EGRE 254
Final Exam
Open book / open notes
5/4/09
Print your name neatly:
__SOLUTIONS_
- For the Karnaugh map shown below, circle all prime implicants.
/ / W / F(W,X,Y,Z)
/ 1 / / 1 / 1
1 / 1 / 1 / Z
Y / / 1
1 / 1 / 1
X
- For the Karnaugh map shown below, circle all essential prime implicants.
/ W / F(W,X,Y,Z)
1 / / 1 / 1
1 / 1 / 1 / Z
Y / 1
1 / 1 / 1
X
- For the Karnaugh map shown below, write F as a minimal sum of products.
/ W / F(W,X,Y,Z)
1 / / 1 / 1
1 / 1 / 1 / Z
Y / 1
1 / 1 / 1
X
ANS:
- For the Karnaugh map shown below, write F as a minimal sum of products, but include any terms necessary to eliminate all static-1 hazards.
/ W / F(W,X,Y,Z)
1 / / 1 / 1
1 / 1 / 1 / Z
Y / 1
1 / 1 / 1
X
ANS:
- For the circuit shown below, plot y on the Karnaugh map.
y(a, b)
1
1 / b
a
ANS:
- Plot the function on the Karnaugh map shown below.
X / F(X,Y,Z)
1 / 1
1 / 1 / Z
Y
ANS:
- Plot the function on the six variable Karnaugh map shown below. Enter only 1’s on the map. Leave blank those cells where F = 0. Note the small numbers in each cell are the minterm values for that cell.
V = 0 / V = 1
W / W
1 / 1 / 1 / 1 / 1 / 1 / U = 0
Z / 1 / 1 / Z
Y / 1 / 1 / 1 / 1 / Y / 1 / 1 / 1 / 1
1 / 1 / 1 / 1 / 1 / 1
X / X
W / W
1 / 1 / 1 / 1 / U = 1
Z / Z
Y / 1 / 1 / 1 / 1 / Y / 1 / 1 / 1 / 1
1 / 1 / 1 / 1
X / X
The truth tables to define the simple decoder and multiplexer used in the problems are shown below.
0 / 0 / 1 / 0 / 0 / 0 / 0 / 0 / D0
0 / 1 / 0 / 1 / 0 / 0 / 0 / 1 / D1
1 / 0 / 0 / 0 / 1 / 0 / 1 / 0 / D2
1 / 1 / 0 / 0 / 0 / 1 / 1 / 1 / D3
Truth table for the decoder. Truth table for the multiplexer.
- Show how to connect the decoder and NOR gate to produce the output F=XÅY..
- Show how to use the multiplexer to implement a two input XOR gate. Do this by connecting either “0”, or “1”to the D inputs so that .
The truth tables to define the simple decoder and multiplexer used in this problem are shown below.
0 / 0 / 1 / 0 / 0 / 0 / 0 / 0 / D0
0 / 1 / 0 / 1 / 0 / 0 / 0 / 1 / D1
1 / 0 / 0 / 0 / 1 / 0 / 1 / 0 / D2
1 / 1 / 0 / 0 / 0 / 1 / 1 / 1 / D3
Truth table for the decoder. Truth table for the multiplexer.
- Plot the function F on the Karnaugh map.
The inputs to the multiplexer are given by:
Using these values gives
W / F1 / 1 / 1 / 1
1 / 1 / Z
Y / 1
1 / 1
X
- Write the minimal sum of product form for the F of the five variable Karnaugh map shown below.
V = 0 / V = 1
W / W / F
1 / 1 / 1 / 1 / 1
1 / 1 / 1 / Z / 1 / 1 / 1 / Z
Y / 1 / Y / 1 / 1 / 1 / 1
1 / 1 / 1 / 1
X / X
- I decided to invent a new type of flip-flop, which I call the JT flip-flop. The dQ map for the JT flip-flop is shown below. Complete the schematic to show the simplest way to build a JT flip-flop using one D flip-flop, one XOR gate, one OR gate, and one AND gate.
J / dQ
0 / D / D / D /
Q / 1 / Ñ / Ñ / 1
T
Using the term, the circuit can be implemented without an inverter.
For the next problems refer to the dQ maps shown below. The dQ maps describe a counter, and CLK is the system clock. Assume JK flip-flops are clocked by the falling edge of the clock.
There are several equally correct answers for some of the J and K inputs. Some of these are shown below. Note: See next page.
Q1
/dQ3
/ Q1 /dQ2
/ /Q1
/dQ1
0 / d / d / 1 / D / d / d / D / D / d / d / Dd / D / 1 / Ñ /
Q3
/ d / Ñ / Ñ / 0 /Q3
/ d / Ñ / 1 / Ñ /Q3
Q2
/Q2
/Q2
- If C1 = CLK, write the equations for J1 and K1.
- If C2 = CLK, write the equations for J2 and K2.
- If C3 = CLK, write the equations for J3 and K3.
- Suppose C3 is connected to Q2 would the counter work if both J3 and K3 are connected to a logic “1”?
Yes No
Q2 can’t be used as clock since it does not provide a transition for the Ñ on the dQ3 map.
- Suppose C3 is connected to Q2 would the counter work if both J3 and K3 are connected to a logic “1”?
Yes No
The above maps should have been labeled as shown below. The actual equations should have Q3 and Q1 interchanged as shown below.
Q3
/dQ3
/ Q3 /dQ2
/ /Q3
/dQ1
0 / d / d / 1 / D / d / d / D / D / d / d / Dd / D / 1 / Ñ /
Q1
/ d / Ñ / Ñ / 0 /Q1
/ d / Ñ / 1 / Ñ /Q1
Q2
/Q2
/Q2
If C1 = CLK, write the equations for J1 and K1.
If C2 = CLK, write the equations for J2 and K2.
If C3 = CLK, write the equations for J3 and K3.
- The state diagram for a certain asynchronous finite state machine is shown below. The state assignment is shown for two states. Enter a valid state assignment for the other states. ANS: One possible solution is shown below. Any solution is correct provided only one state variable changes at a time. This is necessary to prevent a possible race condition.
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