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Writing Efficient Floating-Point Ffts for Adsp-Ts201 Tigersharc®

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Engineer-to-Engineer Note

EE-218

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Writing Efficient Floating-Point FFTs for ADSP-TS201 TigerSHARC® Processors
Contributed by Boris Lerner Rev 2 – March 4, 2004

Introduction
So, you want to write efficient code for the ADSP-TS201 TigerSHARC® processor? Or, maybe, you have come across the optimized example floating-point FFT for this processor and would like to understand how it works and what the author had in mind when writing it. This application note tries to answer both questions by going through that FFT example and all its levels of optimization in detail. This example can be followed in developing other algorithms and code optimized for the ADSPTS201S processor. Generally, most algorithms have several levels of optimization, all of which are discussed in detail in this note. The first and most straightforward level of optimization is paralleling of instructions, as the processor architecture will allow. This is simple and boring. The second level of optimization is loop unrolling and software pipelining to achieve maximum parallelism and to avoid pipeline stalls. Although more complex than the simple parallelism of level one, this can be done in prescribed steps without good understanding of the algorithm and, thus, requires little ingenuity. The third level is to restructure the math of the algorithm to still produce valid results, but so that the new restructured algorithm fits the processor’s architecture better. Being able to do this requires a thorough understanding of the algorithm and, unlike software pipelining, there are no

prescribed steps that lead to the optimal solution. This is where most of the fun in writing optimized code lies. In practical applications it is often unnecessary to go through all of these levels. When all of the levels are required, it is always best to do these levels of optimization in reverse order. By the time the code is fully pipelined, it is too late to try to change the fundamental underlying algorithm. Thus, a programmer would have to think about the algorithm structure first and organize the code accordingly. Then, levels two and one (paralleling, unrolling, and pipelining) are usually done at the same time. The code that this note refers to is supplied by Analog Devices in the form that allows it to be called as either a real or a complex FFT, the last calling parameter of the function defining if real or complex is to be called. The real N-point FFT is obtained from the complex N/2-point FFT with an additional special stage at the end. This note is concerned with code optimization more than the technicalities of the special stage, so it discusses the algorithm for the complex FFT portion of the code only. The last special stage of the real FFT is discussed in detail in the comments of the code.

Standard Radix-2 FFT Algorithm
Figure 1 shows a standard 16-point radix-2 FFT implementation, after the input has been bitreversed. Traditionally, in this algorithm, stages

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1 and 2 are combined together with the required bit reversing into a single optimized loop (since these two stages require no multiplies, only adds and subtracts). Each of the remaining stages is usually done by combining the butterflies that share the same twiddle factors together into groups (so the twiddles have to be fetched only once for each group). Un-optimized assembly source code for a TigerSHARC processor implementing this algorithm is shown in Listing 1. This, with a few tricks that are irrelevant to this discussion, is the way that the 32-bit floating-point FFT code was written when it was targeted to an ADSP-TS101 processor. The benchmarks (in core clock cycles) for this algorithm, including bit reversal, running on an ADSP-TS101 and ADSP-TS201, are shown in Table 1. Note that since the ADSP-TS101 has less memory per memory block than a ADSPTS201, larger point size benchmarks do not apply to the ADSP-TS101. Clearly, as long as the data fits into the ADSP-TS201 cache, it is efficient. Once the data becomes too large for the cache, this FFT implementation becomes extremely inefficient – the cycle count increases from optimal by a factor of five.

Figure 1. Standard Structure of the 16-Point FFT

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//*********************************** Stages ************************************* k10 k11 j12 j10 j11 k20 = = = = = = k31 k31 j31 j31 j31 k31 + + + + + + N;; N/2;; 1;; 2;; 4;; STAGES;; // // // // // twiddles stride butterflies/group groups width of butterfly butterfly stride

_stages_loop: j0 = j31 + j29;; k0 = k31 + k30;; j13 = j31 + 0;; LC1 = j12;; _group_loop: xr1:0 = L[k0 += k10];; j1 = j0 + j10;; LC0 = k11;; _butterfly_loop: xr3:2 = L[j0 += 0];; xr5:4 = L[j1 += 0];; xfr6 = r4 * r0;; xfr7 = r5 * r1;; xfr8 = r6 - r7;; xfr9 = r4 * r1;; xfr10 = r5 * r0;; xfr11 = r9 + r10;; xfr12 = r2 + r8, fr14 = r2 - r8;; xfr13 = r3 + r11, fr15 = r3 - r11;; L[j0 += j11] = xr13:12;; L[j1 += j11] = xr15:14;; if NLC0E, jump _butterfly_loop (NP);; j13 = j13 + 2;; j0 = j29 + j13;; if NLC1E, jump _group_loop (NP);; k10 k11 j12 j10 j11 = = = = = lshiftr k10;; lshiftr k11;; j12 + j12;; j10 + j10;; j11 + j11;; // // // // // // // // // // xr2=Re1, xr3=Im1 xr4=Re2, xr5=Im2 xr6=Re2*cos xr7=Im2*sin xr8=Re(z2*twid) xr6=Re2*sin xr7=Im2*cos xr8=Im(z2*twid) Re(butterfly) Im(butterfly) // xr0=cos, xr1=-sin // j1 -> second input to butterfly // j0 -> internal_buff // k0 -> twiddles

// offset for the next group // // // // // twiddles stride butterflies/group groups width of butterfly butterfly stride

k20 = k20 - 1;; if NKEQ, jump _stages_loop (NP);;

Listing 1. fft32_unoptimized.asm Points N 256 512 1024 2048 4096 8192 16384 32768 65536 2172 4582 9872 21338 46244 99886 215224 NA NA ADSPTS101 ADSP-TS201 Input not in cache 2641 5533 12170 26610 197272 444628 987730 2133220 4720010 ADSP-TS201 Input in cache 2218 4649 9992 22173 NA NA NA NA NA

Optimizing the Structure of the FFT for ADSP-TS201 Processors
To be able to re-structure the algorithm to perform optimally on ADSP-TS201, we have to understand why the performance of large FFTs using the conventional FFT structure is so poor. ADSP-TS201 memory is optimized for sequential reads. Cache is designed to help with algorithms where the reads are not sequential. In the conventional FFT algorithm, each stage’s butterflies stride doubles, so the reads are nonsequential and, with each new stage, the cache is less and less likely to be a hit – the reads are all over the place. The solution lies in re-arranging a stage’s output to ensure that the next stage’s reads are sequential. The structure of the algorithm implementation is shown in Figure 2.

Table 1. Core Clock Cycles for N-point Complex FFT

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Figure 2. Reorganized Structure of the 16-Point FFT

It is simple enough to trace this diagram by hand to see that it is simply a re-ordering of the diagram in Figure 1. Amazingly enough, the final output is in correct order. This can be easily proven for general N = 2k = the number of points in the FFT. Note that the re-ordering is given by the following formula:
⎧n ⎪ , if n is even f (n ) = ⎨ 2 n −1 N ⎪ + , if n is odd 2 ⎩ 2

operation of right 1-bit rotation, which after K = log 2 ( N ) steps returns the original n back. Thus, the output after K stages is in correct order again. Great! We have our new structure. It has sequential reads and we are lucky enough that the output is in the correct order. This should be much more efficient. Right? Let’s write the code for it! Well, before we spend a lot of time writing the code, we should ensure that all of the DSP operations that we are about to do actually fit into our processor architecture efficiently. There may be no reason to optimize data movements if the underlying math suffers.

