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 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 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

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) = 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\)

a = 1

z – transformer of u[n] is given by:

1\)

Concept:

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

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

Application:

Given:

x(n) = -αn u(-n-1)

For the Z-transform to exist,

x(n) = (0.5)ⁿ u(n)

y(n) = δ(n) - 2δ(n – 1)

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

Y(z) = 1 – 2z^-1

Transfer function of the system will be:

= 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

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) = 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:

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

Application:

Given:

x(n) = -αn u(-n-1)

For the Z-transform to exist,

The ROC of  is |z| > 3/4

ROC of  is |z|

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

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) = 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

We have,

From the relation, 

Thus, 

Also, |\frac{1}{3}|\)

Thus, 

x(n) = (0.5)ⁿ u(n)

y(n) = δ(n) - 2δ(n – 1)

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

Y(z) = 1 – 2z^-1

Transfer function of the system will be:

= 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

Concept:

Unit step discrete-time sequence = u[n]

z – transformer of an u[n]:

  a\)

a = 1

z – transformer of u[n] is given by:

1\)

X(z) = cos 2z

Using the time shift property.

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

|_t=0 = 1

|_t=0 = 1

Discrete time Laplace transform is Z- transform.

= |_n=0 = 1