Signal properties (amplitude, frequency, phase) in Signal Processing - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the amplitude of a sine wave.

Signal Processing

import numpy as np

time = np.linspace(0, 1, 500)
signal = 5 * np.sin(2 * np.pi * 10 * time)
amplitude = (np.max(signal) [1] np.min(signal)) / 2
print(amplitude)

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

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition instead of subtraction.

Multiplying max and min values.

Dividing max by min.

The amplitude of a sine wave can be found by (subtracting the minimum value from the maximum value and then dividing by 2).

2fill in blank

medium

Complete the code to calculate the period of a sine wave given the time array.

Signal Processing

import numpy as np

sampling_rate = 1000  # Hz
signal_freq = 50  # Hz
time = np.linspace(0, 1, sampling_rate)
signal = np.sin(2 * np.pi * signal_freq * time)
period = 1 / [1]
print(period)

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

Asampling_rate

Bnp.pi

Ctime

Dsignal_freq

Attempts:

3 left

💡 Hint

Common Mistakes

Using sampling rate instead of signal frequency.

Using pi instead of frequency.

Using the time array directly.

The period of a wave is the inverse of its frequency.

3fill in blank

hard

Fix the error in the code to correctly calculate the phase shift of a sine wave.

Signal Processing

import numpy as np

phase_shift = np.pi / 4
signal = np.sin(2 * np.pi * 5 * np.linspace(0, 1, 500) [1] phase_shift)
print(signal[:5])

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

A*

B-

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of addition.

Subtracting phase shift which shifts wave in opposite direction.

Dividing by phase shift.

The phase shift is added inside the sine function's argument to shift the wave horizontally.

4fill in blank

hard

Fill both blanks to create a dictionary mapping signal properties to their values.

Signal Processing

signal_properties = {"amplitude": [1], "frequency": [2]
print(signal_properties)

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

A5

B10

C50

Dnp.pi

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing amplitude and frequency values.

Using pi as frequency.

Using sampling rate as frequency.

The amplitude is 5 and the frequency is 50 Hz for this signal.

5fill in blank

hard

Fill all three blanks to create a dictionary with amplitude, frequency, and phase shift of a signal.

Signal Processing

signal_info = {"amplitude": [1], "frequency": [2], "phase": [3]
print(signal_info)

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

A7

B60

Cnp.pi / 3

Dnp.pi / 4

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing phase shift values.

Mixing amplitude and frequency values.

Using wrong constants for phase.

The amplitude is 7, frequency is 60 Hz, and phase shift is pi/3 radians.

Practice

1. Which property of a signal describes how tall or strong the wave is?

easy

A. Frequency

B. Amplitude

C. Phase

D. Wavelength

Solution

1. Step 1: Understand amplitude meaning

   Amplitude measures the height or strength of a wave, showing how strong the signal is.

2. Step 2: Compare with other properties

   Frequency is how often the wave repeats, phase shifts the start, wavelength is distance, not height.

3. Final Answer:

   Amplitude -> Option B

4. Quick Check:

   Amplitude = wave height [OK]

Hint: Amplitude means wave height or strength [OK]

Common Mistakes:

• Confusing amplitude with frequency
• Thinking phase affects height
• Mixing wavelength with amplitude

2. Which of the following is the correct way to represent a sine wave with amplitude 3 and frequency 5 Hz in Python using numpy?
t = np.linspace(0, 1, 500)
y = ?

easy

A. 3 + np.sin(2 * np.pi * 5 * t)

B. np.sin(3 * 2 * np.pi * 5 * t)

C. 3 * np.sin(2 * np.pi * 5 * t)

D. 5 * np.sin(2 * np.pi * 3 * t)

Solution

1. Step 1: Recall sine wave formula

   The sine wave formula is amplitude * sin(2π * frequency * time).

2. Step 2: Match formula to options

   3 * np.sin(2 * np.pi * 5 * t) matches amplitude 3 and frequency 5 correctly as 3 * sin(2π * 5 * t).

