Z Transform of Standard Signals MCQ Quiz - Objective Question with Answer for Z Transform of Standard Signals - Download Free PDF

Explanation:

Z-Transform of a Continuous Unit Step Function

Definition: The Z-transform is a powerful mathematical tool used in the analysis and design of discrete-time control systems. It transforms a discrete-time signal, which is a sequence of real or complex numbers, into a complex frequency domain representation. The Z-transform is particularly useful in solving linear, constant-coefficient difference equations.

Continuous Unit Step Function: The continuous unit step function, often denoted as u(t), is defined as:

\[ u(t) = \begin{cases} 0 & \text{for } t < 0 \\ 1 & \text{for } t \ge 0 \end{cases} \]

However, in the context of discrete-time signals, we consider the discrete unit step function, denoted as u[n], which is defined as:

\[ u[n] = \begin{cases} 0 & \text{for } n < 0 \\ 1 & \text{for } n \ge 0 \end{cases} \]

Z-Transform of the Unit Step Function:

The Z-transform of the discrete unit step function u[n] is obtained by applying the definition of the Z-transform:

\[ U(z) = \mathcal{Z}\{u[n]\} = \sum_{n=0}^{\infty} u[n] z^{-n} \]

Since u[n] = 1 for all n ≥ 0, the Z-transform becomes:

\[ U(z) = \sum_{n=0}^{\infty} z^{-n} \]

This is a geometric series with the first term a = 1 and common ratio r = z^-1. The sum of an infinite geometric series is given by:

\[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \]

Applying this formula, we get:

\[ U(z) = \frac{1}{1 - z^{-1}} \]

To express it in a more standard form, we multiply the numerator and the denominator by z:

\[ U(z) = \frac{z}{z - 1} \]

Thus, the Z-transform of the discrete unit step function u[n] is:

\[ U(z) = \frac{z}{z - 1} \]

Correct Option Analysis:

The correct option is:

Option 4: \(\rm X(t)=\frac{z}{z-1}\)

This option correctly represents the Z-transform of the discrete unit step function. The transformation and the properties of the geometric series lead to the result \(\rm \frac{z}{z-1}\), which matches the correct Z-transform of the unit step function.

Additional Information

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

Option 1: \(\rm X(t)=\frac{z}{1-z}\)

This option is incorrect because it does not correctly represent the Z-transform of the unit step function. The denominator should be z - 1, not 1 - z.

Option 2: \(\rm X(t)=\frac{1}{1-z}\)

This option is also incorrect. While it resembles the form of a geometric series, it lacks the factor of z in the numerator, which is necessary for the correct Z-transform expression.

Option 3: \(\rm X(t)=\frac{1}{z-1}\)

This option is incorrect as well. It does not match the correct form of the Z-transform of the unit step function. The numerator should include the factor of z.

Conclusion:

Understanding the Z-transform and its application to discrete-time signals is crucial for analyzing and designing discrete-time control systems. The Z-transform of the discrete unit step function u[n] is correctly given by \(\rm \frac{z}{z-1}\), which matches the correct option 4. Evaluating the other options helps reinforce the proper understanding and application of the Z-transform.

Solution

Some properties of impulse signal are shown below

• x(n) ⋆ δ(n) = x(n)
• x(n) ⋆ δ(n-a) = x(n-a)
• x(n-a) ⋆ δ(n-b) = x(n -a -b)
• x(n)δ(n-a) = x(a)δ(a)
• δ(an) = δ(n)

For a causal LTI system 

h(n) = 0 ; n < 0

Calculation

If x(n) is the input to causal LTI system h(n) ,then the output is y(n)

y(n) = x(n) ⋆ h(n)

Calculation

\(h(n)= \left\{ {\mathop 1\limits_ \uparrow ,2,1,3} \right\}\)

h(n) = δ(n) +2δ(n-1) + δ(n-2) + 3δ(n-3)

x(n) ⋆ h(n) = x(n) ⋆ [δ(n) +2δ(n-1) + δ(n-2) + 3δ(n-3)]

y(n) = x(n) + 2x(n-1) + x(n-2) + 3x(n-3)    

Hence the correct answer is option 2

Concept:

Unit step discrete-time sequence = u[n]

z – transformer of an u[n]:

 \({a^n}u\left[ n \right] = \frac{z}{{z - a}},\;ROC:\left| z \right| > a\)

a = 1

z – transformer of u[n] is given by:

\(u\left[ n \right] \longleftrightarrow \frac{z}{{z - 1}},\;ROC:\left| z \right| > 1\)

Concept:

Z- transform of x(n) is given as:

\(X(z)=\sum_{-\infty}^{\infty} x(n) z^{-n}\)

Region of convergence (ROC) is defined as the region where the Z-transform exists. 

