ddi_compute_wrpll: Rework the logic around r2 and n2 a bit
authorDamien Lespiau <damien.lespiau@intel.com>
Thu, 2 May 2013 15:33:56 +0000 (16:33 +0100)
committerDamien Lespiau <damien.lespiau@intel.com>
Fri, 10 May 2013 14:26:42 +0000 (15:26 +0100)
Let's not use the 2K variants of the frequencies it does not help
in anything here and the explanations are hopefuly more understandable
this way.

On top of that, I noticed that we can just compute the desired min/max
boundaries for r2 and n2, so use that instead of the two tests to
discard out of range values.

Signed-off-by: Damien Lespiau <damien.lespiau@intel.com>
tests/ddi_compute_wrpll.c

index c0fd08c..bdde61c 100644 (file)
 #define P_MAX 64
 #define P_INC 2
 
-#define R2_MIN 5
-#define R2_MAX 256
-#define R2_INC 1
-
-#define N2_MIN 5
-#define N2_MAX 256
-#define N2_INC 1
-
 /* Constraints for PLL good behavior */
 #define REF_MIN 48
 #define REF_MAX 400
@@ -189,33 +181,36 @@ wrpll_compute_rnp(int clock /* in Hz */,
        }
 
        /*
-        * Ref * 2000 = LC_FREQ_2K / R, where Ref is the actual reference input
-        * seen by the WR PLL. In particular:
-        *   REF_MIN <= LC_FREQ_2K / (R * 2000) and
-        *   REF_MAX >= LC_FREQ_2K / (R * 2000).
-        * Eliminating fractions gives:
-        *   2000 * REF_MAX * R >= LC_FREQ_2K >= 2000 * R * REF_MIN, R2 = 2*R
+        * Ref = LC_FREQ / R, where Ref is the actual reference input seen by
+        * the WR PLL.
+        *
+        * We want R so that REF_MIN <= Ref <= REF_MAX.
+        * Injecting R2 = 2 * R gives:
+        *   REF_MAX * r2 > LC_FREQ * 2 and
+        *   REF_MIN * r2 < LC_FREQ * 2
+        *
+        * Which means the desired boundaries for r2 are:
+        *  LC_FREQ * 2 / REF_MAX < r2 < LC_FREQ * 2 / REF_MIN
+        *
         */
-       for (r2 = R2_MIN; r2 <= R2_MAX; r2 += R2_INC) {
-               if (1000 * REF_MAX * r2 < LC_FREQ_2K)
-                       continue;
-
-               if (1000 * REF_MIN * r2 > LC_FREQ_2K)
-                       continue;
+        for (r2 = LC_FREQ * 2 / REF_MAX + 1;
+            r2 <= LC_FREQ * 2 / REF_MIN;
+            r2++) {
 
                /*
-                * VCO is N * Ref, that is: VCO = N * LC_FREQ_2K / (R * 2000).
+                * VCO = N * Ref, that is: VCO = N * LC_FREQ / R
                 *
-                * Once again, VCO_MAX >= N * LC_FREQ_2K / (R * 2000) >= VCO_MIN,
-                * or: VCO_MAX * R * 2000 >= N * LC_FREQ_2K >= VCO_MIN * R * 2000,
-                * where R2=2*R and N2=2*N
+                * Once again we want VCO_MIN <= VCO <= VCO_MAX.
+                * Injecting R2 = 2 * R and N2 = 2 * N, we get:
+                *   VCO_MAX * r2 > n2 * LC_FREQ and
+                *   VCO_MIN * r2 < n2 * LC_FREQ)
+                *
+                * Which means the desired boundaries for n2 are:
+                * VCO_MIN * r2 / LC_FREQ < n2 < VCO_MAX * r2 / LC_FREQ
                 */
-               for (n2 = N2_MIN; n2 <= N2_MAX; n2 += N2_INC) {
-                       if (VCO_MAX * r2 * 2000 < n2 * LC_FREQ_2K)
-                               continue;
-
-                       if (VCO_MIN * r2 * 2000 > n2 * LC_FREQ_2K)
-                               continue;
+               for (n2 = VCO_MIN * r2 / LC_FREQ + 1;
+                    n2 <= VCO_MAX * r2 / LC_FREQ;
+                    n2++) {
 
                        for (p = P_MIN; p <= P_MAX; p += P_INC)
                                wrpll_update_rnp(freq2k, budget,