August 1999
LD201 : LOGIC DESIGN

QUESTION 1 (Compulsory)

Total Marks: 20 Marks

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Question 1

(a) Calculate the value of the expression

(11112 + 234 + B9416 ) ¸ 1610

and give the result in base 8.

[4]

(11112 + 234 + B9416)
¸ 1610
= (1111 + 1011 + 1011 1001 0100)2
¸ 100002
= 1011 1010 11102
¸ 100002
= 1011 1010.11102
= 272.78

 

(b) The following circuit shows a controller for a lightbulb using four switches and a battery.

pic1.gif (4170 bytes)

Give a boolean expression describing the state of the lightbulb, in terms of the settings of the switches W, X, Y, Z. Use 1 for a closed (conducting) switch and 0 for an open switch, and 1 for the lightbulb being on and 0 for it being off.

[2]

X (YZ + W)

 

(c) Express the following circuit symbolically.

pic2.gif (4025 bytes)

Redraw the circuit using a single gate, and express the new circuit symbolically too.

[3]
 
The circuit implements the expression pic7.gif (1213 bytes). It is equivalent to the circuit, which implements the expression Y = A + B + C

                                 pic5.gif (1568 bytes)

 

(d) By constructing a Karnaugh map, find a minimal product-of-sums expression for the function

F (A, B, C, D) = pic3.gif (874 bytes)(2, 4, 5, 6, 7, 10, 12, 13, 14, 15)

[3]
                      
                          pic6.gif (6587 bytes)
                                                                        _   _
so the minimal product-of-sums expression is B(C + D).

 

(e) Complete the diagram below so that the device can act as a 4-bit adder/ subtractor with A/S' as the switch. ( Each full adder FA has inputs bits A and B and carry in Ci, and output bit S and carry out Co. The switch A/S' selects addition if it is high, and subtraction if it is low.) Show all the relevant components and connections to and within the dotted box.

pic4.gif (8359 bytes)

 

[6]
pic8.gif (31431 bytes)

 

(f) Give the truth-table for a 1-to-8 demultiplexor, with a data input A, three selector inputs S0 to S2, and eight data outputs B0 to B7. [2]

S2 S1 S0 A B7 B6 B5 B4 B3 B2 B1 B0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0 0 1
0 0 1 0 0 0 0 0 0 0 0 0

...