Assignment

Assignment

Q:1) Explain the decimal number system with an example. And give the general polynomial representation of a number.

A:) The decimal number system is the most commonly used number system because of its ease of use. We can express any decimal number in units, tens, hundreds, thousands and so on. It has radix 10 (i.e. base).

In decimal number system, the base or radix is 10 i.e. there are ten digits which we can use. They are: 0,1,2,3,4,5,6,7,8,9.

Eg: 59.79 is a decimal number. Its polynomial representation is:

5*10^1+9*10^0+7*10^-1+9*10^-2

In general, polynomial representation of any number in any system is:

N = a(n-1)*r^(n-1)+a(n-2)*r^(n-2)+………+a(1)*r^1+a(0)*r^0+a(-1)*r^-1+………+a(m)*r^(-m)

Where N=number in decimal

a=digit

r=radix or base

n=number of digits in integer portion

m=number of digits in fraction portion

Q:2) Explain the decimal to binary conversion up to 4 binary places. (47.8125)10.

A:) To convert a decimal number whose radix is 10 to binary number of radix 2, we convert the whole number part of the decimal number first to binary by repeated division by 2 in base 10 and keeping track of the remainder.

The fractional part of the decimal number is converted to binary by repeated multiplication by 2 in base 10 and keeping track of resulting whole number part of the answer until required number of binary places.

- To get whole number part, we read binary number upwards.- To get fractional part, we read binary number downward.

Whole no. part: 47

Binary of whole no. part is:

2 47

2 23 1

2 11 1

2 5 1

2 2 1

2 1 0

0 1

Thus (47)10 = (101111)2

Fractional part: 0.8125

2*0.8125=1.625=0.625 carry 1

2*0.625=1.25=0.25 carry 1

2*0.25=0.5=0.5 carry 0

2*0.5=1=0 carry 1

Thus (0.8125)10 = (0.1101)2

Therefore (47.8125)10 = (101111.1101)2

Q:3) Convert binary to:

a) 110101 to decimal110101 = 1*2^5+1*2^4+0*2^3+1*2^2+0*2^1+1*2^0 =32+16+0+4+0+1 =53Thus (110101)2 = (53)10

b) 1110.01101 to octal(1110.01101)2 = (001 110 . 011 010)2 = (1 6 . 3 2)8Thus (1110.01101)2 = (16.32)8

c) 1011101.110111 to hexadecimal(1011101.110111)2 = (0101 1101 . 1101 1100)2 = (5 D . D C)16Thus (1011101.110111)2 = (5D.DC)16

Q:4) Convert octal to decimal

a) 426.73(426.73)8 = 4*8^2+2*8^1+6*8^0+7*8^(-1)+3*8^(-2) = 256+16+6+0.875+0.046875 = (278.921875)10

b) 124.21(124.21)8 = 1*8^2+2*8^1+4*8^0+2*8^(-1)+1*8^(-2) = 64+16+4+0.25+0.015625) = (84.265625)10

c) 0.65(0.65)8 = 0*8^0+6*8^(-1)+5*8^(-2) = 0+0.75+0.078125 =(0.828125)10

Q:5) Convert decimal:

1) To octal:a) (847.951)10

8 8478 105 78 13 18 1 5

0 1(847)10 = (1517)8

And 0.951*8 = 7.608 = 0.608 carry 7

0.608*8 = 4.864 = 0.864 carry 4

0.864*8 = 6.912 = 0.912 carry 6

0.912*8 = 7.296 = 0.296 carry 7 and so on..

Thus (847.951)10 =(1517.7467)8

b) (0.728)100.728*8 = 5.824 = 0.824 carry 50.824*8 = 6.592 = 0.592 carry 60.592*8 = 4.736 = 0.736 carry 40.736*8 = 5.888 = 0.888 carry 5 and so on..Thus (0.728)10 = (0.5645)8

2) To hexadecimala) 2604.10546875

16 260416 162 1216 10 2

0 10(2604)10 = (A2C)16

And 0.10546875*16 = 1.6875 = 0.6875 carry 1

0.6875*16 = 11.00 = 0 carry 11

Thus (2604.10546875)10 = (A2C.1B)16

b) (64026)1016 6402616 4012 1416 250 1216 15 10

0 15Thus (64026)10 = (FACE)16

Q:6) Explain the process of binary subtraction:

a) Subtract ((100.111)2 from (110.101)2 1 1 0 . 1 0 1- 1 0 0 . 1 1 1 0 0 1 . 1 1 0 ans: (1.11)2 (positive)

b) Subtract (1011.11)2 from (101.1)2 1 0 1 1 . 1 1-0 1 0 1 . 1 0 0 0 1 0 . 0 1 ans: (-10.01)2 (as we have subtracted in reverse order)

Binary subtraction can be done by using the following rules:(It is similar to that of decimal numbers. If the borrow is 0, difference is positive, & if borrow is 1, difference is negative).

