Frequency array generation (fftfreq) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of fftfreq with default d=1.0

What is the output of the following code snippet?

SciPy

from scipy.fft import fftfreq
freqs = fftfreq(6)
print(freqs)

A [0. 0.2 0.4 0.6 0.8 1. ]

B [ 0. 0.16666667 0.33333333 0.5 -0.33333333 -0.16666667]

C [ 0. 0.16666667 0.33333333 -0.5 -0.33333333 -0.16666667]

D [ 0. 0.16666667 0.33333333 -0.33333333 -0.16666667 -0.5 ]

💡 Hint

Remember that fftfreq returns frequencies in cycles per unit sample spacing, with negative frequencies after the Nyquist frequency.

fftfreq(n) returns an array of length n with frequencies starting at 0, increasing to the positive Nyquist frequency, then negative frequencies in increasing order. For n=6 and default d=1.0, the output matches option C.

❓ data_output

intermediate

Frequency array with custom sample spacing

Given sample spacing d=0.5 and n=4, what is the output of fftfreq?

SciPy

from scipy.fft import fftfreq
freqs = fftfreq(4, d=0.5)
print(freqs)

A [ 0. 0.25 -1. -0.5]

B [ 0. 0.25 -0.5 -0.25]

C [ 0. 0.5 -0.5 -0.25]

D [ 0. 0.5 -1. -0.5]

💡 Hint

Frequency values scale inversely with sample spacing d.

With n=4 and d=0.5, fftfreq returns frequencies [0, 1/(4*0.5), -1/(2*0.5), -1/(4*0.5)] = [0, 0.5, -1, -0.5].

🧠 Conceptual

advanced

Understanding negative frequencies in fftfreq output

Why does fftfreq return negative frequency values in its output array?

A Negative frequencies represent the direction of rotation in complex exponentials and are necessary for representing real signals in the frequency domain.

B Negative frequencies are errors and should be ignored when analyzing FFT results.

C Negative frequencies indicate frequencies below zero Hertz, which physically do not exist.

D Negative frequencies are placeholders for missing data points in the FFT output.

💡 Hint

Think about how complex sinusoids represent signals in both positive and negative directions.

In Fourier analysis, negative frequencies correspond to complex conjugate components and represent direction of rotation in the complex plane. They are essential for reconstructing real-valued signals.

🔧 Debug

advanced

Identify the error in fftfreq usage

What error will the following code produce?

SciPy

from scipy.fft import fftfreq
freqs = fftfreq(-5)
print(freqs)

A TypeError: fftfreq() missing required positional argument

B ValueError: n must be non-negative

C SyntaxError: invalid syntax

D No error, outputs an empty array

💡 Hint

Check the parameter constraints for fftfreq's n argument.

The fftfreq function requires n to be a non-negative integer. Passing a negative value raises a ValueError.

🚀 Application

expert

Using fftfreq to find the dominant frequency index

Given a signal sampled at 100 Hz with 8 samples, which index corresponds to the dominant frequency 12.5 Hz using fftfreq?

SciPy

from scipy.fft import fftfreq
sample_rate = 100
n = 8
freqs = fftfreq(n, d=1/sample_rate)
print(freqs)

A Index 1

B Index 3

C Index 2

D Index 4

💡 Hint

Calculate frequency bins as k * sample_rate / n for k in [0..n-1], considering fftfreq output order.

With n=8 and sample_rate=100 Hz, fftfreq returns frequencies: [0, 12.5, 25, 37.5, -50, -37.5, -25, -12.5]. The frequency 12.5 Hz is at index 1.

Practice

1. What does the scipy.fft.fftfreq function return when given the number of points and sample spacing?

easy

A. A random array of the same length as input

B. The Fourier transform of the input data

C. The inverse Fourier transform of the input data

D. An array of frequency bins corresponding to the FFT output

Solution

1. Step 1: Understand the purpose of fftfreq

   The fftfreq function generates frequency bins for FFT results, not the transform itself.

2. Step 2: Identify the output type

   It returns an array of frequencies that correspond to FFT output indices, including positive and negative frequencies.

3. Final Answer:

   An array of frequency bins corresponding to the FFT output -> Option D

4. Quick Check:

   fftfreq returns frequency bins [OK]

Hint: fftfreq gives frequency bins, not the transform itself [OK]

Common Mistakes:

• Confusing fftfreq output with FFT transform
• Expecting only positive frequencies
• Thinking it returns time domain data

2. Which of the following is the correct way to import and use fftfreq from scipy to generate frequency bins for 8 points with sample spacing 0.5?

easy

A. from scipy import fftfreq
freqs = fftfreq(8, 0.5)

B. import scipy.fft
freqs = scipy.fft.fftfreq(0.5, 8)

C. from scipy.fft import fftfreq
freqs = fftfreq(8, 0.5)

D. import fftfreq from scipy.fft
freqs = fftfreq(8, 0.5)

Solution

1. Step 1: Check the correct import syntax

   The correct import is from scipy.fft import fftfreq, not directly from scipy or using wrong syntax.

