Time Scaling MCQ Quiz - Objective Question with Answer for Time Scaling - Download Free PDF

Given:

 $X(z)=\frac{4z}{(z-\frac{1}{5})(z-\frac{2}{3})(z-3)}$

Poles of X(z) are located at z = $\frac{1}{5}$, z = $\frac{2}{3}$ and z = 3.

For DTFT to converge, the ROC of Z-transform of x() should contain unit circle.

If x(n) is a right sided sequence then the ROC is |z|>3 which does not include unit circle. So, option (D) and (A) are wrong.

If R.O.C. is $\frac{2}{3}$ < |z| < 3, the R.O.C. includes unit circle. So, option (B) is correct.

If x(n) is a left sided then R.O.C will be |z| < $\frac{1}{5}$ which does not include unit circle. So, option (C) is wrong.

Hence, the correct option is (B).

Concept:

The even part of a signal x(n) is defined as:

$x_e\left(n\right)=\frac{1}{2}\left[x\left(n\right)+x\left(-n\right)\right]$          ---(1)

Property:

If $x\left(n\right)\stackrel{x}{\leftrightarrow }X\left(z\right);\left|z\right|>a$

Then $x\left(-n\right)\stackrel{x}{\leftrightarrow }X\left(\frac{1}{z}\right);\left|\frac{1}{z}\right|>a$

Application:

From Equation (1), the z-transform of the even part of the signal will be:

$X_e\left(z\right)=\frac{1}{2}\left[X\left(z\right)+X\left(\frac{1}{z}\right)\right]$

Given $X\left(z\right)=\frac{z}{2-0.4}$, the above expression becomes:

$X_e\left(z\right)=\underset{I}{\underbrace{\frac{1}{2}\left(\frac{z}{z-0.4}\right)}}+\underset{II}{\underbrace{\frac{1}{2}\left(\frac{\frac{1}{z}}{\frac{1}{z}-0.4}\right)}}$

ROC for the I term will be |z| > 0.4

ROC for the II term will be:

$\left|z\right|<\frac{1}{0.4},i.e.\left|z\right|<2.5$

Now the ROC of X_e(z) will be the intersection of two, i.e.

0.4 < |z| < 2.5

CONCEPT:

The digital frequency (Ω) is related to analog frequency (f) through the sampling frequency f_s.

$\mathrm{\Omega }=\frac{2\pi f}{f_s}\Rightarrow \omega T_s$

PROVING:

Continuous  to discrete frequency converter:

Let continuous signal is x_c(t) and sampling signal is s(t).

$S\left(t\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\delta \left(t-nT\right)\stackrel{\left\{F.T.\right\}}{\to }S\left(J\mathrm{\Omega }\right)\to \frac{2\pi }{T}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\delta J\left(\mathrm{\Omega }-n{\mathrm{\Omega }}_s\right)$

Where Ω_s = sampling frequency

$T=\frac{2\pi }{{\mathrm{\Omega }}_s}=samplinginteraval$

$x_s\left(t\right)=x_c\left(t\right).s\left(t\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{x}_c\left(t\right)\delta \left(t-nT\right)$

$X_s\left(J\mathrm{\Omega }\right)=\frac{1}{2\pi }{X}_c\left(J\mathrm{\Omega }\right)\ast S\left(J\mathrm{\Omega }\right)$

$X_s\left(J\mathrm{\Omega }\right)=\frac{1}{2\pi }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{X}_c\left[J\left(\mathrm{\Omega }-n{\mathrm{\Omega }}_s\right)\right]$

Or we can write

$X_s\left(J\mathrm{\Omega }\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{x}_c\left(nT\right)e^{-J\mathrm{\Omega }Tn}$

If x(nT) = x(n)

$X\left(e^{J\omega }\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left(n\right)e^{-J\mathrm{\Omega }n}$

Concept:

The even part of a signal x(n) is defined as:

$x_e\left(n\right)=\frac{1}{2}\left[x\left(n\right)+x\left(-n\right)\right]$     ---(1)

Property:

If $x\left(n\right)\stackrel{x}{\leftrightarrow }X\left(z\right);\left|z\right|>a$

Then $x\left(-n\right)\stackrel{x}{\leftrightarrow }X\left(\frac{1}{z}\right);\left|\frac{1}{z}\right|>a$

Application:

From Equation (1), the z-transform of the even part of the signal will be:

$X_e\left(z\right)=\frac{1}{2}\left[X\left(z\right)+X\left(\frac{1}{z}\right)\right]$

Given $X\left(z\right)=\frac{z}{2-0.4}$, the above expression becomes:

$X_e\left(z\right)=\underset{I}{\underbrace{\frac{1}{2}\left(\frac{z}{z-0.4}\right)}}+\underset{II}{\underbrace{\frac{1}{2}\left(\frac{\frac{1}{z}}{\frac{1}{z}-0.4}\right)}}$

ROC for the I term will be |z| > 0.4

ROC for the II term will be:

$\left|z\right|<\frac{1}{0.4},i.e.\left|z\right|<2.5$

Now the ROC of X_e(z) will be the intersection of two, i.e.