Thus, if n is even, it is shifted right and if n is odd, it is shifted right and its most significant bit is set. This is, of course, equivalent to the

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The first obvious point to notice is that this structure cannot be done in-place due to its reordering. The stages will have to ping-pong their input/output buffers. This should not be a problem. The ADSP-TS201 processor has a lot of memory on board, but should memory optimization be required (and input does not have to be preserved), we can use the input as one of the two ping-pong buffers. Next, we note that a traditional FFT combines butterflies that share twiddles into the same group to save twiddle fetch cycles. Amazingly, the twiddles of the structure in Figure 2 line up linearly – one group at a time. We are lucky again! Now, what would a butterfly of this new structure consist of? Table 2 lists the operations necessary to perform a single complex butterfly. Since the ADSP-TS201 is a SIMD processor (i.e., it can double all the computes), we will write the steps outlined in Listing 1 in SIMD fashion, so that two adjacent butterflies are computed in parallel, one in the X-Compute block and the other one in the Y-Compute block. Let us analyze the DSP operations in more detail. F1, F2, K2 and F4 fetch a total of four 32-bit words, which on ADSP-TS201 can be done in a single quad fetch into X-Compute block registers. To be able to supply SIMD machine with data, we would also have to perform a second butterfly quad fetch into the Y-Compute block registers. Then, M1, M2, M3, M4, A1, and A2 will perform SIMD operations for both butterflies. The ADSP-TS201 supports a single add/subtract instruction, so A3 and A4 can be combined into a single operation (which is, of course, performed SIMD on both butterflies at once) and similarly A5 and A6 can be combined, as well. Mnemonic
F1 F2 K2 F4 M1 M2 M3 M4 A1 A2 A3 A4 A5 A6 S1 S2 S3 S4

Operation
Fetch Real(Input1) of the Butterfly Fetch Imag(Input1) of the Butterfly Fetch Real(Input2) of the Butterfly Fetch Imag(Input2) of the Butterfly K2 * Real(twiddle) F4 * Imag(twiddle) K2 * Imag(twiddle) F4 * Real(twiddle) M1–M2 = Real(Input2*twiddle) M3+M4 = Imag(Input2*twiddle) F1 + A1 = Real(Output1) F1 - A1 = Real(Output2) F2 + A2 = Imag(Output1) F2 – A2 = Imag(Output2) Store(Real(Output1)) Store(Imag(Output1)) Store(Real(Output2)) Store(Imag(Output2))

Table 2. Single Butterfly Done Linearly – Logical Implementation

Now we run into a problem: S1, S2, S3 and S4 cannot be performed in the same cycle, since S3 and S4 are destined to another place in memory due to our output re-ordering. Instead, we can store S1 and S2 for both butterflies in one cycle (lucky again – these are adjacent!) and S3 and S4 for both butterflies in the next cycle. So far, so good – the new set of operations is summarized in Table 3.

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Mnemonic
F1 F2 M1 M2 M3 M4 A1 A2 A3 A4 S1 S2

Operation
Fetch Input1,2 of the Butterfly1 Fetch Input1,2 of the Butterfly2 Real(Input2) * Real(twiddle) Imag(Input2) * Imag(twiddle) Real(Input2) * Imag(twiddle) Imag(Input2) * Real(twiddle) M1–M2 = Real(Input2*twiddle) M3+M4 = Imag(Input2*twiddle) Real(Input1) +/- A1 = Real(Output1,2) Imag(Input1) +/- A2 = Imag(Output1) Store(Output1, both Butterflies) Store(Output2, both Butterflies)

reserved place in one of the instructions for a loop jump back). Thus, the old rule of fetching twiddles only once per group of butterflies that shares them is no longer necessary – the twiddle fetches come free! And, since the structure of the arrows of Figure 2 is identical at every stage, we may be able to reduce the FFT from the usual three nested loops to only two, provided that we can find a way to correctly fetch the twiddles at each stage (twiddles are the only thing that distinguishes the stages of Figure 2). Figure 2 shows how the twiddles must be fetched at each stage: 1st Stage – all are W0. 2nd Stage – half are W0, next half are WN/4. 3rd Stage – one quarter are W0, the next quarter are WN/8, the next quarter are W2N/8, and the last quarter are W3N/8. And so on… If we keep a virtual twiddle pointer offset, increment it to the next sequential twiddle every butterfly, but AND it with a mask before actually using it in the twiddle fetch, we achieve precisely this order of twiddle fetch. Moreover, this rule is the same for every stage, except that the mask at every stage must be shifted down by one bit (i.e., each stage requires twice as fine a resolution of the twiddles as the previous stage). Here, our unused KALU operations come in very handy. To implement this twiddle fetch, we need to increment the virtual offset, mask it and do a twiddle fetch every butterfly… Oh, no! We are in SIMD (i.e. we are doing two butterflies together) and we do not have the 6 available instruction slots for this! But luck saves us again. We can easily notice that all stages except the last share the twiddles between the SIMD pair of butterflies – so, for these stages, we need only to do the twiddle fetch once per SIMD pair of the butterflies! And the three cycles are precisely what we have to do this. Unfortunately, in the last stage, every butterfly has its own unique twiddle; but in the last stage, we do not have to mask – just step the pointer to the next twiddle every time! It will have to be written separately, but it will optimize completely as well. Table 4 summarizes the latest structure’s steps. Three

Table 3. Single Butterfly Done Linearly – Actual ADSP-TS20x Implementation

Each operation in Table 3 is a single-cycle operation on ADSP-TS201 processor. There is a total of 2 fetches, 4 multiplies, 4 ALU, and 2 store instructions. Since the ADSP-TS201 allows fetches/stores to be paralleled with multiplies and ALUs in a single cycle, loop unrolling, pipelining, and paralleling should yield a 4-cycle execution of these two SIMD butterflies (and we are still efficient in the memory usage!). At this point, we can now be reasonably certain that the above will yield efficient code and we can start developing it. However, careful observation at this point can help us optimize this structure even further. Note that we are only using a total of 4 fetches and stores from a single memory block, say, by using JALU pointer registers. In parallel we can do 3 more fetches/stores/KALU operations without losing any cycles (actually, we can do 4 of them, but we do need one

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a new KALU operations (K1, K2 and K3) have been added to Table 3. Time to write the code? Well, no – let us figure out how to pipeline it first. Mnemonic
K1 K2 K3 F1 F2 M1 M2 M3 M4 A1 A2 A3 A4 S1 S2 K2 -> M1, M2, M3, M4 F1, F2 -> M1, M2, M3, M4, A3, A4 M1, M2 -> A1 M3, M4 -> A2 A1, A2 -> A3, A4

Operation
Virtual Pointer Offset Mask Twiddles Fetch Virtual Pointer Offset Increment Fetch Input1,2 of the Butterfly1 Fetch Input1,2 of the Butterfly2 Real(Input2) * Real(twiddle) Imag(Input2) * Imag(twiddle) Real(Input2) * Imag(twiddle) Imag(Input2) * Real(twiddle) M1–M2 = Real(Input2*twiddle) M3+M4 = Imag(Input2*twiddle) Real(Input1) +/- A1 = Real(Output1,2) Imag(Input1) +/- A2 = Imag(Output1) Store(Output1, both Butterflies) Store(Output2, both Butterflies)

This means that if the operation at the start of the arrow is immediately followed by the operation at the end of that arrow, the result will be correct, but code execution will produce a stall. Thus, to fully optimize the code, operations at the ends of arrows with stalls must be kept more than one instruction line apart.

