SW 4 SW 3 SW 2 SW 1 column. A ооооо -● 0 0 0 1 1 1 1 1 1 1 1 B ● D A (SW4) 2 Apply the inputs shown in Table 38 9 10 0 0 0 0 1 1 1. 1 0 0 0 0 1 1 1 1 B (SW3) tep 6. Output expression: F = LO 5 6 INPUTS 0 0 and record the output state in the right hand. C (SW2) D (SW1) 0 0 1 1 4 0 1 1 0 0 1 1 F 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 L1 OUTPUTS F (L1) Fig. 4-2

Programming Logic & Design Comprehensive
9th Edition
ISBN:9781337669405
Author:FARRELL
Publisher:FARRELL
Chapter5: Looping
Section: Chapter Questions
Problem 4PE
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Question
SW 4
SW 3
SW 2
SW 1
column.
OOO OOO 00
0
0
0
0
0
Apply the inputs shown in Table
0
A
1
1
B
C
1
1
1
1
1
1
D
A (SW4) B (SW3)
2
38
9
3/4 - 7402
оооо
0
0
0
1
1
1
1.
1
0
0
0
0
1
1
1
1
10
Step 6. Output expression: F =
5
6
INPUTS
C (SW2)
and record the output state in the right hand
0
0
1
1
0
COO
0
1
1
LOO
0
0
1
00100
D (SW1)
010
1
OLOHOOOLOL
0
1
0
1
0
1
0
0
1
F
0
1
L1
OUTPUTS
F (L1)
Fig. 4-2
Transcribed Image Text:SW 4 SW 3 SW 2 SW 1 column. OOO OOO 00 0 0 0 0 0 Apply the inputs shown in Table 0 A 1 1 B C 1 1 1 1 1 1 D A (SW4) B (SW3) 2 38 9 3/4 - 7402 оооо 0 0 0 1 1 1 1. 1 0 0 0 0 1 1 1 1 10 Step 6. Output expression: F = 5 6 INPUTS C (SW2) and record the output state in the right hand 0 0 1 1 0 COO 0 1 1 LOO 0 0 1 00100 D (SW1) 010 1 OLOHOOOLOL 0 1 0 1 0 1 0 0 1 F 0 1 L1 OUTPUTS F (L1) Fig. 4-2
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