多通道音頻信號處理 (DTMF decoders)

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1 多通道音頻信號處理 (DTMF decoders)
指導教授:蔡亮宙老師 簡聰富老師 學 生:劉念宗 中華民國 九十 年 九 月 十二 日

2 主題架構 數位振盪器(Digital Oscillators) Goertzel’s algorithm (演算法)
Interrupt(中斷)之資料結構與應用

3 數位振盪器原理推導 利用正弦波 sin x 的指數形式 : 可以得到正弦序列 : 的 z 轉換 設系統的轉移函數 H(z) = Y(z)/X(z) 就是正弦序列 sinkwT的 z 轉換,即

4 Digital Oscillators(數位振盪器)
Digital Sinusoidal Oscillators :  H(z) = [b0*z^(-1)]/[1+a1*z^(-1)+a2*z^(-2)] -----(1) where b0 = Asin(w0) a1 = -2*cos(w0) a2 = 1 w0= 2πfi/fs ;fs：sampling frequency  H(z) = [Asin(w0)*z^(-1)]/[1-2*cos(w0)*z^(-1)+z^(-2)]  P1,2 = [cos(w0)+j sin(w0)]，[cos(w0)-j sin(w0)] = 1∠w0

5 探討Digital Oscillators 與 DTMF decoders
P1,2是一對複數根，其振幅值為1，Phase為w0。振幅值為1的極點對應一個數位振盪器，其頻率由係數b0，a1和a2來決定 。 b0 = Asin(w0) a1 = -2*cos(w0) z-plane a2 = 1 因此，設計一個振盪器，主要在於確定這些係數

6 H(z) = [b0*z^(-1)]/[1+a1*z^(-1)+a2*z^(-2)] -----(1)
由上式(1) : H(z) = [b0*z^(-1)]/[1+a1*z^(-1)+a2*z^(-2)] -----(1) 可得： Y(z)+a1*z^(-1)*Y(z)+a2*z^(-2)*Y(z) = b0*z^(-1)*X(z)  y[n]+a1*y[n-1]+a2*y[n-2] = b0*x[n-1] 或是  y[n] = -a1*y[n-1]-a2*y[n-2]+b0*x[n-1] (2) 上列(2)式，是一個二階差分方程式。 而DTMF decoder 方程式為: Y(n)= X(n)+2cos(coef)×Y(n-1)+(-1)Y(n-2) (3) where coef =360×frequency/sample rate X(n) is the current sample value. Y(n-1) and Y(n-2) are feedback storage elements

7 where coef= 2πfi/fs ; fs：sampling frequency
將上式(2) ---數位振盪器的差分方程式與 上式(3) ---DTMF decoder方程式互相比對，此時 b0 = 1 a1 = -2*cos(coef) a2 = 1 where coef= 2πfi/fs ; fs：sampling frequency 令H(z)的分母[z^2-2*cos(coef)*z+1]=0,求得: P1,2= [cos(coef)+j sin(coef)]，[cos(coef)-j sin(coef)] = 1∠coef 可以明確的知道：DTMF的行(高頻),列(低頻)等八個 頻率之組合是由數位振盪器來提供的。

8 再者，DTMF tone 並非只產生頻率，工作時間間隔 也必須指明。根據AT&T的規格說明，對於按鍵式
撥號電話而言，信號的最大傳輸速率必須為 10digits/sec(或100msec/digit)。 因此，a sample-store interval takes approximately 25 ms. 並且為了解決32 channels在DFT processing時，無法儲 存時間(save time)而使得資料遺失(data loss)的問題 使用Buffer來儲存要傳送給DFT處理的輸入資料流 (input stream)。

9 The buffer uses four levels to store the input data during
a 100msec interval. Buffer Size = N 4 levels = 32 K words where the number of samples, N, is 200 ~ 256.

10 Goertzel’s algorithm (演算法)
X(n) Vk(n) Y(n,k,N) 2×cos(2πk/N) - exp(-j2×π×k/N) -1 <Goertzel’s algorithm(演算法)應用於二階IIR濾波器的結構圖> + + Z^(-1) + Z^(-1)

11 解析運用在DTMF的Goertzel’s algorithm
The DTMF frequency is transformed to a discrete fourier transform (DFT) coefficient. The relationship between the DTMF frequency (fi) and the DFT coefficient (k) is : k = N×fi/fs where fs = Sampling frequency N = Filter length For each k, the state variable, Vk(n) is: Vk(n) = 2×cos(2×π×k/N)×Vk(n-1)-Vk(n-2)+X(n) where n = 0,1,---,N Thereafter, the output, Y(n,k,N) is: Y(n,k,N) = Vk(N)- exp(-j2×π×k/N) ×Vk(N-1)