2. Step 2: Verify function parameters order

   fftfreq takes number of points first, then sample spacing, so fftfreq(8, 0.5) is correct.

3. Final Answer:

   from scipy.fft import fftfreq freqs = fftfreq(8, 0.5) -> Option C

4. Quick Check:

   Correct import and parameter order [OK]

Hint: Import from scipy.fft and pass (n, d) in order [OK]

Common Mistakes:

• Swapping parameters order
• Wrong import statement
• Using scipy.fftfreq instead of scipy.fft.fftfreq

3. What is the output of the following code?

from scipy.fft import fftfreq
freqs = fftfreq(4, 1.0)
print(freqs)

medium

A. [0.0, 0.5, -0.5, -0.25]

B. [0.0, 0.25, -0.5, -0.25]

C. [0.0, 0.25, 0.5, 0.75]

D. [0.0, 0.5, 1.0, 1.5]

Solution

1. Step 1: Understand fftfreq output for n=4, d=1.0

   For 4 points, frequencies are [0, 1/4, -2/4, -1/4] = [0.0, 0.25, -0.5, -0.25].

2. Step 2: Confirm printed array matches expected frequencies

   The printed array matches [0.0, 0.25, -0.5, -0.25] exactly.

3. Final Answer:

   [0.0, 0.25, -0.5, -0.25] -> Option B

4. Quick Check:

   fftfreq(4,1.0) = [0, 0.25, -0.5, -0.25] [OK]

Hint: For even n, negative freqs start at -n/2/n [OK]

Common Mistakes:

• Ignoring negative frequencies
• Miscomputing frequency steps
• Confusing sample spacing effect

4. The code below throws an error. What is the cause and how to fix it?

from scipy import fftfreq
freqs = fftfreq(5)
print(freqs)

medium

A. Import statement is incorrect; use import scipy.fft instead

B. fftfreq requires input data array, not just number of points

C. Missing sample spacing argument; add a second argument like 1.0

D. fftfreq cannot handle odd number of points

Solution

1. Step 1: Identify the cause of the error

   The import from scipy import fftfreq fails because fftfreq is in the scipy.fft submodule.

2. Step 2: Fix by using correct import

   Change to from scipy.fft import fftfreq or import scipy.fft and call scipy.fft.fftfreq(5).

3. Final Answer:

   Import statement is incorrect; use import scipy.fft instead -> Option A

4. Quick Check:

   fftfreq requires scipy.fft import [OK]

Hint: fftfreq is in scipy.fft submodule [OK]

Common Mistakes:

• Importing fftfreq directly from scipy
• Confusing fftfreq with fft function
• Wrong import causing unrelated errors

5. You have a signal sampled at 10 Hz with 10 data points. Using fftfreq, how do you generate the frequency bins array? What is the frequency value at index 3?

hard

A. freqs = fftfreq(10, 0.1); frequency at index 3 is 3 Hz

B. freqs = fftfreq(10, 0.1); frequency at index 3 is 0.3 Hz

C. freqs = fftfreq(10, 10); frequency at index 3 is 3 Hz

D. freqs = fftfreq(10, 1); frequency at index 3 is 0.3 Hz

Solution

1. Step 1: Calculate sample spacing from sampling rate

   Sampling rate is 10 Hz, so sample spacing d = 1/10 = 0.1 seconds.

2. Step 2: Use fftfreq with n=10 and d=0.1

   Calling fftfreq(10, 0.1) generates frequencies: [0, 1, 2, 3, 4, -5, -4, -3, -2, -1] Hz.

3. Step 3: Identify frequency at index 3

   Frequency step df = 1/(n*d) = 1 Hz, so index 3 is 3 Hz.

4. Final Answer:

   freqs = fftfreq(10, 0.1); frequency at index 3 is 3 Hz -> Option A

5. Quick Check:

   Sample spacing = 0.1, freq[3] = 3 Hz [OK]

Hint: Sample spacing = 1 / fs; freq[k] = k / (n * d) [OK]

Common Mistakes:

• Using sampling rate as spacing directly
• Mixing up index and frequency value
• Incorrect parameter order in fftfreq