A B Difference Borrow0 0 0 00 1 1 11 0 1 01 1 0 0

Q:7) Perform binary subtraction using 1’s complement and 2’s complement method for:

1) 101.111 from 10.0101a) I’s complement:

1’s complement of larger number: (101.111)2= (010.000)2 0 1 0 . 0 1 0 1+0 1 0 . 0 0 0 0 1 0 0 . 0 1 0 1 = (100.0101)2

Answer is in one’s complement form:Answer : (-11.1010)2 (1’s complement with negative sign)

b) 2’s complement: 2’s complement of larger number : (101.111)2= (010.000)2 + 1.000= (11)2

1 0 . 0 1 0 1 +1 1 . 0 0 0 0 1 0 1 . 0 1 0 1 Now, answer is in 2’s complement form. Answer : (-011.101)2 (2’s complement wit negative sign)

2) 1101 from 110a) 1’s complement:

1’s complement of larger number: (1101)2= (0010)2

1 1 0 + 1 0 1 0 0 0 Answer in 1’s complement form. Thus answer = (-111)2

b) 2’s complement: 2’s complement of larger number: (1101)2= (10)2 + 1 = (11)2

1 1 0 + 1 1

1 0 0 1 Answer in 2’s complement form. Answer = (-111)2

Q:8)What are binary coded decimal numbers? Represent the following decimal numbers in BCD: (a) 821 (b) 35.67

A:) Binary coded decimal number or BCD, is numeric code in which each digit of a decimal number is represented by a separate group of bits. The most common BCD code is 8-4-2-1 BCD code, in which each digit of decimal number is represented by a 4-bit binary number.

8-4-2-1 BCD code is given as follows:

Decimal Digit BCD Code8 4 2 1

0 0 0 0 01 0 0 0 12 0 0 1 03 0 0 1 14 0 1 0 05 0 1 0 16 0 1 1 07 0 1 1 18 1 0 0 09 1 0 0 1

(positional weights bit 3 has weight 8, bit 2 has weight 4, bit 1 has weight 2 and bit 0 has weight 1)

a) 821 = 1000 0010 0001b) 35.67 = 0011 0101 . 0110 0111

Q:9) Explain the logical AND operation and also the construction of an AND gate using diode.

A:) Logical AND operation can be performed for two or more inputs to get one output. We will get output as high only when all the inputs are high.

Eg: If A & B are inputs and Y is output, then Y is high only when A & B both are high

Truth table:

A B (A.B) Y0 0 00 1 01 0 01 1 1

(Symbol in Boolean for AND is ‘.’)

Symbol used for AND gate is:

Circuit diagram of AND gate using diodes:

We can construct AND gate using diodes. Number of diodes used is equal to the number of inputs used.For two input AND gate. Circuit diagram is shown below:

Working of this gate can be explained as follows: Case 1: A is LOW & B is LOW:

Here, both D1 & D2 diodes are forward biased and both cathodes are grounded. Thus, output obtained Y is LOW.Case 2: A is LOW & B is HIGH:

Here, D1 is forward biased & D2 is reversed biased and cathode of D1 is grounded. It (D1) pulls the output (Vcc) voltage towards it and hence output Y is low.Case 3: A is HIGH & B is LOW:

Here, D1 is reversed biased, and D2 is forward biased and cathode of D2 is grounded. Hence D2 pulls the output (Vcc) voltage towards it and hence output Y is low.Case 4: A is HIGH & B is HIGH:

Here both D1 and D2 are reversed biased. Both diodes will not conduct and hence output Y is high.

Q:10) Explain the construction of a transistor based NOT gate.

A:) A NOT gate can be constructed using a transistor(n-p-n). The truth table for NOT logic is:

A

B Y=A+B

+Vcc

A

BY

Input Output0 11 0

A NOT gate is also called inverter operator. It will reverse the input given to it (as shown in truth table).

Circuit diagram:

Case I: When A is low:

Here resistance drop across resistor R will not be full 5V as no collector current will flow. Hence output Y will be HIGH.

Case II: When A is high:

Here resistance drop across resistor R will be full 5V as collector current flows. Hence, output Y will be LOW.

Symbol for NOT:

A

+Vcc

R

Y

A Y

Q:11) Name the universal gates and why are they called so?

A:) There are two logical universal gates. They are:

(1) NAND gate(NOT-AND gate)(2) NOR gate(NOT-OR gate)

These two gates are called so because these gates can be used to construct any other logic i.e. AND, OR, NOT or any other combination of the given logics.

The truth tables for the two universal gates are:

NAND NOR

A B Y0 0 10 1 11 0 11 1 0

Q:12)Prove the following Boolean identities using Boolean theorems and truth tables.

