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Midterm Answers

1.  Figure 1 Shows a circuit and the input waveforms.  On the diagram, draw the waveforms of the outputs C and D.  Show the number of gate propagation delays relative to the nearest dashed vertical line.

2.  Figure 2 shows a 7 segment DECIMAL DISPLAY.

 (a) Draw the truth table for segment "f".
 (b) Use a Karnaugh map to derive the minimal sum of products.
 (c) Implement the function using only NAND gates (any number of inputs).  In
       answering this question and in part d you may assume that the inverted
       forms of the input variables are available and need not show the
       inverters.
 (d) Use a 8 input multiplexer to form a circuit to implement segment "f".

H

X3

X2

X1

X0

f

H

X3

X2

X1

X0

f

0

0

0

0

0

1

8

1

0

0

0

1

1

0

0

0

1

0

9

1

0

0

1

1

2

0

0

1

0

0

A

1

0

1

0

d

3

0

0

1

1

0

B

1

0

1

1

d

4

0

1

0

0

1

C

1

1

0

0

d

5

0

1

0

1

1

D

1

1

0

1

d

6

0

1

1

0

1

E

1

1

1

0

d

7

0

1

1

1

0

F

1

1

1

1

d

The truth table for segment f is made into a Karnaugh map to minimize the SOP equation.

Implementing this function requires 1 3-input NAND gate, 3 2-input NAND gates, and 1 4-input NAND gate.

The MULTIPLEXER implementation for Segment f is given by:

H

X3

X2

X1

X0

f

H

X3

X2

X1

X0

f

0

0

0

0

0

1

8

1

0

0

0

1
X0#

1

1

0

0

0

1

0

9

1

0

0

1

1

2

0

0

1

0

0

A

1

0

1

0

1

0

d

3

0

0

1

1

0

B

1

0

1

1

1

4

0

1

0

0

1

C

1

1

0

0

1

1

d

5

0

1

0

1

1

D

1

1

0

1

0

6

0

1

1

0

1

E

1

1

1

0

1

X0#

d

7

0

1

1

1

0

F

1

1

1

1

1

 

Please note that although the bottom three inputs of the MUX are shown as not being connected because they will never be selected, it is usually safer to connect them either to Vcc or ground.

3.  A and B are two signed hexadecimal numbers in 2's complement form.
     A = C72 and B = D8E.  See the table of powers of 16 at the bottom of
     the exam.

 (a) Convert A to octal.

       C72 = 1100  0111  0010 = 110  001  110  010 = 6162 Octal

 (b) Subtract B from A and show the result in hexadecimal.

       To subtract B we must add its 2's complement.  D8E = 1101  1000  1110
       Its 2's complement is 0010  0111  0010.  Therefore:

                    1100  0111  0010
                    0010  0111  0010

          =        1110  1110  0100    =   EE4  Hexadecimal

 (c) Show the value of the result in part b in decimal.

       EE4 is a negative number.  To find its decimal value, we take its 2's
       complement.

       0001  0001  1100 = 11C= 1 * 162 + 1 * 161 + 12 * 160 = 256 +16 + 12 = 284.

       The result of the subtraction is therefore -284.

4.  A finite state machine ( FSM ) has 2 inputs, X and R, and a single output Y.  When R is true it resets the machine to its initial state and sets the output to 0.  When R is 0, the machine outputs a 1 only when the last input is the same as the previous one.  Draw the transition diagram for the FSM. 

5. (a)  Draw a circuit diagram for a 3 bit synchronous up/down counter that counts up when its control line is 0 and down when it is 1.  You need not show the gates that make up the flip-flops or multiplexers, but you must label clearly all inputs, outputs, and control lines to the flip-flops and multiplexers. 

    (b) What is the disadvantage in using a ripple counter rather than a
          synchronous one?

          In a ripple counter the clock goes from one stage to the next.  Each
          flip-flop therefore introduces an additional delay from the master clock
          This introduces glitches into the count because the output stages
          change at different times.

 

 

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