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

Z Transform:

The z transform for a discrete signal x[n] is given by:
           X[z] = 

The set of all values of z where X(z) converges to a finite value is called as Radius of Convergence (ROC).
The ROC does not contain any poles.
If x[n] is a finite duration causal sequence or right-sided sequence, then the ROC is entire z-plane except at z = 0.
If x[n] is a finite duration anti-causal sequence or left-sided sequence, then the ROC is the entire z-plane except at z =∞ .

Concept:

The z-transform of a scaled sequence follows the scaling property of z-transforms, which states that if \( Z(u_n) = f(z) \), then \( Z(a^{-n}u_n) = f(az) \). This property is crucial for analyzing discrete-time signals with exponential scaling.

Calculation:

Given:

1. Original z-transform: \(Z(u_n) = f(z) \)
2. Scaled sequence: \(a^{-n}u_n\)

Solution:

1. Apply the scaling property of z-transforms:
\( Z(a^{-n}u_n) = f\left(\frac{z}{a}\right) \)

2. This result comes from the fundamental scaling property where multiplication by \( a^{-n} \) in the time domain corresponds to scaling the z-domain variable by \( \frac{1}{a} \).

Final Answer:

The correct transformed expression is 3) \( f\left(\frac{z}{a}\right) \).

Explanation:

Final Value Theorem in Laplace Transform

Definition: The Final Value Theorem (FVT) is a fundamental concept in control theory and signal processing, which provides a method to determine the steady-state value of a time-varying function as time approaches infinity. It is particularly useful in analyzing the long-term behavior of systems described by differential equations.

Formula: The Final Value Theorem states that for a given function \(f(t)\), if the Laplace transform \(F(s)\) of \(f(t)\) exists, then the final value of \(f(t)\) as \(t\) approaches infinity can be found using the formula:

\(\lim_{t \rightarrow \infty} f(t) = \lim_{s \rightarrow 0} sF(s)\)

This theorem applies under the condition that all poles of \(sF(s)\) (except possibly at \(s = 0\)) are in the left half of the complex plane. This ensures that the function \(f(t)\) approaches a steady-state value as \(t\) approaches infinity.

Working Principle: The Final Value Theorem provides a direct way to find the steady-state value of a function by analyzing its Laplace transform. The theorem essentially relates the behavior of the function in the time domain to its behavior in the frequency domain, allowing for easier computation of the final value.

Example: Consider a function \(f(t)\) with the Laplace transform \(F(s) = \frac{5}{s(s+2)}\). To find the final value of \(f(t)\), we use the Final Value Theorem:

\(\lim_{t \rightarrow \infty} f(t) = \lim_{s \rightarrow 0} s \left(\frac{5}{s(s+2)}\right) = \lim_{s \rightarrow 0} \frac{5}{s+2} = \frac{5}{2} = 2.5\)

Therefore, the final value of \(f(t)\) as \(t\) approaches infinity is 2.5.

Advantages:

• Simplicity in finding the steady-state value without needing to perform inverse Laplace transforms.
• Useful for analyzing the long-term behavior of systems described by differential equations.
• Provides a direct relationship between the time domain and frequency domain representations of a function.

Disadvantages:

• Applicable only if the function \(f(t)\) approaches a steady-state value as \(t\) approaches infinity.
• Requires the existence of the Laplace transform \(F(s)\) of the function \(f(t)\).
• Not applicable if the function has poles on the right half of the complex plane or on the imaginary axis (except at \(s = 0\)).

Correct Option Analysis:

The correct option is:

Option 3: \(\lim_{t \rightarrow \infty} f(t) = \lim_{s \rightarrow 0}sF(s)\)

This option correctly represents the Final Value Theorem. The formula \(\lim_{t \rightarrow \infty} f(t) = \lim_{s \rightarrow 0}sF(s)\) provides a method to find the final value of a time-varying function \(f(t)\) by analyzing its Laplace transform \(F(s)\).

Additional Information

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

Option 1: \(\lim_{t \rightarrow \infty} f(t) = \lim_{s \rightarrow \infty} sF(s)\)

This option is incorrect because it suggests taking the limit as \(s\) approaches infinity, which does not provide the final value of the function \(f(t)\) as \(t\) approaches infinity. The correct limit should be as \(s\) approaches zero.

