Even and odd signal decomposition in Signal Processing - Mini Project: Build & Apply

Even and Odd Signal Decomposition

📖 Scenario: In signal processing, any signal can be split into two parts: an even part and an odd part. This helps us understand the signal better and analyze it in simpler ways.Imagine you have a list of signal values measured at different time points, both positive and negative. We want to break this signal into its even and odd components.

🎯 Goal: You will create a signal as a dictionary with time points as keys and signal values as values. Then, you will write code to find the even and odd parts of this signal using the formulas:Even part at time n = (x[n] + x[-n]) / 2Odd part at time n = (x[n] - x[-n]) / 2Finally, you will print the even and odd parts as dictionaries.

📋 What You'll Learn

Create a dictionary called signal with time points as keys and signal values as values

Create a list called time_points containing all time points from the signal

Use a for loop with variable n to iterate over time_points

Calculate the even part using (signal[n] + signal[-n]) / 2

Calculate the odd part using (signal[n] - signal[-n]) / 2

Store the even parts in a dictionary called even_part

Store the odd parts in a dictionary called odd_part

Print even_part and odd_part

💡 Why This Matters

🌍 Real World

Even and odd decomposition helps engineers analyze signals in communications, audio processing, and control systems by simplifying complex signals into symmetric parts.

💼 Career

Understanding signal decomposition is useful for roles in signal processing, data analysis, and electrical engineering where signal behavior needs to be studied and manipulated.

1

Create the signal dictionary

Create a dictionary called signal with these exact entries: -2: 3, -1: 1, 0: 4, 1: 2, 2: 0

Signal Processing

# Create the signal dictionary with time points and values
# Your code here

Hint

Use curly braces {} to create a dictionary with the given key-value pairs.

2

Create the list of time points

Create a list called time_points containing the keys of signal sorted in ascending order

Signal Processing

signal = {-2: 3, -1: 1, 0: 4, 1: 2, 2: 0}
# Create the list of time points sorted
# Your code here

Hint

Use sorted() on signal.keys() to get the sorted list of time points.

3

Calculate even and odd parts

Create empty dictionaries called even_part and odd_part. Use a for loop with variable n to iterate over time_points. Inside the loop, calculate the even part as (signal[n] + signal[-n]) / 2 and the odd part as (signal[n] - signal[-n]) / 2. Store these values in even_part[n] and odd_part[n] respectively.

Signal Processing

signal = {-2: 3, -1: 1, 0: 4, 1: 2, 2: 0}
time_points = sorted(signal.keys())
# Create even_part and odd_part dictionaries
# Use a for loop to calculate even and odd parts
# Your code here

Hint

Remember to create empty dictionaries before the loop. Use the formulas given to calculate even and odd parts inside the loop.

4

Print the even and odd parts

Print the even_part dictionary and then print the odd_part dictionary.

Signal Processing

signal = {-2: 3, -1: 1, 0: 4, 1: 2, 2: 0}
time_points = sorted(signal.keys())
even_part = {}
odd_part = {}
for n in time_points:
    even_part[n] = (signal[n] + signal[-n]) / 2
    odd_part[n] = (signal[n] - signal[-n]) / 2
# Print even_part and odd_part
# Your code here

Hint

Use two print statements to show the even and odd parts dictionaries.

Practice

1.

What is the main property of an even signal x(t) in signal processing?

easy

A. It satisfies x(t) = -x(-t) for all t.

B. It has no negative values.

C. It is always zero at t=0.

D. It satisfies x(t) = x(-t) for all t.

Solution

1. Step 1: Recall definition of even signal

   An even signal is symmetric about zero, meaning the value at t equals the value at -t.

2. Step 2: Apply the property to the signal

   This means x(t) = x(-t) for every time t.

3. Final Answer:

   It satisfies x(t) = x(-t) for all t. -> Option D

4. Quick Check:

   Even signal = x(t) = x(-t) [OK]

Hint: Even signals mirror values at positive and negative times [OK]

Common Mistakes:

• Confusing even with odd signal definitions
• Thinking even signals must be zero at t=0
• Assuming even signals cannot have negative values

2.

Which of the following is the correct formula to find the odd part x_o(t) of a signal x(t)?

easy

A. x_o(t) = (x(t) + x(-t)) / 2

B. x_o(t) = (x(t) - x(-t)) / 2

C. x_o(t) = x(t) * x(-t)

D. x_o(t) = x(t) / x(-t)

Solution

1. Step 1: Recall formula for odd part of a signal

   The odd part is defined as half the difference between the signal and its time-reversed version: x_o(t) = (x(t) - x(-t)) / 2.