Application:

Given:

x(n) = -αn u(-n-1)

\(X(z)=-\sum_{-\infty}^{-1} \alpha^n z^{-n}\)

\(X(z)=-\sum_{1}^{\infty} (\alpha^{-1} z)^n\)

For the Z-transform to exist,

\(|\alpha ^{-1}z| < 1\)

\({|z|} < |\alpha|\)

\(X(z)=-\frac{\alpha^{-1} z}{1-\alpha^{-1} z}\)

\(X(z)=-(\frac{\frac{z}{\alpha}}{\frac{\alpha-z}{\alpha} }) \)

\(X(z)=\frac{z}{z-\alpha}\)

x(n) = (0.5)ⁿ u(n)

y(n) = δ(n) - 2δ(n – 1)

Taking the z-transform of input and output, we get:

\(X\left( z \right) = \frac{z}{{z - 0.5}}\)

Y(z) = 1 – 2z^-1

\(Y\left( z \right) = \frac{{z - 2}}{z}\)

Transfer function of the system will be:

\(H\left( z \right) = \frac{{Y\left( z \right)}}{{X\left( z \right)}}\)

\(= \left( {\frac{{z - 2}}{z}} \right)\left( {\frac{{z - 0.5}}{z}} \right)\)

\(= \frac{{{z^2} - 2.5\;z + 1}}{{{z^2}}}\)

= 1 – 2.5 z^-1 + z^-2

Taking the inverse z-transform, we get:

h(n) = δ[n] – 2.5 δ[n – 1] + δ[n – 2]

∴ h[1] = - 2.5

Solution

Some properties of impulse signal are shown below

• x(n) ⋆ δ(n) = x(n)
• x(n) ⋆ δ(n-a) = x(n-a)
• x(n-a) ⋆ δ(n-b) = x(n -a -b)
• x(n)δ(n-a) = x(a)δ(a)
• δ(an) = δ(n)

For a causal LTI system 

h(n) = 0 ; n < 0

Calculation

If x(n) is the input to causal LTI system h(n) ,then the output is y(n)

y(n) = x(n) ⋆ h(n)

Calculation

\(h(n)= \left\{ {\mathop 1\limits_ \uparrow ,2,1,3} \right\}\)

h(n) = δ(n) +2δ(n-1) + δ(n-2) + 3δ(n-3)

x(n) ⋆ h(n) = x(n) ⋆ [δ(n) +2δ(n-1) + δ(n-2) + 3δ(n-3)]

y(n) = x(n) + 2x(n-1) + x(n-2) + 3x(n-3)    

Hence the correct answer is option 2

Concept:

Z- transform of x(n) is given as:

\(X(z)=\sum_{-\infty}^{\infty} x(n) z^{-n}\)

Region of convergence (ROC) is defined as the region where the Z-transform exists. 

Application:

Given:

x(n) = -αn u(-n-1)

\(X(z)=-\sum_{-\infty}^{-1} \alpha^n z^{-n}\)

\(X(z)=-\sum_{1}^{\infty} (\alpha^{-1} z)^n\)

For the Z-transform to exist,

\(|\alpha ^{-1}z| < 1\)

\({|z|} < |\alpha|\)

\(X(z)=-\frac{\alpha^{-1} z}{1-\alpha^{-1} z}\)

\(X(z)=-(\frac{\frac{z}{\alpha}}{\frac{\alpha-z}{\alpha} }) \)

\(X(z)=\frac{z}{z-\alpha}\)

The ROC of \({\left( {\frac{3}{4}} \right)^n}u\left( n \right)\) is |z| > 3/4

ROC of \({\left( {\frac{1}{3}} \right)^{n}}u\left( { - n - 1} \right)\) is |z| <1/3

We can see that common area of ROC is zero. So the z- transform doesn't exist

Solution

Some properties of impulse signal are shown below

• x(n) ⋆ δ(n) = x(n)
• x(n) ⋆ δ(n-a) = x(n-a)
• x(n-a) ⋆ δ(n-b) = x(n -a -b)
• x(n)δ(n-a) = x(a)δ(a)
• δ(an) = δ(n)

