Fixed-point arithmetic for embedded DSP in Signal Processing - Mini Project: Build & Apply

Fixed-point Arithmetic for Embedded DSP

📖 Scenario: You are working on a small embedded digital signal processor (DSP) that cannot handle floating-point numbers efficiently. Instead, it uses fixed-point arithmetic to process audio signals. Fixed-point numbers represent real numbers using integers scaled by a fixed factor.For example, if we choose a scale factor of 1000, the number 1.234 is stored as 1234 (an integer). Arithmetic operations must consider this scaling.

🎯 Goal: You will create a simple fixed-point arithmetic system. You will store numbers as integers scaled by 1000, add two fixed-point numbers, and then convert the result back to a readable decimal number.

📋 What You'll Learn

Create a dictionary called fixed_point_numbers with two entries: 'num1' with value 1234 and 'num2' with value 5678.

Create a variable called scale_factor and set it to 1000.

Add the two fixed-point numbers from fixed_point_numbers and store the result in a variable called sum_fixed.

Convert sum_fixed back to a floating-point number by dividing by scale_factor and store it in sum_float.

Print the value of sum_float.

💡 Why This Matters

🌍 Real World

Embedded DSPs in audio devices often use fixed-point arithmetic because they have limited hardware for floating-point calculations. This project shows how to handle such numbers in software.

💼 Career

Understanding fixed-point arithmetic is important for engineers working on embedded systems, audio processing, and real-time signal processing where performance and hardware constraints matter.

1

Create fixed-point numbers dictionary

Create a dictionary called fixed_point_numbers with these exact entries: 'num1': 1234 and 'num2': 5678.

Signal Processing

# Create the fixed_point_numbers dictionary with 'num1' and 'num2'
# Your code here

Hint

Use curly braces {} to create a dictionary with keys 'num1' and 'num2' and their integer values.

2

Set the scale factor

Create a variable called scale_factor and set it to 1000.

Signal Processing

fixed_point_numbers = {'num1': 1234, 'num2': 5678}
# Create the scale_factor variable and set it to 1000
# Your code here

Hint

Just assign the number 1000 to the variable scale_factor.

3

Add fixed-point numbers

Add the two fixed-point numbers from fixed_point_numbers using fixed_point_numbers['num1'] and fixed_point_numbers['num2']. Store the sum in a variable called sum_fixed. Then convert sum_fixed back to a floating-point number by dividing by scale_factor and store it in sum_float.

Signal Processing

fixed_point_numbers = {'num1': 1234, 'num2': 5678}
scale_factor = 1000
# Add the fixed-point numbers and convert to float
# Your code here

Hint

Use the dictionary keys to get the values, add them, and then divide by scale_factor to get the float.

4

Print the floating-point sum

Print the value of sum_float.

Signal Processing

fixed_point_numbers = {'num1': 1234, 'num2': 5678}
scale_factor = 1000
sum_fixed = fixed_point_numbers['num1'] + fixed_point_numbers['num2']
sum_float = sum_fixed / scale_factor
# Print the sum_float value
# Your code here

Hint

Use print(sum_float) to show the result.

Practice

1. What is the main reason to use fixed-point arithmetic in embedded DSP systems?

easy

A. To increase the precision beyond floating-point numbers

B. To perform fast math on devices without floating-point hardware

C. To simplify programming by avoiding scaling

D. To use more memory for storing numbers

Solution

1. Step 1: Understand fixed-point arithmetic purpose

   Fixed-point arithmetic is used to perform calculations efficiently on devices that lack floating-point hardware.

2. Step 2: Compare options with this purpose

   To perform fast math on devices without floating-point hardware correctly states the main reason: fast math on devices without floating-point units. Other options are incorrect because fixed-point does not increase precision beyond floating-point, does not simplify programming by avoiding scaling, and does not use more memory.

3. Final Answer:

   To perform fast math on devices without floating-point hardware -> Option B

4. Quick Check:

   Fixed-point speeds math on limited hardware = A [OK]

Hint: Fixed-point is for speed on limited hardware [OK]

Common Mistakes:

• Thinking fixed-point increases precision beyond floating-point
• Assuming fixed-point uses more memory
• Believing fixed-point avoids scaling complexity

2. Which of the following is the correct way to represent a fixed-point number with 8 fractional bits in C?

easy

A. int16_t fixed = value << 8;

B. int16_t fixed = value >> 8;

C. int16_t fixed = value * 8;

D. int16_t fixed = value / 8;

Solution

1. Step 1: Understand fixed-point scaling

   For 8 fractional bits, multiply the integer value by 2^8 (which is shifting left by 8) to convert to fixed-point.

