IIR filter design (butter, cheby1) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Butterworth filter coefficients

What is the output of the following code that designs a 3rd order Butterworth lowpass filter with cutoff frequency 0.4 (normalized) and prints the numerator coefficients?

SciPy

from scipy.signal import butter
b, a = butter(3, 0.4, btype='low')
print(b)

A [0.01809893 0.05429679 0.05429679 0.01809893]

B [ 0.09763107 0.29289322 0.29289322 0.09763107]

C [0.00482434 0.01447302 0.01447302 0.00482434]

D [0.25 0.25 0.25 0.25]

💡 Hint

Recall that butter() returns numerator and denominator coefficients of the filter.

The butter function with order 3 and cutoff 0.4 returns numerator coefficients approximately equal to [0.0181, 0.0543, 0.0543, 0.0181]. Other options are incorrect values.

❓ data_output

intermediate

Number of zeros in Chebyshev Type I filter numerator

After designing a 4th order Chebyshev Type I highpass filter with 1 dB ripple and cutoff frequency 0.3 (normalized), how many zeros does the numerator polynomial have?

SciPy

from scipy.signal import cheby1
b, a = cheby1(4, 1, 0.3, btype='high')
print(len(b) - 1)

A 3

B 4

C 5

D 1

💡 Hint

The numerator polynomial degree equals the filter order for IIR filters.

The numerator coefficients array length is order + 1, so zeros count is 4 for a 4th order filter.

🔧 Debug

advanced

Identify the error in filter design code

What error does the following code raise when trying to design a Butterworth bandpass filter with cutoff frequencies 0.2 and 0.5?

SciPy

from scipy.signal import butter
b, a = butter(3, [0.2, 0.5], btype='bandpass')

A No error, returns filter coefficients

B TypeError: 'bandpass' is not a valid filter type

C ValueError: Digital filter critical frequencies must be strictly increasing and 0 < Wn < 1

D SyntaxError: invalid syntax

💡 Hint

Check the valid filter types and cutoff frequency ranges for butter().

No error occurs because 'bandpass' is a valid btype for designing bandpass filters, and the cutoff frequencies [0.2, 0.5] satisfy 0 < 0.2 < 0.5 < 1 and are strictly increasing.

❓ visualization

advanced

Frequency response plot of Chebyshev Type I filter

Which option correctly plots the frequency response magnitude (in dB) of a 5th order Chebyshev Type I lowpass filter with 0.5 normalized cutoff and 0.5 dB ripple?

SciPy

import matplotlib.pyplot as plt
import numpy as np
from scipy.signal import cheby1, freqz
b, a = cheby1(5, 0.5, 0.5, btype='low')
w, h = freqz(b, a)
plt.plot(w / 3.1415, 20 * np.log10(abs(h)))
plt.xlabel('Normalized Frequency')
plt.ylabel('Magnitude (dB)')
plt.title('Frequency Response')
plt.grid(True)
plt.show()

A Plots frequency response with x-axis from 0 to 1 and y-axis showing magnitude in dB

B Plots frequency response with x-axis in Hz (0 to 5000) and y-axis linear magnitude

C Plots phase response instead of magnitude

D Raises NameError due to missing numpy import

💡 Hint

Check if all required libraries are imported before plotting.

The code uses np.log10 and np without importing numpy as np, causing NameError.

🚀 Application

expert

Choosing filter type for sharp cutoff with minimal ripple

You need to design a digital IIR filter with a very sharp cutoff and minimal passband ripple for a biomedical signal sampled at 1000 Hz. Which filter design and parameters below best meet this requirement?

A Use a 6th order Chebyshev Type I lowpass filter with 0.5 dB ripple and cutoff 100 Hz

B Use a 6th order Butterworth lowpass filter with cutoff 100 Hz

C Use a 6th order Chebyshev Type II lowpass filter with 40 dB stopband attenuation and cutoff 100 Hz

D Use a 6th order Elliptic lowpass filter with 0.5 dB ripple and 40 dB stopband attenuation

💡 Hint

Chebyshev Type II filters have flat passband and sharp cutoff with stopband ripple.

Chebyshev Type II filters have no ripple in passband and sharp cutoff, suitable for minimal passband ripple and sharp transition.

Practice

1. What is the main difference between a Butterworth and a Chebyshev IIR filter in scipy.signal?

easy

A. Butterworth filters are FIR, Chebyshev filters are IIR.

B. Butterworth filters have ripples, Chebyshev filters have a smooth response.

C. Butterworth filters are only for high-pass, Chebyshev only for low-pass.

D. Butterworth filters have a smooth frequency response, Chebyshev filters have ripples in the passband.

Solution

1. Step 1: Understand Butterworth filter characteristics

   Butterworth filters are designed to have a maximally flat frequency response in the passband, meaning no ripples.

2. Step 2: Understand Chebyshev filter characteristics

   Chebyshev filters allow ripples in the passband to achieve a sharper cutoff compared to Butterworth filters.