0.4 < |z| < 2.5

Given:

 $X(z)=\frac{4z}{(z-\frac{1}{5})(z-\frac{2}{3})(z-3)}$

Poles of X(z) are located at z = $\frac{1}{5}$, z = $\frac{2}{3}$ and z = 3.

For DTFT to converge, the ROC of Z-transform of x() should contain unit circle.

If x(n) is a right sided sequence then the ROC is |z|>3 which does not include unit circle. So, option (D) and (A) are wrong.

If R.O.C. is $\frac{2}{3}$ < |z| < 3, the R.O.C. includes unit circle. So, option (B) is correct.

If x(n) is a left sided then R.O.C will be |z| < $\frac{1}{5}$ which does not include unit circle. So, option (C) is wrong.

Hence, the correct option is (B).

CONCEPT:

The digital frequency (Ω) is related to analog frequency (f) through the sampling frequency f_s.

$\mathrm{\Omega }=\frac{2\pi f}{f_s}\Rightarrow \omega T_s$

PROVING:

Continuous  to discrete frequency converter:

Let continuous signal is x_c(t) and sampling signal is s(t).

$S\left(t\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\delta \left(t-nT\right)\stackrel{\left\{F.T.\right\}}{\to }S\left(J\mathrm{\Omega }\right)\to \frac{2\pi }{T}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\delta J\left(\mathrm{\Omega }-n{\mathrm{\Omega }}_s\right)$

Where Ω_s = sampling frequency

$T=\frac{2\pi }{{\mathrm{\Omega }}_s}=samplinginteraval$

$x_s\left(t\right)=x_c\left(t\right).s\left(t\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{x}_c\left(t\right)\delta \left(t-nT\right)$

$X_s\left(J\mathrm{\Omega }\right)=\frac{1}{2\pi }{X}_c\left(J\mathrm{\Omega }\right)\ast S\left(J\mathrm{\Omega }\right)$

$X_s\left(J\mathrm{\Omega }\right)=\frac{1}{2\pi }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{X}_c\left[J\left(\mathrm{\Omega }-n{\mathrm{\Omega }}_s\right)\right]$

Or we can write

$X_s\left(J\mathrm{\Omega }\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{x}_c\left(nT\right)e^{-J\mathrm{\Omega }Tn}$

If x(nT) = x(n)

$X\left(e^{J\omega }\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left(n\right)e^{-J\mathrm{\Omega }n}$

Concept:

The even part of a signal x(n) is defined as:

$x_e\left(n\right)=\frac{1}{2}\left[x\left(n\right)+x\left(-n\right)\right]$     ---(1)

Property:

If $x\left(n\right)\stackrel{x}{\leftrightarrow }X\left(z\right);\left|z\right|>a$

Then $x\left(-n\right)\stackrel{x}{\leftrightarrow }X\left(\frac{1}{z}\right);\left|\frac{1}{z}\right|>a$

Application:

From Equation (1), the z-transform of the even part of the signal will be:

$X_e\left(z\right)=\frac{1}{2}\left[X\left(z\right)+X\left(\frac{1}{z}\right)\right]$

Given $X\left(z\right)=\frac{z}{2-0.4}$, the above expression becomes:

$X_e\left(z\right)=\underset{I}{\underbrace{\frac{1}{2}\left(\frac{z}{z-0.4}\right)}}+\underset{II}{\underbrace{\frac{1}{2}\left(\frac{\frac{1}{z}}{\frac{1}{z}-0.4}\right)}}$

ROC for the I term will be |z| > 0.4

ROC for the II term will be:

$\left|z\right|<\frac{1}{0.4},i.e.\left|z\right|<2.5$

Now the ROC of X_e(z) will be the intersection of two, i.e.

0.4 < |z| < 2.5

The z-transform of a sequence x(n) is defined as:

$X(Z)=\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}x(n)\cdot z^{-n}$

Now, $Y(z)=X(z^3)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n)(z^3{)}^{-n}$

$\Rightarrow Y(z)=X(z^3)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n)z^{-3n}$

Let 3n = k

$\Rightarrow n=\frac{k}{3}$

$Y(z)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}x\left(\frac{k}{3}\right)z^{-k}$

Thus, $y(n)=x\left(\frac{n}{3}\right)$

$y\left(n\right)=\{\begin{array}{c}x\left(n\right),\\ 0,\end{array}\begin{array}{c}n=0,3,6,--\\ otherwise\end{array}$

So, y(n) at n = 4 is 0.