For a causal LTI system 

h(n) = 0 ; n < 0

Calculation

If x(n) is the input to causal LTI system h(n) ,then the output is y(n)

y(n) = x(n) ⋆ h(n)

Calculation

\(h(n)= \left\{ {\mathop 1\limits_ \uparrow ,2,1,3} \right\}\)

h(n) = δ(n) +2δ(n-1) + δ(n-2) + 3δ(n-3)

x(n) ⋆ h(n) = x(n) ⋆ [δ(n) +2δ(n-1) + δ(n-2) + 3δ(n-3)]

y(n) = x(n) + 2x(n-1) + x(n-2) + 3x(n-3)    

Hence the correct answer is option 2

 \(\rm X(z) =\mathop \sum \limits_{n = - \infty }^{ - 1} {\left( {\frac{1}{3}} \right)^{-n}}{z^{ - n}} + \mathop \sum \limits_0^\infty {\left( {\frac{1}{3}} \right)^n}{z^{ - n}}\)

We have,

\(\rm \mathop \sum \limits_{n = - \infty }^{ - 1} {\left( {\frac{1}{3}} \right)^{-n}}{z^{ - n}} =\mathop \sum \limits_{n = - \infty }^{ - 1} {\left( {3} \right)^{n}}{z^{ - n}} =ZT\{3^nu\left[-n-1\right]\}\)

From the relation, \(\rm \alpha ^nu\left[-u-1\right]\mathop \leftrightarrow^{ZT}-\frac{1}{1-\alpha z^{-1}}\ \ \ \ \ \ |z<|\alpha|\)

Thus, \(\rm 3^nu\left[-n-1\right]\mathop\leftrightarrow^{ZT}-\frac{1}{1-3z^{-1}}\ \ \ \ \ \ \ |z|<|3|\)

Also, \(\rm (\frac{1}{3})^nu\left[n\right]\mathop\leftrightarrow^{ZT}\frac{1}{1-\frac{1}{3}z^{-1}}\ \ \ \ \ \ \ |z|>|\frac{1}{3}|\)

Thus, \(\rm x(n) ={\left( {\frac{1}{3}} \right)^{\left| n \right|}}\mathop\leftrightarrow ^{ZT}\frac{1}{1-\frac{1}{3}z^{-1}}-\frac{1}{1-3z^{-1}}\ \ \ \ \ \ \ \ \ |\frac{1}{3}|<|z|<|3|\)

\(\rm \Rightarrow X(z) =\frac{1-3z^{-1}-1+\frac{1}{3}z^{-1}}{(1-\frac{1}{3}z^{-1})(1-3z^{-1})}\ \ \ \ \ \ \ \ \ |\frac{1}{3}|<|z|<|3|\)

\(\rm \Rightarrow X(z) =\frac{-\frac{8}{3}z^{-1}}{(1-\frac{10}{3}z^{-1}+z^{-2})}\ \ \ \ \ \ \ \ \ |\frac{1}{3}|<|z|<|3|\)

x(n) = (0.5)ⁿ u(n)

y(n) = δ(n) - 2δ(n – 1)

Taking the z-transform of input and output, we get:

\(X\left( z \right) = \frac{z}{{z - 0.5}}\)

Y(z) = 1 – 2z^-1

\(Y\left( z \right) = \frac{{z - 2}}{z}\)

Transfer function of the system will be:

\(H\left( z \right) = \frac{{Y\left( z \right)}}{{X\left( z \right)}}\)

\(= \left( {\frac{{z - 2}}{z}} \right)\left( {\frac{{z - 0.5}}{z}} \right)\)

\(= \frac{{{z^2} - 2.5\;z + 1}}{{{z^2}}}\)

= 1 – 2.5 z^-1 + z^-2

Taking the inverse z-transform, we get:

h(n) = δ[n] – 2.5 δ[n – 1] + δ[n – 2]