12 Goertzel’s algorithm與DTMF之能量 轉換
The DTMF frequency at which the incoming signal has maximum energy is the detected frequency. This energy calculation is : 經Goertzel’s algorithm運算後的頻譜 |Y(n,k,N)|^2 = Vk(N)^2+Vk(N-1)^2 –2*cos(2πk/N)* Vk(N)^2*Vk(N-1)^2

13 DTMF decoders與Goertzel’s algorithm
(1).使用Goertzel’s演算法執行離散傅立葉轉換(discrete Fourier transform ,DFT)。 每個頻率只需要一個實數係數就可以決定數值的 振幅。 (2).Goertzel’s algorithm只要獲得任何取樣就可以立刻處 理，不必全數獲得取樣之後才進行處理。 (3).Goertzel’s algorithm運用於DTMF解碼上是非常適合 的，其優點為: A.步驟少、執行速度快。 B.所需常數較少，可節省記憶體空間。 DTMF decoders與Goertzel’s algorithm

14 Y(n)= X(n)+2cos(coef)×Y(n-1)+(-1)Y(n-2)
(4) The DTMF decoder equation is: Y(n)= X(n)+2cos(coef)×Y(n-1)+(-1)Y(n-2) where coef =360×frequency/sample rate X(n) is the current sample value. Y(n-1) and Y(n-2) are feedback storage elements for the frequency point ,k. The value of k is : k = N × Freq/(Sample Rate) The decoder equation uses the initial data : Y(-1) = 0, Y(-2) = 0, Y(-3) = 0,…for n =0,1,2,…,n-1 The sample rate is 8 KHz.

15

16 <Tone Decoder ---16 DFT Transforms >

17 <Modified DTMF Tone Decoder >

18 系統流程圖 No Yes No Yes Yes Find Max Channel No Initialize
Below Threshold Initialize Signal Strength? Above Threshold Pointer to Next 2 Channels Below Threshold Relative Peak? Above Threshold Get Channel Samples and Store to Buffer Channel row.4 DFTs and Find Max Channel Row Energy No Channel Column 4 DFTs Twist & Level Yes No Nth Sample of DFT Channel-2nd Harmonic DFT of Max Column and Max Row Yes Yes Find Max Channel Column Energy 2nd Harmonic Signal too Strong? No

19 to Pointer to Next 2 Channels
系統流程圖(續) to Pointer to Next 2 Channels to Find Max Channel Column Energy No Same as 2nd to Last Digit? Yes No No Same as Last Digit Yes Valid Tone Output No Channel Finished? Yes

20 System Implementation
RecvInt4 Main Program (主程式) INT2 ISR ,turn on IMR Recv int No Overflow? Yes RecvInt5 RecvInt Cout0Flg(= -1時,205點收到) RecvInt6 PQueFlg(Recv int use Nth queue block buffer) No Overflow? PProcFlg(n)(Block(n)資料 已經填滿時 = -1) Yes RecvInt7

21 Interrupt Set up the ISR vector INT2 INT2 ISR ( Frame 0 INT10 ) INT11
Receive int subroutine RecvInt RecvInt4 RecvInt5 RecvInt6 RecvInt7

22 Set up the ISR vector .sect "vectors" b Main b Int1 ;02 INT1 interrupt
b Int ;04 INT2 interrupt(Frame 0) b Int ;06 INT3 interrupt .space 2 * ;08 TINT interrupt  Rint: b RecvInt ;0A Serial port receive interrupt RI XInt: b TransInt ;0C Serial port transmit interrupt X TRnt: b TRecvInt ;0E TDM receive interrupt TXnt: b TTransInt ;10 TDM transmit interrupt .space 18 * ;12

23 Ch0 Ch1 Ch2 Ch3 F Channel us 8kHz Tchannel = 125us/32 = 3.9 us

24

25 INT2 ISR ( Frame 0 ) Int2: ;turn on IMR Recv int
;FilCout(取205點)/=0,執行Int10 ;Cout0Flg /=0,執行Int11 ;否則,read DRR data Int10: ;FilCout – 1 ;set ar5CoefBase3c,ar6F32Ch0 ; ar4PQue(B1QueCh0) & PQue+1 CoefBase3c,ar6F32Ch0, ar4PQue(B1QueCh0) & PQue+1

26 ;caculate Filter result Block address ;Block move F11Ch31 BnF11Ch31
Int11: ;stop Recv Int ;caculate Filter result Block address ;Block move F11Ch31 BnF11Ch31 ;以ar5指向Block(n)(第n個Block) ;Clear F11Ch31~F32Ch0 ;set Block(0 ~ 3)資料填滿時PProcFlg(n) = -1 ;set Block number(0 ~ 3),並以PQueFlg表示 ;caculate PQue address