3. Final Answer:

   3 * np.sin(2 * np.pi * 5 * t) -> Option C

4. Quick Check:

   Amplitude * sin(2π * frequency * t) [OK]

Hint: Amplitude multiplies sine, frequency inside sine argument [OK]

Common Mistakes:

• Multiplying frequency by amplitude inside sine
• Adding amplitude instead of multiplying
• Swapping amplitude and frequency

3. Given the code:
import numpy as np
t = np.linspace(0, 1, 5)
y = 2 * np.sin(2 * np.pi * 1 * t + np.pi/2)
print(np.round(y, 2))
What is the output?

medium

A. [ 2. 0. -2. 0. 2. ]

B. [ 2. 1.41 0. -1.41 -2. ]

C. [ 0. 1.41 2. 1.41 0. ]

D. [ 1.41 2. 1.41 0. -1.41]

Solution

1. Step 1: Calculate time points

   t = [0, 0.25, 0.5, 0.75, 1]

2. Step 2: Compute y values

   y = 2 * sin(2π * 1 * t + π/2) = 2 * cos(2π * t) due to phase shift π/2.
   Values: 2*cos(0)=2, 2*cos(π/2)=0, 2*cos(π)=-2, 2*cos(3π/2)=0, 2*cos(2π)=2
   np.round(y, 2) prints [ 2. 0. -2. 0. 2. ]

3. Final Answer:

   [ 2. 0. -2. 0. 2. ] -> Option A

4. Quick Check:

   Phase π/2 shifts sine to cosine [OK]

Hint: Phase π/2 shifts sine to cosine wave [OK]

Common Mistakes:

• Ignoring phase shift
• Rounding errors
• Mixing sine and cosine values

4. Identify the error in this code snippet that tries to create a sine wave with amplitude 4 and frequency 2 Hz:
import numpy as np
t = np.linspace(0, 1, 100)
y = 4 * np.sin(2 * np.pi * t) * 2

medium

A. The time vector t should be integers, not floats

B. Amplitude should be added, not multiplied

C. np.linspace should use 0 to 2 instead of 0 to 1

D. The frequency multiplier is in the wrong place; it should be inside the sine argument as 2 * np.pi * 2 * t

Solution

1. Step 1: Analyze sine argument order

   The sine argument is 2π * t (frequency 1 Hz), but *2 is outside sin(), incorrectly scaling the amplitude instead of frequency.

2. Step 2: Correct placement for frequency

   Frequency multiplier must be inside sine: 2 * np.pi * 2 * t.

3. Final Answer:

   The frequency multiplier is in the wrong place; it should be inside the sine argument as 2 * np.pi * 2 * t -> Option D

4. Quick Check:

   Frequency must multiply time inside sine [OK]

Hint: Write frequency right after 2π inside sine [OK]

Common Mistakes:

• Multiplying amplitude incorrectly
• Using wrong time vector type
• Changing linspace range unnecessarily

5. You have a signal defined as y = 5 * np.sin(2 * np.pi * 3 * t + π/4). If you want to double the frequency and halve the amplitude while keeping the phase shift the same, which expression is correct?

hard

A. y = 2.5 * np.sin(2 * np.pi * 6 * t + π/4)

B. y = 10 * np.sin(2 * np.pi * 1.5 * t + π/4)

C. y = 2.5 * np.sin(2 * np.pi * 3 * t + π/2)

D. y = 5 * np.sin(2 * np.pi * 6 * t)

Solution

1. Step 1: Adjust amplitude and frequency

   Original amplitude is 5, halved is 2.5. Original frequency is 3, doubled is 6.

2. Step 2: Keep phase shift same

   Phase shift π/4 remains unchanged in the new expression.

3. Final Answer:

   y = 2.5 * np.sin(2 * np.pi * 6 * t + π/4) -> Option A

4. Quick Check:

   Amplitude halved, frequency doubled, phase same [OK]

Hint: Multiply frequency by 2, amplitude by 0.5, keep phase [OK]

Common Mistakes:

• Changing phase shift incorrectly
• Multiplying amplitude instead of halving
• Swapping frequency and amplitude values