Real FFT (rfft) in SciPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to compute the real FFT of the signal array.

SciPy

import numpy as np
from scipy.fft import [1]
signal = np.array([0, 1, 0, -1])
result = [1](signal)

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

Arfft

Bfft

Cifft

Dirfft

Attempts:

3 left

💡 Hint

Common Mistakes

Using fft instead of rfft causes output to be complex and longer.

Using ifft or irfft is for inverse transforms, not forward.

The function rfft computes the real FFT of a real-valued input signal.

2fill in blank

medium

Complete the code to compute the frequencies corresponding to the rfft output.

SciPy

import numpy as np
from scipy.fft import rfft, [1]
signal = np.array([0, 1, 0, -1])
sample_rate = 4
freqs = [1](len(signal), 1/sample_rate)

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

Afftfreq

Bfftshift

Cirfft

Drfftfreq

Attempts:

3 left

💡 Hint

Common Mistakes

Using fftfreq returns frequencies for full FFT, not real FFT.

Using irfft or fftshift does not return frequencies.

rfftfreq returns the sample frequencies for the output of rfft.

3fill in blank

hard

Fix the error in the code to correctly compute the inverse real FFT.

SciPy

import numpy as np
from scipy.fft import rfft, [1]
signal = np.array([0, 1, 0, -1])
transformed = rfft(signal)
recovered = [1](transformed)

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

Aifft

Birfft

Crfft

Dfft

Attempts:

3 left

💡 Hint

Common Mistakes

Using ifft instead of irfft may cause shape mismatch.

Using rfft or fft again does not invert the transform.

irfft computes the inverse of rfft, recovering the original real signal.

4fill in blank

hard

Fill both blanks to create a dictionary of frequencies and their corresponding rfft magnitudes.

SciPy

import numpy as np
from scipy.fft import rfft, rfftfreq
signal = np.array([0, 1, 0, -1])
sample_rate = 4
freqs = [1](len(signal), 1/sample_rate)
magnitudes = np.abs([2](signal))
result = dict(zip(freqs, magnitudes))

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

Arfftfreq

Brfft

Cfft

Dirfft

Attempts:

3 left

💡 Hint

Common Mistakes

Using fft instead of rfft causes mismatch in frequencies and magnitudes.

Using irfft instead of rfft for magnitudes is incorrect.

rfftfreq gives frequencies for rfft output; rfft computes the transform.

5fill in blank

hard

Fill all three blanks to compute the normalized power spectrum of a real signal.

SciPy

import numpy as np
from scipy.fft import rfft, rfftfreq
signal = np.array([0, 1, 0, -1])
sample_rate = 4
freqs = [1](len(signal), 1/sample_rate)
transform = [2](signal)
power = np.square(np.abs(transform)) / [3](signal)

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

Arfftfreq

Brfft

Cfloat

Dlen

Attempts:

3 left

💡 Hint

Common Mistakes

Using float(len(signal)) instead of len(signal) causes type issues.

Using fftfreq instead of rfftfreq causes frequency mismatch.

rfftfreq gets frequencies; rfft computes transform; len(signal) normalizes power.

Practice

1. What does the scipy.fft.rfft function compute when applied to a real-valued signal?

easy

A. The full complex Fourier transform including negative frequencies

B. The positive frequency components of the signal's Fourier transform

C. The inverse Fourier transform of the signal

D. The time-domain signal filtered by a low-pass filter

Solution

1. Step 1: Understand the input type

   rfft is designed for real-valued input signals.

2. Step 2: Analyze the output of rfft

   It returns only the positive frequency components because the negative frequencies are mirror images for real signals.

3. Final Answer:

   The positive frequency components of the signal's Fourier transform -> Option B

4. Quick Check:

   Real input -> half spectrum output [OK]

Hint: Real signals -> use rfft for positive frequencies only [OK]

Common Mistakes:

• Confusing rfft with full fft output
• Thinking rfft returns time-domain data
• Assuming rfft returns negative frequencies

2. Which of the following is the correct way to import and use rfft from SciPy's FFT module?

easy

A. import scipy.fft as rfft
result = rfft.rfft(signal)

B. import scipy.rfft
result = scipy.rfft(signal)

C. from scipy.fft import rfft
result = rfft(signal)

D. from scipy import rfft
result = rfft.fft(signal)

Solution

1. Step 1: Recall correct import syntax

   The rfft function is in scipy.fft, so import it directly from there.

2. Step 2: Check usage

   Calling rfft(signal) after import is the correct usage.