27 Int2: ldp #FilCout ;#FilCout=#205 lmmr IMR,#D16h ;turn on IMR Recv int, lacc FilCout,0 ;FilCout =?0 bcnd Int10,neq lacc Cout0Flg ;Cout0Flg complement cmpl sacl Cout0Flg bcnd Int11,neq lamm DRR ;read DRR data lacc #205

28 INT10 INT10: sub One ;FilCout - 1 sacl FilCout,0
lar ar5,#CoefBase3c ;set Ar5 - Ar6 lar ar6,#F32Ch0 lar ar4,PQue ;set ar4 = PQue & PQue + 1 lacc PQue add One sacl PQue rete

29 Int11 ;stop Recv Int ;caculate Filter result Block addr ;Block move F11Ch31 BnF11Ch31 ;Clear F11Ch31 ~ F32Ch0 ;set PProcFlg(n) = -1,即表示Block資料已填滿 ;caculate PQue addr

30 Int11 Int11: lmmr IMR,#D06h ;stop Recv Int,RINT=0 ;Stop Recv Int ;caculate Filter result Block addr lt PQueFlg ;(PQueFlg)TREG0 ;PQueFlg:Block number mpy #192 ;(TREG0)*#192PREG pac ;(PREG)ACC adds PB1F11Ch31 ;(ACC)+(PB1F11Ch31) ;B1F11Ch31ACC sacl TempInt ;16 LSBs of (ACC)TempInt ;TempInt=[(PQueFlg)*#192+B1F11Ch31]

31 mar *,ar5 ;Block move F11Ch31 BnF11Ch31
lar ar5,TempInt ;ar5負責做選取4個Block的哪一個 ;Block(Bn) rpt # ;#191RPTC,6*32=192 bldd #F11Ch31,*+ lar ar5,#F11Ch31 ;Clear F11Ch31 ~ F32Ch0 zap ;0ACC,0PREG rpt #191 sacl *+ lacc PQueFlg ;set PProcFlg(n) = -1 ;表示由ar5選取到的Block資料已經完 ;全填完 adds PPProcFlg0 sacl TempInt

32 lacc PQueFlg ;PQueFlg+1 & #3 PQueFlg add One and Three ;Block number
lar ar5,TempInt zap ;0ACC,0PREG sub One sacl * lacc PQueFlg ;PQueFlg+1 & #3 PQueFlg add One and Three ;Block number ;(Bn0~3,總共4個Block) sacl PQueFlg

33 ;caculus PQue addr lt PQueFlg ;(PQueFlg)TREG0 mpy #205*32 ;(TREG0)*#205*32PREG pac ;PREGACC ;(ACC)=[(PQueFlg)*#205*32] adds PB1QueCh0 ;(ACC)+(PB1QueCh0) ;B1QueCh0ACC sacl PQue ;16 LSBs of (ACC)PQue rete

34 lmmr IMR,#D16h (1)lmmr : Load Memory-Mapped Register
(1) (2) (3) (1)lmmr : Load Memory-Mapped Register Syntax Direct: [label] lmmr dma,#addr Indirect: [label] lmmr {ind},#addr [,next ARP]

35 (src,addressed by PFC)(dst,specified by lower
承上(1)lmmr Execution PFC  MCS (PC)+2  PC 1k  PFC While (repeat counter/=0): (src,addressed by PFC)(dst,specified by lower 7 bits of dma) (PFC)+1  PFC (repeat counter)-1 repeat counter MCS  PFC

36 (2)IMR: Interrupt Mask Register
A 16-bit memory-mapped register used to mask interrupts. (3)#D16h :以D16h為位址,其內容值為 16h = (2) RINT=1,start Recv Int Reserved /INT4 TXNT TRNT XINT RINT TINT /INT3 /INT2 /INT1

37 lamm DRR (1)lamm : Load Accumulator With Memory - Mapped Register
(1) (2) (1)lamm : Load Accumulator With Memory - Mapped Register Syntax Direct: [label] lamm dma Indirect: [label] lamm {ind} [,next ARP] Execution (PC)+1  PC (dma)  ACC , Not affected by SXM

38 (2)DRR : Data Receive Register

39 Burst Mode Serial Port Receive Operation

40 Burst Mode Serial Port Receive Operation & 16 bits 資料傳送/接收的必備條件
(1).If FSM=1 ,every transfer and the mode is referred to as burst mode. (2).If FO=0,data is transmitted and/or received as 16- bits words. (3).在主程式(Main Program)中的Initialization: splk #00c8h,SPC;定義SPC中FSM=1,FO=0

