Fixed-point arithmetic for embedded DSP in Signal Processing

Introduction

Fixed-point arithmetic helps embedded devices do math fast using limited memory. It is useful when floating-point is too slow or unavailable.

When programming small devices like microcontrollers that process sound or signals.

When you need fast math but have no floating-point hardware.

When saving memory is important in your embedded system.

When working with real-time audio or sensor data on limited hardware.

When you want predictable, repeatable math results in your DSP code.

Syntax

Signal Processing

Qm.n format: number has m bits for integer part and n bits for fractional part.
Example: Q1.15 means 1 bit integer, 15 bits fraction.

Fixed-point numbers are stored as integers but represent fractions.

Scaling and shifting bits are used to do multiplication and division.

Examples

This shows how a fractional number is stored as an integer.

Signal Processing

Q1.15 format stores numbers like 0.5 as 16384 (because 0.5 * 2^15 = 16384).

Shifting right adjusts the result back to the fixed-point scale.

Signal Processing

To multiply two Q1.15 numbers: multiply integers, then shift right by 15 bits to keep scale.

Addition is simpler because scale is the same.

Signal Processing

Adding two Q1.15 numbers is just integer addition.

Sample Program

This code multiplies two fixed-point numbers in Q1.15 format and converts the result back to float to check correctness.

Signal Processing

def fixed_point_multiply(a, b, fractional_bits=15):
    product = a * b
    # Shift right to keep scale
    result = product >> fractional_bits
    return result

# Example: multiply 0.5 * 0.25 in Q1.15
# 0.5 in Q1.15 is int(0.5 * 2**15) = 16384
# 0.25 in Q1.15 is int(0.25 * 2**15) = 8192

a = 16384
b = 8192
result = fixed_point_multiply(a, b)

# Convert back to float for display
float_result = result / (2**15)
print(f"Fixed-point multiply result (int): {result}")
print(f"Fixed-point multiply result (float): {float_result}")

Important Notes

Be careful with overflow: fixed-point math can overflow if numbers get too big.

Always keep track of the number of fractional bits to interpret results correctly.

Fixed-point math is faster but less flexible than floating-point.

Summary

Fixed-point arithmetic stores fractions as scaled integers for fast math on small devices.

Multiplication requires shifting bits to keep the scale correct.

Use fixed-point when floating-point hardware is not available or too slow.

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