( {n + 3} \right) \leftrightarrow \frac{{{{\left( {\frac{3}{2}} \right)}^3} \cdot {e^{j6\pi f}}}}{{1 - \left( {\frac{2}{3}} \right){e^{ - j2\pi f}}}}\)

Comparing this with the given transform, we get:

\(A = {\left( {\frac{3}{2}} \right)^3} = {\left( {1.5} \right)^3}\)

A = 3.375

FIR Design Methods include:

• Impulse response truncation – the simplest design method, has undesirable frequency domain-characteristics.
•  Windowing design method – simple and convenient but not optimal, i.e. order achieved is not minimum possible.
• There are much less ripples for the Hanning window but than the transition width has increased.