Magnitude and phase spectrum in Signal Processing - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of magnitude spectrum calculation

What is the output of the following Python code that computes the magnitude spectrum of a signal?

Signal Processing

import numpy as np
signal = np.array([1, 2, 3, 4])
fft_result = np.fft.fft(signal)
magnitude = np.abs(fft_result)
print(np.round(magnitude, 2))

A [10. 2.83 2. 2.83]

B [10. 2.83 2.83 2.83]

C [10. 2. 2. 2. ]

D [10. 3. 2. 3. ]

💡 Hint

Recall that magnitude is the absolute value of the FFT complex output.

The FFT of [1,2,3,4] is [10+0j, -2+2j, -2+0j, -2-2j]. The magnitudes are [10, sqrt(2^2+2^2), 2, sqrt(2^2+2^2)] = [10, 2.83, 2, 2.83].

❓ data_output

intermediate

Phase spectrum values from FFT

Given the signal array and its FFT, what is the phase spectrum output (in radians) from the code below?

Signal Processing

import numpy as np
signal = np.array([0, 1, 0, -1])
fft_result = np.fft.fft(signal)
phase = np.angle(fft_result)
print(np.round(phase, 2))

A [ 0. -1.57 3.14 1.57]

B [ 0. 0. 0. 0. ]

C [ 0. 1.57 3.14 -1.57]

D [ 0. -3.14 1.57 3.14]

💡 Hint

Phase is the angle of the complex FFT output, use np.angle.

The FFT of [0,1,0,-1] is [0+0j, 0-2j, 0+0j, 0+2j]. The phases are [0, -pi/2, pi, pi/2] which rounds to [0, -1.57, 3.14, 1.57].

❓ visualization

advanced

Identify the correct magnitude spectrum plot

Which plot correctly shows the magnitude spectrum of a 1-second 5 Hz sine wave sampled at 50 Hz?

Signal Processing

import numpy as np
import matplotlib.pyplot as plt
fs = 50
f = 5
t = np.arange(0, 1, 1/fs)
signal = np.sin(2 * np.pi * f * t)
fft_result = np.fft.fft(signal)
magnitude = np.abs(fft_result) / len(signal)
freq = np.fft.fftfreq(len(signal), 1/fs)
plt.stem(freq[:len(freq)//2], magnitude[:len(magnitude)//2])
plt.xlabel('Frequency (Hz)')
plt.ylabel('Magnitude')
plt.title('Magnitude Spectrum')
plt.show()

A A flat plot with no peaks

B A plot with a single peak at 5 Hz and zero elsewhere

C A plot with peaks at 5 Hz and 45 Hz

D A plot with multiple peaks evenly spaced every 5 Hz

💡 Hint

Remember FFT frequency bins include negative frequencies represented as high positive frequencies.

The FFT of a real sine wave shows two peaks: one at the positive frequency (5 Hz) and one at the negative frequency (which appears at 45 Hz due to FFT frequency wrapping).

🧠 Conceptual

advanced

Effect of zero-padding on magnitude spectrum resolution

What is the main effect of zero-padding a time-domain signal before computing its FFT magnitude spectrum?

A It changes the actual frequency content of the signal

B It removes noise from the signal

C It decreases the amplitude of all frequency components

D It increases the frequency resolution by interpolating the spectrum

💡 Hint

Think about how zero-padding affects the FFT length and frequency bins.

Zero-padding adds extra zeros to the signal, increasing FFT length and producing more frequency bins, which interpolates the spectrum and improves frequency resolution visually without changing actual frequency content.

🔧 Debug

expert

Identify the error in phase spectrum calculation code

What error will the following code produce when calculating the phase spectrum of a signal?

Signal Processing

import numpy as np
signal = np.array([1, 0, -1, 0])
fft_result = np.fft.fft(signal)
phase = np.angle(fft_result.real)
print(phase)

A SyntaxError due to incorrect np.angle usage

B The phase array will be all zeros, losing phase information

C TypeError because fft_result.real is not a complex array

D Correct phase spectrum output

💡 Hint

Check what np.angle expects as input and what fft_result.real contains.

np.angle expects complex numbers to compute phase. Passing only the real part (fft_result.real) removes imaginary parts, so phase is zero everywhere, losing phase information.

Practice

1. What does the magnitude spectrum represent in signal processing?

easy

A. The strength or amplitude of different frequency components

B. The timing or phase shift of frequency components

C. The original time-domain signal values

D. The noise level in the signal

Solution

1. Step 1: Understand magnitude spectrum meaning

   The magnitude spectrum shows how strong or intense each frequency is in a signal.

2. Step 2: Differentiate from phase spectrum

   The phase spectrum shows timing, not strength, so it is not the magnitude.