Option 2: \(\lim_{t \rightarrow \infty} f(t) = \lim_{t \rightarrow 0} tf(t)\)

This option is incorrect because it involves the time domain limit as \(t\) approaches zero, which does not relate to finding the final value as \(t\) approaches infinity. The Final Value Theorem specifically deals with the limit as \(t\) approaches infinity.

Option 4: \(\lim_{t \rightarrow \infty} f(t) = \lim_{t \rightarrow 0} f(t)\)

This option is incorrect because it again involves the time domain limit as \(t\) approaches zero, which does not help in finding the final value as \(t\) approaches infinity. The correct approach involves the Laplace transform and the limit as \(s\) approaches zero.

Conclusion:

Understanding the Final Value Theorem is essential for analyzing the steady-state behavior of time-varying functions. The correct application of the theorem allows for the determination of the final value of a function by evaluating its Laplace transform. This method simplifies the process and provides a direct relationship between the time domain and frequency domain representations of the function.

Explanation:

Load Flow Analysis

Definition: Load flow analysis, also known as power flow analysis, is a fundamental tool used in power system engineering to determine the voltage, current, real power (P), and reactive power (Q) at various points in an electrical power system under steady-state conditions. It helps in planning and operating power systems efficiently and safely.

Load Flow Equations: The load flow study involves solving a set of nonlinear algebraic equations known as power flow equations. These equations are derived from Kirchhoff's laws and the power balance equations at each bus in the power system. The primary goal is to find the bus voltages and the corresponding power flows in the network.

Y-Bus Matrix: The Y-bus (admittance) matrix is a key component in load flow analysis. It is a sparse matrix that represents the admittance (inverse of impedance) between the buses (nodes) in the power system. The elements of the Y-bus matrix are derived from the network's impedance data.

Correct Option Analysis:

The correct option is:

Option 2: Both Statement 1 and Statement 2 are false.

Statement 1: "The diagonal element of each node is the negated admittance between the nodes."

This statement is false. In the Y-bus matrix, the diagonal elements represent the sum of the admittances of all the branches connected to that particular bus (node). These elements are not the negated admittance between the nodes. The correct representation of the diagonal element is the sum of the admittances of the branches connected to the node.

Statement 2: "The off-diagonal element is the sum of the admittances connected to it."

This statement is also false. In the Y-bus matrix, the off-diagonal elements represent the negative of the admittance of the branch directly connecting the two nodes. It is not the sum of the admittances connected to the node but rather the negative of the admittance of the branch between the two nodes.

Y-Bus Matrix Formation:

• Diagonal Elements (Y_ii): These elements are calculated as the sum of the admittances of all the branches connected to bus i. Mathematically, it can be expressed as:

  Y_ii = ∑ y_ik for all k ≠ i

  where y_ik is the admittance of the branch between bus i and bus k.
• Off-Diagonal Elements (Y_ij): These elements represent the negative of the admittance of the branch connecting bus i and bus j. Mathematically, it can be expressed as:

  Y_ij = -y_ij

  where y_ij is the admittance of the branch between bus i and bus j.

Additional Information

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

Option 1: "Statement 1 is true and Statement 2 is false."

This option is incorrect because Statement 1 is false. The diagonal elements of the Y-bus matrix are not the negated admittance between the nodes but the sum of the admittances of the branches connected to the node.

Option 3: "Both Statement 1 and Statement 2 are true."

This option is incorrect because both statements are false. The correct representation of the diagonal and off-diagonal elements of the Y-bus matrix has been explained above.

Option 4: "Statement 2 is true and Statement 1 is false."

This option is incorrect because Statement 2 is false. The off-diagonal elements of the Y-bus matrix represent the negative of the admittance of the branch connecting the two nodes, not the sum of the admittances connected to the node.

Conclusion:

Understanding the formation of the Y-bus matrix is crucial for accurate load flow analysis in power systems. The diagonal elements represent the sum of the admittances of the branches connected to the bus, while the off-diagonal elements represent the negative of the admittance of the branch connecting the two buses. Both statements provided in the question are false, making Option 2 the correct choice.

Explanation:

Z-Transform

Definition: The Z-transform is a mathematical tool used in the field of signal processing and control systems to analyze discrete-time signals and systems. It transforms a discrete-time signal, which is a sequence of real or complex numbers, into a complex frequency domain representation.

Working Principle: The Z-transform of a discrete-time signal \( x[n] \) is defined as:

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

where \( z \) is a complex variable. This transformation allows the analysis of the signal in the z-domain, providing insights into its frequency components and system behavior.