Table 4. Single Butterfly Done Linearly – Modified ADSP-TS20x Implementation

Pipelining of the Algorithm
Figure 3 shows the algorithm’s operations from Table 4 with arrows showing the dependencies. The arrows of the dependencies indicate that the result of the operation at the start of the arrow is used by the operation at the end of that arrow and, thus, must be completed first to ensure correct data. Some arrows have a stall associated with them, specifically:
Figure 3. Reorganized Structure’s Dependencies

A quick observation of the dependencies in Figure 3 is sufficient to analyze the level of pipelining and the number of compute block registers needed to do it.

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State
K1 K2 K3 F1

Dependency To States
K2 M1,M2,M3,M4 K1 M1,M2,M3,M4, A1,A2 M1,M2,M3,M4, A1,A2 A1 A1 A2 A2 A3,A4 A3,A4 S1,S2 S1,S2 none none

Max Dep Cycles
1 5 1 10

Compute Block Registers Needed
0 4*([5/4]+1)=8 0 4*([10/4]+1)=16

many output registers as M1, M2, M3, M4, A1 and A2 since A3 and A4 are add/subtracts. The resulting requirement to fully pipeline this code is 60 compute block registers, out of 64 total – just barely made it! Cycle/ Operation
1 2

JALU
F1 F2 S1--S2--F1+ F2+ S1-S2-F1++ F2++ S1S2F1+++ F2+++ S1 S2

KALU
K1 K2

MAC
M4-M2M3-

ALU
A3--A4--A2-A1A3-A4-A2A1 A3A4A2 A1+ A3 A4 A2+ A1++

F2 M1 M2 M3 M4 A1 A2 A3 A4 S1 S2

10 2 2 2 2 2 2 1 1 0 0

4*([10/4]+1)=16 2([2/4]+1)=2 2([2/4]+1)=2 2([2/4]+1)=2 2([2/4]+1)=2 2([2/4]+1)=2 2([2/4]+1)=2 4([1/4]+1)=4 4([1/4]+1)=4 0 0 60

3 4 5 6 7 8 9 10 11 12 13 14 15 16

K3 K1+ K2+

M1 M4M2 M3

K3+ K1++ K2++

M1+ M4 M2+ M3+

K3++ K1+++ K2+++

M1++ M4+ M2++ M3++

Total Regs
Table 5. Number of Compute Block Registers Required to Pipeline the Butterflies

Full pipelining, as mentioned earlier, would give a 4-cycle SIMD pair of butterflies. Thus,
Pipelined_CB_Registers_Per_State_Output = Unpipelined_CB_Registers_Per_State_Output * ([Maximum_Dependency_Cycles/4]+1)

K3+++

M1+++

Table 6. Pipelined Butterflies

Here, [x] denotes the integer part of the number x. We can therefore determine the number of compute block registers needed, as shown in Table 5. Note that A3 and A4 require twice as

We pipeline this fully symbolically, using the mnemonics of Table 4 and Figure 3. The pipelining is shown in Table 6, in which “+” in the operation indicates the operation that corresponds to the next set of the butterflies and “-” corresponds to the operation in the previous set of the butterflies.
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All instructions are paralleled, there are no stalls, and there is a place to put the jump to the top of the loop (actually, four places, but this is only because the pipeline is 4 pairs of butterflies deep, each iteration of the loop in Table 6 will actually do 4 pairs of butterflies). and doing them separately – they do not really require a complex multiply and can be done faster. Also, bit-reversal is incorporated into the first two stages, as well. Now, for the bottom line – how much did the cycle count improve? In Table 7 we repeat Table 1 with additional columns for the benchmarks for the new algorithm. The cycle count for the larger-thancache FFTs improved by a factor greater than 3! Moreover, the cycle count for FFTs that fit into the cache is better than it was on the original ADSP-TS101 processor, which had no cache or memory latency issues of any kind. The reason for this is that the new architecture allows the code to be written in two nested loops instead of three and, thus, has significantly less overhead. This code, ported to the ADSP-TS101, improves its benchmarks, too – as shown in Table 7.

The Code
Now, writing the code is trivial. The ADSPTS201 is so flexible that it takes all the challenge right out of it. Just follow the pipeline of Table 6 and the code is done. The resulting code for the stages other than last is shown in Listing 2. Outside of this inner loop is a stage loop that ping-pongs input/output buffers and shifts the twiddle modifier mask. Pretty simple! Additional optimization is done by breaking the first two stages away from the main of the code
.align_code 4; _BflyLoop: q[j2+=4]=r27:26; yr3:0=q[j0+=4]; xr3:0=q[j0+=4]; q[j1+=4]=r25:24; q[j2+=4]=r27:26; yr11:8=q[j0+=4]; xr11:8=q[j0+=4]; q[j1+=4]=r25:24; q[j2+=4]=r27:26; k5=k5+k9; k3=k5 and k4; r5:4=l[k7+k3]; k5=k5+k9; fr6=r30*r12; fr15=r23*r4; fr7=r31*r13; fr14=r30*r13; fr6=r2*r4;

fr16=r6-r7;; fr24=r8+r18, fr26=r8-r18;; fr25=r9+r19, fr27=r9-r19;; fr17=r14+r15;; fr18=r6-r7;;

// S2----,K3-, // // // // // // // // // // // // F1, K1, F2, K2, S1---, S2---, K3, F1+, F2+, S1--, S2--, F1++, F2++, S1-, S2-, K1+, K2+, K3+, K1++, K2++, K3++,

M1-, M4--, M2-, M3-, M1, M4-, M2, M3, M1+, M4, M2+, M3+, M1++,

A1-A3--A4--A2-A1A3-A4-A2A1 A3A4A2 A1+ A3 A4 A2+

k3=k5 and k4; fr15=r31*r12; fr24=r20+r16, fr26=r20-r16;; r13:12=l[k7+k3]; fr7=r3*r5; fr25=r21+r17, fr27=r21-r17;; fr14=r2*r5; fr19=r14+r15;; k5=k5+k9; fr6=r10*r12; fr16=r6-r7;; fr15=r3*r4; fr7=r11*r13; fr14=r10*r13; fr6=r22*r4; fr24=r28+r18, fr26=r28-r18;; fr25=r29+r19, fr27=r29-r19;; fr17=r14+r15;; fr18=r6-r7;; fr26=r0-r16;; fr27=r1-r17;;

yr23:20=q[j0+=4]; k3=k5 and k4; xr23:20=q[j0+=4]; r5:4=l[k7+k3]; q[j1+=4]=r25:24; q[j2+=4]=r27:26; k5=k5+k9;

yr31:28=q[j0+=4]; k3=k5 and k4; fr15=r11*r12; fr24=r0+r16, xr31:28=q[j0+=4]; r13:12=l[k7+k3]; fr7=r23*r5; fr25=r1+r17, .align_code 4; if NLC0E, jump _BflyLoop; q[j1+=4]=r25:24; fr14=r22*r5; fr19=r14+r15;;

// F1+++, K1+++, M4+, // F2+++, K2+++, M2++, // S1, M3++,

Listing 2. fft32.asm - fragment

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ADSP-TS101 ADSP-TS101
New structure

