digital tutotrials
TRANSCRIPT
-
7/29/2019 Digital Tutotrials
1/29
Digital Tutotrials
R.K.Tiwary
-
7/29/2019 Digital Tutotrials
2/29
(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.
-
7/29/2019 Digital Tutotrials
3/29
(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.
-
7/29/2019 Digital Tutotrials
4/29
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
-
7/29/2019 Digital Tutotrials
5/29
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.
-
7/29/2019 Digital Tutotrials
6/29
Q. Express the following binary numbers as Gray codes (a) 101010011 (b) 10001110110
(a) Binary number is 101010011
Gray code is 111111010
(b) Binary number is 10001110110
Gray code is 11001001101
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
-
7/29/2019 Digital Tutotrials
7/29
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
-
7/29/2019 Digital Tutotrials
8/29
Q. Simplify the Boolean function F = AB + (AC) + ABC(AB + C).
Solution
-
7/29/2019 Digital Tutotrials
9/29
Q. Simplify the Boolean function F = ((XY + XYZ) + X(Y + XY)).
Q. Simplify the Boolean function F = XYZ + XYZ + XYZ.
-
7/29/2019 Digital Tutotrials
10/29
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
-
7/29/2019 Digital Tutotrials
11/29
Obtain the canonical sum of product form of the following function.
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.
-
7/29/2019 Digital Tutotrials
12/29
Obtain the canonical sum of product form of the following function.
Solution
-
7/29/2019 Digital Tutotrials
13/29
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.
-
7/29/2019 Digital Tutotrials
14/29
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
-
7/29/2019 Digital Tutotrials
15/29
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.
-
7/29/2019 Digital Tutotrials
16/29
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:
-
7/29/2019 Digital Tutotrials
17/29
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
-
7/29/2019 Digital Tutotrials
18/29
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
-
7/29/2019 Digital Tutotrials
19/29
Realize the following function by NAND gates only, F = B(A + CD) + AC.
Realization by AND, OR, and NOT gates.
-
7/29/2019 Digital Tutotrials
20/29
AND and OR gates are replaced by equivalent NAND gates.
NAND gate realization after two cascaded inverters are removed.
-
7/29/2019 Digital Tutotrials
21/29
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.
-
7/29/2019 Digital Tutotrials
22/29
Implementation by NOR gates after two cascaded inverters are removed.
-
7/29/2019 Digital Tutotrials
23/29
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.
-
7/29/2019 Digital Tutotrials
24/29
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.
-
7/29/2019 Digital Tutotrials
25/29
-
7/29/2019 Digital Tutotrials
26/29
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
-
7/29/2019 Digital Tutotrials
27/29
-
7/29/2019 Digital Tutotrials
28/29
-
7/29/2019 Digital Tutotrials
29/29