Properties of DTFT MCQ Quiz in मराठी - Objective Question with Answer for Properties of DTFT - मोफत PDF डाउनलोड करा

Concept:

\(x\left( n \right)u\left( n \right)\overset{DTFT}{\longleftrightarrow }x\left( \omega \right)\)

The time-shifting property states that:

\(x\left( {n\; - \;6} \right)u\left( {n\; - \;6} \right)\overset{DTFT}{\longleftrightarrow } x\left( \omega \right){e^{ - j6\omega }}\)

Application:

\({0.8^n}\;u\left( n \right)\overset{DTFT}{\longleftrightarrow } \frac{1}{{1 - 0.8{e^{ - j\omega }}}}\)

Now, we wish to calculate the DTFT of

\(x\left( n \right) = \left\{ {\begin{array}{*{20}{c}} {{{\left( {0.8} \right)}^n},}&{for\;0 \le n \le 5}\\ {0,}&{Otherwise} \end{array}} \right.\)

This can be written as:

x(n) = (0.8)ⁿ [u(n) - u(n - 6)]

x(n) = (0.8)ⁿ u(n) – (0.8)ⁿ u(n - 6)

x(n) = (0.8)ⁿ u(n) – (0.8)⁶ (0.8)^{n – 6} u(n - 6)

Using time-shifting property, the DTFT of x(n) will be:

\(X\left( \omega \right) = \frac{1}{{1 - 0.8{e^{ - j\omega }}}} - \frac{{{{\left( {0.8} \right)}^6}{e^{ - j6\omega }}}}{{1 - 0.8{e^{ - j\omega }}}}\)

\(X\left( \omega \right) = \frac{{1 - {{0.8}^6}\;{e^{ - j6\omega }}}}{{1 - 0.8\;{e^{ - j\omega }}}}\)

Alternate Method (Option checking):

1) Option (3) is ruled out because we can see that DTFT of x(n) will consist of two terms as the interval 0 ≤ n ≤ 5 and can be defined by:

\(\underbrace {u\left( n \right)}_{one\;term} - \underbrace {u\left( {n - 6} \right)}_{another\;term}\)

2) Option (4) is ruled out since there is a positive sign in the denominator that comes only when (-a)ⁿ u(n) is given. But we have 0.8 which does not contain a negative sign in the given question.

3) Option (2) is also ruled out since DTFT of both terms u(n) and u(n - 6) components are getting added. Instead, they must have been subtracted.

So the correct answer is Option (1).

Tips: In the early stage of learning, try to eliminate the options. Eventually, with practice, one can get used to it to solve all the easy questions in minimum time.

Concept:

The DTFT and Inverse DTFT is defined as:

\(X\left( {{e}^{j\omega }} \right)=\underset{n=-\infty }{\overset{\infty }{\mathop \sum }}\,x\left( n \right){{e}^{-j\omega n}}\)

\(x\left( n \right)=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\mathop \int }}\,X\left( {{e}^{j\omega }} \right){{e}^{j\omega n}}d\omega\)

Application:

If \(x\left( n \right) = {\left( { - a} \right)^{n - 1}}u\left( {n - 1} \right)\)

The Fourier transform will be:

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

x²(n) → (a²)^n-1 u(n – 1)

∴ DTFT of x²(n) will be:

\({x^2}\left( n \right) = \mathop \sum \nolimits_{n = - \infty }^\infty {\left( {{a^2}} \right)^{n - 1}}u\left( {n - 1} \right){e^{ - j\omega n}}\)

\( = \frac{1}{{{a^2}}}\mathop \sum \nolimits_{n = 1}^\infty {\left( {{a^2}{e^{ - j\omega }}} \right)^n}\)

\(= \frac{1}{{{a^2}}}\frac{{{a^2}{e^{ - j\omega }}}}{{1 - {a^2}{e^{ - j\omega }}}}\)

\(= \frac{{{e^{ - j\omega }}}}{{1 - {a^2}{e^{ - j\omega }}}} = \frac{1}{{{e^{j\omega }} - {a^2}}}\)

Concept:

Any discrete signal is said to be periodic if:

\(\frac{{{\omega _0}}}{{2\pi }}\) is a rational number \(\left( {\frac{M}{N}} \right)\)

Calculation:

Given signal is \({e^{j{\omega _0}n}}\), ω₀ = 1 rad / sec

\(\frac{{{\omega _0}}}{{2\pi }} = \frac{1}{{2\pi }}\) 

Since this is not a rational number, so it is not periodic at all.

Note:

Continuous sinusoidal and complex sinusoids are periodic for every value of ‘ω₀’, but discrete-time signals are periodic only if \(\frac{{{\omega _0}}}{{2\pi }}\) is a rational number \(\frac{M}{N}\).