ADSP-TS201
Old structureInput not in cache 2641 5533 12170 26610 197272 444628 987730 2133220 4720010

ADSP-TS201
New structureInput not in cache 2402 5192 11662 25316 69924 147628 313292 662614 1397544

ADSP-TS201 ADSP-TS201
Old structureInput in cache New structureInput in cache

N
256 512 1024 2048 4096 8192 16384 32768 65536

Old structure

2172 4582 9872 21338 46244 99886 215224 NA NA

1958 4276 9410 20688 45278 98540 213243 NA NA

2218 4649 9992 22173 NA NA NA NA NA

1963 4283 9419 20699 NA NA NA NA NA

Table 7. Core Clock Cycles for N-point Complex FFT, New versus Old Structure

Usage Rules
The C-callable complex FFT routine is called as
FFT32(&(input), &(ping_pong_buffer1), &(ping_pong_buffer2), &(output), N, F);

where input -> FFT input buffer, output -> FFT output buffer, ping_pong_bufferx are the ping pong buffers, N=Number of complex points, F=0 if FFT is real and 1 if FFT is complex. As mentioned earlier, due to data re-ordering, stages cannot be done in-place and have to pingpong. Thus, ping_pong_buffer1 and ping_pong_buffer2 have to be two distinct buffers. However, depending on the routine’s user requirements, some memory optimization is possible. Ping_pong_buffer1 can be made the

same as input if input does not need to be preserved. Also, if Log2(N) is even, output can be made the same as ping_pong_buffer2 and if Log2(N) is odd, output can be made the same as ping_pong_buffer1. Below are two examples of the routine usage with minimal use of memory:
FFT32(&(input), &( input), &( output), &(output), 1024, 1); FFT32(&(input), &(input), &( ping_pong_buffer2), &( input), 2048, 1);

To eliminate memory block access conflicts, input must reside in a different memory block than ping_pong_buffer2 and twiddle factors must reside in a different memory block than the pingpong buffers. Of course, all code must reside in a block that is different from all the data buffers, as well. Ping-pong buffers can share a memory block, however – there is no instruction that accesses both ping-pong buffers in the same cycle.

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Appendix
Complete Source Code of the Optimized FFT
/* fft32.asm Prelim rev. October 19, 2003 - BL Rev. 1.0 - added real inputs case - PM This is assembly routine for the Complex radix-2 C-callable FFT on TigerSHARC family of DSPs. I. Description of Calling. 1. Inputs: j4 -> input (ping-pong buffer 1) j5 -> ping-pong buffer 1 j6 -> ping-pong buffer 2 j7 -> output j27+0x18 -> N = Number of points j27+0x19 -> REAL or COMPLEX 2. C-Calling Example: fft32(&(input), &(ping_pong_buffer1), &(ping_pong_buffer2), &(output), N, COMPLEX); 3. Limitations: a. All buffers must be aligned on memory boundary which is a multiple of 4. b. N must be between 32 and MAX_FFT_SIZE. c. If memory space savings are required and input does not have to be preserved, ping_pong_buffer1 can be the same buffer as input. d. If memory space savings are required, output can be the same buffer as ping_pong_buffer2 if the number of FFT stages is even (i.e. Log2(N) is even) or the same as ping_pong_buffer1 if the number of FFT stages is odd (i.e. Log2(N) is odd). 4. MAX_FFT_SIZE can be selected via #define. Larger values allow for more choices of N, but its twiddles will occupy more memory. 5. This C - callable function can process up to 64K blocks of data on TS201 (16K blocks on TS101) because C environment itself necessitates memory. Therefore, if more input points are necessary, assembly language development may become a must. On TS201, a block of memory is 128K words long, so maximum N is 128K real points or 64K complex points. TS101 contains only 2 blocks of data memory of 64K words and 4 buffers must be accommodated. Therefore, maximum N is 32K real words or 16K complex words. II. Description of the FFT algorithm. 1. The input data is treated as complex interleaved N-point. 2. Due to re-ordering, no stage can be done in-place. 3. The bit reversal and the first two stages are combined into a single loop. This loop takes data from input and stores it in the ping-pong buffer1. 4. Each subsequent stage ping-pongs the data between the two ping-pong buffers. The last stage uses FFT output buffer for its output. 5. Although the FFT is designed to be called with any point size N ------------------------>--------> F(n) \ +/ \ +/ \/ \/ /\ /\ conj / -\ exp(-2*pi*i*n)*i / -\ conj G(N/2-n) -----> conj(G(N/2-n))------>------------------------>--------> F(N/2-n) This is very efficient on the TigerSHARC architecture due to the add/subtract instruction. IV. For all additional details regarding this algorithm and code, see EE-218

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a application note, available from the ADI web site. */ //************************************ Includes ********************************** #include "FFTDef.h" #include "defts201.h" //************************* Externs ************************************* .extern _twiddles; //********************************* FFT Routine ********************************* .section program; .global _FFT32; _FFT32: //********************************** Prologue *********************************** mENTER mPUSHQ(xR31:28) mPUSHQ(xR27:24) mPUSHQ(yR31:28) mPUSHQ(yR27:24) //************************************ Setup ************************************* j17 = [j27 + 0x18];; //j17 = N j11 = [j27 + 0x19];; // j11=COMPLEX or REAL, off the stack comp(j11,COMPLEX);; if jeq, jump _FFTStages1and2;; j17=ashiftr j17;; // Complex or Real? // if Real, half N

//******************************************************************************** _FFTStages1and2: j11 = j31 + j17;; xr3=j11; k7=k31+_twiddles;; k1=j11; j8=lshiftr j11;; j9=lshiftr j8; xr0=MAX_FFT_SIZE; xr3=LD0 r3;; k8=lshiftr k1; xr0=LD0 r0; xr1=j11;; k8=lshiftr k8; xr1=LD0 r1; xr2=(31-3);; k0=j4; k10=lshiftr k8; xr1=r1-r0; xr0=lshift r0 by -32;; k10=lshiftr k10; xr0=bset r0 by r1; xr30=r2-r3;; k10=k10-1; xr0=lshift r0 by 2; LC1=xr30;; k9=xr0; k4=k31+(MAX_FFT_SIZE/4-1);; k4=not k4; j10=lshiftr j9;; // j11=N // k1=N, j8=N/2 // j9=N/4, compute the twiddle stride // // // // k8=N/4, Compute Stages-3 k0->input, xr1=bit difference between MAX and N k10=N/16, xr30=Stages-3 k10=N/16-1, LC1=Stages-3