(a) A+BC = (A+B)(A+C)LHS : A+BC =A(1+C)+BC (As 1+C=1) =A+AC+BC (Distributive property) =A(1+B)+AC+BC (As 1+B=1) =A+AB+AC+BC (Distributive property) =AA+AB+AC+BC (AA=A) =A(A+B)+C(A+B) =(A+C)(A+B) (Distributive property) = RHS Hence proved

-Truth table:

A B C BC A+BC A+B A+C (A+B)(A+C)0 0 0 0 0 0 0 00 0 1 0 0 0 1 00 1 0 0 0 1 0 00 1 1 1 1 1 1 11 0 0 0 1 1 1 11 0 1 0 1 1 1 1

A B Y0 0 10 1 01 0 01 1 0

1 1 0 0 1 1 1 11 1 1 1 1 1 1 1

(b) ABC+AB’C+ABC’ = AB+ACLHS = ABC+AB’C+ABC’ =AC(B+B’)+ABC’ (Distributive property) =AC(1)+ABC’ (As B+B’=1) =AC+ABC’ (As AC.1=AC) =A(C+BC’) (Distributive property) =A(C+B) (As C+BC’=C+B) =AB+AC (Distributive property) = RHS Hence proved

-Truth table:

A B C B’ C’ ABC AB’C ABC’ (ABC+AB’C+ABC’) AB AC (AB+AC)0 0 0 1 1 0 0 0 0 0 0 00 0 1 1 0 0 0 0 0 0 0 00 1 0 0 1 0 0 0 0 0 0 00 1 1 0 0 0 0 0 0 0 0 01 0 0 1 1 0 0 0 0 0 0 01 0 1 1 0 0 1 0 1 0 1 11 1 0 0 1 0 0 1 1 1 0 11 1 1 0 0 1 0 0 1 1 1 1

Q:13) Simplify and realize the following Boolean expressions using logic gates:

(a) Y= ((BC)’+(AD)’((AB)’+(CD)’))’=(BC)’’.((AD)’((AB)’+(CD)’))’ (De Morgan’s theorem)=BC.((AD)’’+((AB)’+(CD)’)’) (De Morgan’s theorem)=BC.(AD+((AB)’’.(CD)’’) (De Morgan’s theorem)=BC.(AD+(AB.CD))=ABCD+(ABCD)(BC) (Distributive property)=ABCD+ABCD (As ABCD.BC=ABCD)=ABCD

Using logical gates:

(b) Y= AB+A’C+BC(Cannot be simplified further)Using logical gates:

(c) Y=(A+B’+C’)(A+B’+C)=(A+B’+C’)A+(A+B’+C’)B’+(A+B’+C’)C (Distributive property)=AA+AB’+AC’+AB’+B’B’+C’B’+AC+B’C+C’C (Distributive property)=A+AB’+AC’+B’+B’C’+AC+B’C+0=A(1+B’)+A(C’+C)+B’(1+C’)+B’C=A+A+B’+B’C=A+B’(1+C)=A+B’

Using logic gates:

ABC

D

Y=ABCD

Y

A

B

C

A’

AB

A’C

BC

Q:14) Simplify the expression and implement using only NAND gates and only NOR gates:

(a) XYZ+YZ+Z’=(X+1)YZ+Z’ (Distributive property)=1.YZ+Z’=YZ+Z’ (as 1.YZ=YZ)=Y+Z’ (as YZ+Z’=Y+Z’)

Using NAND gates:

Using NOR gates:

A

B

C

Y

O

X

Y

Z

Y’

(b) ((A+B’+C)(A’+B+C)(A+B))’=(((A+B’+C)A’+(A+B’+C)B+(A+B’+C)C)(A+B))’ (Distributive property)=((AA’+A’B’+A’C+AB+B’B+BC+AC+B’C+CC)(A+B))’ (Distributive property)=((0+A’B’+A’C+AB+0+BC+AC+B’C+C)(A+B))’=((A’B’+AB+C(A’+A)+C(B+B’)+C)(A+B))’ (Distributive property)=((A’B’+AB+C)(A+B ))’ (as A+A’=1, C.1=C & B+B’=1)=((A’B’+AB+C)A+(A’B’+AB+C)B)’ (Distributive property)=(A’B’.A+AB.A+AC+A’B’.B+AB.B+BC)’=(0+AB+AC+0+AB+BC)’=(AB+BC+AC)’ -> For NOR=(AB+BC)’.(AC)’=(AB)’(BC)’(AC)’ -> For NAND

Using NAND gates:

Using NOR gates:

X

Y

Z Z’

O

Y

(AC)’

(BC)’

(AB)’A

B

C

Q:15) Draw the output waveform and truth table of logic circuit for input waveform:

A

B

C

A:) Given logic circuit has equivalent Boolean expression:

Y=((AB)’+C)’=(A’+B’+C)=(A’+B’)’.C’=A.B.C’

Truth table :

A B C C’ Y=A.B.C’0 0 0 1 01 1 0 1 10 1 1 0 01 0 1 0 00 0 0 1 0

A

B

C

A

B

C

AB

BC

AC

Y

B

CY

A

1 1 0 1 10 1 1 0 0

Output timing diagram:

A

B

C

Y

Q:16) Explain the operation of half adder and full adder.

