Which of the following Verilog code(s) correctly multiply the contents of Q by 2? Don't worry about the overflow issue. Q[3] is the most significant bit and Q[O] is the lowest bit. Choose all that apply. always@(posedge clock) begin Q[0] =Q[1]: Q[1]-Q[2]; Q[2]-Q[3]: Q[3]=0; end always@(posedge clock) begin Q[0]=0; Q=Q<<1; end endmodule always@(posedge clock) begin Q={Q[2:01,0); and endmodule always@(posedge clock) begin Q[1] <=Q[0]; Q[2] <=Q[1]; Q[3] <-Q[2]: 9[0]<=0;
Which of the following Verilog code(s) correctly multiply the contents of Q by 2? Don't worry about the overflow issue. Q[3] is the most significant bit and Q[O] is the lowest bit. Choose all that apply. always@(posedge clock) begin Q[0] =Q[1]: Q[1]-Q[2]; Q[2]-Q[3]: Q[3]=0; end always@(posedge clock) begin Q[0]=0; Q=Q<<1; end endmodule always@(posedge clock) begin Q={Q[2:01,0); and endmodule always@(posedge clock) begin Q[1] <=Q[0]; Q[2] <=Q[1]; Q[3] <-Q[2]: 9[0]<=0;
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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![Which of the following Verilog code(s) correctly multiply the contents of Q by 2? Don't worry about the overflow issue.
Q[3] is the most significant bit and Q[0] is the lowest bit.
Choose all that apply.
always@(posedge clock)
begin
Q[0] =Q[1];
Q[1] =Q[2];
Q[2] =Q[3]:
Q[3] =0;
end
always@(posedge clock)
begin
Q[0]=0;
Q=Q<<1;
end
endmodule
always@(posedge clock)
begin
Q={Q[2:0],0};
end
endmodule
always@(posedge clock)
begin
Q[1] <=Q[0];
Q[2] <=Q[1];
Q[3] <=Q[2];
Q[0] <=0;
end](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F83b97d9f-eac5-4c59-98f4-44d1d3519004%2Fdcce2f45-e77f-4dc9-87ed-9e911d96350f%2Fqw54qzb_processed.png&w=3840&q=75)
Transcribed Image Text:Which of the following Verilog code(s) correctly multiply the contents of Q by 2? Don't worry about the overflow issue.
Q[3] is the most significant bit and Q[0] is the lowest bit.
Choose all that apply.
always@(posedge clock)
begin
Q[0] =Q[1];
Q[1] =Q[2];
Q[2] =Q[3]:
Q[3] =0;
end
always@(posedge clock)
begin
Q[0]=0;
Q=Q<<1;
end
endmodule
always@(posedge clock)
begin
Q={Q[2:0],0};
end
endmodule
always@(posedge clock)
begin
Q[1] <=Q[0];
Q[2] <=Q[1];
Q[3] <=Q[2];
Q[0] <=0;
end
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