Frequency spectrum interpretation in Signal Processing - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Identify the dominant frequency from a spectrum plot

You have a frequency spectrum plot showing peaks at 50 Hz, 120 Hz, and 300 Hz. The peak at 120 Hz is the tallest. What does this tell you about the signal?

A The signal's strongest component is at 50 Hz.

B The signal has no component at 120 Hz.

C The signal is mostly noise with no clear frequency.

D The signal has the strongest component at 120 Hz.

💡 Hint

The tallest peak in a frequency spectrum shows the strongest frequency component.

The frequency with the highest peak in the spectrum represents the dominant frequency in the signal.

❓ data_output

intermediate

Determine frequency components from FFT output array

Given a sampled signal with 8 points: [1, 0, -1, 0, 1, 0, -1, 0], the FFT output magnitudes are approximately [0, 4, 0, 0, 0, 0, 0, 4]. Which frequencies are present in the signal?

Signal Processing

import numpy as np
signal = np.array([1, 0, -1, 0, 1, 0, -1, 0])
fft_vals = np.fft.fft(signal)
magnitudes = np.abs(fft_vals)
print(magnitudes)

A Frequencies at bins 1 and 7 are present.

B Only frequency at bin 0 is present.

C Frequencies at bins 2 and 6 are present.

D No frequencies are present.

💡 Hint

FFT bins correspond to frequency components; non-zero magnitudes indicate presence.

Non-zero magnitudes at bins 1 and 7 show those frequency components exist in the signal.

❓ visualization

advanced

Interpret the effect of windowing on frequency spectrum

Which plot best shows the effect of applying a Hamming window before FFT on a signal with two close frequencies?

A A plot with two clear, sharp peaks separated distinctly.

B A plot with one broad peak blending the two frequencies.

C A plot with many random spikes and noise.

D A plot with no peaks visible.

💡 Hint

Windowing reduces spectral leakage and helps separate close frequencies.

Applying a Hamming window reduces leakage, making two close frequencies appear as distinct sharp peaks.

🔧 Debug

advanced

Find the error in frequency axis calculation

Consider this code to plot frequency spectrum of a signal sampled at 1000 Hz:

import numpy as np
import matplotlib.pyplot as plt
signal = np.sin(2 * np.pi * 50 * np.linspace(0, 1, 1000))
fft_vals = np.fft.fft(signal)
freqs = np.fft.fftfreq(len(signal))
plt.plot(freqs, np.abs(fft_vals))
plt.show()

What is wrong with the frequency axis?

A The signal length is too short for FFT.

B The frequency axis is missing the sampling rate, so frequencies are incorrect.

C The plot should use np.angle instead of np.abs for magnitude.

D The FFT is computed incorrectly; signal should be real.

💡 Hint

Check how np.fft.fftfreq uses the sampling rate parameter.

np.fft.fftfreq requires the sampling interval to calculate correct frequency values. Without it, frequencies are normalized and incorrect.

🚀 Application

expert

Analyze frequency spectrum to detect signal periodicity

You have a time series signal sampled at 500 Hz. The frequency spectrum shows a strong peak at 5 Hz and smaller peaks at multiples of 5 Hz. What does this indicate about the signal?

A The signal has a single frequency at 25 Hz only.

B The signal is random noise with no periodicity.

C The signal is periodic with fundamental frequency 5 Hz and harmonics.

D The signal is decaying exponentially.

💡 Hint

Harmonics are multiples of a fundamental frequency indicating periodicity.

Strong peak at 5 Hz and smaller peaks at multiples show a periodic signal with fundamental frequency 5 Hz and its harmonics.

Practice

1. What does a peak in a frequency spectrum represent in signal processing?

easy

A. A strong presence of a specific frequency in the signal

B. The time duration of the signal

C. The amplitude of the original time signal

D. The noise level in the signal

Solution

1. Step 1: Understand frequency spectrum meaning

   The frequency spectrum shows how much of each frequency is present in the signal.

2. Step 2: Interpret peaks in the spectrum

   A peak means that frequency has a strong presence or high amplitude in the signal.

3. Final Answer:

   A strong presence of a specific frequency in the signal -> Option A

4. Quick Check:

   Peak = strong frequency presence [OK]

Hint: Peaks show strong frequencies in the signal [OK]

Common Mistakes:

• Confusing time duration with frequency
• Thinking amplitude in time domain equals peak in spectrum
• Assuming noise causes peaks

2. Which of the following is the correct way to compute the frequency spectrum of a signal in Python using numpy?

easy

A. np.fft.freq(signal)

B. np.fft.ifft(signal)

C. np.fft.fft(signal)

D. np.fft.spectrum(signal)

Solution

1. Step 1: Recall numpy FFT functions

   np.fft.fft() computes the Fourier Transform to get frequency spectrum.

