UCR CS 122B
Winter 2004
Prof. Frank Vahid

Quiz 2


1. What is the approximate fastest sampling rate, in seconds, of a 128-tap 
   FIR filter dealing with 12-bit data samples for the following processing
   technologies. Assume an infinitely-fast A2D converter. Show your
   work for each, state your assumptions, and give the speedups of
   (b) and (c) compared to (a).

      (a) A 200 MHz 32-bit microprocessor
      (b) A 2 GHz 32-bit microprocessor
      (c) An FPGA running at 200 MHz (briefly explain the design you 
          are assuming).
















2. We want to sample any signal every 10 microseconds and perform a
   a 32-tap FIR filtering on that signal. Using the above implementation 
   technologies, compute the number of different signals that we can
   simultaneously process. Assume unlimited pins are available for each
   technology. Again, show your work and state all assumptions clearly.




















3. Optimize the following code for **performance**. Explain why you chose
   to optimize what you did, and GIVE AN ESTIMATE OF THE OVERALL SPEEDUP 
   you obtain. ASSUME that the code was originally written for readability
   and maintainability, and thus you should only make changes if they
   really make a substantial difference. Full credit will be given as long 
   as *most* of the available speedup is obtained -- getting the absolute 
   maximum speedup isn't required.

   #define MAX1 1000
   #define MAX2 2000
   int M[MAX1], N[MAX2];
   int total=0;
   int minM=32000, minN=32000;

   for (i=0; i < MAX1; i++)
   {
      M[i] = 0; // initialize M
   }

   for (j=0; j < MAX2; 2++)
   {
      N[j] = 0; // initialize N
   }

   for (i=0; i < MAX1; i++)
   {
      cin>> M[i]; // fill M
   }

   for (j=0; j < MAX2; j++)
   {
      cin>> N[j]; // fill N
   }

   for (i=0; i < MAX1; i++)
   {
      for (j=0; j < MAX2; j++)
      {
         if (M[i] < minM) { // find smallest in M
            minM = M[i];
         }
         if (N[j] < minN) { // find smallest in N
            minN = N[j];
         }  
      }
   }

   for (i=0; i < MAX1; i++)
   {
      total = total + (minM * minN * i);
   }