Fixed-point arithmetic for embedded DSP in Signal Processing - Time & Space Complexity

Time Complexity: Fixed-point arithmetic for embedded DSP

O(n)

Understanding Time Complexity

When working with fixed-point arithmetic in embedded DSP, it is important to understand how the time to perform calculations changes as the data size grows.

We want to know how the number of operations grows when processing more data points.

Scenario Under Consideration

Analyze the time complexity of the following fixed-point multiply-accumulate operation over an input array.

// Fixed-point multiply-accumulate on input array
int32_t fixed_point_mac(int16_t *input, int16_t *coeff, int n) {
    int32_t acc = 0;
    for (int i = 0; i < n; i++) {
        acc += (int32_t)input[i] * (int32_t)coeff[i];
    }
    return acc;
}
    

This code multiplies each input value by a coefficient and sums the results using fixed-point arithmetic.

Identify Repeating Operations

• Primary operation: The for-loop that multiplies and accumulates each pair of input and coefficient values.
• How many times: Exactly n times, where n is the size of the input arrays.

How Execution Grows With Input

Each additional data point adds one multiply and one add operation, so the total work grows directly with the number of data points.

Input Size (n)	Approx. Operations
10	10 multiply-accumulate steps
100	100 multiply-accumulate steps
1000	1000 multiply-accumulate steps

Pattern observation: The number of operations increases in a straight line as input size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the fixed-point multiply-accumulate grows directly in proportion to the number of data points.

Common Mistake

[X] Wrong: "The multiply-accumulate operation takes constant time regardless of input size because it is a simple calculation."

[OK] Correct: While one multiply-accumulate is quick, the code repeats it for every data point, so total time grows with the number of points.

Interview Connect

Understanding how loops over data affect time is key in embedded DSP tasks, showing you can reason about performance in real systems.

Self-Check

"What if we added a nested loop inside the multiply-accumulate to process pairs of coefficients for each input? How would the time complexity change?"

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