Question

Two reservoirs R 1 and R2, the level of each of which is controlled by a detector of
high level (a for R1, b for R2) and a low level detector (c for R1, d for R2). When these sensors detect the presence of the liquid they are at level 1.
 
There are three LEDs V1, V2, V3, which operate under the following conditions:
V1 = 1 if both tanks are full.
V2 = 1 if both tanks are empty.
V3 = 1 in all other cases (tank half full or full one empty ...).
 
A number of combinations are technologically impossible, the outputs V1, V2,
V3, will take in these cases an indifferent value (X).
 
1. Establish the truth table of this system.
2. Determine simplified logic equations using Karnaugh's tables.
3. Give the V1, V2, V3 circuits with doors.

Answer 1

Given a,b indicates the high level of system and c,d indicate the low level of system for R1and R2 reservoir respectively.

V1,V2,V3 are the LEDs that are used to know the condition of condition of detectors.

Let's the write the combinations. As we have 4 sensors completely we get 16 different combinations of abcd and V1,V2,V3 are to be represented for these combinations. We know all four sensors won't react at a time i.e, abcd =1111/0000 is impossible then we will consider it as X as per given. Similarly both ac , bd react at a time is not possible as one represents full and the other represent empty. Hence these conditions are also taken as X.

V1 reacts when both ab are 11, V2 reacts when both cd are 11 and V3 for other conditions.

i) Truth table of system

a	b	c	d	V1	V2	V3
0	0	0	0	X	X	X
0	0	0	1	0	0	1
0	0	1	0	0	0	1
0	0	1	1	0	1	0
0	1	0	0	0	0	1
0	1	0	1	X	X	X
0	1	1	0	0	0	1
0	1	1	1	X	X	X
1	0	0	0	0	0	1
1	0	0	1	0	0	1
1	0	1	0	X	X	X
1	0	1	1	X	X	X
1	1	0	0	1	0	0
1	1	0	1	X	X	X
1	1	1	0	X	X	X
1	1	1	1	X	X	X

ii) Using Karnaugh's table the simplified expression for V1,V2, V3 is obtained as follows.

For V1

ab/cd	00	01	11	10
00	X	0	0	0
01	0	X	X	0
11	1	X	X	X
10	0	0	X	X

As we have only one at 1100 , to minimise the equation we are considering all don't care conditions in that row as 1. Hence we get V1= ab from the above k map.

For V2

ab/cd	00	01	11	10
00	X	0	1	0
01	0	X	X	0
11	0	X	X	X
10	0	0	X	X

As we have only one 1 at 0011 am considering all don't cares of that column as 1's to get simplified equation as V2=cd by using k map.

For V3

ab/cd	00	01	11	10
00	X	1	0	1
01	1	X	X	1
11	0	X	X	X
10	1	1	X	X

We got totally six 1's. Hence ones due to 0001 and 0100 are taken as quad using 0000,0101 don't cares as 1. We get it as a'c'.(complement are written as a' here)

Similarly 1's corresponding to 0010 and 0110 as a quad considering the column don'tcares as 1. We get it as cd'

Similarly 1's corresponding to 1000 and 1001 as a quad considering the row don't cares as 1. We get it as ab'.

Thus V3 is the combination of all 3 expression. We get V3= a'c'+cd'+ab'

Hence V1=ab

V2=cd

V3= a'c'+cd'+ab'