41 PQue( B1QueCh0 ~ B1QueCh31) 特寫

42 ;Receive int subroutine ;Processing 2 channels in single Interrupt
RecvInt: ;DRR u_law Expanding ;(1).(DRR)ACCL(Ch1 Ch0) ;(2).取出Ch0(低8Bits) ;(3).經查表(u_Tab)解壓縮後 ; 存入B1QueCh0~B1QueCh31

43 RecvInt4 (2)計算X(n)+(-1)*Y(n-2)
(3)計算cos(coef)*Y(n-1) ;Y(n)=X(n)+2*cos(coef)*Y(n-1)+(-1)*Y(n-2) ; (ar4) (1)計算Column係數 ;(4)計算X(n)-Y(n-2)+cos(coef)*Y(n-1) ;(5)再計算X(n)+2*cos(coef)*Y(n-1)+(-1)*Y(n-2) ;4 frequency IIR ;ar4PQue(B1QueCh0 ~ B1QueCh31;X(n)) ;ar5CoefBase3c,2c,1c(cos(coef)) ;ar6F32Ch0(Y(n-2)),F31Ch0(Y(n-1))

44 RecvInt5: ;Get high byte data RecvInt6: ;4 frequency IIR of 2nd channel

45 Receive Interrupt RecvInt: ldp #TempInt lamm DRR sacl TempInt
mar *,ar4 ;*ar4:PQue ;(B1QueCh0 ~ B1QueCh31) and #0ffh ;ACC and 0ffh(保留低8位元) add #uTab ;(ACC)+#uTabACC tblr *,ar5 ;next *ar5:#CoefBase3c lar ar7,# ;4 Freq of 1st channel

46 Indirect: [label] tblr {ind} [,next ARP]
tblr *,ar5 tblr : Table Read Syntax Direct: [label] tblr dma Indirect: [label] tblr {ind} [,next ARP] Execution (PC)+1 PC (PFC)  MCS (ACC(15-0)) PFC

47 承上tblr 所述: Then ( pma,addressed by PFC ) dma
If ( repeat counter )/=0: Then ( pma,addressed by PFC ) dma Modify AR(ARP) and ARP as specified, ( PFC )+1 PFC (repeat counter )-1  repeat counter. Else ( pma,addressed by PFC )  dma, Modify AR(ARP) and ARP as specified. ( MCS ) PFC

48 Example

49 RecvInt4 RecvInt4 (2)計算X(n)+(-1)*Y(n-2) (3)計算cos(coef)*Y(n-1)
;Y(n)=X(n)+2*cos(coef)*Y(n-1)+(-1)*Y(n-2) ; (ar4) (1)計算Column係數 ;(4)計算X(n)-Y(n-2)+cos(coef)*Y(n-1) ;(5)再計算X(n)+2*cos(coef)*Y(n-1)+(-1)*Y(n-2) ;4 frequency IIR ;ar4PQue(B1QueCh0 ~ B1QueCh31;X(n)) ;ar5CoefBase3c,2c,1c(cos(coef)) ;ar6F32Ch0(Y(n-2)),F31Ch0(Y(n-1))

50

51 lt *-,ar4 ;ar5(CoefBase3c,2c,1c)TREG0,ar5=ar5-1
;(CoefBase2c,1c,4r),next *PQue ---- (1) lacc *,12,ar6 ;ar4*2^12ACC,next *ar6 sub *-, ;(ACC)-[(F32Ch0~F11Ch31)*2^16]ACC ;ar6F31Ch (2) mpy * ;(TREG0)*(ar6)PREG (3) ltd * ;(ar6)TREG0,(ar6)ar (4) ;(ACC)+PREGACC apac ;(ACC)+PREGACC

52 RecvInt5: ;Get high byte data
sach *-,0,ar ;(ACC)(ar6+1),ar6-1, ;next *ar7(3) bcnd RecvInt5,ov ;if(Overflow Flag)=1, ;跳至RecvInt5執行 banz RecvInt4,*-,ar5 ;if ar7/=0,跳至RecvInt4執行 ov = 1 RecvInt5: ;Get high byte data

53 ltd * Previous Product,and Move Data Syntax Direct: [label] ltd dma
ltd : Load TREG0,Accumulate Previous Product,and Move Data Syntax Direct: [label] ltd dma Indirect: [label] {ind}[,next ARP]

54 承上ltd所述: Execution (PC) +1 PC (dma) TREG0 (dma)  dma+1
(ACC) + (shifted P register)  ACC Affected by OVM ,PM,and TRM;affects C and OV.