2. Step 2: Match with options

   int16_t fixed = value << 8; shifts left by 8 bits, correctly scaling the value. int16_t fixed = value >> 8; shifts right (wrong direction), A divides by 8 (wrong), C multiplies by 8 (not 256).

3. Final Answer:

   int16_t fixed = value << 8; -> Option A

4. Quick Check:

   Shift left by fractional bits to scale = D [OK]

Hint: Shift left by fractional bits to convert to fixed-point [OK]

Common Mistakes:

• Shifting right instead of left
• Multiplying/dividing by wrong powers of two
• Confusing fractional bits with decimal places

3. Given two fixed-point numbers with 8 fractional bits: a = 0x0100 (1.0) and b = 0x0200 (2.0), what is the fixed-point result of c = (a * b) >> 8;?

medium

A. 0x0400 (4.0)

B. 0x0300 (3.0)

C. 0x0200 (2.0)

D. 0x0100 (1.0)

Solution

1. Step 1: Multiply fixed-point values

   a * b = 0x0100 * 0x0200 = 256 * 512 = 131072 (decimal)

2. Step 2: Adjust scale by shifting right 8 bits

   c = 131072 >> 8 = 131072 / 256 = 512 decimal = 0x0200

3. Step 3: Recalculate carefully

   Wait, 0x0100 = 256 decimal, 0x0200 = 512 decimal. Multiplying: 256 * 512 = 131072 decimal. Shifting right 8 bits divides by 256: 131072 / 256 = 512 decimal = 0x0200. So c = 0x0200 (2.0).

4. Final Answer:

   0x0200 (2.0) -> Option C

5. Quick Check:

   (1.0 * 2.0) = 2.0 fixed-point = 0x0200 [OK]

Hint: Multiply then shift right by fractional bits to rescale [OK]

Common Mistakes:

• Forgetting to shift right after multiplication
• Misinterpreting hex values as decimal
• Incorrect bit shift amount

4. Identify the error in this fixed-point multiplication code snippet with 8 fractional bits:

int16_t a = 0x0100; // 1.0
int16_t b = 0x0200; // 2.0
int16_t c = a * b;

medium

A. Missing right shift to rescale after multiplication

B. Variables should be float, not int16_t

C. Should shift left before multiplication

D. No error, code is correct

Solution

1. Step 1: Understand fixed-point multiplication scaling

   Multiplying two fixed-point numbers doubles the fractional bits, so the result must be shifted right by the fractional bit count to rescale.

2. Step 2: Analyze given code

   The code multiplies a and b but does not shift right by 8 bits, so the scale is incorrect.

3. Final Answer:

   Missing right shift to rescale after multiplication -> Option A

4. Quick Check:

   Shift right after multiply to fix scale = C [OK]

Hint: Always shift right after fixed-point multiply [OK]

Common Mistakes:

• Not shifting right after multiplication
• Using float instead of fixed-point on embedded
• Shifting left instead of right

5. You have a fixed-point format with 12 fractional bits. You want to multiply two numbers and keep the result in the same format without overflow. Which approach is best?

hard

A. Multiply, then shift left by 12 bits, and store in a 16-bit integer

B. Shift right by 12 bits before multiplying, store in 16-bit integer

C. Multiply without shifting, store in 16-bit integer

D. Multiply, then shift right by 12 bits, and store in a 32-bit integer

Solution

1. Step 1: Understand fixed-point multiplication with 12 fractional bits

   Multiplying two fixed-point numbers doubles fractional bits (24 bits), so shifting right by 12 bits rescales back to 12 fractional bits.

2. Step 2: Prevent overflow by using wider integer

   Multiplying two 16-bit numbers can overflow 16 bits, so use a 32-bit integer to hold the intermediate product safely.

3. Final Answer:

   Multiply, then shift right by 12 bits, and store in a 32-bit integer -> Option D

4. Quick Check:

   Shift right after multiply and use wider int = B [OK]

Hint: Use wider int and shift right after multiply [OK]

Common Mistakes:

• Not using wider integer causing overflow
• Shifting left instead of right after multiply
• Shifting before multiply losing precision