Build an XOR gate three ways, using an assign statement, a combinational always block, and a clocked always block. Note that the clocked always block produces a different circuit from the other two: There is a flip-flop so the output is delayed. Module Declaration // synthesis verilog_inpu...
Build an AND gate using both an assign statement and a combinational always block. (Since assign statements and combinational always blocks function identically, there is no way to enforce that you're using both methods. But you're here for practice, right?...) Module Declaration // synthesis...
The simulation runs smoothly, but there is now a dual performance penalty because both theassignstatement and thealwaysblock are sensitive to changes in A. As a result, the proceduralassignstatement is executed repeatedly, replacing the same RHS. Solution 4: Assign, a continuous assignment statement...
both y1 and y2 will take on the value of 1. If the second always block executes first after a reset, both y1 and y2 will take on the value 0. This clearly represents a Verilog race condition.
Only reg nets can be assigned in an always block inputwirea;inputwirec;outputwireb;regb_out;//always@(*)beginb_out = ~a;endassignb = b_out;// if-else statementsalways@(*)beginif(a) b_out = c;elseb_out = ~c;end// case statementalways@(*)begincase(a)0: b_out = c;1:...
// Wait 1 time unitb=a;// Assign the value of reg a to reg bendalways@(aorb)// Any time a or b CHANGE, run the processbeginif(a)c=b;elsed=~b;end// Done with this block, now return to the top (i.e. the @ event-control)always@(posedgea)// Run whenever reg a has a...
I'm getting this error for my verilog code, “Illegal, The if needs to be inside an always block; otherwise, the simulator thinks you are using a generate, which expects state_check_dgo to be a constant (like a parameter). In this case, you would not have an assign inside an always...
its initial value is x. The register type variable in Verilog is different from the real hardware register. It refers to a variable that stores a value. If you want to assign a variable in a process (initial process or always process), the variable must be of the register type. There ...
When the reset is not active, then the rising edge of the clock signal has triggered the always_ff block. We use the else branch of the first if statement to capture this condition. We use a second if statement to model the behaviour of the multiplexor circuit. This is an example of ...
Verilog generate block, that do not have a name label, creates a hierarchy that is only visible within the block, and within the sub-tree formed by this block—and nowhere else. Therefore it’s good practice to always name Verilog generate blocks so all identifiers can be referenced through...