Let the signal in frequency be represented by:

To occupy the entire region from -π to π, x[n] must be down sample by factor 14/3

Since it is not possible to down sample by non-integer factor, up sample single first by 3 (= L).

Then down sample signal by 14 (= N).

Thus, L = 3

N = 14

\(\begin{array}{l} X\left( {{e^{i\omega }}} \right) = \frac{1}{{1 - \frac{1}{4}{e^{ - j\left( {\omega - 0} \right)}}}} + \frac{{\frac{1}{2}}}{{1 - \frac{1}{4}{e^{ - j\left( {\omega - \frac{\pi }{2}} \right)}}}} + \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{1 - \frac{1}{4}{e^{ - j\left( {\omega - \frac{{2\pi }}{2}} \right)}}}} + \frac{{{{\left( {\frac{1}{2}} \right)}^3}}}{{1 - \frac{1}{4}{e^{ - j\left( {\omega - \frac{{3\pi }}{2}} \right)}}}}\\ x\left[ n \right] = {\left( {\frac{1}{4}} \right)^n}u\left[ n \right] + \frac{1}{2}{e^{\frac{{j\pi }}{{2n}}}}{\left( {\frac{1}{4}} \right)^n}u\left[ n \right] + {\left( {\frac{1}{2}} \right)^2}{e^{j\pi 4}}{\left( {\frac{1}{4}} \right)^n}u\left[ n \right] + {\left( {\frac{1}{2}} \right)^3}{\left( {\frac{1}{4}} \right)^n}{e^{\frac{{j3\pi }}{{2n}}}}u\left[ n \right]\\ = {\left( {\frac{1}{4}} \right)^{n}}u\left[ n \right]\left[ {1 + \frac{1}{2}{e^{\frac{{j\pi }}{{2n}}}} + {{\left( {\frac{1}{2}} \right)}^2}{e^{j\pi n}} + {{\left( {\frac{1}{2}} \right)}^3}{e^{\frac{{j3\pi }}{{2n}}}}} \right] \end{array}\)

= g[n].q[n]

 g[n] is of form αⁿ u[n]

⇒ α = ¼

Period of q[n] is 4

Hence the product αN = ¼ × 4 = 1

We have

\({\rm{x}}\left[ {\rm{n}} \right]\mathop \leftrightarrow \limits^{{\rm{FT}}} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)\) 

From properties of Fourier transform we have,

\({\rm{nx}}\left[ {\rm{n}} \right]\mathop \leftrightarrow \limits^{{\rm{FT}}} {\rm{j}}\frac{{{\rm{dX}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)}}{{{\rm{d\omega }}}}\) 

Rewriting it we have

\({\rm{j}}\frac{{{\rm{dX}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)}}{{{\rm{d\omega }}}} = \mathop \sum \limits_{{\rm{n}} = - \infty }^\infty {\rm{nx}}\left[ {\rm{n}} \right] \cdot {{\rm{e}}^{ - {\rm{j\omega n}}}}\) 

Putting ω = 0, we have

\({\rm{j\;}}{\left. {\frac{{{\rm{dX}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)}}{{{\rm{d\omega }}}}} \right|_{{\rm{\omega }} = 0}} = \mathop \sum \limits_{{\rm{n}} = - \infty }^\infty {\rm{nx}}\left[ {\rm{n}} \right]\) 

Likewise,

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

Again at ω = 0

\({\rm{X}}{\left. {\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)} \right|_{{\rm{\omega }} = 0}} = \mathop \sum \limits_{{\rm{n}} = - \infty }^\infty {\rm{x}}\left[ {\rm{n}} \right]\) 

Since, for calculating α, we need value only at \({\rm{\omega }} = 0\). So, simply, we express \({\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)\) mathematically for interval \(\left( { - \frac{{\rm{\pi }}}{6},\frac{{\rm{\pi }}}{6}} \right)\)

So,

\(\begin{array}{l} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right) = 1 + {\rm{j}}\left( {\frac{{ - 4}}{{\frac{{2{\rm{\pi }}}}{3}}}} \right){\rm{\omega \;\;\;\;\;\;\;\;\;}}-\frac{{\rm{\pi }}}{6} < {\rm{\omega }} < \frac{{\rm{\pi }}}{6}\\ \Rightarrow {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right) = 1 - {\rm{j}}\left( {\frac{6}{{\rm{\pi }}}} \right){\rm{\omega }} \end{array}\)

Thus, \({\rm{X}}{\left. {\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)} \right|_{{\rm{\omega }} = 0}} = 1\) 

and \(\frac{{{\rm{dX}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)}}{{{\rm{d\omega }}}} = - {\rm{j}}\left( {\frac{6}{{\rm{\pi }}}} \right)\)