// initial twiddles pointer mask, j10=N/8

//****************** Bit Reverse and Stages 1 & 2 ****************************** k5=lshiftr k1;; j0=j31+j6; k6=k6-k6;; j1=j0+j9; LC0=k10;; j2=j1+j9; k1=k0+k5;; j3=j2+j9; k2=k1+k5;; j12=j3+j9; k3=k2+k5;; j13=j12+j9; k5=lshiftr k5;; j14=j13+j9; r1:0=q[k0+k6];; j15=j14+j9; r3:2=q[k2+k6];; r5:4=q[k1+k6];; r7:6=q[k3+k6];; k6=k6+k5 (br); fr0=r0+r2, fr20=r0-r2;; r9:8=q[k0+k6]; fr2=r1+r3, fr29=r1-r3;; r11:10=q[k2+k6]; fr4=r4+r6, fr21=r4-r6;; r13:12=q[k1+k6]; fr5=r5+r7, fr28=r5-r7;; r15:14=q[k3+k6]; fr18=r8+r10, fr22=r8-r10;; k6=k6+k5 (br); fr19=r9+r11, fr31=r9-r11;; fr26=r12+r14, fr23=r12-r14;; fr27=r13+r15, fr30=r13-r15;; fr20=r20+r28, fr29=r29+r21, fr22=r22+r30, fr31=r31+r23, fr28=r20-r28;; fr21=r29-r21;; fr30=r22-r30;; fr23=r31-r23;; // // // // // // // // // k5=N/2 j0->ping_pong_buffer2 j1->ping_pong_buffer2+N/4, LC0=N/16-1 j2->ping_pong_buffer2+N/2, k1->input+N/2 j3->ping_pong_buffer2+3N/4, k2->input+N j12->ping_pong_buffer2+N, k3->input+3N/2 j13->ping_pong_buffer2+5N/4, k5=N/4 j14->ping_pong_buffer2+3N/2 j15->ping_pong_buffer2+7N/4

.align_code 4; _Stages1and2Loop: r1:0=q[k0+k6]; r3:2=q[k2+k6]; r5:4=q[k1+k6]; r7:6=q[k3+k6];

q[j2+=4]=yr23:20; fr16=r0+r4, fr24=r0-r4;; q[j3+=4]=xr23:20; fr17=r2+r5, fr25=r2-r5;; q[j14+=4]=yr31:28; fr18=r18+r26, fr26=r18-r26;; q[j15+=4]=xr31:28; fr19=r19+r27, fr27=r19-r27;;

k6=k6+k5 (br); q[j0+=4]=yr19:16; fr0=r0+r2, fr20=r0-r2;; r9:8=q[k0+k6]; q[j1+=4]=xr19:16; fr2=r1+r3, fr29=r1-r3;; r11:10=q[k2+k6]; q[j12+=4]=yr27:24; fr4=r4+r6, fr21=r4-r6;; r13:12=q[k1+k6]; q[j13+=4]=xr27:24; fr5=r5+r7, fr28=r5-r7;; r15:14=q[k3+k6]; fr18=r8+r10, fr22=r8-r10;; k6=k6+k5 (br); fr19=r9+r11, fr31=r9-r11;; fr26=r12+r14, fr23=r12-r14;; fr27=r13+r15, fr30=r13-r15;; fr20=r20+r28, fr28=r20-r28;; fr29=r29+r21, fr21=r29-r21;; fr22=r22+r30, fr30=r22-r30;;

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a
.align_code 4; if NLC0E, jump _Stages1and2Loop; fr31=r31+r23, fr23=r31-r23;; q[j2+=4]=yr23:20; fr16=r0+r4, fr24=r0-r4;; q[j3+=4]=xr23:20; fr17=r2+r5, fr25=r2-r5;; q[j14+=4]=yr31:28; fr18=r18+r26, fr26=r18-r26;; q[j15+=4]=xr31:28; fr19=r19+r27, fr27=r19-r27;; q[j0+=4]=yr19:16;; q[j1+=4]=xr19:16;; q[j12+=4]=yr27:24;; q[j13+=4]=xr27:24;; //************************ Stages 3 to Log2(N)-1 j0=j31+j6; k5=k31+0;; .align_code 4; _StageLoop: yr3:0=q[j0+=4]; xr3:0=q[j0+=4]; LC0=k10; yr11:8=q[j0+=4]; xr11:8=q[j0+=4]; j1=j31+j5; k3=k5 and k4;; r5:4=l[k7+k3];; k5=k5+k9;; k3=k5 and k4; fr6=r2*r4;; r13:12=l[k7+k3]; fr7=r3*r5;; k5=k5+k9; fr14=r2*r5;; fr6=r10*r12; fr16=r6-r7;; fr15=r3*r4;; fr7=r11*r13;; fr14=r10*r13; fr17=r14+r15;; fr6=r22*r4; fr18=r6-r7;; // F1, // F2, // // F1+, // F2+, // // F1++, // F2++, // // K1 K2 K3, K1+ K2+, // K3+, K1++, K2++, K3++, M1 M2 M3 M1+, M4 M2+ M3+, M1++, A1 *******************************

yr23:20=q[j0+=4]; k3=k5 and k4; xr23:20=q[j0+=4]; r5:4=l[k7+k3]; j2=j1+j11; k5=k5+k9;

A2 A1+ A3 A4 A2+

yr31:28=q[j0+=4]; k3=k5 and k4; fr15=r11*r12; fr24=r0+r16, fr26=r0-r16;; xr31:28=q[j0+=4]; r13:12=l[k7+k3]; fr7=r23*r5; fr25=r1+r17, fr27=r1-r17;; q[j1+=4]=r25:24; fr14=r22*r5; fr19=r14+r15;; .align_code 4; _BflyLoop: q[j2+=4]=r27:26; yr3:0=q[j0+=4]; xr3:0=q[j0+=4]; q[j1+=4]=r25:24; q[j2+=4]=r27:26; yr11:8=q[j0+=4]; xr11:8=q[j0+=4]; q[j1+=4]=r25:24; q[j2+=4]=r27:26; k5=k5+k9; k3=k5 and k4; r5:4=l[k7+k3]; k5=k5+k9; fr6=r30*r12; fr15=r23*r4; fr7=r31*r13; fr14=r30*r13; fr6=r2*r4; fr16=r6-r7;; fr24=r8+r18, fr26=r8-r18;; fr25=r9+r19, fr27=r9-r19;; fr17=r14+r15;; fr18=r6-r7;;

// F1+++, K1+++, M4+, // F2+++, K2+++, M2++, // S1, M3++,

// S2----,K3-, // // // // // // // // // // // // F1, K1, F2, K2, S1---, S2---, K3, F1+, F2+, S1--, S2--, F1++, F2++, S1-, S2-, K1+, K2+, K3+, K1++, K2++, K3++,

M1-, M4--, M2-, M3-, M1, M4-, M2, M3, M1+, M4, M2+, M3+, M1++,

A1-A3--A4--A2-A1A3-A4-A2A1 A3A4A2 A1+ A3 A4 A2+

k3=k5 and k4; fr15=r31*r12; fr24=r20+r16, fr26=r20-r16;; r13:12=l[k7+k3]; fr7=r3*r5; fr25=r21+r17, fr27=r21-r17;; fr14=r2*r5; fr19=r14+r15;; k5=k5+k9; fr6=r10*r12; fr16=r6-r7;; fr15=r3*r4; fr7=r11*r13; fr14=r10*r13; fr6=r22*r4; fr24=r28+r18, fr26=r28-r18;; fr25=r29+r19, fr27=r29-r19;; fr17=r14+r15;; fr18=r6-r7;; fr26=r0-r16;; fr27=r1-r17;;

yr23:20=q[j0+=4]; k3=k5 and k4; xr23:20=q[j0+=4]; r5:4=l[k7+k3]; q[j1+=4]=r25:24; q[j2+=4]=r27:26; k5=k5+k9;

