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標(biāo)題: Verilog編程問(wèn)題 [打印本頁(yè)]

作者: eqgyzgs 時(shí)間: 2012-7-18 23:43
標(biāo)題: Verilog編程問(wèn)題
我在編寫(xiě)FFT內(nèi)核的時(shí)候在大循環(huán)里總是出現(xiàn)下面的錯(cuò)誤
Error (10119): Verilog HDL Loop Statement error at Test.v(72): loop with non-constant loop condition must terminate within 250 iterations
Verilog不能支持250層以上的循環(huán)么？求高手指教。
有沒(méi)有現(xiàn)成的FFT內(nèi)核，希望能參考一下。

一下為代碼
/*
FFT code
*/
module Test(CLK_50M,REST,CS,ADCDAT,FULL,BUSY,EXT,OUTDAT);
input CLK_50M,REST,CS,EXT;
input [ADCBit-1:0] ADCDAT;
output FULL,BUSY;
output [63:0] OUTDAT;//計(jì)算結(jié)果是浮點(diǎn)型
//reg [ADCBit-1:0] ADCDAT;
reg [31:0] OUTDAT;
reg FULL,BUSY;
reg [15:0] CNT;
reg [15:0] N;
reg CLK;
reg [ADCBit-1:0] ADCMEMORY[FFTN-1:0];//輸入為實(shí)數(shù)
reg [31:0] DOREMEMORY[FFTN-1:0],DOIMMEMORY[FFTN-1:0];//輸出為復(fù)數(shù)
reg [15:0] i,j,k;//中間過(guò)程變量寄存器
reg [15:0] f,m,l,le,lei,ip;
reg [ADCBit-1:0] temp;
reg [31:0] u[1:0],v[1:0],w[1:0];//x[0]=real,x[1]=image
parameter FFTN=16'd256,//FFT點(diǎn)數(shù)
Pi=3.14159,//pi
ADCBit=4'd8,//8bit ADC
CLKDiv=16'd50;//1MHz

always @(posedge CLK_50M or negedge REST)
if(!REST) CNT <= 16'h0000;
else if((!CS)&&(CNT!=CLKDiv)) CNT <= CNT+1'b1;
else CNT <= 16'h0000;
always @(posedge CLK_50M or negedge REST)
if(!REST) CLK <= 1'b0;
else if(CNT==CLKDiv) CLK <= ~CLK;

always @(posedge CLK or negedge REST)//ACD數(shù)據(jù)輸入
if(!REST) N <= 16'h0000;
else if(!CS) begin ADCMEMORY[N] <= ADCDAT;N <= N+1'b1; end

always @(N)
if(N==FFTN-1) FULL <= 1'b1;
else FULL <= 1'b0;

always @(posedge CLK_50M or negedge REST)
if(!REST) begin BUSY <= 1'b0;j <= 16'h0000;f <= FFTN/16'd2; end
else if(N==FFTN-1)
  begin
BUSY <= 1'b1;
for(i=16'h0000;i begin
   if(i    begin
   temp <= ADCMEMORY[j];
   ADCMEMORY[j] <= ADCMEMORY[i];
   ADCMEMORY[i] <= temp;
   end
   k <= FFTN/16'd2;
   while(k<=j)
   begin
   j <= j-k;
   k <= k/16'd2;
   end
   j <= j+k;
end
for(i=16'h0000;i begin
   DOREMEMORY[i] <= ADCMEMORY[i];
   DOIMMEMORY[i] <= 0;
end
for(l=16'd1;f!=16'd1;l=l+1'b1) f <= f/2;
for(m=16'd1;m<=l;m=m+1'b1)
begin
   le <= 16'd2<<(m-1);
   lei <= le/2;
   u[0] <= 16'd1;
   u[1] <= 16'd0;
   w[0] <= cos(Pi/lei);
   w[1] <= -sin(Pi/lei);
   for(j=0;j<=lei-1;j=j+1'b1)
   begin
   ip <= i+lei;
   EE(DOREMEMORY[ip],DOIMMEMORY[ip],u[0],u[1],v[0],v[1]);
   DOREMEMORY[ip] <= DOREMEMORY[i]-v[0];
   DOIMMEMORY[ip] <= DOIMMEMORY[ip]-v[1];
   DOREMEMORY[i] <= DOREMEMORY[i]+v[0];
   DOIMMEMORY[i] <= DOIMMEMORY[i]+v[1];
   end
   EE(u[0],u[1],w[0],w[1],u[0],u[1]);
end
BUSY <= 1'b0;
  end

always @(posedge CLK or posedge EXT or negedge BUSY or negedge CS)
if((!CS)&&(!BUSY)&&EXT)
  begin
OUTDAT <= {DOIMMEMORY[N],DOREMEMORY[N]};
N <= N-1'b1;
  end
//task EE(ARE,AIM,BRE,BIM,CRE,CIM);
task EE;
input [31:0] ARE,AIM,BRE,BIM;
output [31:0] CRE,CIM;
begin
CRE=ARE*BRE-AIM*BIM;
CIM=ARE*BIM+AIM*BRE;
end
endtask
endmodule

作者: whb_fei 時(shí)間: 2012-9-5 08:55
在FPGA中編程，最好多分模塊，這樣不容易出錯(cuò)，檢查時(shí)也方便。。。

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