Applying filters (lfilter, sosfilt) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of lfilter with simple coefficients

What is the output array after applying lfilter with given coefficients on the input signal?

SciPy

from scipy.signal import lfilter
import numpy as np

b = [0.5, 0.5]
a = [1]
x = np.array([1, 2, 3, 4, 5])
y = lfilter(b, a, x)
print(y)

A [0.5 1.5 2.5 3.5 4.5]

B [0.5 1. 1.5 2. 2.5]

C [1. 1.5 2.5 3.5 4.5]

D [0.5 1. 2. 3. 4. ]

💡 Hint

Remember that lfilter applies the filter coefficients as a convolution-like operation starting from the first element.

The filter coefficients b = [0.5, 0.5] average the current and previous input values. The output at each step is 0.5 * x[n] + 0.5 * x[n-1], with x[-1] assumed zero at start.

❓ data_output

intermediate

Number of output samples from sosfilt

Given a second-order sections filter and an input signal of length 8, how many samples does sosfilt return?

SciPy

from scipy.signal import sosfilt
import numpy as np

sos = np.array([[1, 0, -1, 1, -1, 0]])
x = np.arange(8)
y = sosfilt(sos, x)
print(len(y))

A 7

B 8

C 9

D 6

💡 Hint

The output length of sosfilt matches the input length.

sosfilt applies the filter to the input signal and returns an output array of the same length as the input.

🔧 Debug

advanced

Identify the error in lfilter usage

What error will this code raise when run?

SciPy

from scipy.signal import lfilter
b = [1, -1]
a = [1, -0.9]
x = [1, 2, 3]
y = lfilter(b, a)
print(y)

A TypeError: lfilter() missing 1 required positional argument: 'x'

B ValueError: Input arrays must have the same length

C IndexError: list index out of range

D No error, prints filtered output

💡 Hint

Check the function call arguments carefully.

The lfilter function requires three arguments: b, a, and the input signal x. Omitting x causes a TypeError.

🧠 Conceptual

advanced

Effect of filter coefficients on output shape

Which statement about the output length of lfilter and sosfilt is correct?

A <code>lfilter</code> output length is always shorter than input length; <code>sosfilt</code> output length equals input length

B <code>lfilter</code> output length is always longer than input length; <code>sosfilt</code> output length equals input length

C <code>lfilter</code> output length equals input length; <code>sosfilt</code> output length is always longer

D Both <code>lfilter</code> and <code>sosfilt</code> output lengths equal input length

💡 Hint

Think about how these filters process signals sample by sample.

Both lfilter and sosfilt produce output arrays with the same length as the input signal.

🚀 Application

expert

Filter a noisy signal and identify the filtered output mean

Given a noisy signal x and a low-pass filter defined by second-order sections sos, what is the mean of the filtered output y?

SciPy

import numpy as np
from scipy.signal import sosfilt

np.random.seed(0)
x = np.sin(np.linspace(0, 2*np.pi, 100)) + 0.5 * np.random.randn(100)
sos = np.array([[0.2929, 0.5858, 0.2929, 1.0000, -0.0000, 0.1716]])
y = sosfilt(sos, x)
print(round(np.mean(y), 3))

A -0.005

B -0.050

C 0.002

D 0.050

💡 Hint

The filter smooths the signal, so the mean should be close to zero.

The filtered signal mean is close to zero because the low-pass filter removes noise but preserves the sine wave's zero-centered nature.

Practice

1. What is the main purpose of using lfilter in signal processing with SciPy?

easy

A. To generate random noise signals

B. To apply a digital filter using numerator and denominator coefficients

C. To visualize the frequency response of a filter

D. To compute the Fourier transform of a signal

Solution

1. Step 1: Understand the role of lfilter

   lfilter applies a digital filter to a signal using given numerator (b) and denominator (a) coefficients.

2. Step 2: Compare with other options

   Options A, B, and D describe different signal processing tasks unrelated to filtering with coefficients.

3. Final Answer:

   To apply a digital filter using numerator and denominator coefficients -> Option B

4. Quick Check:

   lfilter = apply filter with b, a coefficients [OK]

Hint: Remember: lfilter uses b, a coefficients to filter signals [OK]

Common Mistakes:

• Confusing lfilter with Fourier transform functions
• Thinking lfilter generates signals instead of filtering
• Mixing up lfilter with visualization tools

2. Which of the following is the correct way to import and use lfilter from SciPy to filter a signal array named data with coefficients b and a?

easy

A. from scipy.signal import lfilter; filtered = lfilter(data, b, a)

B. import scipy.signal as signal; filtered = signal.lfilter(data, b, a)

C. from scipy.signal import lfilter; filtered = lfilter(b, a, data)

D. import lfilter from scipy.signal; filtered = lfilter(b, a, data)

Solution

1. Step 1: Recall the correct import syntax

   The correct import is from scipy.signal import lfilter.

2. Step 2: Check the function call order

   lfilter takes arguments in order: numerator coefficients (b), denominator coefficients (a), then the signal (data).