yr31:28=q[j0+=4]; k3=k5 and k4; fr15=r11*r12; fr24=r0+r16, xr31:28=q[j0+=4]; r13:12=l[k7+k3]; fr7=r23*r5; fr25=r1+r17, .align_code 4; if NLC0E, jump _BflyLoop; q[j1+=4]=r25:24; fr14=r22*r5; q[j2+=4]=r27:26; j0=j31+j5; j5=j31+j6; q[j1+=4]=r25:24; q[j2+=4]=r27:26; j6=j31+j0; q[j1+=4]=r25:24; q[j2+=4]=r27:26; j0=j31+j6; q[j1+=4]=r25:24; k5=k31+0;; .align_code 4; if NLC1E, jump _StageLoop; q[j2+=4]=r23:22; k4=ashiftr k4;; //******************************* Last stage ********************************* k9 = ashiftr k9;;//in this manner any MAX_FFT_SIZE can be used yr3:0=q[j0+=4]; yr5:4 = l[k7+=k9];; xr3:0=q[j0+=4]; xr5:4=l[k7+=k9];; j1=j31+j7; fr6=r2*r4; LC0=k10;; yr11:8=q[j0+=4]; xr11:8=q[j0+=4]; j2=j1+j11; yr13:12=l[k7+=k9];; xr13:12=l[k7+=k9]; fr7=r3*r5;; fr14=r2*r5;; fr6=r10*r12; fr19=r14+r15;; fr6=r30*r12; fr15=r23*r4; fr16=r6-r7;; fr24=r8+r18,

// F1+++, K1+++, M4+, // F2+++, K2+++, M2++, // S1, // S2----, M3++,

M1-, A1--

fr26=r8-r18;;

fr7=r31*r13; fr25=r9+r19, fr27=r9-r19;; fr14=r30*r13; fr17=r14+r15;; fr18=r6-r7;; fr15=r31*r12; fr24=r20+r16, fr26=r20-r16;; fr25=r21+r17, fr27=r21-r17;; fr19=r14+r15;; fr24=r28+r18, fr22=r28-r18;; fr25=r29+r19, fr23=r29-r19;;

// M4--, A3--// swap ping-pong pointers // M2-, A4--// S1---, M3-, A2-// S2---, A1// // // S1--, // S2-// // S1// S2-, shift the mask M4-, A3-A4-A2A3A4-

// F1, // F2, // // F1+ // F2+, // // // F1++, // F2++, // //

K2 M1 K2+, M2 M3 M1+, M4 M2+ M3+, M1++,

fr16=r6-r7;;

A1

yr23:20=q[j0+=4]; yr5:4=l[k7+=k9]; xr23:20=q[j0+=4]; xr5:4=l[k7+=k9];

fr15=r3*r4;; fr7=r11*r13;; fr14=r10*r13; fr17=r14+r15;; fr6=r22*r4; fr18=r6-r7;;

K2++,

A2 A1+

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a yr31:28=q[j0+=4]; yr13:12=l[k7+=k9];fr15=r11*r12; fr24=r0+r16, fr26=r0-r16;; xr31:28=q[j0+=4]; xr13:12=l[k7+=k9]; fr7=r23*r5; fr25=r1+r17, fr27=r1-r17;; q[j1+=4]=r25:24; fr14=r22*r5; fr19=r14+r15;; .align_code 4; _BflyLastLoop: q[j2+=4]=r27:26; yr3:0=q[j0+=4]; xr3:0=q[j0+=4]; q[j1+=4]=r25:24; q[j2+=4]=r27:26; yr11:8=q[j0+=4]; xr11:8=q[j0+=4]; q[j1+=4]=r25:24; q[j2+=4]=r27:26; yr5:4=l[k7+=k9]; xr5:4=l[k7+=k9]; fr6=r30*r12; fr15=r23*r4; fr7=r31*r13; fr14=r30*r13; fr6=r2*r4; fr16=r6-r7;; fr24=r8+r18, fr26=r8-r18;; fr25=r9+r19, fr27=r9-r19;; fr17=r14+r15;; fr18=r6-r7;; fr24=r20+r16, fr26=r20-r16;; fr25=r21+r17, fr27=r21-r17;; fr19=r14+r15;; fr16=r6-r7;; fr24=r28+r18, fr26=r28-r18;; fr25=r29+r19, fr27=r29-r19;; fr17=r14+r15;; fr18=r6-r7;; fr26=r0-r16;; fr27=r1-r17;; // S2----, // // // // // // // // // // // // F1, F2, K2, S1---, S2---, F1+, F2+, S1--, S2--, F1++, F2++, S1-, S2-, K2+, M1-, M4--, M2-, M3-, M1, M4-, M2, M3, M1+, M4, M2+, M3+, M1++, A1-A3--A4--A2-A1A3-A4-A2A1 A3A4A2 A1+ A3 A4 A2+ // F1+++, M4+, // F2+++, K2+++, M2++, // S1, M3++, A3 A4 A2+

yr13:12=l[k7+=k9];fr15=r31*r12; xr13:12=l[k7+=k9];fr7=r3*r5; fr14=r2*r5; fr6=r10*r12; fr15=r3*r4; fr7=r11*r13; fr14=r10*r13; fr6=r22*r4;

yr23:20=q[j0+=4]; yr5:4=l[k7+=k9]; xr23:20=q[j0+=4]; xr5:4=l[k7+=k9]; q[j1+=4]=r25:24; q[j2+=4]=r27:26;

K2++,

yr31:28=q[j0+=4];yr13:12=l[k7+=k9]; fr15=r11*r12; fr24=r0+r16, xr31:28=q[j0+=4];xr13:12=l[k7+=k9]; fr7=r23*r5; fr25=r1+r17, .align_code 4; if NLC0E, jump _BflyLastLoop; q[j1+=4]=r25:24; fr14=r22*r5; fr19=r14+r15;; q[j2+=4]=r27:26; q[j1+=4]=r25:24; q[j2+=4]=r27:26; q[j1+=4]=r25:24; q[j2+=4]=r27:26;; q[j1+=4]=r25:24;; q[j2+=4]=r27:26;; j11=[j27+0x19];; comp(j11,COMPLEX);; .align_code 4; if jeq, jump _FFTEpilogue;; fr6=r30*r12; fr15=r23*r4; fr7=r31*r13; fr14=r30*r13;

// F1+++, M4+, // F2+++, K2+++, M2++, // S1, M3++,

fr16=r6-r7;; fr24=r8+r18, fr26=r8-r18;; fr25=r9+r19, fr27=r9-r19;; fr17=r14+r15;; fr18=r6-r7;; fr15=r31*r12; fr24=r20+r16, fr26=r20-r16;; fr25=r21+r17, fr27=r21-r17;; fr19=r14+r15;; fr24=r28+r18, fr26=r28-r18;; fr25=r29+r19, fr27=r29-r19;;

// S2----, M1-, A1-// M4--, A3--// M2-, A4--// S1---, M3-, A2-// S2---, A1// M4-, A3-// A4-// S1--, A2// S2-// A3// A4// S1// S2// j11=COMPLEX or REAL, off the stack // Complex or Real? // If Complex, done