55 RecvInt5 RecvInt5 adrk #4 ;ar5+4 pointer proces lar ar7,#3
lacl TempInt ;Get high byte data ;(TempInt)ACC(15-0) bsar ;(ACC)/2^8ACC ----取出高8Bits ;ACCL( Ch1,3,5,..31) mar *,ar ;next *ar4(B1QueCh0) adrk # ;ar4+#205ar4 add #uTab ;(ACC)+#uTabACC tblr *,ar ;rear table,(ACC(15-0))(ar4)

56 以陣列形式表示Channel的資料結構 以陣列形式可將“Channel”與“點數”表示為 [Channel][點數] [0] [0]
[0~31][0~204] : [0][204] 即 [0][0] [1] [0] : : : [1][204] [31][204] : [31] [0] : [31][204]

57 RecvInt6 RecvInt6 lt *-,ar4 ;4 frequency IIR of 2nd channel
lacc *,12,ar6 sub *-,16 mpy * ltd * apac sach *-,0,ar7 bcnd RecvInt7,ov banz RecvInt6,*-,ar5

58 RecvInt7 ---告知已經Overflow
adrk # ;pointer process,ar5+4 ;adrk # ;pointer process mar *,ar ;next *ar4,PQue ;(B1QueCh0 ~ B1QueCh31) adrk # ;ar4+#205ar4 lacc #0ffffh ;output level 0 to phone samm DXR ;(ACC)  DXR rete ;Enable Interrupts and From Interrupt

59 補充資料 補充: sammStore Accumulator in Memory-Mapped Register
lmmr Load Memory-Mapped Register ; Ex: lmmr dma,#addr DXR Data transmit register DRR Data receive register XSR Transmit shift register (不能存取,但可 允許雙緩衝功能) RSR Receive shift register (不能存取,但可 IMR Interrupt mask register(16Bits)

60 The flow chart of INT2

61 The flow chart of the Check Peak level and Twist

62 Main Program ; Initialization Main: setc intm ;Set Interrupt Mode Bit in ST0 setc ovm ;Set Overflow Mode Bit in ST0 ldp #0 ;Load data memory pointer=0 opl #003ch,PMST ;ndx=1, ram=ovly=1 splk #00c8h,SPC ;soft=free=0, rrst=xrst=1, txm=0 ;mcm=0,fsm=1, fo=0 splk #00c8h,TSPC ;same SPC

63 Initialization spm 0 ;No shift of multiplier output (PM=00 in ST1)
setc sxm ;sxm=1,produces sign extension on data as it is ;passed into the accumulator through the scaling shifter. clrc cnf ;B1 = data memory lmmr INDX,#Two ;indx=2 zap ;0-->ACC,0-->PREG sacl CWSR ;CWSR=PDWSR=IOWSR=0 sacl PDWSR sacl IOWSR splk #06h,IMR ;Series Recv Int on & Int2 & Int3

64 Block Move ; block move mar *,ar ;table move,*ar1 lar ar1,#CoefBase1r;CoefBase1rar1 rpt #Len ;rpt #43 blpd #Table,* ;#Table=#0800h,#Table(ar1) ; Table ar1 CoefBase1r Base1r CoefBase2r : : : Harm1r : : : Len PB1QueCh0

65 lar ar1,#F11Ch31;clear filter result buffer6*32=192
rpt #191 sacl *+ clrc intm ldp #CoefBase1r ;check if need to process MainChk0: lar ar0,PGropSave0 ;PGropSave0(0)-->ar0 lar ar1,PGropSave1 ;PGropSave1(PProcFlg0)-->ar1 lar ar2,PGropSave2 ;PGropSave2(3)-->ar2

66 MainChk1 lacc * ;ar1(PProcFlg0)ACC bcnd MainProc0,neq ;PProcFlg(n)=? –1(Block(n)資料已經 ;填滿) mar *+,ar ; mar *+,ar ; banz MainChk1,*-,ar1; lar ar0,# ; lar ar1,#PProcFlg0 ; lar ar2,# ; b MainChk ;

67 MainProc(n) MainProc0: zap ;0ACC,0PREG sacl * ;0 ar1(PProcFlg(n))
sar ar0,PGropSave0 sar ar1,PGropSave1 sar ar2,PGropSave2 lacc # ;set PChSave=#31(32ch) sacl PChSave

68 MainProc1 MainProc1 lt PGropSave0 ;Group start then,PGropSave0TREG0
mpy # ;1.PFilC31b,PFilC31b pac adds PB1F31Ch31 ;PB1F31Ch31 sacl PFilC31b add One sacl PFilC32b

