I realize that this may be very easy to fix but I can't find a way to make it work.
My question is how can I get visual signals of CE2 and CEO in this case? I know by looking on RTL Scheme CE2 and CEO isn't connected to pins. And I just can't connect them.
CE2 should be ON when first counter reach 9 but on waveform its always as X. And CEO should be ON when Q is 9 but on waveform its always Z.
This circuit is just for self learning.
Testbench
Circuit Scheme
TOP MODULE:
`timescale 1ns / 1ps
module top(
input CLK,
input CLR,
input CE,
input CE2,
output reg [3:0] Q,
output reg [3:0] Q2,
output wire CEO,
output CEO2
);
wire CLK;
wire CLR;
wire CENABLE;
wire CE2;
wire Q;
wire Q2;
wire CEO;
wire CEO2;
licznik licznik(.CLK(CLK),.CLR(CLR),.CE(CE),.CEO(CENABLE),.Q(Q));
licznik2 licznik2(.CLK(CLK),.CLR(CLR),.CE2(CENABLE),.Q2(Q2),.CEO2(CEO2));
endmodule
TESTBENCH:
`timescale 1ns / 1ps
module testbench;
reg CLK;
reg CLR;
reg CE;
reg CE2;
wire [3:0] Q;
wire [3:0] Q2;
wire CEO;
wire CEO2;
top UUT (
.CLK(CLK),
.CLR(CLR),
.CE(CE),
.CE2(CE2),
.Q(Q),
.Q2(Q2),
.CEO(CEO),
.CEO2(CEO2)
);
initial CLK=1'b0;
always #5 CLK=~CLK;
initial
begin
CLR = 1'b1;
CE= 1'b1;
#18 CLR = 1'b0;
end
endmodule
FIRST MODULE:
`timescale 1ns / 1ps
module licznik(
input CLK,
input CLR,
input CE,
output reg [3:0] Q,
output CEO
);
always #(posedge CLK or posedge CLR)
if(CLR)
Q <= 4'd0;
else begin
if(CE) begin
if(Q != 4'd9)
Q <= Q + 1;
else
Q <= 4'd0;
end
end
assign CEO = CE & (Q == 4'd9);
endmodule
SECOND MODULE:
`timescale 1ns / 1ps
module licznik2(
input CLK,
input CLR,
input CE2,
output reg [3:0] Q2,
output CEO2
);
always #(posedge CLK or posedge CLR)
if(CLR)
Q2 <= 4'd0;
else begin
if(CE2) begin
if(Q2 != 4'd9)
Q2 <= Q2 + 1;
else
Q2 <= 4'd0;
end
end
assign CEO2 = CE2 & (Q2 == 4'd9);
endmodule
I ran your code on 2 simulators, and I got compile errors on both. Try your code on EDAPlayground.
To fix the compile errors, I removed the wire declarations in module top. To fix the problem with Z on CEO, I replaced CENABLE with CEO. Here is the new top module:
module top(
input CLK,
input CLR,
input CE,
input CE2,
output reg [3:0] Q,
output reg [3:0] Q2,
output wire CEO,
output CEO2
);
licznik licznik(.CLK(CLK),.CLR(CLR),.CE(CE),.CEO(CEO),.Q(Q));
licznik2 licznik2(.CLK(CLK),.CLR(CLR),.CE2(CEO),.Q2(Q2),.CEO2(CEO2));
endmodule
CE2 is X because it is an undriven input. I think you can just delete it.
Related
I am building a shift-unit that is capable of arithmetic and logical right shift, and logical left shift depending on the control signals given to it. However, the arithmetic right shift operator output generates output similar to that logical right shift operator, i.e. sign extension does not occur.
