1. Implementing real-time convolution
Circular buffer
Implementing real-time convolution

2. Problem Statement
Implement the 4-th order FIR filter shown in Figure 3.2 on page 28.
n = 4
x[] is the buffer for the input
b[] is the coefficients vector
the newest input will be multiplied by b[0] while the oldest will be multiplied by b[4]
y is the output to the DAC
sn denotes the value of the n-th input sample

3. Initially
The system is relaxed, that is, the input buffer is filled with 0.
x[0] = x[1] = … = x[4] = 0
We need an index idx to denote where is the latest input.
Initially, idx = 0. that means x[0] will hold the current input at t = 0. Note that’s different from what we did in problem 2 where x[n] ALWAYS stores the latest input.

4. t = 0 idx = 0 x[] 1 2 3 4 s0 b[] b[0] b[1] b[2] b[3] b[4]1 2 3 4 s0
b[]
b[0]
b[1]
b[2]
b[3]
b[4]
y = x[idx] * b[0] = x[0] * b[0]

5. t = 1 idx= 0+1=1 x[] 1 2 3 4 s0 s1 b[] b[0] b[1] b[2] b[3] b[4]1 2 3 4 s0 s1
b[]
b[0]
b[1]
b[2]
b[3]
b[4]
y = x[idx] * b[0] + x[idx-1] * b[1] = x[1] * b[0] + x[0] * b[1]

6. t = 4 idx = 3+1 = 4 x[] 1 2 3 4 s0 s1 s2 s3 s4 b[] b[0] b[1] b[2] b[3]1 2 3 4 s0 s1 s2 s3 s4
b[]
b[0]
b[1]
b[2]
b[3]
b[4]
y = x[i] * b[0] + x[i-1] * b[1] + x[i-2] * b[2] + x[i-3] * b[3] + x[i-4] * b[4]
= x[4] * b[0] + x[3] * b[1] + x[2] * b[2] + x[1] * b[3] + x[0] * b[4]

7. t = 5 idx = 4 +1 = 5 = 0 x[] 1 2 3 4 s5 s1 s2 s3 s4 b[] b[0] b[1] b[2]1 2 3 4 s5 s1 s2 s3 s4
b[]
b[0]
b[1]
b[2]
b[3]
b[4]
y = x[idx] * b[0] + x[idx-1] * b[1] + x[idx-2] * b[2] + x[idx-3] * b[3]
+ x[idx-4] * b[4]
= x[0] * b[0] + x[-1] * b[1] + x[-2] * b[2] + x[-3] * b[3] + x[-4] * b[4]
= x[0] * b[0] + x[4] * b[1] + x[3] * b[2] + x[2] * b[3] + x[1] * b[4]

8. Mod operator
b[0] should be always multiplied by the latest data while b[4] times the oldest one.
Also notice the negative indexes in the equation. Since it’s a “circular” buffer, x[-1] stands for x[4]
But how to mathematically convert -1 to 4?
Mod operation: (k + n) mod n = k mod n
(3+5) mod 5 = 3 mod 5 =3
(-1+5) mod 5 = 4 mod 5 = 4
In C/C++, % is the mod operator

9. Solution #define N 4 int idx = 0; int i; float y;
Interrupt void McBSP_Rx_ISR() { …
idx = (idx+1) % (N+1);
y = 0.0;
for (i=0; i<=N;i++)
y = y + x[(idx-i+N+1)%(N+1)] * b[i]; }