69 MainProc1(續) lt PGropSave0 ;2.PB1QueCh31 mpy #205*32 pac bd MainProc2
adds PB1QueCh ;PB1QueCh31 sacl PQueBuf

70 MainProc3 MainProc3: sar ar1,PChSave ;channel change then
lacc PQueBuf, ;1.PQueBuf-205,channel-1 sub #205 sacl PQueBuf lacc PFilC31b ;2.PFilC31b+6&PFilC32b+6 add #6 sacl PFilC31b add One sacl PFilC32b MainProc2: lmmr INDX,#Two ;indx=2

71 ; Caculate Energy at Col 1,2,3,4 and
; Find Max Energy of Col 1,2,3,4 ;ENERGY = [(Y(n-2)-Y(n-1))/2]^2 - Y(n-1)*Y(n-2)*(coef-1) EnergyCol: lar ar0,Two ;2ar0 lar ar1,PFilC32b ;(PFilC32b)ar1 lar ar2,#CoefBase3c ;(CoefBase3c)ar2 EnergyCol0: ;next * ar2 call EnergySr,*,ar ;ar2-1,[(ACC)*2^1]ar2 sach *-,1,ar ;next * ar0 banz EnergyCol0,*-,ar1 ;if ar0/=0,跳至EnergyCo10 ;執行,ar0-1,next * ar1

72

73 Find Column 1,2,3 Energy Max

74 Check Column 1,2,3,4 level and Relative peak
ChkLev: lacc ColMaxDat ;ColMaxDat must >2^4,(acc) = (ColMaxDat) sub One,4 bcnd NotCode,lt ;if (ACC<0),執行NotCode ; Check col1,2,3 Relative ratio ChkRelCol: ; mar *,ar : lar ar0,Two ;ar0= ar2One lar ar1,PFilC31b ;ar1=PFilC31b : lar ar2,One ;ar2=1 ar0Two lt ColMaxDat ;P=ColMaxDat*683 mpy #683 ar1PFilC31b PFilC32b 2 1

75 ChkRelCol0 lacc *0-,12,ar0 ;(ar1) - ColMaxDat*683 & ar1-2 spac bcnd ChkRelCol1,lt mar *,ar ;ar2 - 1 mar *-,ar0 ChkRelCol1 banz ChkRelCol0,*-,ar1 ;ar0 =?0 mar *,ar2 banz NotCode,*,ar ;ar2=?0

76 Clear Row & Harmonic Filter result
ClearFil: lar ar1,#FilR11b zap rpt #7 sacl *+ lar ar1,#FilR01h rpt #3

77 Process DFT of ROW 1,2,3,4 ;********************************************************* ; Process DFT of Row 1,2,3,4 DFTRow: lar ar0,BaseLen ;ar0 = 204 lar ar1,#FilR12b lar ar2,PQueBuf lar ar3,#CoefBase1r lacc Four sacl FreqCout4 call DFTCall,*,ar3

78 Caculate Energy at ROW 1,2,3,4 ;********************************************************* ; Caculate Energy at Row 1,2,3,4 and ; Find Max Energy of Row 1,2,3,4 ;ENERGY = [(Y(n-2)-Y(n-1))/2]^2 - Y(n-1)*Y(n-2)*(coef-1) EnergyRow: lar ar0,Three lar ar1,#FilR42b lar ar2,#CoefBase4r EnergyRow0: call EnergySr,*,ar ;21 sach *-,1,ar ;23..ar1 - 1 (Y(n-2) banz EnergyRow0,*-,ar1 ;24..ar0(FreqCount)=?0 ar0-1

79 ;---------------------- Find Row 1,2,3,4 Energy Max --------------------------
;RowMaxDat = row peak data RowMaxNo= row peak number FindRowMax: ; mar *,ar1 lacc Three ;RowMaxNo=3 sacl RowMaxNo FindRow lar ar2,Two ;ar2=2 lar ar1,#FilR41b ;RowMaxDat=(#FilR41b)&ar1=#FilR31b lacc *0- sacl RowMaxDat

80 FindRow0 lacc RowMaxDat sub * bcnd FindRow1,geq ;RowMax - (ar1) sar ar2,RowMaxNo lacc * sacl RowMaxDat FindRow1 mar *0-,ar2 banz FindRow0,*-,ar1

81 ;Check Peak level and Twist
TwistLevel: lacc ColMaxDat ;(ColMaxDat)-(RowMaxDat)ACC sub RowMaxDat bcnd TwistLevel0,geq ;if (ACC >=0)執行TwistLevel0 TwistRev lac ColMaxDat ;row > col sub One,4 bcnd NotCode,lt lac RowMaxDat ;row_max - col_max*12 sub ColMaxDat, ;8dB = 12 sub ColMaxDat,2