Inverse Z-Transform: The inverse Z-transform is used to convert a signal back from the z-domain to the time domain. One common method to find the inverse Z-transform is the partial fraction expansion method. This involves expressing the Z-transform as a sum of simpler fractions, each of which can be inverted individually.

Advantages:

• Facilitates the analysis of discrete-time systems and signals in the frequency domain.
• Helps in the design and analysis of digital filters and control systems.
• Provides a convenient way to solve linear difference equations.

Disadvantages:

• Complexity in mathematical manipulation, especially for large systems.
• Requires a good understanding of complex variable theory.

Applications: The Z-transform is widely used in digital signal processing, control system design, and telecommunications to analyze and design discrete-time systems and filters.

Correct Option Analysis:

The correct option is:

Option 1: Both Statement 1 and Statement 2 are true.

This option correctly identifies the validity of both statements. Let's analyze the given statements in detail:

Statement 1: Inverse Z-transforms can be obtained by partial fraction expansion methods.

This statement is true. One of the common methods to find the inverse Z-transform is by using partial fraction expansion. By expressing the Z-transform as a sum of simpler fractions, the inverse can be determined for each fraction, which are then summed to obtain the original discrete-time signal.

Statement 2: Z-transform can be used for the solution of linear difference equations.

This statement is also true. The Z-transform is a powerful tool for solving linear difference equations. By transforming the difference equation into the z-domain, it becomes an algebraic equation that is easier to solve. The solution can then be transformed back to the time domain using the inverse Z-transform.

Additional Information

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

Option 2: Statement 1 is true and Statement 2 is false.

This option is incorrect because both statements are true. The inverse Z-transform can indeed be obtained by partial fraction expansion methods, and the Z-transform is also used for solving linear difference equations.

Option 3: Both Statement 1 and Statement 2 are false.

This option is incorrect because both statements are true. The inverse Z-transform can be obtained by partial fraction expansion, and the Z-transform is used for solving linear difference equations.

Option 4: Statement 2 is true and Statement 1 is false.

This option is incorrect because both statements are true. The Z-transform is used for solving linear difference equations, and the inverse Z-transform can be obtained by partial fraction expansion methods.

Conclusion:

Understanding the Z-transform and its applications is crucial in the fields of signal processing and control systems. The Z-transform provides a powerful method for analyzing discrete-time signals and systems, and its inverse can be determined using partial fraction expansion. It is also a valuable tool for solving linear difference equations, making it an essential concept for engineers and researchers in related fields.

Definition:

Z transform is defined as

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

Properties:

Differentiation in z domain:

If X(z) is a z transform of x(n), then the z transform of n x(n) is,

\(nx\left( n \right) \leftrightarrow - z\frac{d}{{dz}}\left( {X\left( z \right)} \right)\)

Time-shifting:

If X(z) is a z transform of x(n), then the z transform of x(n – n₀) is,

\(x\left( {n - {n_0}} \right) \leftrightarrow {z^{ - {n_0}}}X\left( z \right)\)

Application:

Let x(n) = 1

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

Now, by applying the property of differentiation in the z domain,

\(x\left( n \right) = n \leftrightarrow - z\frac{d}{{dz}}\left( {\frac{z}{{z - 1}}} \right) = \frac{z}{{{{\left( {z - 1} \right)}^2}}}\)

Now, by applying the property of differentiation in the z domain,

\(x\left( n \right) = {n^2} \leftrightarrow - z\frac{d}{{dz}}\left( {\frac{z}{{{{\left( {z - 1} \right)}^2}}}} \right) = \frac{{z\left( {z + 1} \right)}}{{{{\left( {z - 1} \right)}^3}}}\)

Now, by applying the property of time-shifting,

\(x\left( n \right) = {\left( {n + 1} \right)^2} \leftrightarrow z.\frac{{z\left( {z + 1} \right)}}{{{{\left( {z - 1} \right)}^3}}} = \frac{{{z^2}\left( {z + 1} \right)}}{{{{\left( {z - 1} \right)}^3}}}\)

Concept:

Final value theorem:

It states that:

\(x\left( \infty \right) = \mathop {\lim }\limits_{z \to 1} \left( {1 - {z^1}} \right)X\left( z \right)\)

Conditions:

1. It is valid only for causal systems. 

2. Pole of (1 – z-1) X(z) must lie inside the unit circle.

Calculation:

The final value theorem for z-transform is:

\( = \mathop {\lim }\limits_{z \to 1} \frac{1}{4}\frac{{\left( {1 - {z^{ - 1}}} \right){z^{ - 1}}\left( {1 - {z^{ - 4}}} \right)}}{{{{\left( {1 - {z^{ - 1}}} \right)}^2}}}\)

\(= \mathop {\lim }\limits_{z \to 1} \frac{1}{4}\;\frac{{\left( {{z^2} - 1} \right)\left( {{z^2} + 1} \right)}}{{{z^4}\left( {z - 1} \right)}}\)

\( = \mathop {\lim }\limits_{z \to 1} \frac{1}{4}\;\frac{{\left( {z + 1} \right)\left( {{z^2} + 1} \right)}}{{{z^4}}} \)

= 1/4 × 1 × 2 × 2 = 1

Concept:

The z-transform of a unit impulse function or Kronecker delta δ [n] ↔ 1

The time-shifting affects the z-transform as:

x[n - n0] = z -n0 X(z)

Application:

Given:

h1[n] = δ[n - 1] + δ[n + 1] 

h2[n] = δ[n] + δ[n - 1] 

If h₁[n]and h₂[n] are cascaded connected then h[n] = h₁[n] * h₂[n]

Where '*' denotes convolution.

h[n] = h1[n] * h2[n]

Taking z-transform both side

H[z] = H₁[z] ⋅ H₂[z]

H[z] = (z^-1 + z) ⋅ (1 + z^-1) = (z^-1 + z-2 + z + 1 )

Taking inverse z-transform both side

h[n] = δ[n-1] + δ[n-2] + δ[n+1] + δ[n]

∴ Impulse response of the cascaded system is δ[n - 2] + δ[n - 1] + δ[n] + δ[n + 1]

Concept:

Z transform:

The Z transform of x(t) is, denoted by X(z), is defined as:

\(X\left( z \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left( n \right){z^{ - n}}\)

\({a^n}u\left( n \right)\mathop \to \limits^{ZT} \frac{1}{{1 - a{z^{ - 1}}}}\); ROC:|z|>|a|

Shifting property:

If \(x\left( n \right)\mathop \to \limits^{ZT} x\left( z \right)\) ; ROC: R

Then \(x\left( {n - {n_0}} \right)\mathop \to \limits^{ZT} {z^{ - {n_0}}}x\left( z \right)\); ROC: R

The time-shifting will not affect ROC.

Calculation:

Given that, \(x\left( n \right) = {\left( {\frac{1}{2}} \right)^n}1\left[ n \right]\)

Z Transform (ZT) of x(n) will be,

\(x\left( z \right) = \frac{z}{{z - \frac{1}{2}}} = \frac{1}{{1 - \frac{1}{2}{z^{ - 1}}}}\) ; ROC: \(\left| z \right| > \frac{1}{2}\)

Concept:

Causal Signals:

• Causal Signals are signals that are zero for all negative time.
• Causality in a system determines whether a system relies on future information of a signal x [n+1].
• Talking about “causality” in signals, it means whether they are zero to the left of t = 0 or zero to the right of t = 0.
• A causal signal is zero for t < 0.
• A system is said to be causal if its output depends upon present and past inputs, and does not depend upon future input. 

Calculation:

Given a casual signal,

X (Z) = z2 (z - a)-2

\(X(Z) = \frac{{{z^2}}}{{{{\left( {a - z} \right)}^2}}}\)

From standard Z-Transform,

\(n{\left( a \right)^n}u\left( n \right) \leftrightarrow \frac{{az}}{{{{\left( {z - a} \right)}^2}}}\)

\(n{\left( a \right)^{n - 1}}u\left( n \right) \leftrightarrow \frac{z}{{{{\left( {z - a} \right)}^2}}}\)

From the time-shifting property,

\(x\left( {n - {n_0}} \right) \leftrightarrow {z^{ - {n_0}}}X\left( z \right)\)

\(\left( {n + 1} \right){\left( a \right)^{n + 1 - 1}}u\left( {n + 1} \right) \leftrightarrow z\left[ {\frac{z}{{{{\left( {z - a} \right)}^2}}}} \right]\)

\(\left( {n + 1} \right){\left( a \right)^n}u\left( {n + 1} \right) \leftrightarrow \frac{{{z^2}}}{{{{\left( {z - a} \right)}^2}}}\)

