Digital Tutotrials

7/29/2019 Digital Tutotrials

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Digital Tutotrials

R.K.Tiwary


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(a) Decimal number

is

4 5

BCD code is 0100 0101

Hence the BCD coded form of 4510 is 0100 0101

(b) Decimal number

is

2 7 3 9 8

BCD code is 0010 0111 0011 1001 1000

Hence the BCD coded form of 273.9810 is 0010 0111 0011.1001 1000

(c) Decimal numberis

6 2 9 0 5

BCD code is 0110 0010 1001 0000 0101

Hence the BCD coded form of 62.90510 is 0110 0010.1001 0000 0101

Encode the following decimal numbers in BCD code:a.45 b.273.98 c.62.905Solution.


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(a) BCD code is 1 0010 1001

By padding up

the first numberwith 3 zeros

0001 0010 1001

Decimal number

is

1 2 9

Hence the decimal number is 129.(b) BCD code is 1000 1001 0011

Decimal number

is

8 9 3

Hence the decimal number is 893.

(c) BCD code is 011 1000 1001 1001 0010

By padding up

the first number

with 1 zero

0011 1000 1001 1001 0010

Decimal number

is

3 8 9 9 2

Hence the decimal number is 389.92.

Write down the decimal numbers represented by the following BCDcodes:a.100101001 b.100010010011 c.01110001001.10010010Solution.


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Encode the following decimal numbers to Excess-3 code:

a.38 b. 471.78(a) Decimal

number is

3 8

BCD code is 0011 1000

Now adding 3 +0011 +0011

Excess-3 code

is

0110 1011

Hence the Excess-3 coded form of 3810 is 0110 1011

(b) Decimal

number is

4 7 1 7 8

BCD code is 0100 0111 0001 0111 1000

Now adding 3 +0011 +0011 +0011 +0011 +0011

Excess-3 code

is

0111 1010 0100 1010 1011

Hence the Excess-3 coded form of 471.7810 is 0111 1010 0100.1010 1011

S l i


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Encode the following decimal numbers to Gray codes: (a) 61 (b) 83

(a) Decimal number is 61

Binary code is 111101

Gray code is 100011

(b) Decimal number is 83

Binary code is 1010011

Gray code is 1111010

Solution.


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Q. Express the following binary numbers as Gray codes (a) 101010011 (b) 10001110110

(a) Binary number is 101010011


(b) Binary number is 10001110110


Q. Express the following Gray codes as binary numbers: (a) 100111100 (b) 10110010101

(a) Gray code is 100111100

Binary number is 111010111

(b) Gray code is 10110010101

Binary number is 11011100110


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Q. Simplify the Boolean function F=AB+ BC + BC.Solution

Q. Simplify the Boolean function F= A + AB.Solution.

Q. Simplify the Boolean function F= ABC + ABC + AB.

Solution


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Q. Simplify the Boolean function F = AB + (AC) + ABC(AB + C).

Solution


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Q. Simplify the Boolean function F = ((XY + XYZ) + X(Y + XY)).

Q. Simplify the Boolean function F = XYZ + XYZ + XYZ.


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Obtain the canonical sum of product form of the following function.

Solution.The given function contains two variables A and B.

The variable B is missing from the first term of the expression and the

variable A is missing from the second term of the expression.

Therefore, the first term is to be multiplied by (B + B) and the second

term is to be multiplied by (A + A) as demonstrated below.

Hence the canonical sum of the product expression of the given function is


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Hence the canonical sum of the product expression of the given function is

Solution. Here neither the first term nor the second term is minterm.

The given function contains three variables A, B, and C.

The variables B and C are missing from the first term of the expressionand the variable A is missing from the second term of the expression.

Therefore, the first term is to be multiplied by (B + B) and (C + C).

The second term is to be multiplied by (A + A).

This is demonstrated below.


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Solution


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Obtain the canonical product of the sum form of the following function

Solution. In the above three-variable expression, C is missing from the

first term, A is missing from the second term, and B is missing from the

third term.

Therefore, CC is to be added with first term, AA is to be added with the

second, and BB is to be added with the third term. This is shown below.


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Obtain the canonical product of the sum form of the following function.

Solution. In the above three-variable expression, the function is given at sum of the product

form. First, the function needs to be changed to product of the sum form by applying the

distributive law as shown .

Now, in the above expression, C is missing from the first term and B is missing from the second

term. Hence CC is to be added with the first term and BB is to be added with the second term

as shown below.

It can be clearly noted that the following relation holds true


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Conversion between Canonical Forms

This has the complement that can be expressed as

Now, if we take complement of F by DeMorgans theorem, we obtain F as

It can be clearly noted that the following relation holds true :

It can be clearly noted that the following relation holds true

that is, the maxterm with subscriptjis a complement of the minterm with the

same subscriptj, and vice versa.


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Realize the function Y = BDE + BF + CDE + CF + A with a multilevelnetwork.

Solution: By the straightforward method, the function can be realized

in a two-level AND-OR network as shown in Figure:


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However, the expression may be factored into a different form as below

The same function can be realized as a multilevel gate network as shown in Figure


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Hence, from the above examples, we observe that the multilevel network has

distinct advantages over the two-level network, which may be summarized as below.

1. Multilevel networks use less number of literals or inputs,

thus reducing the number of wires for connection.

2. Sometimes the multilevel network reduces the number of

gates.

3. It reduces the variety type of gates and hence the number ofICs

4. Multilevel gate networks can be very easily converted to

universal gates realization In that case the switching network

can be implemented by less variety of the logic gates.

However, the biggest disadvantage of the multilevel network is that

it increases the propagation delay. The propagation delay is the

inherent characteristics of any logic gate, and it increases with the

increase of number of levels

R li th f ll i f ti b NAND t l F B(A CD) AC


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Realize the following function by NAND gates only, F = B(A + CD) + AC.

Realization by AND, OR, and NOT gates.


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AND and OR gates are replaced by equivalent NAND gates.

NAND gate realization after two cascaded inverters are removed.


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Realize the following function by NOR gates only, F = A(B + CD) + BC.

Circuit realization by AND-OR

gates.

AND and OR gates are replaced by NOR gates.


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Implementation by NOR gates after two cascaded inverters are removed.


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Realize the function F = BC + AC + AB by (i) basic gates, (ii) NANDgates only, (iii) NOR gates

only.

Solution.

The function is realized basic gates as in Figure

For the NAND realization, at the first step, each of the gates are converted to NAND gates as

in Figure

Figure represents the conversion of each gate to a NOR gate and next Figure is the logic

diagram for the given function realized with NOR gates only.


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a. Realize the function F = A + BCD using NAND gates only.

b. Realize the function F = (A + C)(A + D) (A + B + C) using NOR gates only.

Solution.

a.


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I II III IV

Decimal equivalent Binary equivalent

A B C D BCD ABCD

1 0 0 0 1 1,3 001 1,3,9,11 01

4 0 1 0 0 1,5 001 8,9,10,11 10

8 1 0 0 0 1,9 001

4,5 010

4,6 010

8,9 100 8,10 100

3 0 0 1 1 3,11 011

5 0 1 0 1 9,11 101

6 0 1 1 0 10,11 101

9 1 0 0 1 1 0 1 0 1 0

11 1 0 1 1


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Digital Tutotrials