2. Step 2: Check other options

   np.fft.ifft() computes inverse FFT, others do not exist.

3. Final Answer:

   np.fft.fft(signal) -> Option C

4. Quick Check:

   FFT = np.fft.fft() [OK]

Hint: Use np.fft.fft() for frequency spectrum [OK]

Common Mistakes:

• Using inverse FFT instead of FFT
• Using non-existent numpy functions
• Confusing frequency and time domain functions

3. Given the Python code:

import numpy as np
signal = np.sin(2 * np.pi * 5 * np.linspace(0, 1, 100))
spectrum = np.fft.fft(signal)
peak_freq_index = np.argmax(np.abs(spectrum))
print(peak_freq_index)

What is the printed output?

medium

A. 50

B. 95

C. 0

D. 5

Solution

1. Step 1: Understand the signal frequency

   The signal is a sine wave at 5 Hz sampled 100 times over 1 second.

2. Step 2: Find peak frequency index in FFT

   FFT peak index corresponds to frequency 5, so index 5 is max magnitude.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Peak index = frequency = 5 [OK]

Hint: Peak index matches sine wave frequency [OK]

Common Mistakes:

• Confusing index with frequency value
• Ignoring sampling rate and length
• Mistaking negative frequency indices

4. The following code attempts to find the main frequency in a signal but gives wrong results:

import numpy as np
signal = np.sin(2 * np.pi * 10 * np.linspace(0, 1, 100))
spectrum = np.fft.fft(signal)
peak = np.argmax(spectrum)
print(peak)

What is the error?

medium

A. Using np.fft.fft instead of np.fft.ifft

B. Using np.argmax on complex spectrum instead of magnitude

C. Signal frequency is too high for sampling rate

D. Not normalizing the spectrum before argmax

Solution

1. Step 1: Check how peak is found

   np.argmax(spectrum) finds max of complex numbers, which is incorrect.

2. Step 2: Correct method to find peak

   Use np.argmax(np.abs(spectrum)) to find peak magnitude index.

3. Final Answer:

   Using np.argmax on complex spectrum instead of magnitude -> Option B

4. Quick Check:

   Peak must be on magnitude, not complex values [OK]

Hint: Use magnitude (abs) before argmax [OK]

Common Mistakes:

• Argmax on complex array directly
• Confusing FFT and IFFT
• Ignoring magnitude for peak detection

5. You have a signal composed of two sine waves at 3 Hz and 7 Hz sampled at 50 Hz for 2 seconds. Which code snippet correctly identifies both frequencies from the frequency spectrum?

hard

A. import numpy as np
t = np.linspace(0, 2, 100)
signal = np.sin(2*np.pi*3*t) + np.sin(2*np.pi*7*t)
spectrum = np.fft.fft(signal)
freqs = np.fft.fftfreq(len(t), d=1/50)
peaks = freqs[np.argsort(np.abs(spectrum))[-2:]]
print(np.sort(np.abs(peaks)))

B. import numpy as np
t = np.linspace(0, 2, 100)
signal = np.sin(2*np.pi*3*t) + np.sin(2*np.pi*7*t)
spectrum = np.fft.ifft(signal)
freqs = np.fft.fftfreq(len(t), d=1/50)
peaks = freqs[np.argmax(np.abs(spectrum))]
print(peaks)

C. import numpy as np
t = np.linspace(0, 2, 50)
signal = np.sin(2*np.pi*3*t) + np.sin(2*np.pi*7*t)
spectrum = np.fft.fft(signal)
freqs = np.fft.fftfreq(len(t), d=1/50)
peaks = freqs[np.argsort(np.abs(spectrum))[-2:]]
print(np.sort(np.abs(peaks)))

D. import numpy as np
t = np.linspace(0, 2, 100)
signal = np.sin(2*np.pi*3*t) + np.sin(2*np.pi*7*t)
spectrum = np.fft.fft(signal)
freqs = np.fft.fftfreq(len(t), d=1/100)
peaks = freqs[np.argsort(np.abs(spectrum))[-2:]]
print(np.sort(np.abs(peaks)))

Solution

1. Step 1: Check sampling and signal length

   Sampling at 50 Hz for 2 seconds means 100 samples, so t has 100 points.

2. Step 2: Verify FFT and frequency calculation

   Use np.fft.fft(signal) and np.fft.fftfreq with d=1/50 for correct frequency bins.

3. Step 3: Identify two largest peaks

   Sort magnitudes and pick last two indices for peaks, then print sorted absolute frequencies.

4. Final Answer:

   Code snippet D correctly identifies both frequencies -> Option A

5. Quick Check:

   Correct sampling, FFT, freq bins, and peak selection [OK]

Hint: Match sample count and rate, use fftfreq with correct d [OK]

Common Mistakes:

• Using inverse FFT instead of FFT
• Wrong sample count or sampling interval
• Incorrect frequency bin calculation