Titel: Circuit Analysis and Design
Objectives:
- Know how to apply the basic Boolean laws
- Able to create digital circuits with the help of the sum of products method
Time |
Method |
Topic |
Way |
Remark | |
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* Review Lesson 2 |
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* Introduction |
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* Boolean laws and theorems |
HO |
Handout No. 2 (Boolean Algebra Theorems) | |
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- Basic laws |
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- Boolean relation about OR operations | |
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- Boolean relations about AND operations | |
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* Sum of products method | | | |
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- Fundamental products | | |
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- Sum of products equations | | |
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* Review Exercise |
WS |
Worksheet No. 3 | |
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S: Speech | |
B: Boardscript | |
Circuit Analysis and Design
Basic laws
(see also Handout No. 2)
Commutative law:
A + B = B + AA B = B A
Fig. 3-1: Commutative law
Associative law:
A + (B + C) = (A + B) + CA (B + C) = (A B) C
Fig. 3-2: Associative law
Distributive law:
A (B + C) = AB + AC
Fig. 3-3: Distributive law
Boolean relations about OR operations
A + 0 = A
Proof:
when A is 0
0 + 0 = 0
when A is 1
1 + 0 = 1
A + A = A
Proof:
when A is 0
0 + 0 = 0
when A is 1
1 + 1 = 1
A + 1 = 1
Proof:
when A is 0
0 + 1 = 1
when A is 1
1 + 1 = 1
A + A = 1
If one input is high, the output is high no matter what the other input is.
Boolean relations about AND operations
A · 1 = A
A · A = A
A · 0 = 0
HO: Check the equations above in the same way as we did it before.
A · A = 0
If one input is low, the output is low no matter what the other input is.
Double inversion
De Morgan's theorems
see also Lesson 1
Duality theorem
1. Change each OR sign to an AND sign
2. Change ech AND sign to an OR sign
3. Complement any 0 or 1 appearing in the expression
Ex: | |
A + 0 = A |
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® |
A 1 = A |
Ex: | |
A (B + C) = AB + AC |
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A + B C = (A + B) (A + C) |
TIP: Proof it with a truth table
Ex: Simplify the following Boolean equation
Solution: (see also Handout No. 2)
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Handout 2 |
No. 3a |
X = A (1) | |
No. 8a |
X = A | |
No. 7b |
HO: Simplify the following Boolean equation
X = B + B
X = B
Fig. 3-4: Example truth table with fundamental products
A |
B |
X |
Fundamental Products |
0 |
0 |
0 |
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0 |
0 |
1 |
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0 |
1 |
0 |
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0 |
1 |
1 |
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1 |
0 |
0 |
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1 |
0 |
1 |
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1 |
1 |
0 |
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1 |
1 |
1 |
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Sum of products equation
Given is the following truth table:
Fig. 3-5: Example truth table
A |
B |
C |
X | |
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0 |
0 |
0 |
0 | | |
0 |
0 |
1 |
0 | | |
0 |
1 |
0 |
0 | | |
0 |
1 |
1 |
1 |
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1 |
0 |
0 |
0 | | |
1 |
0 |
1 |
1 |
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1 |
1 |
0 |
1 |
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1 |
1 |
1 |
1 |
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We have to locate each output 1 in the truth table and write down the fundamental product.
The next step is to OR the fundamental products:
Now we can derive the corresponding logic circuit:
Fig. 3-6: Logic circuit
HO: What is the sum of product circuit for the given truth table?
Fig. 3-7: Truth table
A |
B |
C |
X |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
Boolean Algebra Theorems
No |
Theorem |
Name |
1a |
A + B = B + A |
Commutative law |
2a |
(A + B) + C = A + (B + C) |
Associative law |
3a |
A (B + C) = AB + AC |
Distributive law |
4a |
A + A = A |
Identity law |
5a |
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Negation |
6a |
A + AB = A |
Redundancy |
7a |
0 + A = A | |
8A |
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9a |
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10a |
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De Morgan's laws |
No. 1 Simplify the Boolean equation and discribe the logic circuit:
No. 2 Simplify the following Boolean expressions:
No. 3 A digital system has a 4-bit input from 0000 to 1111. Design a logic circuit that produces a high output whenever the equivalent decimal input is greater than 13.
No. 4 In a heating plant the burner X has to be switched on, when the circulating pump A is actuated and the temperature probe B for the warm water supply or the room temperature probe C respond.
a) Develop the truth table
b) Write down the sum of products equation
c) Draw the logic circuit
d) Use Boolean algebra to simplify the equation
e) Draw the corresponding logic circuit.