//******************************* Real re-combine ******************************** k8=k31+_twiddles; j0=j31+j7;; k9=ashiftr k9; j10=j31+j7;; j14=j17+j17;; j14=j14-4;; j1=j0+j14;; j14=j10+j14;; j29 = ashiftr j17;; k15=k31+MAX_FFT_SIZE/4; j30=ashiftr j29;; j30 = ashiftr j30;; k8=k8+k9; r0=l[j7+j17];; j0=j0+2; k12=k8+k15;; LC0=j30; fr0=r0+r0; j2=j0+j29;; j3=j1-j29;; xfr0=-r0; j10=j10+2; k10=yr0;; j12=j10+j29;; if LC0E; j13=j14-j29; k11=xr0;; yr3:0=DAB q[j0+=4];; xr3:0=DAB q[j2+=4];; yr3:0=DAB q[j0+=4];; xr3:0=DAB q[j2+=4];; yr7:4=q[j1+=-4]; xr9:8=l[k12+=k9];; xr7:4=q[j3+=-4]; xr11:10=l[k12+=k9];; if LC0E; fr16=r0+r6, fr20=r0-r6; yr9:8=l[k8+=k9];; fr18=r2+r4, fr22=r2-r4; yr11:10=l[k8+=k9];; fr24=r20*r9; fr21=r1+r7, fr17=r1-r7;; fr26=r22*r11; fr23=r3+r5, fr19=r3-r5; xr3:0=DAB q[j2+=4];; //j17=N/2, j7=output // k8->twiddles, j0->internal buffer // k9=twiddle stride, j10->internal buffer // j14=N (N/2 complex values) // j14=N-4 real=N/2-2 complex // j1->internal buffer+(N/2-2) // j14->internal buffer+(N/2-2) // j29=N/4 // k15=N/4*twiddle_stride, j30=N/8 // N/16 // k8->twiddles+1, get G(N/4) // j0->internal buffer+1, k12->twiddles+N/8+1 // LC0=N/16, compute F(N/4)=2*conj(G(N/4)), // j2->internal buffer+1+N/8 // j3->internal buffer+3N/8-2 // j10->internal buffer+1, k10=Im(F(N/4)) // j12->internal buffer+N/8+1 // LC0=N/16-1, j13->internal buffer+3N/8-2, k11=Re(F(N/4)) // Prime the DAB // Prime the DAB // yr0=Re(G(n)), yr1=Im(G(n)), yr2=Re(G(n+1)), yr3=Im(G(n+1)) // xr0=Re(G(n+N/8)), xr1=Im(G(n+N/8)) // xr2=Re(G(n+1+N/8)), xr3=Im(G(n+1+N/8)) // yr4=Re(G(N/2-(n+1))), yr5=Im(G(N/2-(n+1))) // yr6=Re(G(N/2-n)), yr7=Im(G(N/2-n)) // twiddles(n+N/8) - want to mult by sin(x)-icos(x) // xr4=Re(G(N/2-(n+1+N/8))), xr5=Im(G(N/2-(n+1+N/8))) // xr6=Re(G(N/2-(n+N/8))), xr7=Im(G(N/2-(n+N/8))) // twiddles(n+1+N/8) // LC0=N/16-2, r16=Re(G(n)+conj(G(N/2-n))), // r20=Re(G(n)-conj(G(N/2-n))) // twiddles(n) // r18=Re(G(n+1)+conj(G(N/2-(n+1)))), // r22=Re(G(n+1)-conj(G(N/2-(n+1)))) // twiddles(n+1) // r24=s(n)*Re(G(n)-conj(G(N/2-n))) // r17=Im(G(n)+conj(G(N/2-n))), r21=Im(G(n)-conj(G(N/2-n))) // r26=s(n+1)*Re(G(n+1)-conj(G(N/2-(n+1)))) // r19=Im(G(n+1)+conj(G(N/2-(n+1)))), // r23=Im(G(n+1)-conj(G(N/2-(n+1)))) // xr3:0=next G(n+2+N/8), G(n+3+N/8) // r25=c(n)*Im(G(n)-conj(G(N/2-n))), // yr3:0=next G(n+2), G(n+3) // r27=c(n+1)*Im(G(n+1)-conj(G(N/2-(n+1)))) // r24=Re(-i*exp(2*pi*i*n)(G(n)-conj(G(N/2-n)))) // r13=s(n)*Im(G(n)-conj(G(N/2-n))) // yr7:4=next G(N/2-(n+2)), G(N/2-(n+3)) // r26=Re(-i*exp(2*pi*i*(n+1))(G(n+1)-conj(G(N/2-(n+1)))))

fr25=r21*r8;

yr3:0=DAB q[j0+=4];;

fr27=r23*r10;; fr24=r24+r25; fr25=r21*r9; yr7:4=q[j1+=-4];; fr26=r26+r27; fr27=r23*r11; xr7:4=q[j3+=-4];;

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a fr13=r20*r8; fr12=r16+r24, fr30=r16-r24;; fr15=r22*r10; fr14=r18+r26, fr28=r18-r26;; fr13=r25-r13; xr9:8=l[k12+=k9];; fr15=r27-r15; xr11:10=l[k12+=k9];; .align_code 4; _combine_stage: fr16=r0+r6, fr20=r0-r6; yr9:8=l[k8+=k9];; fr18=r2+r4, fr22=r2-r4; yr11:10=l[k8+=k9];; fr13=r13+r17, fr31=r13-r17;; fr15=r15+r19, fr29=r15-r19; l[j12+=2]=xr13:12;; fr24=r20*r9; fr21=r1+r7, fr17=r1-r7; q[j14+=-4]=yr31:28;; // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // r27=s(n+1)*Im(G(n+1)-conj(G(N/2-(n+1)))) xr7:4=next G(N/2-(n+2+N/8)), G(N/2-(n+3+N/8)) r13=c(n)*Re(G(n)-conj(G(N/2-n))), r12=Re(F(n)), r30=Re(F(N/2-n)) r15=c(n+1)*Re(G(n+1)-conj(G(N/2-(n+1)))) r14=Re(F(n+1)), r28=Re(F(N/2-(n+1))) r13=Im(-i*exp(2*pi*i*x)(G(n)-conj(G(N/2-n)))), next twiddles(n+2+N/8) r15=Im(-i*exp(2*pi*i*x)(G(n+1)-conj(G(N/2-(n+1))))) next twiddles(n+3+N/8) r16=Re(G(n+2)+conj(G(N/2-(n+2)))), r20=Re(G(n+2)-conj(G(N/2-(n+2)))) next twiddles(n+2) r18=Re(G(n+3)+conj(G(N/2-(n+3)))), r22=Re(G(n+3)-conj(G(N/2-(n+3)))) next twiddles(n+3) r13=Im(F(n)), r31=Im(F(N/2-n)) r15=Im(F(n+1)), r29=Im(F(N/2-(n+1))), store F(n+N/8) r24=s(n+2)*Re(G(n+2)-conj(G(N/2-(n+2)))) r21=Im(G(n+2)+conj(G(N/2-(n+2)))), r17=Im(G(n+2)-conj(G(N/2-(n+2)))) store F(N/2-n), F(N/2-(n+1)) r26=s(n+3)*Re(G(n+3)-conj(G(N/2-(n+3)))) r23=Im(G(n+3)+conj(G(N/2-(n+3)))), r19=Im(G(n+3)-conj(G(N/2-(n+3)))) xr3:0=next G(n+4+N/8), G(n+5+N/8) r25=c(n+2)*Im(G(n+2)-conj(G(N/2-(n+2)))) yr3:0=next G(n+4), G(n+5) r27=c(n+3)*Im(G(n+3)-conj(G(N/2-(n+3)))) store F(N/2-(n+N/8)), F(N/2-(n+1+N/8)) r24=Re(-i*exp(2*pi*i*x)(G(n+2)-conj(G(N/2-(n+2))))) r25=s(n+2)*Im(G(n+2)-conj(G(N/2-(n+2)))), store F(n) r26=Re(-i*exp(2*pi*i*x)(G(n+3)-conj(G(N/2-(n+3))))) r27=s(n+3)*Im(G(n+3)-conj(G(N/2-(n+3)))), store F(n+1) r13=cos(n+2)*Re(G(n+2)-conj(G(N/2-(n+2)))) r12=Re(F(n+2)), r30=Re(F(N/2-(n+2))), store F(n+1+N/8) r15=cos(n+3)*Re(G(n+3)-conj(G(N/2-(n+3)))) r14=Re(F(n+3)), r28=Re(F(N/2-(n+3))) xr7:4=next G(N/2-(n+4+N/8)), G(N/2-(n+5+N/8)) r13=Im(-i*exp(2*pi*i*x)(G(n+2)-conj(G(N/2-(n+2))))) next twiddles(n+4+N/8) yr7:4=next G(N/2-(n+4)), G(N/2-(n+5))