Main code
`timescale 1ns / 1ps
module shift_unit(
input [15:0] a,
input [3:0] b,
input clk,
input isLSL,
input isLSR,
input isASR,
output reg [15:0] result
);
wire [15:0] LSL_result, LSR_result, ASR_result;
LSL lsl(a, b, clk, isLSL, LSL_result);
LSR lsr(a, b, clk, isLSR, LSR_result);
ASR asr(a, b, clk, isASR, ASR_result);
always#(posedge clk) begin
case({isLSL, isLSR, isASR})
3'b001: result <= ASR_result;
3'b010: result <= LSR_result;
3'b100: result <= LSL_result;
endcase
end
endmodule
LSL code:
`timescale 1ns / 1ps
module LSL(
input [15:0] a,
input [3:0] b,
input clk,
input isLSL,
output [15:0] out
);
reg [15:0] result;
always#(posedge clk) begin
if(isLSL) result = a << b;
end
assign out = result;
endmodule
LSR code:
`timescale 1ns / 1ps
module LSR(
input [15:0] a,
input [3:0] b,
input clk,
input isLSR,
output [15:0] out
);
reg [15:0] result;
always#(posedge clk) begin
if(isLSR) result = a >> b;
end
assign out = result;
endmodule
ASR code:
`timescale 1ns / 1ps
module ASR(
input [15:0] a,
input [3:0] b,
input clk,
input isASR,
output [15:0] out
);
reg [15:0] result;
always#(posedge clk) begin
if(isASR) result = a >>> b;
end
assign out = result;
endmodule
And finally, the testbench:
`timescale 1ns / 1ps
module shift_unit_test;
reg [15:0] a;
reg [3:0] b;
reg clk;
reg isLSL;
reg isLSR;
reg isASR;
wire [15:0] result;
shift_unit uut (
.a(a),
.b(b),
.clk(clk),
.isLSL(isLSL),
.isLSR(isLSR),
.isASR(isASR),
.result(result)
);
always #5 clk = ~clk;
initial begin
clk = 1'b0;
a = 16'b1100101011001010;
b = 4;
{isLSL, isLSR, isASR} = 3'b100; #100;
{isLSL, isLSR, isASR} = 3'b010; #100;
{isLSL, isLSR, isASR} = 3'b001; #100;
end
endmodule
The above code has been modelled using Xilinx ISE 14.7.
Any help would be greatly appreciated.
You need to be working with signed signals to get sign extension.
module ASR(
input wire signed [15:0] a,
input [3:0] b,
input clk,
input isASR,
output reg signed [15:0] out
);
always#(posedge clk) begin
if(isASR) out = a >>> b;
end
endmodule
I'm trying to design a 4-bit counter with T-flipflop, here's what i did:
1- From a D-flipflop to T-flipflop:
module T_FlipFlop( clk,T, Q);
input wire clk;
input wire T;
output reg Q;
wire D;
initial
begin
Q<=1'b0;
end
assign D= T ^ Q;
always #(negedge clk)
begin
Q<=D;
end
endmodule
with RTL shematic :
following this "D_ff to T_ff" conversion:
2- Then, i instantiated 4 T-flipflops in the top module and connected the output of each flipflop to the clk of the next one:
module Counters_FreqDividers( sysclk,Q1,Q2,Q3,Q4);
input sysclk;
output wire Q1;
output wire Q2;
output wire Q3;
output wire Q4;
T_FlipFlop num_1(.clk(sysclk),.T(1'b1),.Q(Q1));
T_FlipFlop num_2(.clk(Q1),.T(1'b1),.Q(Q2));
T_FlipFlop num_3(.clk(Q2),.T(1'b1),.Q(Q3));
T_FlipFlop num_4(.clk(Q3),.T(1'b1),.Q(Q4));
endmodule
with RTL schematic :
to follow this diagram:
We know that T-flipflop is just a JK-flipflop with J and K connected to each other and that's what we have here, so consider them as T-flipflops.
3-The simulation:
4- Finally, my questions:
1) why Q1 is the ONLY output that operates properly?
2) Why Q2, Q3, Q4 starts with 1 although i have initialized them as 0?
I can't figure out what's missing, i tried to play around but nothing worked and i'm stuck here!
