Applying filters (lfilter, sosfilt) in SciPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to apply a filter using lfilter.

SciPy

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

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

Ab

Ba

Csignal

D[1, 0, 0, 0, 0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using the denominator coefficients a as the first argument instead of numerator b.

Passing the signal as the first argument.

The lfilter function applies the filter defined by coefficients b and a to the input signal. Here, b is the numerator coefficient array.

2fill in blank

medium

Complete the code to apply a second-order sections filter using sosfilt.

SciPy

from scipy.signal import sosfilt
import numpy as np
sos = np.array([[1, 2, 1, 1, -0.5, 0.25]])
signal = [1, 2, 3, 4, 5]
filtered = sosfilt([1], signal)
print(filtered)

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

Asignal

Bsos

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

D[1, 0, 0, 0, 0]

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the signal as the first argument instead of the filter coefficients.

Using lfilter instead of sosfilt for second-order sections.

The sosfilt function applies a filter defined by second-order sections stored in sos to the input signal.

3fill in blank

hard

Fix the error in the code to correctly apply lfilter with numerator and denominator coefficients.

SciPy

from scipy.signal import lfilter
b = [0.1, 0.1, 0.1]
a = [1, -0.8]
signal = [1, 2, 3, 4, 5]
filtered = lfilter(b, [1], signal)
print(filtered)

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

Aa

Bb

Csignal

D[1]

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping numerator and denominator coefficients.

Passing the signal as the second argument.

The second argument to lfilter is the denominator coefficients a. This defines the recursive part of the filter.

4fill in blank

hard

Fill both blanks to create a dictionary of filtered signals for words longer than 3 characters.

SciPy

from scipy.signal import lfilter
b = [0.2, 0.2, 0.2, 0.2, 0.2]
a = [1]
words = ['data', 'ai', 'science', 'ml']
filtered_signals = {word: lfilter([1], [2], [1, 2, 3, 4, 5]) for word in words if len(word) > 3}
print(filtered_signals)

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

Ab

Ba

C[1]

Dsignal

Attempts:

3 left

💡 Hint

Common Mistakes

Using the signal or [1] as filter coefficients.

Swapping numerator and denominator coefficients.

To filter signals, lfilter needs numerator b and denominator a coefficients. Here, we use b and a for each word longer than 3 characters.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension filtering signals with condition on values.

SciPy

from scipy.signal import lfilter
b = [0.1, 0.1, 0.1]
a = [1, -0.8]
data = {'x': 0.5, 'y': 1.5, 'z': 2.5}
result = {k[1]: lfilter(b, a, [1, 2, 3]) for k, v in data.items() if v [2] 1 and k != [3]
print(result)

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

A.upper()

B>

C'y'

D.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using .upper() instead of .lower() for key modification.

Using '<' instead of '>' for filtering values.

Not quoting the key string in the condition.

The dictionary comprehension uses .lower() on keys, filters values greater than 1, and excludes key 'y'.

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