Concept:

The even part of a signal x(n) is defined as:

$x_e\left(n\right)=\frac{1}{2}\left[x\left(n\right)+x\left(-n\right)\right]$          ---(1)

Property:

If $x\left(n\right)\stackrel{x}{\leftrightarrow }X\left(z\right);\left|z\right|>a$

Then $x\left(-n\right)\stackrel{x}{\leftrightarrow }X\left(\frac{1}{z}\right);\left|\frac{1}{z}\right|>a$

Application:

From Equation (1), the z-transform of the even part of the signal will be:

$X_e\left(z\right)=\frac{1}{2}\left[X\left(z\right)+X\left(\frac{1}{z}\right)\right]$

Given $X\left(z\right)=\frac{z}{2-0.4}$, the above expression becomes:

$X_e\left(z\right)=\underset{I}{\underbrace{\frac{1}{2}\left(\frac{z}{z-0.4}\right)}}+\underset{II}{\underbrace{\frac{1}{2}\left(\frac{\frac{1}{z}}{\frac{1}{z}-0.4}\right)}}$

ROC for the I term will be |z| > 0.4

ROC for the II term will be:

$\left|z\right|<\frac{1}{0.4},i.e.\left|z\right|<2.5$

Now the ROC of X_e(z) will be the intersection of two, i.e.

0.4 < |z| < 2.5

Given:

 $X(z)=\frac{4z}{(z-\frac{1}{5})(z-\frac{2}{3})(z-3)}$

Poles of X(z) are located at z = $\frac{1}{5}$, z = $\frac{2}{3}$ and z = 3.

For DTFT to converge, the ROC of Z-transform of x() should contain unit circle.

If x(n) is a right sided sequence then the ROC is |z|>3 which does not include unit circle. So, option (D) and (A) are wrong.

If R.O.C. is $\frac{2}{3}$ < |z| < 3, the R.O.C. includes unit circle. So, option (B) is correct.

If x(n) is a left sided then R.O.C will be |z| < $\frac{1}{5}$ which does not include unit circle. So, option (C) is wrong.

Hence, the correct option is (B).

CONCEPT:

The digital frequency (Ω) is related to analog frequency (f) through the sampling frequency f_s.

PROVING:

Continuous  to discrete frequency converter:

Let continuous signal is x_c(t) and sampling signal is s(t).

Where Ω_s = sampling frequency

Or we can write

If x(nT) = x(n)

CONCEPT:

The digital frequency (Ω) is related to analog frequency (f) through the sampling frequency f_s.

$\mathrm{\Omega }=\frac{2\pi f}{f_s}\Rightarrow \omega T_s$

PROVING:

Continuous  to discrete frequency converter:

Let continuous signal is x_c(t) and sampling signal is s(t).

$S\left(t\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\delta \left(t-nT\right)\stackrel{\left\{F.T.\right\}}{\to }S\left(J\mathrm{\Omega }\right)\to \frac{2\pi }{T}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\delta J\left(\mathrm{\Omega }-n{\mathrm{\Omega }}_s\right)$

Where Ω_s = sampling frequency

$T=\frac{2\pi }{{\mathrm{\Omega }}_s}=samplinginteraval$

$x_s\left(t\right)=x_c\left(t\right).s\left(t\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{x}_c\left(t\right)\delta \left(t-nT\right)$

$X_s\left(J\mathrm{\Omega }\right)=\frac{1}{2\pi }{X}_c\left(J\mathrm{\Omega }\right)\ast S\left(J\mathrm{\Omega }\right)$

$X_s\left(J\mathrm{\Omega }\right)=\frac{1}{2\pi }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{X}_c\left[J\left(\mathrm{\Omega }-n{\mathrm{\Omega }}_s\right)\right]$

Or we can write

$X_s\left(J\mathrm{\Omega }\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{x}_c\left(nT\right)e^{-J\mathrm{\Omega }Tn}$

If x(nT) = x(n)

$X\left(e^{J\omega }\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left(n\right)e^{-J\mathrm{\Omega }n}$

$\begin{array}{l}X\left(z\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left(n\right)z^{-n}\\ =\sum _{n=0}^{\mathrm{\infty }}{\left(-\frac{1}{4}\right)}^n{z}^{-n}\\ Y\left(z\right)=X\left(z^3\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left(n\right){\left(z^3\right)}^{-n}\\ =\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left(n\right){\left(z\right)}^{-3n}\end{array}$

Taking 3n = k, we get

$\begin{array}{l}Y\left(z\right)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}x\left(\frac{k}{3}\right)z^{-k}\\ \therefore y\left(n\right)=x\left(\frac{n}{3}\right)=\{\begin{array}{cc}{\left(-\frac{1}{4}\right)}^{\frac{n}{3}}& n=0,3,6,\dots \\ 0& otherwise\end{array}\end{array}$

∴ y(2) = 0