Download Digital Electronics Tutorial - 2

Digital Electronics Tutorial
2
13th February, 2006
Indian Institute of Technology,
Kharagpur, India.
Tutors : Daibashish, Subhodeep, Narayanan and Saurabh,
5th Year Dual Degree, Dept. of E &ECE. IIT Kharagpur.
Digital Electronics Class Test -1
Solutions
•
Question 1.
– Compute the logic equations for a BCD
to 7 segment code converter.
– For example if the input <x3,x2,x1,x0> =
<0,0,1,1> i.e. decimal equivalent of 3 then
LEDs a,b,c,d and g turn on. For input
=<1,0,0,0> all LEDs turn on.
Write the logic equations for a,b, d and g
LEDs
Solution
for Q1.
Fig. A 7 Segment Display
a = X3’ ( X1 + (X2X0)’ ) + X3 X2’X1’
b = X3’ ( X2’ + (X1X0)’ ) + X3 X2’X1’
c = X3’ ( X2+X1’+X0) + X3 X2’X1’
d = X3’ ( X2(X1X0) +X2’(X1+X0’) )+X3 X2’X1’
e = X3’ X0’ ( X1 + X2’ ) +X3 X2’X1’X0’
f = X3’ (X2 X1’ X0’) +X3 X2’X1’
g = X3’(X1X2) + X3’X2 X1 X0’ +X3 X2’X1’
NO need to optimize.
Just straight 4-variable
equation required
Question 2.
•
Let
F(A,B,C,D)= f(A,B,C,D) + g(A,B,C,D) .
Given f =B’D’ + A’C’D’ + AB’C’ , find
g(A,B,C,D) such that F is self-dual.
Also find the minimal sop and pos
expressions for F.
[ Note : A function is self dual if
F(a’,b’,c’,d’)=F’ (a,b,c,d) ]
Solution to Q2.
• A 4 variable, self dual function will have
opposite entries in locations marked n and
15 – n . We observe that location pairs
(12,3) and (14,1) are the only
complementary locations which do not
have 1 in them. So, to make F self dual set
locations { 12 and 14 } to 1.
• (Note : We may also set { 12 and 1 } or { 3
and 14 } or { 3 and 1} to 1 )
• This makes G(A,B,C,D) =∑(12,14) =ABD’
• Minimal Sop expression of F(A,B,C,D) =
B’D’ + C’D’ + AD’ + AB’C’
• Minimal Pos expression of F(A,B,C,D) =
(B’+D’) (C’+D’)(A+D’) ( A+B’+C’)
Question 3.
Design a magnitude comparator M101 which
compares the 3 bit numbers A=<a2,a1,a0> and
B=<b2,b1,b0> taking three additional inputs
Ein, Gin and Lin, that represent the equal,
greater than and lesser than signals from a
previous more significant stage. Write the
logic equations for Eout , Gout and Lout for
this module in terms of its inputs.
[ Use two M101 modules to compare two 5-bit
numbers viz.
X=<x4,x3,x2,x1,x0> and Y= <y4,y3,y2,y1,y0>]
Solution for Question 3.
• Let ci = ( ai xor bi )’ for I = 0,1,2.
Then :
 Eout = Ein . c0.c1 .c2
 Gout = Gin + Ein ( a2 . b2’ + c2 a1 b1’ +
c2 c1 a0 b0’ )
 Lout = Lin + Ein ( a2’ .b2 + c2 a1’ b1 +
c2 c1 a0’ b0 )
Fig. Recommended design.
Question 4.
• What is the minimum number of 2 to 1
multiplexers required to determine the
square of a 3 bit number <x2,x1,x0> ?
Solution for Q4.
Fig. Shows the Truth Table of the Squaring circuit
From above it’s quite clear that
• Y0=X0 , Y1=0 , Y2 = X2’ X1 X0’ + X2 X1 X0’
= X1 X0’
• Y3 = X2’ X1 X0 + X2 X1’ X0
= X0 ( X1 X2)
• Y4 = X2 X1’ X0’ + X2 X1’ X0 + + X2 X1 X0
= X2 ( X0 + X1’)
• Y5 = X2 X1
• So Y0 and Y1 do not require muxes.
Y2 requires 1 mux .
Y3 requires 3 muxes : 2 for Xor & 1 for And. Y4
requires 2 muxes : 1 for Or & 1 for And.
Y5 requires 1 mux.
Hence a total of 7 2-to-1 multiplexers are needed.
• Please Submit your homework sheets
from the previous Tutorial (the QM Problems)
******** The End.********
Have a Good time at the
Mid-Semester Examinations!