fr26=r22*r11; fr23=r3+r5, fr19=r3-r5; xr3:0=DAB q[j2+=4];;

fr25=r21*r8; yr3:0=DAB q[j0+=4];; fr27=r23*r10; q[j13+=-4]=xr31:28;; fr24=r24+r25; fr25=r21*r9; l[j10+=2]=yr13:12;; fr26=r26+r27; fr27=r23*r11; l[j10+=2]=yr15:14;; fr13=r20*r8; fr12=r16+r24, fr30=r16-r24; l[j12+=2]=xr15:14;; fr15=r22*r10; fr14=r18+r26, fr28=r18-r26; xr7:4=q[j3+=-4];; fr13=r25-r13; xr9:8=l[k12+=k9]; yr7:4=q[j1+=-4];; .align_code 4;

if NLC0E, jump _combine_stage(P); fr15=r27-r15; xr11:10=l[k12+=k9];;// r15=Im(-i*exp(2*pi*i*x)(G(n+3)-conj(G(N/2-(n+3))))) // next twiddles(n+5+N/8) fr16=r0+r6, fr20=r0-r6; yr9:8=l[k8+=k9];; fr18=r2+r4, fr22=r2-r4; yr11:10=l[k8+=k9];; fr13=r13+r17, fr31=r13-r17;; fr15=r15+r19, fr29=r15-r19;; fr24=r20*r9; fr21=r1+r7, fr17=r1-r7; yr1:0=l[j31+j7];; // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // // r16=Re(G(n+4)+conj(G(N/2-(n+4)))), r20=Re(G(n+4)-conj(G(N/2-(n+4)))) next twiddles(n+4) r18=Re(G(n+5)+conj(G(N/2-(n+5)))), r22=Re(G(n+5)-conj(G(N/2-(n+5)))) next twiddles(n+5) r13=Im(F(n+2)), r31=Im(F(N/2-(n+2))) r15=Im(F(n+3)), r29=Im(F(N/2-(n+3))) r24=s(n+4)*Re(G(n+4)-conj(G(N/2-(n+4)))) r21=Im(G(n+4)+conj(G(N/2-(n+4)))), r17=Im(G(n+4)-conj(G(N/2-(n+4)))) yr0=Re(G(0)), yr1=Im(G(0)) r26=s(n+5)*Re(G(n+5)-conj(G(N/2-(n+5)))) r23=Im(G(n+5)+conj(G(N/2-(n+5)))), r19=Im(G(n+5)-conj(G(N/2-(n+5)))) r25=cos(x)*Im(G(n)-conj(G(N/2-n))) r27=cos(x)*Im(G(n)-conj(G(N/2-n))) store F(n+2+N/8) yr0=Re(G(0))+Im(G(0)), yr1=0=Im(F(0)) store F(N/2-(n+2)), F(N/2-(n+3)) yr0=Re(F(0)) store F(N/2-(n+2+N/8)), F(N/2-(n+3+N/8)) r24=Re(-i*exp(2*pi*i*x)(G(n+4)-conj(G(N/2-(n+4))))) r25=s(n+4)*Im(G(n+4)-conj(G(N/2-(n+4)))), store F(n+2) r26=Re(-i*exp(2*pi*i*x)(G(n+5)-conj(G(N/2-(n+5))))) r27=s(n+5)*Im(G(n+5)-conj(G(N/2-(n+5)))), store F(n+3) r13=c(n+4)*Re(G(n+4)-conj(G(N/2-(n+4)))) r12=Re(F(n+4)), r30=Re(F(N/2-(n+4))), store F(n+3+N/8) r15=c(n+5)*Re(G(n+5)-conj(G(N/2-(n+5)))) r14=Re(F(n+5)), r28=Re(F(N/2-(n+5))) store F(0) r13=Im(-i*exp(2*pi*i*x)(G(n+4)-conj(G(N/2-(n+4))))) store F(N/4) r15=Im(-i*exp(2*pi*i*x)(G(n+5)-conj(G(N/2-(n+5))))) r13=Im(F(n+4)), r31=Im(F(N/2-(n+4))) r15=Im(F(n+5)), r29=Im(F(N/2-(n+5))), store F(n+4+N/8) store F(N/2-(n+4)), F(N/2-(n+5)) store F(N/2-(n+4+N/8)), F(N/2-(n+5+N/8)) store F(n+4) store F(n+5) store F(n+5+N/8)

fr26=r22*r11; fr23=r3+r5, fr19=r3-r5;; fr25=r21*r8;; fr27=r23*r10; l[j12+=2]=xr13:12;; yfr0=r1+r0; yr1=lshift r1 by -32; q[j14+=-4]=yr31:28;; yfr0=r0+r0; q[j13+=-4]=xr31:28;; fr24=r24+r25; fr25=r21*r9; l[j10+=2]=yr13:12;; fr26=r26+r27; fr27=r23*r11; l[j10+=2]=yr15:14;; fr13=r20*r8; fr12=r16+r24, fr30=r16-r24; l[j12+=2]=xr15:14;; fr15=r22*r10; fr14=r18+r26, fr28=r18-r26; l[j31+j7]=yr1:0;; fr13=r25-r13; l[j7+j17]=k11:10;; fr15=r27-r15;; fr13=r13+r17, fr31=r13-r17;; fr15=r15+r19, fr29=r15-r19; l[j12+=2]=xr13:12;; q[j14+=-4]=yr31:28;; q[j13+=-4]=xr31:28;; l[j10+=2]=yr13:12;; l[j10+=2]=yr15:14;; l[j12+=2]=xr15:14;; //******************************** Epilogue ********************************** _FFTEpilogue: mPOPQ(yR27:24) mPOPQ(yR31:28) mPOPQ(xR27:24) mPOPQ(xR31:28) mRETURN

Writing Efficient Floating-Point FFTs for ADSP-TS201 TigerSHARC® Processors (EE-218)

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a
//********************* End Label For Statistical Profiling ****************** _FFT32.end:

Listing 3. fft32.asm

References
[1] ADSP-TS201 TigerSHARC Processor Programming Reference. Revision 0.1, June 2003. Analog Devices, Inc.

Document History
Revision Rev 2 – March 04, 2004 by Boris Lerner Rev 1 – December 18, 2003 by Boris Lerner Description Added mention of the real stage and updated the calling examples appropriately.

Initial Release

Writing Efficient Floating-Point FFTs for ADSP-TS201 TigerSHARC® Processors (EE-218)

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