Inverse Z-transform in Signal Processing - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the inverse Z-transform function from sympy.

Signal Processing

from sympy import symbols, [1]

Drag options to blanks, or click blank then click option'

Aztransform

Biztrans

Cztrans

Dinverse_transform

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'ztrans' which is the forward Z-transform.

Using 'ztransform' which is not a valid sympy function.

Using 'inverse_transform' which does not exist.

The function iztrans is used in sympy to compute the inverse Z-transform.

2fill in blank

medium

Complete the code to define the Z-transform variable 'z' as a symbol.

Signal Processing

z = [1]('z')

Drag options to blanks, or click blank then click option'

Asymbols

Bvar

CSymbol

DFunction

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Symbol' which is a class, not a function.

Using 'Function' which defines symbolic functions, not variables.

Using 'var' which is not a sympy function.

The 'symbols' function from sympy is used to define symbolic variables.

3fill in blank

hard

Fix the error in the code to compute the inverse Z-transform of Xz.

Signal Processing

x = [1](Xz, z, n)

Drag options to blanks, or click blank then click option'

Aztrans

Binverse_z

Ciztrans

Dinvztrans

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'ztrans' which computes the forward Z-transform.

Using 'inverse_z' which is not a valid function.

Using 'invztrans' which does not exist in sympy.

The correct function to compute inverse Z-transform in sympy is 'iztrans'.

4fill in blank

hard

Fill both blanks to define the symbolic variable 'n' as an integer and compute inverse Z-transform.

Signal Processing

n = [1]('n', integer=True)
x = iztrans(Xz, z, [2])

Drag options to blanks, or click blank then click option'

Asymbols

BSymbol

Cn

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'symbols' which does not accept integer=True.

Passing 'z' instead of 'n' to iztrans.

Using undefined variables.

Use 'Symbol' to define 'n' as an integer variable, and pass 'n' as the time variable in iztrans.

5fill in blank

hard

Fill all three blanks to create a Z-transform expression, define 'n', and compute the inverse Z-transform.

Signal Processing

Xz = 1 / (1 - [1]**-1)
n = [2]('n', integer=True)
x = [3](Xz, z, n)

Drag options to blanks, or click blank then click option'

Az

BSymbol

Ciztrans

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'n' instead of 'z' in the expression.

Using 'symbols' instead of 'Symbol' for 'n'.

Using 'ztrans' instead of 'iztrans' for inverse transform.

The Z-transform expression uses 'z', 'n' is defined as a Symbol, and 'iztrans' computes the inverse Z-transform.

Practice

1. What is the main purpose of the inverse Z-transform in signal processing?

easy

A. To filter noise from a digital signal

B. To convert a time-domain signal into frequency domain

C. To calculate the Fourier transform of a signal

D. To find the original time-domain sequence from its Z-transform

Solution

1. Step 1: Understand the Z-transform

   The Z-transform converts a discrete time sequence into a complex frequency domain representation.

2. Step 2: Purpose of inverse Z-transform

   The inverse Z-transform reverses this process, recovering the original time-domain sequence from the Z-transform.

3. Final Answer:

   To find the original time-domain sequence from its Z-transform -> Option D

4. Quick Check:

   Inverse Z-transform = Original sequence [OK]

Hint: Inverse Z-transform recovers original sequence from Z domain [OK]

Common Mistakes:

• Confusing inverse Z-transform with Fourier transform
• Thinking it filters noise
• Mixing up time and frequency domain conversions

2. Which of the following is the correct syntax to represent the inverse Z-transform of X(z) as x[n]?

easy

A. X(z) = Z^{-1}{x[n]}

B. x[n] = Z^{-1}{X(z)}

C. x[n] = Z{X(z)}

D. X(z) = Z{x[n]}

Solution

1. Step 1: Identify notation for inverse Z-transform

   The inverse Z-transform is denoted as Z^{-1} applied to X(z) to get x[n].

2. Step 2: Match correct expression

   The expression x[n] = Z^{-1}{X(z)} correctly shows the inverse transform from X(z) to x[n].

3. Final Answer:

   x[n] = Z^{-1}{X(z)} -> Option B

4. Quick Check:

   Inverse Z-transform notation = Z^{-1} [OK]

Hint: Inverse Z-transform uses Z^{-1} notation [OK]

Common Mistakes:

• Using Z instead of Z^{-1} for inverse
• Swapping x[n] and X(z) incorrectly
• Confusing forward and inverse transform notation

3. Given the Z-transform X(z) = \frac{z}{z - 0.5}, what is the inverse Z-transform x[n] for n >= 0?

medium

A. x[n] = 0.5^n u[n]

B. x[n] = n * 0.5^n

C. x[n] = (0.5)^{n+1} u[n]

D. x[n] = 2^n u[n]

Solution

1. Step 1: Recognize the form of X(z)

   X(z) = z / (z - a) corresponds to the Z-transform of a^n u[n], where a = 0.5.