\(\Rightarrow {\rm{j}}{\left. {\frac{{{\rm{dX}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)}}{{{\rm{d\omega }}}}} \right|_{{\rm{\omega }} = 0}} = \left( {\frac{6}{{\rm{\pi }}}} \right)\) 

Now,

\({\rm{\alpha }} = \frac{{\mathop \sum \nolimits_{{\rm{n}} = - \infty }^\infty {\rm{nx}}\left[ {\rm{n}} \right]}}{{\mathop \sum \nolimits_{{\rm{n}} = - \infty }^\infty {\rm{x}}\left[ {\rm{n}} \right]}}\) 

Substituting the Fourier equivalents we have,

\(\begin{array}{l} {\rm{\alpha }} = \frac{{{\rm{j}}{{\left. {\frac{{{\rm{dX}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)}}{{{\rm{d\omega }}}}} \right|}_{{\rm{\omega }} = 0}}}}{{{{\left. {{\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)} \right|}_{{\rm{\omega }} = 0}}}}\\ \Rightarrow {\rm{\alpha }} = \frac{{\left( {\frac{6}{{\rm{\pi }}}} \right)}}{1} = \frac{6}{{\rm{\pi }}} \end{array}\)

Using the property that \(\rm{c[n]\ast\delta[n-n_\circ]=c[n-n_\circ]}\),

\({\rm{x}}\left[ {\rm{n}} \right]\) can be rewritten, as

\({\rm{x}}\left[ {\rm{n}} \right] = {\left( {\frac{1}{2}} \right)^{\rm{n}}}{\rm{u}}\left[ {\rm{n}} \right]{\rm{*}}\mathop \sum \limits_{{\rm{k}} = - \infty }^\infty {\rm{\delta }}\left[ {{\rm{n}} - 4{\rm{k}}} \right]\)

Let, \({\left( {\frac{1}{2}} \right)^{\rm{n}}}{\rm{u}}\left[ {\rm{n}} \right] = {\rm{p}}\left[ {\rm{n}} \right]\)and \(\mathop \sum \limits_{{\rm{k}} = - \infty }^\infty {\rm{\delta }}\left[ {{\rm{n}} - 4{\rm{k}}} \right] = {\rm{q}}\left[ {\rm{n}} \right]\).

Now, \({\rm{y}}\left[ {\rm{n}} \right]\) can be obtained by first passing \({\rm{q}}\left[ {\rm{n}} \right]\) through the filter with response \({\rm{H}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)\) and the convolving the result with \({\rm{p}}\left[ {\rm{n}} \right]\).

Now, \({\rm{q}}\left[ {\rm{n}} \right]\) is periodic with period \(4\), the Fourier series coefficient \({{\rm{a}}_{\rm{k}}}\) of \({\rm{q}}\left[ {\rm{n}} \right]\) are

\({{\rm{a}}_{\rm{k}}} = \frac{1}{4}\) for all \({\rm{k}}\left(\because {{\rm{for}}\mathop \sum \limits_{{\rm{k}} = - \infty }^\infty {\rm{\delta }}\left[ {{\rm{n}} - {\rm{pk}}} \right]\mathop \leftrightarrow \limits^{{\rm{DFS}}} {{\rm{a}}_{\rm{k}}} = \frac{1}{{\rm{p}}}{\rm{for\ all\ k}}} \right)\)

Now, output \({\rm{r}}\left[ {\rm{n}} \right] = {\rm{q}}\left[ {\rm{n}} \right]{\rm{*h}}\left[ {\rm{n}} \right]\)

As, the signal is periodic, taking the convolution over one period we have,

\(\begin{array}{l} {\rm{r}}\left[ {\rm{n}} \right] = \mathop \sum \limits_{{\rm{k}} = 0}^3 {{\rm{a}}_{\rm{k}}}{{\rm{e}}^{{\rm{jk}}\left( {\frac{{2{\rm{\pi }}}}{4}} \right){\rm{n}}}} \cdot {\rm{H}}\left( {{{\rm{e}}^{{\rm{jk}}\left( {\frac{{2{\rm{\pi }}}}{4}} \right)}}} \right)\\ \Rightarrow {\rm{r}}\left[ {\rm{n}} \right] = {{\rm{a}}_0}{{\rm{e}}^{{\rm{j}}0}}{\rm{H}}\left( {{{\rm{e}}^{{\rm{j}}0}}} \right) + {{\rm{a}}_1}{{\rm{e}}^{\frac{{{\rm{j\pi n}}}}{2}}}{\rm{H}}\left( {{{\rm{e}}^{{\rm{j}}\frac{{\rm{\pi }}}{2}}}} \right) + {{\rm{a}}_2}{{\rm{e}}^{{\rm{j\pi n}}}} \cdot {\rm{H}}\left( {{{\rm{e}}^{{\rm{j\pi }}}}} \right) + {{\rm{a}}_3}{{\rm{e}}^{{\rm{j}}\frac{{3{\rm{\pi n}}}}{2}}}{\rm{H}}\left( {{{\rm{e}}^{{{\rm{e}}^{\frac{{{\rm{j}}3{\rm{\pi }}}}{2}}}}}} \right) \end{array}\)

