> Some further information. The specgrm2 FFT program has 30 instructions
> in the basic operation, including 4 multiplies, 2 adds, and 2 subtracts
> on the actual data, which gives order 20 Gflops, or 100 Gips; the processor
> clock cycle count will obviously be several times this.
So, let me turn this question around, to answer the question more directly.
You let me know if I have this right. You are saying that approximately 300
Gips is neccessary to FFT a 100 mhz bandwidth. Lets say a 300 Mhz pentium II
processor has only about 10 percent of that capacity. Therefore, the upper
limit of FFT capacity of a 300 Mhz pentium II should be about 10 mhz. This
would leave no extra cpu cycles for other things. So, maybe 5 mhz of bandwidth
would use about half of the CPU cycles of a pentium II. Do I have this
correct? It seems plausible to me.
John Marcus MD