Why Z-transform is used in DSP in Signal Processing - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the library used for Z-transform in Python.

Signal Processing

from sympy import [1]

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

Aintegrate

Bfft

Cztrans

Dplot

Attempts:

3 left

💡 Hint

Common Mistakes

Importing unrelated functions like fft or integrate.

The ztrans function from sympy is used to compute the Z-transform symbolically.

2fill in blank

medium

Complete the code to define a discrete signal x[n] as a list.

Signal Processing

x = [1, 2, 3, 4, [1]]

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

A5

B-1

C0

D6

Attempts:

3 left

💡 Hint

Common Mistakes

Adding zero or negative numbers which break the sequence.

The signal is defined as a list of values; adding 5 continues the sequence naturally.

3fill in blank

hard

Fix the error in the code to compute the Z-transform of x[n].

Signal Processing

from sympy import symbols, [1]
n, z = symbols('n z')
x = 1
Z = [1](x, n, z)

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

Adiff

Bfft

Cintegrate

Dztrans

Attempts:

3 left

💡 Hint

Common Mistakes

Using fft or integrate which are unrelated to Z-transform.

The ztrans function computes the Z-transform symbolically.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps n to x[n] for n from 0 to 4.

Signal Processing

x = {n: [1] for n in range([2])}

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

An+1

B5

C4

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong range or wrong value expression.

This creates a dictionary where each key n maps to n+1, for n from 0 to 4 (5 values).

5fill in blank

hard

Fill all three blanks to compute the Z-transform of a signal and print the result.

Signal Processing

from sympy import symbols, [1]
n, z = symbols('n z')
x = [2]
Z = [1](x, n, [3])
print(Z)

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

Aztrans

Bn+1

Cz

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong function names or variables.

The ztrans function computes the Z-transform of the signal n + 1 with respect to z.

Practice

1. Why is the Z-transform used in digital signal processing (DSP)?

easy

A. To convert discrete signals into a simpler form for easier analysis

B. To convert analog signals into digital signals

C. To increase the speed of signal transmission

D. To compress the signal data for storage

Solution

1. Step 1: Understand the purpose of Z-transform in DSP

   The Z-transform converts discrete-time signals into a complex frequency domain representation, making analysis easier.

2. Step 2: Compare options with the purpose

   Only To convert discrete signals into a simpler form for easier analysis correctly states that Z-transform simplifies discrete signals for analysis, unlike other options which describe unrelated processes.

3. Final Answer:

   To convert discrete signals into a simpler form for easier analysis -> Option A

4. Quick Check:

   Z-transform simplifies discrete signals for analysis [OK]

Hint: Z-transform simplifies discrete signals for analysis [OK]

Common Mistakes:

• Confusing Z-transform with analog-to-digital conversion
• Thinking it compresses data
• Assuming it speeds up transmission

2. Which of the following is the correct mathematical expression for the Z-transform of a discrete signal x[n]?

easy

A. Z\{x[n]\} = \sum_{n=-\infty}^{\infty} x[n] z^n

B. Z\{x[n]\} = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt

C. Z\{x[n]\} = \sum_{n=0}^{\infty} x[n] z^{-n}

D. Z\{x[n]\} = \sum_{n=0}^{N} x[n] e^{j2\pi fn/N}

Solution

1. Step 1: Recall the Z-transform formula

   The Z-transform is defined as the sum from n=0 to infinity of x[n] times z to the power of negative n.

2. Step 2: Match the formula with options

   Z\{x[n]\} = \sum_{n=0}^{\infty} x[n] z^{-n} matches the correct formula. Z\{x[n]\} = \sum_{n=-\infty}^{\infty} x[n] z^n has incorrect power sign and limits, the integral is the Fourier transform, the finite sum with exponentials is the DFT formula.

3. Final Answer:

   Z\{x[n]\} = \sum_{n=0}^{\infty} x[n] z^{-n} -> Option C

4. Quick Check:

   Z-transform formula = Z\{x[n]\} = \sum_{n=0}^{\infty} x[n] z^{-n} [OK]

Hint: Z-transform sums x[n] times z to negative n from 0 to infinity [OK]

Common Mistakes:

• Using integral instead of sum (confusing with Fourier transform)
• Wrong power sign on z
• Wrong summation limits

3. Given the discrete signal x[n] = {1, 2, 3} for n=0,1,2 and zero elsewhere, what is the Z-transform X(z)?

medium

A. X(z) = 1 + 2z + 3z^2

B. X(z) = 1 + 2z^{-1} + 3z^{-2}

C. X(z) = 1z^{-1} + 2z^{-2} + 3z^{-3}

D. X(z) = 1 + 2 + 3

Solution

1. Step 1: Apply Z-transform formula to given x[n]

   Using X(z) = sum x[n] z^{-n}, calculate: 1*z^0 + 2*z^{-1} + 3*z^{-2}.

2. Step 2: Write the expression

   This gives X(z) = 1 + 2z^{-1} + 3z^{-2}, matching X(z) = 1 + 2z^{-1} + 3z^{-2}.

3. Final Answer:

   X(z) = 1 + 2z^{-1} + 3z^{-2} -> Option B

4. Quick Check:

   Sum x[n]z^{-n} = X(z) = 1 + 2z^{-1} + 3z^{-2} [OK]

Hint: Use powers of z as negative n for each x[n] term [OK]

Common Mistakes:

• Using positive powers of z instead of negative
• Starting powers from z^{-1} instead of z^{0}
• Summing values without z terms

4. Identify the error in this Z-transform calculation code snippet for x = [1, 2, 3]:

def z_transform(x, z):
    Xz = 0
    for n in range(len(x)):
        Xz += x[n] * z**n
    return Xz

medium

A. The power of z should be negative n, not positive n

B. The loop should start from n=1, not 0

C. The initial value of Xz should be 1, not 0

D. The function should return x[n] instead of Xz

Solution

1. Step 1: Review Z-transform formula

   The Z-transform uses z to the power of negative n, so the code should use z**(-n).

2. Step 2: Check the code's power usage

   The code uses z**n, which is incorrect and will give wrong results.

3. Final Answer:

   The power of z should be negative n, not positive n -> Option A

4. Quick Check:

   Power sign in Z-transform = negative n [OK]

Hint: Z-transform uses z to negative n power [OK]

Common Mistakes:

• Using positive powers of z
• Starting loop at wrong index
• Wrong initial sum value

5. In DSP, why is the Z-transform preferred over the Fourier transform when analyzing system stability?

hard

A. Because Fourier transform cannot handle complex numbers

B. Because Fourier transform only works for continuous signals

C. Because Z-transform is faster to compute than Fourier transform

D. Because Z-transform can analyze signals outside the unit circle, capturing stability better

Solution

1. Step 1: Understand stability analysis in DSP

   Stability depends on poles inside the unit circle in the Z-plane; Z-transform represents this well.

2. Step 2: Compare Z-transform and Fourier transform

   Fourier transform evaluates on the unit circle only, missing behavior outside it, so it cannot fully analyze stability.

3. Final Answer:

   Because Z-transform can analyze signals outside the unit circle, capturing stability better -> Option D

4. Quick Check:

   Z-transform analyzes poles outside unit circle for stability [OK]

Hint: Z-transform analyzes poles beyond unit circle for stability [OK]

Common Mistakes:

• Thinking Fourier transform works only for continuous signals
• Assuming Z-transform is faster to compute
• Believing Fourier transform can't handle complex numbers