Now since all \({\rm{H}}\left( {{{\rm{e}}^{{\rm{j}}0}}} \right),{\rm{H}}\left( {{{\rm{e}}^{\frac{{{\rm{j\pi }}}}{2}}}} \right),{\rm{H}}\left( {{{\rm{e}}^{{\rm{j\pi }}}}} \right)\) and \({\rm{H}}\left( {{{\rm{e}}^{\frac{{{\rm{j}}3{\rm{\pi }}}}{2}}}} \right)\) are 0

\(\Rightarrow {\rm{r}}\left[ {\rm{n}} \right] = 0\) for all \({\rm{n}}\)

Now, \({\rm{y}}\left[ {\rm{n}} \right] = {\rm{p}}\left[ {\rm{n}} \right]{\rm{*r}}\left[ {\rm{n}} \right] = 0\) for all \({\rm{n}}\)

DTFT (x[n]) = X(e^jω)

\(\begin{array}{l} = \mathop \sum \limits_{n = - \infty }^\infty {a^{\left| n \right|}}{e^{ - j\omega n}}\\ = \mathop \sum \limits_{n = - \infty }^{ - 1} {a^{ - n}}{e^{ - j\omega n}} + \mathop \sum \limits_{n = 0}^\infty {a^n}{e^{ - j\omega n}} \end{array}\)

In first part, we take m = -n then we get

\(\begin{array}{l} X\left( {{e^{j\omega }}} \right) = \mathop \sum \limits_{m = 1}^\infty {a^m}{e^{j\omega m}} + \mathop \sum \limits_{n = 0}^\infty {a^n}{e^{ - j\omega n}}\\ = \mathop \sum \limits_{m = 1}^\infty {\left( {a{e^{j\omega }}} \right)^m} + \mathop \sum \limits_{n = 0}^\infty {\left( {a{e^{ - j\omega }}} \right)^n}\\ = \frac{{a{e^{j\omega }}}}{{1 - a{e^{j\omega }}}} + \frac{1}{{1 - a{e^{ - j\omega }}}}\\ = \frac{{a{e^{j\omega }}\left( {1 - a{e^{ - j\omega }}} \right) + 1 - a{e^{j\omega }}}}{{\left( {1 - a{e^{j\omega }}} \right)\left( {1 - a{e^{ - j\omega }}} \right)}}\\ = \frac{{1 - {a^2}}}{{1 - a{e^{j\omega }} - a{e^{ - j\omega }} + {a^2}}}\\ = \frac{{1 - {a^2}}}{{1 - 2a\cos \omega + {a^2}}} \end{array}\)

x[n] = u [n – 2] – u [n – 5]

= δ[n – 2] + δ[n – 3] + δ[n - 4] 

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

= e^-j2Ω + e^-j3Ω + e^-j4Ω

\(\begin{array}{l} H\left( {{e^{j\omega }}} \right) = \frac{1}{{1 - \alpha {e^{ - j\omega }}}}\\ X\left( {{e^{j\omega }}} \right) = \frac{1}{{1 - \beta {e^{ - j\omega }}}} \end{array}\)

So that

\(\begin{array}{l} Y\left( {{e^{j\omega }}} \right) = H\left( {{e^{j\omega }}} \right)X\left( {{e^{j\omega }}} \right) = \frac{1}{{\left( {1 - \alpha {e^{ - j\omega }}} \right)\left( {1 - \beta {e^{ - j\omega }}} \right)}}\\ \Rightarrow Y\left( {{e^{j\omega }}} \right) = \frac{A}{{\left( {1 - \alpha {e^{ - j\omega }}} \right)}} + \frac{B}{{\left( {1 - \beta {e^{ - j\omega }}} \right)}} \end{array}\)

Now,

\(\begin{array}{l} A = {\left. {\frac{1}{{\left( {1 - \beta {e^{ - j\omega }}} \right)}}} \right|_{{e^{j\omega }} = \alpha }}\\ \Rightarrow A = \frac{1}{{1 - \frac{\beta }{\alpha }}} = \frac{\alpha }{{\alpha - \beta }} \end{array}\)