mac_5x9x8-Revised.md

 We have defined a KxMxN MAC unit as a device which takes K signed M-bit numbers ($a_0, a_1, ..., a_{K-1}$) and K signed N-bit numbers ($b_0, b_1, ..., b_{K-1}$), and computes $$a_0 \ast b_0 + a_1 \ast b_1 + ... + a_{K-1} \ast b_{K-1}$$
 
 For making an KxMxN MAC unit, the number of output bits will be
-$$\lceil\log_2(K * 2^M * 2^N)\rceil = \lceil\log_2 K\rceil + M + N$$
+$$B = \lceil\log_2(K * 2^M * 2^N)\rceil = \lceil\log_2 K\rceil + M + N$$
 
 For our specialized case of a 5x9x8 MAC unit, $\lceil\log_2 5\rceil + 9 + 8 = 20$ bits.
 
@@ -24,7 +24,33 @@ p3[4:0] = a[4:0] * b[3]
   P[8]    P[7]     P[6]     P[5]     P[4]     P[3]     P[2]     P[1]     P[0]
 ```
 
+Now, to sum K products together, these products must be sign-extended to B bits, then summed.
+Simply adding 1's to the constant addition correctly sign-extends the result to as many bits as needed.
 
+In this case, if this multiply was a part of a 5x4x5 MAC unit, $\lceil\log_2(5)\rceil= 3$ ones will need to be prepended to the constant term.
+
+Since there are no more sign extensions required, we can rewrite the entire 5x4x5 MAC unit as a single large array addition:
+
+```
+p00[4:0] = a0[4:0] * b0[0]
+p01[4:0] = a0[4:0] * b0[1]
+p02[4:0] = a0[4:0] * b0[2]
+p03[4:0] = a0[4:0] * b0[3]
+
+p00[4:0] = a0[4:0] * b0[0]
+p01[4:0] = a0[4:0] * b0[1]
+p02[4:0] = a0[4:0] * b0[2]
+p03[4:0] = a0[4:0] * b0[3]
+
+
+                                    ~p0[4]    p0[3]    p0[2]    p0[1]    p0[0]
+                           ~p0[4]    p0[3]    p0[2]    p0[1]    p0[0]
+                  ~p0[4]    p0[3]    p0[2]    p0[1]    p0[0]
+          p0[4]   ~p0[3]   ~p0[2]   ~p0[1]   ~p0[0]
+ + 1       0        0        0        0        1        0        0        0
+-------------------------------------------------------------------------------
+  P[8]    P[7]     P[6]     P[5]     P[4]     P[3]     P[2]     P[1]     P[0]
+```