Basic Electronics and its fundamentals needed for a fresh start or a quick revision, Study notes of Basic Electronics

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UNIT-

DIGITAL SYSTEMS

i

¥1,213,456,718, IE ÷i Hexadecimal 0,112,3,^ 4,516,718,9^ ,^

A ,B, c. DIE, t^16

Number

system

Used to^

represent information Babe 1 Radix^ Cr)

Total no .

of digits

used in the

number

system

91 Convert^ 5362, to binary Z#¥ = (^1101012)

2. 13
   • (^) O la. TEO 2 ¥ 0.62×2= O -^24 ×2=0. 0.48× 2 = (^) 0- = 1001112 0.96× 2 =^ 1. O - 92 ×2=1. 53.62-0.84×2=1.68110101.

2. Decimal^ to^ octal ① If (^) given decimal no.^ is^ less^ than 8 , octal no. is same ② If decimal no^.^77 , divide mo by 8 ③ Repeat steps^ till quotient is less than^8 ⑨ Write^ Remainder^ in^ reverse

• Eos^ fractional part^

Multiply by^ 8

3- Decimal^ to^ Hexadecimal ④ Divide^ no. by 16 ② White^ remainder ③ Repeat until result^ is^ O ⑥ Hex value^

is

sequence of remainders from last to^ first

• For^ fractional part^ A

• to B-^11

Multiply by^

16 c- 12 D - 13 E-If E-IS

Ox convert (^) 444.456,0 to^ Hexadecimal iE÷± ' Bc 0.456× 16 =^ 7296

• 7 0296 × 16 =^ 4.^
• q O - 74 B O -^736 ×^16 = 11.
• (^11) 494.456 =^ IBC^ . 74 B

Binary to decimal^

Use the formula 28 -^256 I -^128 26 -^64 25 - 32 I-^16 23 -^8 22 -^4 21 -^ Z

Binary to (^) decimal Of convert 101111. 01012 to^ decimal D= as × (^25) tax It azx (^23) t (^) a 22 t

a

,

x 2

' t (^) Ao x 20 t a

x 2-

t a

X 2-

• (^) a -3×2-

t a

• (^) ex 2- = (^1) × 25 t^ O^ X 24 t^1 ×^23 t^ 1 × 271 × 2 ' t (^) l (^) X 20 t O X 2-

I

t 1 ×2-2^ t^ 0 ×2-

• Ix^ 2- = 47. 312510

6- (^) Hexadecimal to^ Decimal Oy DZA^

• SC

to decimal D= (^) Az x^ (^162) t a , X 16

t (^) aox 160

at X 16

t (^) a-

z

x 16- = 13 × 162 t^ 2x^16

t 10x^160 t^ (^5) X 16

• (^) I t 12 X^ lb^ - Z = 3370. 35937510

Ky

. (^) Other than^ decimal^ conversion 7.^ octal^ to Binary to (^) each octal^ digit , write three^ bit^ binary O -^000 1

• 001 2

O l O 3 -^ O^ l^ l 4 - l O^ O S - 101 6 - l l^ O 7 -^ l^ l^ l

9. Hexadecimal^ to Binary - For each hexadecimal^ digit , write a^ bit binary

I O

• 0000 9
• (^) I 001 1- 0001 A -^ I^ O^ l^ O z -^0010 B
• (^) l O (^) l l 3 - O O^ l^ l 4 -^ O^ l^

O O

C -^ l^ l^ O^ O 5 - O l O I b

• (^) l (^) l 01 6 - O^ l^ l^ O E
• l l (^) l O F

l l^ l^ l 7 - O (^) l l l 8 - 1000

19 Binary to (^) Hexadecimal Oy l l O^ l^ l^ l^ l^01.^ I^010 l^ l^ l^ z to Hexadecimal 000 l (^) l O (^) l l^ l (^) l O l (^). I O^ l^ O^ l l^ l O 1 B^ D A^ E = 1 B^ D^.^ A^ E 16

① One's^ complement of

a - ne

binary no. is the BITWISE (^) NOT applied to it^. 43, = 0 lolol^ ! 322216 8 4 21 so " ' oo (^) I 0 10 l (^) l

• 43, = (^) l l^ O 10100 I^ μ A- ③ Two's^ complement o
• ne nos^ rep by bit pattern which (^) is one greater than one's^ complement of^ the (^) value 43 , o = 00101 o^ l^ l^ "t HIM

I

oto=O

• 43

, o

= I^10 I^0 100 Cone's^ complement)

•  Gmo's^ complement 

9. i) 85-

4=

• 42=85+1- ) (^85) in to bit 000101010742 in (^) to bit (^0000101010) -42 in^1 's complement : (^1111010101)

• I

'

• 42 1111010110 in 2 's^ complement .

85 =^0001010101

• 42 = 10000101011 = y= 0000101011 7=