82 bcnd NotCode,geq ;row_max > =col_max*12執行NotCode
b DFTHarmRow TwistLevel0: lac RowMaxDat ;col > row sub One,4 bcnd NotCode,lt lac ColMaxDat ;col_max - row_max*3 sub RowMaxDat ;4dB = 3 sub RowMaxDat,1 bcnd NotCode,geq ;col_max >= row_max*3執行NotCode

83 ;------- Process 2nd Harmonic DFT of Max Row Energy-------
DFTHarmRow: mar *,ar3 lmmr INDX,#RowMaxNo ;INDX=3 lar ar0,HarmLen ;ar0=200 lar ar1,#FilR02h lar ar2,PQueBuf lar ar3,#CoefHarm1r mar *0+,ar3 lt *,ar ;1..T = Cos(*)

84 DFTHarmRow0之資料結構

85 DFTHarmRow0 lacc *+,12,ar ;3..X(n) sub *-, ;X(n)-Y(n-2) --> Acc mpy * ;Cos(*) * Y(n-1) --> P ltd *,ar ;X(n)+ Cos(*) * Y(n-1)-Y(n-2) ;Y(n-1)-->Y(n-2) lta *,ar ;X(n)+2Cos(*) * Y(n-1)-Y(n-2) apac apac ;X(n)+2*2Cos(*)*Y(n-1)-Y(n-2) sach *+,0,ar ;2..--> Y(n-1) Q30 bcnd HarmRowLev,ov ;check overflow banz DFTHarmRow0,*-,ar2 ;4..ar0(len)=?0

86 ;--------------- Check Row Harmonic Energy -------------
HarmRowLev: ; mar *,ar2 lar ar2,#CoefHarm1r;ar2 = harmonic coefficent Pointer mar *0+,ar2 ;= #coeff + row_max*2 lar ar1,#FilR02h ;ar1 = harmonic data pointer ;= #dat + row_mxp*2 call EnergySr,*,ar2 lt RowMaxDat ;harmonic energy row_max/4 mpy #4096 spac bcnd NotCode,geq

87 ;*********************************************************
; Process 2nd Harmonic DFT of Max Col Energy and ; Check Col Harmonic Energy ;*********************************************************DFTHarmCol: mar *,ar3 lmmr INDX,#ColMaxNo ;INDX=1 lar ar0,HarmLen ;ar0=200 lar ar1,#FilC02h lar ar2,PQueBuf lar ar3,#CoefHarm1c mar *0+,ar3 lt *,ar ;1..T = Cos(*)

88 DFTHarmCol0 lacc *+,12,ar ;3..X(n) sub *-, ;X(n)-Y(n-2)--> Acc mpy * ;Cos(*) * Y(n-1)--> P ltd *,ar ;X(n)+ Cos(*) * Y(n-1)-Y(n-2) ;Y(n-1)-->Y(n-2) lta *,ar ;X(n)+2Cos(*) * Y(n-1)-Y(n-2) apac apac ;X(n)+2*2Cos(*)*Y(n-1)-Y(n-2) sach *+,0,ar ;2..--> Y(n-1) Q30 bcnd HarmColLev,ov ;check overflow banz DFTHarmCol0,*-,ar2 ;4..ar0(len)=?0

89 DFTHarmCol0之資料結構

90 ;----------------- Check Col Harmonic Energy ------------
HarmColLev: mar *,ar2 lar ar2,#CoefHarm1c ;ar2 = harmonic coefficent pointer mar *0+,ar ;= #Coef + ColMax*2 lar ar1,#FilC02h ;ar1=harmonic data pointer = #dat ;ColMax*2 call EnergySr,*,ar2 lt ColMaxDat ;harmonic energy ColMax/4 mpy #4096 spac bcnd NotCode,geq

91 ;*********************************************************
; Check for valid number ; Compare number with last number and 2nd to last number ;********************************************************* ; load recognized number mar *,ar1 lt PChSave ;ch*3 + #Last1Ch0 mpy Three ;(PChSave)*3PREG lacc #Last0Ch0 ;ACC= #Last0Ch0 apac ;(ACC)+(PREG)ACC sacl TempMain,0 lar ar1,TempMain

92 mar *+,ar1 ;Last0Ch0+1 = Last1Ch0
dmov * ;last1 --> last2 dmov * ;last0 --> last1

93 ;----------------------- Check for new number ------------
ChkNewCode: lacc RowMaxNo,2 ;row_mxp*2^2 + col_mxp + Ch*2^8 ;--> last0 add ColMaxNo add PChSave,8 sacl * ;--> last0 mar *+,ar1 sub * ;last0 =? last2 bcnd NotCode1,eq ;ACC(last0-last2)=0,執行NotCode1