2. Step 2: Verify the formula matches the odd signal property

   This formula ensures x_o(-t) = -x_o(t), which is the odd signal condition.

3. Final Answer:

   x_o(t) = (x(t) - x(-t)) / 2 -> Option B

4. Quick Check:

   Odd part = (x(t) - x(-t))/2 [OK]

Hint: Odd part uses subtraction of reversed signal, then divide by 2 [OK]

Common Mistakes:

• Using addition instead of subtraction for odd part
• Multiplying signals instead of subtracting
• Dividing by wrong number or not dividing at all

3.

Given a discrete signal x = [2, 3, 4, 3, 2] indexed from -2 to 2, what is the even part x_e of this signal?

Use the formula x_e[n] = (x[n] + x[-n]) / 2.

medium

A. [2, 3, 4, 3, 2]

B. [2, 3.5, 4, 3.5, 2]

C. [2, 3, 4, 4, 2]

D. [2, 3.5, 4, 3, 2]

Solution

1. Step 1: Identify signal values and indices

   Signal x indexed from -2 to 2 is: x[-2]=2, x[-1]=3, x[0]=4, x[1]=3, x[2]=2.

2. Step 2: Calculate even part using formula

   Calculate each x_e[n] = (x[n] + x[-n]) / 2: - n=-2: (x[-2]+x[2])/2 = (2+2)/2=2 - n=-1: (x[-1]+x[1])/2 = (3+3)/2=3 - n=0: (x[0]+x[0])/2 = (4+4)/2=4 - n=1: (x[1]+x[-1])/2 = (3+3)/2=3 - n=2: (x[2]+x[-2])/2 = (2+2)/2=2 But note the signal is symmetric, so values are same as original except at n=1 and n=-1 where average is 3.

3. Final Answer:

   [2, 3, 4, 3, 2] -> Option A

4. Quick Check:

   Even part averages signal and its mirror [OK]

Hint: Average signal values at n and -n for even part [OK]

Common Mistakes:

• Not averaging values correctly
• Mixing indices n and -n
• Using odd part formula instead

4.

Consider the Python code below to compute the odd part of a signal x:

def odd_part(x):
    return [(x[i] - x[-i]) / 2 for i in range(len(x))]

What is the main error in this code?

medium

A. The range should start from 1, not 0.

B. The formula uses addition instead of subtraction.

C. Indexing with negative indices -i is incorrect for this context.

D. The division by 2 is missing.

Solution

1. Step 1: Understand indexing in Python lists

   Python negative indices count from the end, so -i does not correspond to -n time index but to list end positions.

2. Step 2: Identify correct indexing for signal symmetry

   For signal indexed from negative to positive, indexing must map i to -i properly, which this code does not do because list indices are 0-based and -i accesses wrong elements.

3. Final Answer:

   Indexing with negative indices -i is incorrect for this context. -> Option C

4. Quick Check:

   Python negative indices ≠ signal negative time [OK]

Hint: Python negative indices count from list end, not negative time [OK]

Common Mistakes:

• Assuming -i indexes negative time correctly
• Ignoring zero-based indexing
• Confusing addition/subtraction in formula

5.

You have a signal x(t) = t^3 + 2t. Decompose it into even and odd parts. What is the even part x_e(t)?

hard

A. x_e(t) = 0

B. x_e(t) = t^3

C. x_e(t) = 2t

D. x_e(t) = t^3 + 2t

Solution

1. Step 1: Identify even and odd nature of terms

   The term t^3 is an odd function because (-t)^3 = -t^3. The term 2t is also odd because 2(-t) = -2t.

2. Step 2: Calculate even part using formula

   Even part is x_e(t) = (x(t) + x(-t)) / 2. Substitute: x_e(t) = (t^3 + 2t + (-t)^3 + 2(-t)) / 2 = (t^3 + 2t - t^3 - 2t) / 2 = 0.

3. Final Answer:

   x_e(t) = 0 -> Option A

4. Quick Check:

   Sum of odd functions is odd, so even part is zero [OK]

Hint: Sum of odd functions has zero even part [OK]

Common Mistakes:

• Assuming polynomial terms are even by default
• Mixing odd and even function properties
• Forgetting to apply formula correctly