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

Concept:

For a linear time-invariant (LTI) discrete-time system, stability is determined by the location of the poles (roots of the characteristic equation). The system is:

• Stable if all poles lie inside the unit circle (|z| < 1).
• Unstable if any pole lies outside the unit circle (|z| > 1).
• Marginally stable if poles lie on the unit circle and are non-repeating.

Given:

Difference equation:

y(n) = 1.8y(n - 1) - 0.72y (n - 2) + x (n) + 0.5x (n - 1)

Characteristic equation (from homogeneous part):

\(H(z): z^2 - 1.8z + 0.72 = 0 \)

Calculation:

Use the quadratic formula to find roots:

 \(\frac{1.8 \pm \sqrt{(1.8)^2 - 4 \cdot 0.72}}{2}\)

Since one root is greater than 1, the system is unstable.

Explanation:

Minimum Phase LTI System

Definition: A Linear Time-Invariant (LTI) system is considered to be a minimum phase system if all its poles and zeros are located within the unit circle in the Z-plane. The unit circle is defined as the set of points in the complex plane where the magnitude of the complex number is equal to 1. For a minimum phase system, both stability and minimum phase conditions are satisfied.

Key Characteristics of a Minimum Phase System:

• All poles of the system must lie strictly inside the unit circle. This ensures the system is stable.
• All zeros of the system must also lie strictly inside the unit circle. This ensures the system has the minimum phase property.
• The system's phase response is minimized, which is crucial in many control and signal processing applications where phase linearity or minimal phase shift is required.

Explanation of the Correct Option:

The correct option is:

Option 4: All poles and zeros are inside the unit circle.

This is the correct condition for a minimum phase LTI system. For a system to be classified as minimum phase, it is not sufficient for just the poles or just the zeros to be inside the unit circle—both must satisfy this condition. This ensures that the system is stable and has the minimum phase property.

Explanation:

Analysis of the Given Problem:

The problem involves cascading two systems with impulse responses h₁[n] = A(b₁)ⁿu[n] and h₂[n] = A(b₂)ⁿu[n]. The overall transfer function of the cascaded system is given as:

H(z) = 4 / (4 - z^-2).

We need to determine the possible values of A, b₁, and b₂ that satisfy the given transfer function. Let us solve step-by-step:

Step 1: Analyze the impulse response and transfer function of individual systems

The impulse response of the first system is:

h₁[n] = A(b₁)ⁿu[n],

where u[n] is the unit step function. The Z-transform of h₁[n] is:

H₁(z) = A / (1 - b₁z^-1) for |z| > |b₁|.

Similarly, the impulse response of the second system is:

h₂[n] = A(b₂)ⁿu[n].

The Z-transform of h₂[n] is:

H₂(z) = A / (1 - b₂z^-1) for |z| > |b₂|.

Step 2: Combine the two systems

When the two systems are cascaded, the overall transfer function is the product of the transfer functions of the individual systems:

H(z) = H₁(z) × H₂(z).

Substituting the expressions for H₁(z) and H₂(z):

H(z) = [A / (1 - b₁z^-1)] × [A / (1 - b₂z^-1)].

H(z) = A² / [(1 - b₁z^-1)(1 - b₂z^-1)].

Step 3: Match the given H(z)

The given H(z) is:

H(z) = 4 / (4 - z^-2).

To match this with the derived expression, rewrite the denominator of the given H(z):

4 - z^-2 = (2 - z^-1)(2 + z^-1).

So, the given H(z) can be expressed as:

H(z) = 4 / [(2 - z^-1)(2 + z^-1)].

Comparing this with the derived H(z), we identify:

b₁ = 1/2, b₂ = -1/2, and A² = 1.

Thus, A = 1 (since A must be positive), b₁ = 1/2, and b₂ = -1/2.

Correct Option: Option 3 (A = 1, b₁ = 1/2, b₂ = -1/2)

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) \( f\left(\frac{z}{a}\right) \).

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

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