94 lac * ;last0 =? last1 sub *,0,ar1 bcnd ChkNewCode0,neq ;last0/=last1,執行ChkNewCode0 out *, ;Output Last0 Digital data ChkNewCode0 lar ar1,PChSave banz MainProc3,*-,ar1 ;PChSave=?0 & PChSave-1 b MainChk0 ;Load ar0,ar1,ar2

95 Not New Code -當(1)row_max-col_max >=8dB (2)col_max-row_max>=4dB 時執行
NotCode: lt PChSave ;ch*3 + #Last0Ch0 mpy Three lacc #Last0Ch0 apac ;ch*3 + #Last0Ch0ACC sacl TempMain, ;ACCTempMain NotCode1 lar ar1,TempMain mar *,ar1 lacc #0ffh bd ChkNewCode0 dmov * ;last0 --> last1 sacl *

96 ;********************************************************
; DFT Subroutine DFTCall: lt *,ar ;1..T = Cos(*) ,next *ar ;*ar3(CoefBase1r,2r,3r,4r) DFTCall0 lacc *+,12,ar ;3..X(n) sub *-, ;X(n)-Y(n-2)--> Acc mpy * ;Cos(*) * Y(n-1) --> P ltd *,ar ;X(n)+ Cos(*) * Y(n-1)-Y(n-2) ;Y(n-1)-->Y(n-2) lta *,ar ;X(n)+2Cos(*) * Y(n-1)-Y(n-2) apac apac ;X(n)+2*2Cos(*)*Y(n-1)-Y(n-2) sach *+,0,ar ;2..--> Y(n-1) Q30 bcnd DFTCall1,ov ;check overflow banz DFTCall0,*-,ar2 ;4..ar0(len)=?0

97 DFT Subroutine之資料結構

98 DFTCall1 lar ar0,BaseLen ;ar0=204 lar ar2,PQueBuf mar *,ar1 mar *0+,ar3 mar *+,ar3 lacc FreqCout4 ;FreqCout4-1 sub One sacl FreqCout4 bcnd DFTCall,neq ;ACC(FreqCout4-1)/=0,跳至DFTCall執行 ret

99 Energy Subroutine ar2=(coeff-1),負責指到Coefficent ar1Y(n-2),ar1-1Y(n-1)
CoefBase1r ~CoefHarm4c ar2=(coeff-1),負責指到Coefficent ar1Y(n-2),ar1-1Y(n-1) 運算式:[(Y(n-2)-Y(n-1))/2]^2-Y(n-1)*Y(n-2)*(coef-1) 執行步驟:(1).(coef-1)/2 (2).Y(n-2)*(coef-1)/2 (3).Y(n-1)*Y(n-2)*(coef-1)/2 (4).ABS[Y(n-2)-Y(n-1)]/2 (5).{[Y(n-2)-Y(n-1)]/2} ^2 (6).{[Y(n-2)-Y(n-1)]/2}^2-Y(n-1)*Y(n-2)*(coef-1)

100 ;------------------- ENERGY Subroutine--------------------
EnergySr: lac NegMax,15 ;(coef-1)/2 add *-,15,ar1 ;ar2=&coeff-1 sach TempMain ;[(Coef-1)/2]TempMain lt * ;Y(n-2)*(coef-1)/2,Y(n-2)TREG0 mpy TempMain ;(TREG0)*(TempMain)PREG pac ;[Y(n-2)*(coef-1)/2]ACC sach TempMain,1 lt * ;Y(n-1)*Y(n-2)*(coef-1)/2,Y(n-1)Y(n-2) mpy TempMain

101 pac sach TempMain,1 lac *-, ;ABS[Y(n-2)-Y(n-1)]/2,Y(n-1)Y(n-2) sub *,15 ;*Y(n-1) abs sach * lt * ;[Y(n-2)-Y(n-1)]/2 ^2 mpy * pac ;[(Y(n-2)-Y(n-1))/2]^2-Y(n-1)*Y(n-2)*(coef-1) sub TempMain,15 ret

102 Energy Subroutine ;caculate Energy at column 1,2,3:
;ar0=2,ar1PFilC32b,ar2CoefBase3c

103 Energy Subroutine(續) ;caculate Energy at Row 1,2,3,4: ;ar0=3,ar1FilR42b,ar2CoefBase4r

104 Energy Subroutine(續) ;check column Harmonic Energy
;ar0HarmLen(200),ar1FilC02h,ar2CoefHarm1c

105 Energy Subroutine(續) ;check Row Harmonic Energy: ;ar0HarmLen(200),ar1FilR02h,ar2CoefHarm1r

106 謝謝各位指導!! 學生:劉念宗 製作