Since,

\(X\left( z \right) = \frac{{{z^2}}}{{{{\left( {z - a} \right)}^2}}}\)

Therefore,

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

The given signal is causal, therefore,

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

Important Points

Z-Transform basic functions:

Concept:

Z transform of \(\dfrac {Z}{Z-a} \space \rightarrow \space a^n u(n ), \space \space \space |Z| \space > \space a \space\)

Analysis:

\(X(Z) \space = \space \dfrac {(Z-1) \space (Z+0.8)} {(Z-0.5)(Z+0.2)} \space \)

Divide by Z on both side

\(\dfrac{X(Z)}{Z} \space = \space \dfrac {(Z-1) \space (Z+0.8)} {Z(Z-0.5)(Z+0.2)} \space \)

By partial fraction 

\(\dfrac{X(Z)}{Z} \space = \space \dfrac {8}{Z} \space \space -\space \dfrac{1.857}{Z-0.5} \space - \dfrac {5.142}{Z+0.2} \space \)

\({X(Z)}\space = \space {8} \space \space -\space \dfrac{\space Z\space (1.857) }{Z-0.5} \space - \dfrac {Z\space (5.142)}{Z+0.2} \space \)

Taking inverse Z transform

x(n) = 8 δ(n) - 1.857 (0.5)ⁿ u(n) - 5.142 (-0.2)ⁿ u(n)

Concept:

For a causal signal x(n), the initial value theorem states that:

\(x\left( 0 \right) = \mathop {\lim }\limits_{z \to \infty } X\left( z \right)\)

For a causal signal x(n), the final value theorem states that:

\(x\left( \infty \right) = \mathop {\lim }\limits_{z \to 1} \left[ {z - 1} \right]X\left( z \right)\)

Calculation:

Given:

\(H\left( z \right) = \frac{{{z^2} + 1}}{{\left( {z + 0.5} \right)\left( {z - 0.5} \right)}}\)

∴ Poles = 0.5 and -0.5

Zeros = ±j

Hence all the poles are lying inside the unit circle. Therefore, the system is stable.

Now by using the initial value theorem, we get

\(h\left( 0 \right) = \mathop {\lim }\limits_{z \to \infty } H\left( z \right) = \mathop {\lim }\limits_{z \to \infty } \frac{{{z^2} + 1}}{{\left( {z + 0.5} \right)\left( {z - 0.5} \right)}} = 1\)

Final Value theorem:

\(h\left( \infty \right) = \mathop {\lim }\limits_{z \to 1 } (z-1)H\left( z \right) \)

\(h\left( \infty \right) = \mathop {\lim }\limits_{z \to 1 } \frac{(z-1){{z^2} + 1}}{{\left( {z + 0.5} \right)\left( {z - 0.5} \right)}}=0\)

Hence, statement C is also correct.

Concept:

The initial value theorem is given by:

In time domain:

\(Lt\overset{t=0}{\rightarrow}x(t) \)

In the Laplace domain:

\(Lt\overset{s=∞}{\rightarrow}sx(s) \)

The final value theorem is given by:

In time domain:

\(Lt\overset{t=∞}{\rightarrow}x(t) \)

In the Laplace domain:

Calculation:

Concept:

The Z-transform is defined by

\(X\left( z \right) = \mathop \sum \limits_{k = 0}^\infty x\left[ k \right]{z^{ - k}}\)

Time-shifting:

If X(z) is a z transform of x(n), then the z transform of x(n – n0) is,

\(x\left( {n - {n_0}} \right) \leftrightarrow {z^{ - {n_0}}}X\left( z \right)\)

\(x\left( {n + {n_0}} \right) \leftrightarrow {z^{ {n_0}}}X\left( z \right)\)

Z.T{δ(n)} = 1

Analysis:

Given:

X(z) = 5z2 +4z-1 + 3; 0 < |z| < ∞

Taking the inverse z transform, we get

x(n) = 5δ(n + 2) + 4δ(n - 1) + 3δ 

Concept:

For all-pass systems poles must lie on the left and zeroes on the mirror image of the pole that is on the right.

Calculation:

For all pass system,

If pole is at  z = p, then zero will be at z = 1/p*

Pole of H(z) is at z = a

∴zero of H(z) should be at \(z = \frac{1}{{{a^*}}}\)           

zero of H(z) is z = 1/b

\( \Rightarrow \frac{1}{b} = \frac{1}{{{a^*}}} \Rightarrow b = {a^*}\)