∴ h[1] = - 2.5

\(\eqalign{ & x\left[ n \right] = \cos \left( {\frac{{\pi n}}{4}} \right)u\left( n \right) = \left( {{e^{j\frac{\pi }{4}n}} + {e^{ - j\frac{\pi }{4}n}}} \right)u\left( 0 \right) \cr & = \frac{1}{2}{\left( {{e^{j\frac{\pi }{4}}}} \right)^n}u\left( n \right) + \frac{1}{2}{\left( {{e^{ - j\frac{\pi }{4}}}} \right)^n}u\left( n \right) \cr & x\left( z \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{z^{ - n}} = \mathop \sum \limits_{n = - \infty }^\infty \frac{1}{2}\left[ {{{\left( {{e^{j\frac{\pi }{4}}}} \right)}^n} + {{\left( {{e^{ - j\frac{\pi }{4}}}} \right)}^n}} \right]u\left( n \right){z^{ - n}} \cr & = \frac{1}{2}\mathop \sum \limits_{n = - 0}^\infty {\left( {{e^{j\frac{\pi }{4}}}} \right)^n}{z^{ - n}} + \frac{1}{2}\mathop \sum \limits_{n = - 0}^\infty {\left( {{e^{ - j\frac{\pi }{4}}}} \right)^n}{z^{ - n}} \cr & = \frac{1}{2}\left[ {\frac{1}{{1 - {e^{j\frac{\pi }{4}{z^{ - 1}}}}}} + \frac{1}{{1 - {e^{ - j\frac{\pi }{4}{z^{ - 1}}}}}}} \right] \cr & = \frac{1}{2}\left[ {\frac{{2 - \left( {{e^{j\frac{\pi }{4}}} + {e^{ - j\frac{\pi }{4}}}} \right){z^{ - 1}}}}{{1 + {z^{ - 2}} - {z^{ - 1}}\left( {{e^{j\frac{\pi }{4}}} + {e^{ - j\frac{\pi }{4}}}} \right)}}} \right] \cr & = \left[ {\frac{{1 - \left( {cos\frac{\pi }{4}} \right){z^{ - 1}}}}{{1 + {z^{ - 2}} - {z^{ - 1}}2\left( {{{cos\frac{\pi }{4}}}} \right)}}} \right] = \frac{{1 - \frac{1}{{\sqrt 2 }}{z^{ - 1}}}}{{1 + {z^{ - 2}} - {z^{ - 1}}{{\sqrt 2 }}}} \cr & x\left( z \right) = \frac{{1 - {{\left( {\sqrt 2 z} \right)}^{ - 1}}}}{{1 + {z^{ - 2}} - {{\sqrt 2z}^{ - 1}}}} \cr}\)

Concept:

Unit step discrete-time sequence = u[n]

z – transformer of an u[n]:

 \({a^n}u\left[ n \right] = \frac{z}{{z - a}},\;ROC:\left| z \right| > a\)

a = 1

z – transformer of u[n] is given by:

\(u\left[ n \right] \longleftrightarrow \frac{z}{{z - 1}},\;ROC:\left| z \right| > 1\)

X(z) = cos 2z

\(\cos \theta = 1 - \frac{{{\theta ^2}}}{{2!}} + \frac{{{\theta ^4}}}{{4!}} - \frac{{{\theta ^6}}}{{6!}} + \ldots \)

\(\cos \theta = \mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}\frac{{{\theta ^{2k}}}}{{\left( {2k} \right)!}}\)

\(\cos 2z = \mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}\frac{{{2^{2k}}{z^{2k}}}}{{\left( {2k} \right)!}}\)

\( = \mathop \sum \limits_{k = 0}^\infty {\left( { - 4} \right)^k}\frac{{{z^{2k}}}}{{\left( {2k!} \right)}}\)

Using the time shift property.

\(\delta \left( {n + k} \right)\mathop \leftrightarrow \limits^z {z^{ + k}}\)

\(\delta \left( {n + 2k} \right) \leftrightarrow {z^{ + 2k}}\)

\(x\left( n \right) = \mathop \sum \limits_{k = 0}^\infty \frac{{{{\left( { - 4} \right)}^k}}}{{\left( {2k} \right)!}}\delta \left( {n + 2k} \right)\)

Comparing with the equation given, we get a = -4

Impulse function is exist only at t= 0 and the area under impulse function is unity \(\mathop \smallint \limits_{ - \infty }^\infty \delta \left( t \right).dt = 1\)

\(L\left( {\delta \left( t \right)} \right) = \mathop \smallint \limits_{ - \infty }^\infty \delta \left( t \right).{e^{ - st}}.dt = {e^{ - st}}\)|_t=0 = 1

\(F\left( {\delta \left( t \right)} \right) = \mathop \smallint \limits_{ - \infty }^\infty \delta \left( t \right).{e^{ - j\omega t}}.dt = {e^{ - j\omega t}}\)|_t=0 = 1

Discrete time Laplace transform is Z- transform.