3. Final Answer:

   The strength or amplitude of different frequency components -> Option A

4. Quick Check:

   Magnitude spectrum = strength [OK]

Hint: Magnitude = strength of frequencies, phase = timing [OK]

Common Mistakes:

• Confusing magnitude with phase
• Thinking magnitude shows time values
• Mixing magnitude with noise level

2. Which Python function is used to get the phase spectrum from a Fourier Transform result stored in variable F?

easy

A. np.abs(F)

B. np.angle(F)

C. np.fft(F)

D. np.real(F)

Solution

1. Step 1: Recall phase extraction function

   In Python's NumPy, np.angle() returns the phase angles of complex numbers.

2. Step 2: Differentiate from magnitude and other functions

   np.abs() returns magnitude, np.fft() is not a valid function, and np.real() returns real parts only.

3. Final Answer:

   np.angle(F) -> Option B

4. Quick Check:

   Phase spectrum = np.angle() [OK]

Hint: Use np.angle() for phase, np.abs() for magnitude [OK]

Common Mistakes:

• Using np.abs() instead of np.angle() for phase
• Trying to use np.fft() which is invalid
• Confusing real part with phase

3. Given the code:

import numpy as np
x = np.array([1, 0, -1, 0])
F = np.fft.fft(x)
magnitude = np.abs(F)
phase = np.angle(F)
print(np.round(magnitude, 2))
print(np.round(phase, 2))

What is the printed magnitude and phase arrays?

medium

A. [4.00 0.00 0.00 0.00]
[0.00 0.00 0.00 0.00]

B. [0.00 2.00 0.00 2.00]
[0.00 0.00 0.00 0.00]

C. [2.00 2.00 2.00 2.00]
[ 0.00 -1.57 3.14 1.57]

D. [2.00 0.00 2.00 0.00]
[0.00 1.57 3.14 -1.57]

Solution

1. Step 1: Compute FFT of x

   The FFT of [1, 0, -1, 0] is [0+0j, 2+0j, 0+0j, 2+0j].

2. Step 2: Calculate magnitude and phase arrays

   Magnitude is [2.00 2.00 2.00 2.00]. Phase angles are [0.00 -1.57 3.14 1.57].

3. Final Answer:

   [2.00 2.00 2.00 2.00] [ 0.00 -1.57 3.14 1.57] -> Option C

4. Quick Check:

   Magnitude = [2 2 2 2], Phase = approx [-pi/2, pi, pi/2] [OK]

Hint: FFT magnitude is abs(), phase is angle() in radians [OK]

Common Mistakes:

• Confusing magnitude with real part
• Not rounding phase correctly
• Mixing up signs of phase angles

4. The following code attempts to compute magnitude and phase but has an error:

import numpy as np
x = np.array([1, 2, 3, 4])
F = np.fft.fft(x)
magnitude = np.angle(F)
phase = np.abs(F)
print(magnitude)
print(phase)

What is the error?

medium

A. No error; code runs correctly

B. np.fft.fft() is used incorrectly; should be np.fft.ifft()

C. The input array x must be complex numbers

D. Magnitude and phase are swapped; np.angle() should be for phase, np.abs() for magnitude

Solution

1. Step 1: Check function usage for magnitude and phase

   Magnitude is the absolute value, so np.abs() should be used for magnitude, not np.angle(). Phase is angle, so np.angle() is for phase.

2. Step 2: Identify swapped assignments

   The code assigns magnitude = np.angle(F) and phase = np.abs(F), which is reversed.

3. Final Answer:

   Magnitude and phase are swapped; np.angle() should be for phase, np.abs() for magnitude -> Option D

4. Quick Check:

   Magnitude = abs(), Phase = angle() [OK]

Hint: Remember: abs() for magnitude, angle() for phase [OK]

Common Mistakes:

• Swapping magnitude and phase functions
• Thinking FFT input must be complex
• Confusing FFT and IFFT usage

5. You have a noisy signal and want to keep only frequencies with magnitude above 5. Which steps correctly apply this filter using FFT results stored in F?

hard

A. Calculate magnitude = np.abs(F), create mask = magnitude > 5, then set F_filtered = F * mask

B. Calculate phase = np.angle(F), create mask = phase > 5, then set F_filtered = F * mask

C. Calculate magnitude = np.abs(F), create mask = magnitude < 5, then set F_filtered = F * mask

D. Calculate magnitude = np.abs(F), create mask = magnitude > 5, then set F_filtered = np.abs(F) * mask

Solution

1. Step 1: Compute magnitude and create mask

   Calculate magnitude with np.abs(F). Create a boolean mask where magnitude > 5 to keep strong frequencies.

2. Step 2: Apply mask to original FFT data

   Multiply the complex FFT array F by the mask to zero out weak frequencies, preserving phase and magnitude of strong ones.

3. Final Answer:

   Calculate magnitude = np.abs(F), create mask = magnitude > 5, then set F_filtered = F * mask -> Option A

4. Quick Check:

   Filter by magnitude > 5, multiply mask with F [OK]

Hint: Filter frequencies by magnitude mask, multiply mask with FFT [OK]

Common Mistakes:

• Filtering by phase instead of magnitude
• Using magnitude < 5 to keep frequencies
• Multiplying mask with magnitude only, losing phase info