3. Final Answer:

   from scipy.fft import rfft\nresult = rfft(signal) -> Option C

4. Quick Check:

   Correct import and call syntax [OK]

Hint: Import rfft directly from scipy.fft for simple use [OK]

Common Mistakes:

• Trying to import rfft from scipy directly
• Using wrong module names
• Calling rfft as a method of scipy.fft without import

3. What is the output of the following code?

import numpy as np
from scipy.fft import rfft
signal = np.array([1, 0, -1, 0])
result = rfft(signal)
print(np.round(result, 2))

medium

A. [0.+0.j 1.+0.j 0.+0.j]

B. [0.+0.j 0.+0.j 0.+0.j]

C. [0.+0.j -2.+0.j 0.+0.j]

D. [0.+0.j 2.+0.j 0.+0.j]

Solution

1. Step 1: Understand the input signal

   The signal is [1, 0, -1, 0], a simple 4-point real array.

2. Step 2: Compute rfft manually or recall pattern

   rfft returns N/2+1 = 3 complex values for length 4 input. The values correspond to DC, positive frequency, and Nyquist frequency components. The result is approximately [0.+0.j, 2.+0.j, 0.+0.j].

3. Final Answer:

   [0.+0.j 2.+0.j 0.+0.j] -> Option D

4. Quick Check:

   rfft length 4 -> 3 outputs, middle value 2 [OK]

Hint: rfft output length = input//2 + 1 [OK]

Common Mistakes:

• Expecting output length same as input
• Ignoring complex number format
• Misreading the signal values

4. Identify the error in this code snippet using rfft and fix it:

import numpy as np
from scipy.fft import rfft
signal = [1, 2, 3, 4]
result = rfft(signal, n=2)
print(result)

medium

A. The parameter n is smaller than signal length; use n=4 instead

B. The input signal must be a NumPy array, not a list

C. rfft does not accept the parameter n

D. The import statement is incorrect; use import scipy.fftpack

Solution

1. Step 1: Check the n parameter usage

   The n parameter sets the length of the FFT. Here, n=2 is less than the signal length 4, so the signal is truncated.

2. Step 2: Fix the parameter to match or exceed signal length

   Using n=4 ensures the full signal is transformed without truncation.

3. Final Answer:

   The parameter n is smaller than signal length; use n=4 instead -> Option A

4. Quick Check:

   FFT length must cover signal length [OK]

Hint: Set n ≥ signal length to avoid truncation [OK]

Common Mistakes:

• Passing list instead of array (allowed but less efficient)
• Ignoring effect of n parameter truncation
• Wrong import module for rfft

5. You have a real sensor signal sampled at 100 Hz with 1000 points. You want to find the frequency with the highest amplitude using rfft. Which code snippet correctly computes this frequency?

hard

A. import numpy as np
from scipy.fft import rfft, rfftfreq
signal = np.random.rand(1000)
fft_vals = rfft(signal)
freqs = rfftfreq(len(signal), 1/100)
peak_freq = freqs[np.argmax(np.abs(fft_vals))]
print(peak_freq)

B. import numpy as np
from scipy.fft import fft
signal = np.random.rand(1000)
fft_vals = fft(signal)
freqs = np.fft.fftfreq(len(signal), 1/100)
peak_freq = freqs[np.argmax(np.abs(fft_vals))]
print(peak_freq)

C. import numpy as np
from scipy.fft import rfft
signal = np.random.rand(1000)
fft_vals = rfft(signal)
peak_freq = np.argmax(np.abs(fft_vals))
print(peak_freq)

D. import numpy as np
from scipy.fft import rfft, rfftfreq
signal = np.random.rand(1000)
fft_vals = rfft(signal)
freqs = rfftfreq(len(signal), 100)
peak_freq = freqs[np.argmax(np.abs(fft_vals))]
print(peak_freq)

Solution

1. Step 1: Use rfft and rfftfreq correctly

   For real signals, rfft computes the positive frequencies, and rfftfreq gives the corresponding frequency bins. The sampling interval is 1/100 seconds.

2. Step 2: Find the peak frequency

   Calculate the absolute values of FFT coefficients, find the index of the max, then map to frequency using freqs.

3. Final Answer:

   Code snippet A correctly computes the peak frequency -> Option A

4. Quick Check:

   Use rfft + rfftfreq with correct sampling interval [OK]

Hint: Use rfftfreq with 1/sampling_rate for correct frequencies [OK]

Common Mistakes:

• Using fft and fftfreq for real signals (less efficient)
• Wrong sampling interval in rfftfreq
• Not mapping peak index to frequency