1. Processor Architecture Needed to handle FFT algoarithm M. Smith

2. Tackled already this term Three types of DSP algorithms Long loops, multiplication and addition intensive, regular (simple) memory accesses – e.g. 300 taps in FIR algorithms Short loops involving multiplications and additions – e.g. 3 stages in IIR algorithms DSP Introduction, M. Smith, ECE, University of Calgary, Canada 2

3. 3 Comparing IIR and FIR filters Infinite Impulse Response filters – few operations to produce output from input for each IIR stage 3 – 7 stages Finite Impulse Response filters – many operations to produce output from input. Long FIFO buffer which may require as many operations As FIR calculation itself. Easy to optimize

4. Discrete Fourier Transform  FIR and IIR algorithms directly manipulate the data in “the time domain”.  FIR -- Process M data points using N point FIR filter – involves M * (N-1) additions M * N multiplications M * N * 2 + M memory accesses Algorithm takes a time of Order (M * N)  Very slow if manipulating large amount of data DSP Introduction, M. Smith, ECE, University of Calgary, Canada 4

5. Frequency domain analysis  Apply discrete Fourier transform (implemented via FFT)  Transform to frequency domain takes time Order (M log M)  Perform FIR in frequency domain takes time Order (M)  Transform back to time-domain takes time Order (M log M)  FFT (Order (M log M) is orders of magnitude faster that FIR (Order (M log N) DSP Introduction, M. Smith, ECE, University of Calgary, Canada 5

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8. 4 point DFT to show concepts DSP Introduction, M. Smith, ECE, University of Calgary, Canada 8

9. Simplify using special complex exponential properties DSP Introduction, M. Smith, ECE, University of Calgary, Canada 9

10. Running FFT on data stored in array DSP Introduction, M. Smith, ECE, University of Calgary, Canada 10

11. 8 point FFT with log 8 (= 3) stages  3 stages – with N / 2 butterflies / stage Order (N log N) in time DSP Introduction, M. Smith, ECE, University of Calgary, Canada 11

12. Architectural characteristics needed to handle FFT efficiently DSP Introduction, M. Smith, ECE, University of Calgary, Canada 12

13. Add / subtract in one instruction  The following instruction is illegal as a single instruction XFR4 = R2 + R3, XFR5 = R6 + R7;; Note: comma and NOT semi-colon, means “one instruction” using 6 registers; Not enough data paths to get data into ALU (4 in -- 2 out) XFR4 = R2 + R3; XFR5 = R6 + R7;; ILLEGAL  FFT Butterfly add is special instruction XFR4 = R2 + R3, XFR5 = R2 – R3;; Uses only “4 registers”, 2 in, 2 out DSP Introduction, M. Smith, ECE, University of Calgary, Canada 13

14. Memory accesses  Stage 1 Fetch X data at location k and k + N /2 Store X data at location k and k + N /2  Stage 2 Fetch X data at location k and k + N /4 Store X data at location k and k + N /4  Stage 3 -- Final stage Fetch X data at location k and k + N /8 Store X data at bit-reversed location k and k + N /4 DSP Introduction, M. Smith, ECE, University of Calgary, Canada 14

15. First issue – how do you store complex numbers?  One option Use 16-bit values Store real part in top 16-bits Store imaginary part in bottom 16 bits Access data on J-bus Access complex sinusoids on J-bus Access both components (R and I) in one cycle TigerSHARC has the ability to do 16-bit complex additions and multiplications as specific instructions – INTEGER only Can Use both X and Y compute blocks DSP Introduction, M. Smith, ECE, University of Calgary, Canada 15

16. Integer operations a pain – tend to overflow  Option 2 – floating point Store Real component in location X and imaginary component in location Y Use R1:0 = Q[J4 += 4];; Store first imaginary number in X0 and Y0 Store second imaginary number in X1 and Y1 FR3 = R1 + R0;; – performs complex floating point addition in single cycle L[J5] = R3;; stores complex answer back DSP Introduction, M. Smith, ECE, University of Calgary, Canada 16

17. Integer operations a pain – tend to overflow  Option 3 – floating point Access Real component along J- bus from “data1” and Imaginary component along K- bus from “data 2” Use XR3:0 = Q[J4 += 4]; YR3:0 = Q[K4 += 4]; ; Store first imaginary number in X0 and Y0 Store second, third and fourth imaginary number in XR1, YR1;; XR2, YR2;; XR3, YR3 Which option is best? Depends? How handle bring in complex sinusoids DSP Introduction, M. Smith, ECE, University of Calgary, Canada 17

18. Bit reverse addressing DSP Introduction, M. Smith, ECE, University of Calgary, Canada 18

19. Bit reverse addressing – Check manual for “accurate details” before MII  Only possible with J0, J1, J2 and J3 registers (also K0, K1, K2, K3)  You must start the array on a N aligned boundary otherwise it does not work J0 = address pointer JB0 = base register – point to start of array JL0 = length of array register JM0 = special circular buffer modify register ???? XR4 = BR [J0 += 1]; Bit-reverse addressinbg only works on POST-MODIFY (permits next address to be calculated in parallel) DSP Introduction, M. Smith, ECE, University of Calgary, Canada 19

20. Issues handling “FFT Butterfly DSP Introduction, M. Smith, ECE, University of Calgary, Canada 20

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24. Wrong again  This is using the “Radix 2” form of the algorithm – breaks down into 2-pt DFT  There is also a Radix 4 form of the algorithm – which is faster again FFT Introduction, M. Smith, ECE, University of Calgary, Canada 24

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32. Many special TigerSHARC features to handle FFT FFT Introduction, M. Smith, ECE, University of Calgary, Canada 32