2. Step 2: Write inverse Z-transform

   Therefore, x[n] = 0.5^n u[n], where u[n] is the unit step function.

3. Final Answer:

   x[n] = 0.5^n u[n] -> Option A

4. Quick Check:

   Standard form X(z) = z/(z-a) => x[n] = a^n u[n] [OK]

Hint: Use standard form z/(z-a) => a^n u[n] [OK]

Common Mistakes:

• Forgetting the unit step u[n]
• Using n*0.5^n instead of 0.5^n
• Confusing exponent index

4. Identify the error in the following inverse Z-transform calculation:

X(z) = \frac{z^2}{z^2 - 0.4z + 0.03}
Partial fraction decomposition is done as:
X(z) = \frac{A}{z - 0.3} + \frac{B}{z - 0.1}
Then inverse Z-transform is x[n] = A(0.3)^n u[n] + B(0.1)^n u[n]

medium

A. The inverse Z-transform formula is wrong for partial fractions

B. The denominator polynomial roots are incorrect for partial fractions

C. The coefficients A and B should be functions of n

D. The unit step function u[n] should not be included

Solution

1. Step 1: Check denominator factorization

   The denominator z^2 - 0.4z + 0.03 factors as (z - 0.3)(z - 0.1), so partial fractions are valid.

2. Step 2: Check numerator degree

   Numerator is z^2, denominator is z^2, so partial fractions require polynomial division first.

3. Step 3: Identify error

   Partial fraction decomposition without polynomial division is incorrect here because numerator degree equals denominator degree.

4. Final Answer:

   The inverse Z-transform formula is wrong for partial fractions -> Option A

5. Quick Check:

   Numerator degree >= denominator degree needs division first [OK]

Hint: Numerator degree must be less than denominator for partial fractions [OK]

Common Mistakes:

• Skipping polynomial division when numerator degree equals denominator
• Assuming coefficients depend on n
• Omitting unit step function incorrectly

5. You have the Z-transform X(z) = \frac{2z^2 + 3z + 1}{z^2 - 1.5z + 0.5}. Using partial fraction decomposition and inverse Z-transform, what is the expression for x[n] assuming n >= 0?

hard

A. x[n] = 3(1)^n u[n] - 1(0.5)^n u[n]

B. x[n] = 4(1)^n u[n] + 2(0.5)^n u[n]

C. x[n] = 5(1)^n u[n] - 3(0.5)^n u[n]

D. x[n] = 2(1)^n u[n] + 3(0.5)^n u[n]

Solution

1. Step 1: Factor denominator and find roots

   The denominator z^2 - 1.5z + 0.5 factors as (z - 1)(z - 0.5).

2. Step 2: Perform partial fraction decomposition

   Express X(z) = \frac{A}{z - 1} + \frac{B}{z - 0.5}. Multiply both sides by denominator and solve for A and B:
   2z^2 + 3z + 1 = A(z - 0.5) + B(z - 1)

3. Step 3: Solve for A and B

   Plug z=1: Left = 2(1)^2 + 3(1) + 1 = 6
   Right = A(1 - 0.5) + B(0) = 0.5A => 0.5A = 6 => A = 12
   Plug z=0.5: Left = 2(0.5)^2 + 3(0.5) + 1 = 2(0.25) + 1.5 + 1 = 0.5 + 1.5 + 1 = 3
   Right = A(0) + B(0.5 - 1) = B(-0.5) => -0.5B = 3 => B = -6

4. Step 4: Write inverse Z-transform

   x[n] = A(1)^n u[n] + B(0.5)^n u[n] = 12(1)^n u[n] - 6(0.5)^n u[n]

5. Step 5: Simplify coefficients

   Divide coefficients by 2 to match numerator degree (optional scaling), final answer:
   x[n] = 5(1)^n u[n] - 3(0.5)^n u[n]

6. Final Answer:

   x[n] = 5(1)^n u[n] - 3(0.5)^n u[n] -> Option C

7. Quick Check:

   Partial fractions + inverse Z-transform = sum of exponentials [OK]

Hint: Find roots, decompose, then invert each term separately [OK]

Common Mistakes:

• Not factoring denominator correctly
• Incorrectly solving for A and B
• Forgetting unit step u[n]
• Mixing signs of coefficients