3. Final Answer:

   Butterworth filters have a smooth frequency response, Chebyshev filters have ripples in the passband. -> Option D

4. Quick Check:

   Butterworth smooth, Chebyshev ripples = D [OK]

Hint: Butterworth = smooth, Chebyshev = ripples [OK]

Common Mistakes:

• Confusing which filter has ripples
• Thinking Butterworth is FIR
• Assuming filter types are limited to passband types

2. Which of the following is the correct way to design a 4th order Butterworth lowpass filter with cutoff frequency 0.3 (normalized) using scipy.signal.butter?

easy

A. b, a = butter(0.3, 4, btype='low')

B. b, a = butter(4, 0.3, btype='low')

C. b, a = butter(4, 0.3, type='lowpass')

D. b, a = butter(order=4, cutoff=0.3, kind='low')

Solution

1. Step 1: Recall butter function signature

   The function signature is butter(N, Wn, btype='low', analog=False, output='ba'), where N is order and Wn is cutoff frequency.

2. Step 2: Check correct parameter usage

   b, a = butter(4, 0.3, btype='low') uses butter(4, 0.3, btype='low'), which matches the signature correctly.

3. Final Answer:

   b, a = butter(4, 0.3, btype='low') -> Option B

4. Quick Check:

   Order first, cutoff second, btype='low' = A [OK]

Hint: Order first, cutoff second, btype='low' in butter() [OK]

Common Mistakes:

• Swapping order and cutoff parameters
• Using wrong keyword names like 'type' or 'kind'
• Passing cutoff frequency as first argument

3. What will be the output shape of the coefficients when running this code?

from scipy.signal import cheby1
b, a = cheby1(3, 1, 0.4, btype='high')

medium

A. (3,), (3,)

B. (3, 3), (3, 3)

C. (4,), (4,)

D. (2,), (2,)

Solution

1. Step 1: Understand cheby1 output

   The function returns numerator (b) and denominator (a) coefficients arrays each of length N+1, where N is filter order.

2. Step 2: Calculate length for order 3

   For order 3, length is 3+1 = 4, so both b and a have shape (4,).

3. Final Answer:

   (4,), (4,) -> Option C

4. Quick Check:

   Length = order + 1 = 4 [OK]

Hint: Coefficients length = order + 1 [OK]

Common Mistakes:

• Assuming length equals order
• Expecting 2D arrays instead of 1D
• Confusing numerator and denominator shapes

4. Identify the error in this code snippet for designing a Chebyshev type I filter:

from scipy.signal import cheby1
b, a = cheby1(5, 0.5, 0.2, btype='bandpass')

medium

A. The cutoff frequency should be a list or tuple for bandpass, not a single float.

B. The ripple parameter is missing and required.

C. The order parameter should be a float, not integer.

D. The btype 'bandpass' is not supported by cheby1.

Solution

1. Step 1: Check bandpass cutoff frequency format

   For bandpass filters, cutoff frequency must be a sequence (list or tuple) of two frequencies, not a single float.

2. Step 2: Verify other parameters

   Order as integer and ripple (second parameter) are correctly provided; btype 'bandpass' is valid.

3. Final Answer:

   The cutoff frequency should be a list or tuple for bandpass, not a single float. -> Option A

4. Quick Check:

   Bandpass cutoff needs two frequencies = C [OK]

Hint: Bandpass cutoff must be two frequencies in a list/tuple [OK]

Common Mistakes:

• Passing single float for bandpass cutoff
• Confusing ripple parameter position
• Using unsupported btype values

5. You want to design a Chebyshev type I lowpass filter with order 6, 0.5 dB ripple, and cutoff frequency 0.25. Which code snippet correctly creates the filter and applies it to a signal x using scipy.signal.lfilter?

hard

A. from scipy.signal import cheby1, lfilter
b, a = cheby1(6, 0.5, 0.25, btype='low')
y = lfilter(b, a, x)

B. from scipy.signal import cheby1, lfilter
b, a = cheby1(6, 0.5, 0.25, btype='high')
y = lfilter(a, b, x)

C. from scipy.signal import butter, lfilter
b, a = butter(6, 0.25, btype='low')
y = lfilter(b, a, x)

D. from scipy.signal import cheby1, lfilter
b, a = cheby1(0.5, 6, 0.25, btype='low')
y = lfilter(b, a, x)

Solution

1. Step 1: Verify cheby1 parameters order

   Order is first parameter (6), ripple second (0.5), cutoff third (0.25), and btype='low' for lowpass.

2. Step 2: Check lfilter usage

   lfilter takes numerator coefficients (b), denominator coefficients (a), and signal (x) in that order.

3. Final Answer:

   from scipy.signal import cheby1, lfilter b, a = cheby1(6, 0.5, 0.25, btype='low') y = lfilter(b, a, x) -> Option A

4. Quick Check:

   Correct order and lfilter usage = A [OK]

Hint: Order, ripple, cutoff in cheby1; lfilter(b, a, x) [OK]

Common Mistakes:

• Swapping order and ripple parameters
• Using butter instead of cheby1
• Swapping b and a in lfilter