- We can use a number of
multiplexor circuits to
create a shifter circuit.
Recall the functionality of a 2×1 multiplexor:
For convenience, we draw the Mux circuit rotated 90-degrees:
- How to construct
a 4 bits shifter circuit using
2x1 multiplexors:
- We need to use 4 (four)
2x1 multiplexors:
And make c control all the multiplexors:
- We know that when c = 0,
all the multiplexors will
switch the
right input to its
output:
The circuit must (pretend) to shift d3 d2 d1 d0 one bit to the left:
- output z3 z2 z1 z0 = d2 d1 d0 0
To obtain the output z3 z2 z1 z0 = d2 d1 d0 0, we connect the appropriate input signals to the activate inputs of the multiplexors:
- Next...
We know that when c = 1,
all the multiplexors will
switch the
left input to its
output:
The circuit must (pretend) to shift d3 d2 d1 d0 one bit to the right:
- output z3 z2 z1 z0 = 0 d3 d2 d1
To obtain the output z3 z2 z1 z0 = 0 d3 d2 d1, we connect the appropriate input signals to the activate inputs of the multiplexors:
- We need to use 4 (four)
2x1 multiplexors:
- The shifter circuit using
2×1 multiplexors
(the mux is drawn rotated 90 degrees) is:
EDiSim macro definition for this circuit using EDiSim's built-in Mux component:
Define Shifter c d3 d2 d1 d0 | z3 z2 z1 z0; /* Bit 3 */ Mux aa c | ZERO d2 | z3; /* Bit 2 */ Mux aa c | d3 d1 | z2; /* Bit 1 */ Mux aa c | d2 d0 | z1; /* Bit 0 */ Mux aa c | d1 ZERO | z0; Endef;
- Example Program:
(Demo above code)
- Prog file: /home/cs355001/demo/circuits/shifter
How to run the program:
/home/cs355001/bin/cs355sim /home/cs355001/demo/circuits/shifter