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

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)

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).

 Properties of convolution of two signals:

→ Convolution of an odd and even function is an odd function.

→ Convolution of two odd functions is an even function.

→ Convolution of two even function is an even function

→ Convolution of two causal functions is causal function.

→ Convolution of two Anti causal function is Anticausal function.

→ Convolution of two unequal length rectangular pulse is trapezium.

→ Convolution of two equal length rectangular pulses is triangle.

→ Convolution of signal with periodic train of impulses is periodic representation of signal.

Therefore option (4) statement is not correct.

Concept:

• The overall impulse response of systems connected in cascade in the time domain is given by the convolution of the two impulse responses, i.e. convolution h1 (t) and h2 (t).
• In the frequency domain, the overall impulse response is the multiplication of the impulse responses in the frequency domain.

Given two impulse responses are h1(t) and h2(t).

These are cascaded, so the resulting system response will be a convolution of these two systems.

Properties:

Commutative property:

 x(t) ∗ h(t) = h(t) ∗ x(t)

Associative property: 

x(t) ∗ [h1(t) ∗ h2(t)] = [x(t) ∗ h1(t)] ∗ h2(t)

Distributive property:

x(t) ∗ [h1(t) + h2(t)] = x(t) ∗ h1(t) + x(t) ∗  h2(t)

Convolution with impulse property:

 x(t) ∗ δ (t) = x(t)

Width property: if the durations(widths) of x(t) and h(t) are finite and given by Wx and Wh,

Then the duration(width) of the x(t) ∗ h(t) is Wx + Wh

Important points:

Convolution of two equal width rectangles will give resultant as a triangular profile.

Convolution of two unequal width rectangles will give resultant as a trapezoidal profile.

Concept:

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

$x\left(0\right)=\underset{z\to \mathrm{\infty }}{lim}X\left(z\right)$

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

$x\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left[z-1\right]X\left(z\right)$

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

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

Final Value theorem:

$h\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}(z-1)H\left(z\right)$

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

Hence, statement C is also correct.

Concept:

The z-transform of a unit impulse function x[n] = δ [n] is given as:

X(z) = 1

Also, the time-shifting affects the z-transform as:

x[n - n₀] = z ^-n₀ X(z)

Application:

Given:

x[n] = δ [n - 3] + 2δ [n - 5] 

Taking the z transform, we get:

X(z) = z^-3 + 2 z^-5

Replacing z by -z we get:

$\mathrm{Y}\left(\mathrm{z}\right)=\mathrm{X}\left(-\mathrm{z}\right)$

$={\left(-\mathrm{z}\right)}^{-3}+2{\left(-\mathrm{z}\right)}^{-5}$

$=-{\mathrm{z}}^{-3}-2{\mathrm{z}}^{-5}$

$Y(z)=-\left({\mathrm{z}}^{-3}+2{\mathrm{z}}^{-5}\right)$

We observe that:

$\mathrm{Y}\left(\mathrm{z}\right)=-\mathrm{X}\left(\mathrm{z}\right)$

Taking inverse z-transfrom of the above, we get:

Definition:

Z transform is defined as

$X\left(z\right)=\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}x\left[n\right]z^{-n}$

Time-shifting:

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

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

Analysis:

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

Discrete-Time System:

The convolution of two a signal with a system with impulse response h(n) is represented as:

y[n] = x[n] ∗ h[n]

Continuous Time System:

The convolution of two a signal with a system with impulse response h(t) is represented as:

y(t) = x(t) ∗ h(t)

The z-transform for 1/n does not exist for n=0, but it exists for n > 0 and for n < 0

Let n > 0

X(z) = $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}{z}^{-n}$

Differentiating both sides w.r.t z we get,

$\frac{dX(z)}{dz}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}(-n)z^{-n-1}$

$\frac{dX(z)}{dz}=-\sum _{n=1}^{\mathrm{\infty }}{z}^{-n-1}$

 $\frac{dX(z)}{dz}=-\frac{z^{-2}}{1-z^{-1}}$  |z-1| < 1 or |z| > 1

Integrating both sides, we get

$X(z)=-\int \frac{z^{-2}}{1-z^{-1}}dz$

$-X(z)=\int \frac{1}{z^2-z}dz$

$-X(z)=\int (\frac{-1}{z}+\frac{1}{z-1})dz$

$-X(z)=-log(z)+log(z-1)$

$\mathbf{\text{X(z)}}=\mathbf{\text{log}}(\frac{\mathbf{\text{z}}}{\mathbf{\text{z-1}}})$ for |z| > 1

