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)

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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)

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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'.