3. Final Answer:

   from scipy.signal import lfilter; filtered = lfilter(b, a, data) -> Option C

4. Quick Check:

   Import and call lfilter(b, a, data) [OK]

Hint: Import lfilter then call with (b, a, data) order [OK]

Common Mistakes:

• Swapping the order of arguments in lfilter
• Incorrect import syntax causing errors
• Passing data as first argument instead of third

3. Given the code below, what will be the output of filtered_signal?

from scipy.signal import lfilter
b = [0.2, 0.2, 0.2, 0.2, 0.2]
a = [1]
data = [1, 2, 3, 4, 5]
filtered_signal = lfilter(b, a, data)
print(list(filtered_signal))

medium

A. [0.2, 0.6, 1.2, 2.0, 3.0]

B. [1, 2, 3, 4, 5]

C. [0.2, 0.2, 0.2, 0.2, 0.2]

D. [0, 0, 0, 0, 0]

Solution

1. Step 1: Understand the filter coefficients

   The numerator coefficients b average the last 5 samples equally (each 0.2), denominator a is 1 (no feedback).

2. Step 2: Calculate filtered values step-by-step

   Filtered signal is moving average: - y[0] = 0.2*1 = 0.2 - y[1] = 0.2*1 + 0.2*2 = 0.6 - y[2] = 0.2*1 + 0.2*2 + 0.2*3 = 1.2 - y[3] = 0.2*1 + 0.2*2 + 0.2*3 + 0.2*4 = 2.0 - y[4] = 0.2*1 + 0.2*2 + 0.2*3 + 0.2*4 + 0.2*5 = 3.0

3. Final Answer:

   [0.2, 0.6, 1.2, 2.0, 3.0] -> Option A

4. Quick Check:

   Moving average filter output = [0.2, 0.6, 1.2, 2.0, 3.0] [OK]

Hint: Sum weighted inputs by b coefficients for each output [OK]

Common Mistakes:

• Ignoring the filter coefficients and outputting original data
• Confusing denominator coefficients effect
• Calculating only single sample instead of cumulative

4. What is wrong with the following code snippet that attempts to filter a signal using sosfilt?

from scipy.signal import sosfilt
sos = [[0.2929, 0.5858, 0.2929, 1, -0.0000, 0.1716]]
data = [1, 2, 3, 4, 5]
filtered = sosfilt(sos, data)
print(filtered)

medium

A. The sos array should have denominator coefficients in positions 3, 4, 5, but here the second denominator coefficient is zero, which is invalid

B. The denominator coefficients in sos must be normalized; here they are not

C. The sos array should be a 2D array with float values, but the given sos is a list of lists which is invalid

D. The sos array must have shape (n_sections, 6), but here it has shape (1, 6) which is correct

Solution

1. Step 1: Check the sos array format

   The sos array must be a 2D array with each row having 6 coefficients: [b0, b1, b2, a0, a1, a2].

2. Step 2: Validate denominator coefficients

   Denominator coefficients (a0, a1, a2) must be valid. Here, a0=1 (correct), a1=-0.0000 (effectively zero), a2=0.1716. Having a1 near zero is unusual and can cause instability or errors.

3. Final Answer:

   The sos array should have denominator coefficients in positions 3, 4, 5, but here the second denominator coefficient is zero, which is invalid -> Option A

4. Quick Check:

   sos denominator coefficients must be valid and non-zero [OK]

Hint: Check sos denominator coefficients are valid and non-zero [OK]

Common Mistakes:

• Assuming list of lists is invalid input for sos
• Ignoring denominator coefficients causing filter errors
• Confusing numerator and denominator positions in sos

5. You want to apply a stable Butterworth low-pass filter of order 4 with cutoff frequency 0.3 (normalized) to a noisy signal data. Which approach is best to ensure numerical stability and why? A) Use lfilter with numerator and denominator coefficients from butter. B) Use sosfilt with second-order sections from butter. C) Use lfilter with only numerator coefficients. D) Use sosfilt with random coefficients.

hard

A. Use lfilter with numerator and denominator coefficients from butter

B. Use sosfilt with random coefficients

C. Use lfilter with only numerator coefficients ignoring denominator

D. Use sosfilt with second-order sections from butter for better stability

Solution

1. Step 1: Understand filter stability issues

   High order filters can be unstable when applied directly with numerator and denominator coefficients due to numerical errors.

2. Step 2: Recognize sosfilt advantage

   Using second-order sections (sos) breaks the filter into smaller stable parts, improving numerical stability.

3. Step 3: Evaluate options

   Use sosfilt with second-order sections from butter for better stability (recommended stable method). Options C and D are incorrect or nonsensical.

4. Final Answer:

   Use sosfilt with second-order sections from butter for better stability -> Option D

5. Quick Check:

   Use sosfilt with sos for stable filtering [OK]

Hint: Use sosfilt with butter sos for stable high-order filters [OK]

Common Mistakes:

• Using lfilter directly for high order filters causing instability
• Ignoring denominator coefficients in filtering
• Using random coefficients causing wrong results