Let n < 0

X(z) = $\sum _{n=-\mathrm{\infty }}^{-1}\frac{1}{n}{z}^{-n}$

Differentiating both sides w.r.t z we get,

$\frac{dX(z)}{dz}=\sum _{n=-\mathrm{\infty }}^{-1}\frac{1}{n}(-n)z^{-n-1}$

$\frac{dX(z)}{dz}=-\sum _{n=-\mathrm{\infty }}^{-1}{z}^{-n-1}$

Let n = -p

$\frac{dX(z)}{dz}=-\sum _{p=\mathrm{\infty }}^1{z}^{p-1}$

 $\frac{dX(z)}{dz}=-\frac{1}{1-z}$    |z| < 1

Integrating both sides, we get

$\mathbf{\text{X(z)}}=\mathbf{\text{log(z-1)}}$ for |z| < 1

∴ By looking at the options the correct answer is 3 (given for n > 0)

Concept:

Time reversal property of Z-transform

X[n] ↔ X(z)

X[-n] ↔ X(z^-1)

Calculation:

Given that x[n] = x[-n]

⇒ X(z) = X(z^-1)

Zero of X(z) = 1 + j2

Then another zero will be:

$\frac{1}{1+j2}=\frac{1-j2}{5}$

$=0.2-0.4j$

Concept:

• The overall impulse response of systems connected in cascade in the time domain is given by the convolution of the two impulse responses, i.e. convolution h1 (t) and h2 (t).
• In the frequency domain, the overall impulse response is the multiplication of the impulse responses in the frequency domain.

Given two impulse responses are h1(t) and h2(t).

These are cascaded, so the resulting system response will be a convolution of these two systems.

Properties:

Commutative property:

 x(t) ∗ h(t) = h(t) ∗ x(t)

Associative property: 

x(t) ∗ [h1(t) ∗ h2(t)] = [x(t) ∗ h1(t)] ∗ h2(t)

Distributive property:

x(t) ∗ [h1(t) + h2(t)] = x(t) ∗ h1(t) + x(t) ∗  h2(t)

Convolution with impulse property:

 x(t) ∗ δ (t) = x(t)

Width property: if the durations(widths) of x(t) and h(t) are finite and given by Wx and Wh,

Then the duration(width) of the x(t) ∗ h(t) is Wx + Wh

Important points:

Convolution of two equal width rectangles will give resultant as a triangular profile.

Convolution of two unequal width rectangles will give resultant as a trapezoidal profile.

 Properties of convolution of two signals:

→ Convolution of an odd and even function is an odd function.

→ Convolution of two odd functions is an even function.

→ Convolution of two even function is an even function

→ Convolution of two causal functions is causal function.

→ Convolution of two Anti causal function is Anticausal function.

→ Convolution of two unequal length rectangular pulse is trapezium.

→ Convolution of two equal length rectangular pulses is triangle.

→ Convolution of signal with periodic train of impulses is periodic representation of signal.

Therefore option (4) statement is not correct.

${\mathrm{a}}^{\mathrm{n}}\mathrm{u}(\mathrm{n})=\frac{\mathrm{z}}{\mathrm{z}-\mathrm{a}}$ ;  ROC = |z| > a

$-a^{n}u(-n-a)=\frac{z}{z-a}$ ; ROC = |z| < a

From property of linearity:

x₁(n) → X₁(z);  ROC: R₁​​

_​x₂(n) → X₂(Z);  ROC: R₂

_​If g(n) = x₁(n) + x₂(n) then ROC of g(n) is R₁ intersection R₂​​​_​

If R₁ intersection R₂ is nullity then that expression's Z transform does not exist. 

Analysis:

g(n) = 2ⁿu(n) - 3ⁿ(n) u(-n-1)

$G(z)=\frac{z}{z-2}+\frac{z}{z+3}$

ROC: |z| > 2 intersection |z| < 3

∴ The resultant ROC will be 2 < |z| < 3

Concept:

Let,

y_o(n) = x(n) * y(n)

Then to find Y_o(z)

$Y_o(z)=\sum _{n=-\mathrm{\infty }}^{n=+\mathrm{\infty }}{y}_o(n)z^{-n}$

In terms of Z transform we can write:

x(n) * y(n) = X(z) × Y(z)

The convolution of two discrete time signals gives the ROC as the intersection of respective ROCs. 

Additional Information

Properties of the region of convergence:

The properties of the ROC depend on the nature of the signal. Assuming that the signal has a finite amplitude and that the z-transform is a rational function.

• The ROC is a ring or disk in the z-plane, centred on the origin (0 < rR < |z| < rL ≤ ∞)
• The Fourier transform of x[n] converges absolutely if and only if the ROC of the z-transform includes the unit circle.
• The ROC cannot contain any poles.
• If x[n] is finite duration, then the ROC is the entire z-plane except perhaps at z = 0 or z = ∞
• If x[n] is a right-sided sequence then the ROC extends outward from the outermost finite pole to infinity.
• If x[n] is left-sided then the ROC extends inward from the innermost nonzero pole to z = 0
• A two-sided sequence (neither left nor right-sided) has a ROC consisting of a ring in the z-plane, bounded on the interior and exterior by a pole (and not containing any poles).
• The ROC is a connected region.