Edit: my testbench:
module test;
// Inputs
reg sysclk;
// Outputs
wire Q1;
wire Q2;
wire Q3;
wire Q4;
// Instantiate the Unit Under Test (UUT)
Counters_FreqDividers uut (
.sysclk(sysclk),
.Q1(Q1),
.Q2(Q2),
.Q3(Q3),
.Q4(Q4)
);
initial begin
// Initialize Inputs
sysclk <= 1'b1;
#200 $finish();
end
always #5 sysclk=~sysclk;
endmodule
I am using the below mentioned module and testbench to use MAC megawizard in quartus, can anyone tell me how can I use floating point numbers for the same megawizard?
Testbench
`timescale 1ns/1ps
module projecttry2_tb;
reg [15:0] A, B;
wire [31:0] P;
reg clk;
projecttry2 M(.A(A),.B(B),.P(P),.clk(clk));
initial
begin
clk = 1;
forever #25 clk = ~clk;
end
initial
begin
A=3008;
B=255;
#50
A=5859;
B=255;
#50
A=1133;
B=255;
#50
A=0;
B=0;
end
endmodule
Design Module
module projecttry2(A,B,P,clk);
input [15:0] A,B;
output [31:0] P;
input clk;
mult_acc mult_acc_inst (
.clock0(clk),
.dataa(A),
.datab(B),
.result (P)
);
endmodule
ALTFP_MULT is available and documented on page 40 of Floating-Point Megafunctions.
I cannot get a T-Flipflop from a D flipflop to work in Modelsim even after it came directly from class notes. It must be something simple I'm just overlooking.
module D_FF (q, Clk, reset_n,d);
output q;
input Clk, reset_n, d;
reg q;
always #(posedge reset_n or negedge Clk)
if (~reset_n)
q <= 1'b0;
else
q <= d;
endmodule
module T_ff (q, Clk, reset_n);
output q;
input Clk, reset_n;
wire d;
D_FF DFF0 (q, Clk, reset_n, Vcc);
not n1 (d,q);
endmodule
Hi there are two problems:
1. Your reset is active low, so it should be sensitive to the falling edge of the clock.
2. What is that VCC? you should use d there.
Here is the correct version
D_FF:
module D_FF (q, Clk, reset_n,d);
output q;
input Clk, reset_n, d;
reg q;
always #(negedge reset_n or posedge Clk)
if (~reset_n)
q <= 1'b0;
else
q <= d;
endmodule
T_FF:
module T_FF (q, Clk, reset_n);
output q;
input Clk, reset_n;
wire d;
D_FF DFF0 (q, Clk, reset_n, d);
not n1 (d,q);
endmodule
This my Verilog code to convert the number x into form x=a0*R+a1 ,e.g 51 = 5*10 +1. My code does not work, it cannot enter the always loop.
`timescale 1ns / 1ps
module poly(
input [15:0] r,
input [15:0] x,
output reg[15:0] a1,
output reg [15:0] a0,
output finish,
input clk,
input reset
);
reg [15:0] sum;
assign finish =(sum > x);
always# (posedge clk )
begin
if(reset)
begin
a0 <=0;
sum <=0;
end
else if (!finish)
begin
a0 <=a0+1;
sum <= sum+r;
end
else
a1<=x-sum;
end
initial begin
$monitor ( "a1=%b,a0=%b,finish=%b,reset=%b",a1,a0,finish,reset);
end
endmodule
testbench
`timescale 1ns / 1ps
module tb_p;
reg [15:0] r;
reg [15:0] x;
wire[15:0] a1;
wire [15:0] a0;
wire finish;
reg clk;
reg reset;
initial clk=0;
always #5 clk=!clk;
poly m1(r,x,a1,a0,finish,clk,reset);
initial begin
r<=10;
x <=17;
#1 reset<=1;
#2 reset<=0;
end
endmodule
Since your reset signal is synchronous to the clock, you need to extend it so that it is high for at least one posedge of the clock:
initial begin
r<=10;
x <=17;
#1 reset<=1;
#20 reset<=0;
#500 $finish;
end
Note that I added $finish just so my simulation would end.