A:) Half Adder:-

The logic circuit which can perform addition of two (significant) bits is called half adder. It needs two inputs and it can give two outputs i.e. sum and carry. Truth table for the relation of the input and output can be given below:

Inputs OutputsA B Carry Sum0 0 0 00 1 0 11 0 0 11 1 1 0

It’s K-map diagram is as follows:

-for carry -for sum

A\B 0 10 0 01 0 1

Hence its logic and bloc diagram are as follows:

A\B 0 10 0 11 1 0

Half adder has limited use as it can add only two bits and for addition of three bits becomes more difficult.

Full adder:-

It is a combinational circuit that forms the arithmetic sum of three input bits. It consists of 3 inputs and two outputs. Inputs are denoted by A, B and Cin. Truth table is as follows:

Inputs OutputsA B Cin Carry Sum0 0 0 0 00 0 1 0 10 1 0 0 10 1 1 1 01 0 0 0 11 0 1 1 01 1 0 1 01 1 1 1 1

It’s K-map can be written as follows:

For Cout (Carry)-

A\BCin 00 01 11 100 0 0 1 01 0 1 1 1

Cout=AB+ACin+BCin

For Sum-

Sum=A’B’Cin+A’BCin’+AB’Cin’+ABCin

Half Adder

A

BSum

Carry

A

B

Sum

Carry

A\BCin 00 01 11 100 0 1 0 11 1 0 1 0

=ABCin

Hence it’s logic and block diagram is as follows:

Full adder can be used also to add n-bits when used as parallel binary adder.

CATHODE RAY OSCILLOSCOPE

The cathode-ray oscilloscope (CRO) is a common laboratory instrument that provides accurate time and amplitude measurements of voltage signals over a wide range of frequencies. Its reliability, stability, and ease of operation make it suitable as a general purpose laboratory instrument. The heart of the CRO is a cathode-ray tube .

Main parts of the Cathode ray tube are

Electron gun, Deflection system, Fluorescent screen, Glass tube, Base.

Cin

AB Sum

Cout

A

B

Cin

Sum

Cout

Full Adder

The cathode ray is a beam of electrons which are emitted by the heated cathode (negative electrode) and accelerated toward the fluorescent screen. The assembly of the cathode, intensity grid, focus grid, and accelerating anode (positive electrode) is called an electron gun. Its purpose is to generate the electron beam and control its intensity and focus. Between the electron gun and the fluorescent screen are two pair of metal plates - one oriented to provide horizontal deflection of the beam and one pair oriented ot give vertical deflection to the beam. These plates are thus referred to as the horizontal and vertical deflection plates. The combination of these two deflections allows the beam to reach any portion of the fluorescent screen. Wherever the electron beam hits the screen, the phosphor is excited and light is emitted from that point.

(Refer to Godse page no.381 to 384 for mainparts in detail)

 The linear deflection or sweep of the beam horizontally is accomplished by use of a sweep generator that is incorporated in the oscilloscope circuitry. The voltage output of such a generator is that of a sawtooth wave as shown in Fig. 2. Application of one cycle of this voltage difference, which increases linearly with time, to the horizontal plates causes the beam to be deflected linearly with time across the tube face. When the voltage suddenly falls to zero, as at points (a) (b) (c), etc...., the end of each sweep - the beam flies back to its initial position. The horizontal deflection of the beam is repeated periodically, the frequency of synchronization of the two deflections, the pattern on the tube face repeats itself and hence appears to remain stationary. this periodicity is adjustable by external controls.

Front panel controls of Simple C.R.O

These are divided into 4 groups:

1.BASIC CONTROLS:

ON-OFF:This switch turns ON or OFF the CRO.

Intensity:This controls the intensity or the brightness of the light produced by beam spot.It actually controls the number of electrons bombarding the screen per second.The high intensity spot shorten the life of cathoderay tube.Potentiometer varies the control grid voltage and is the intensity contol.

Focus:This controls the sharpness of the spot.Focussing of the spot is obtained by vaying the voltage applied to the focusing anodes of the cathode ray tube.

Asigmatism:With the help of focus control and astigmatism control,a sharp spot can be obtained both in the centre and also at the edges of the screen.

Scale illumination:The screen infront of the CRT has grid engraved on it appears similar to graph paper.This is called graticule.This control illuminates thescreenand hence lines can be seen easily.

VERTICAL AMPLIFIER SECTION

Position:  Controls vertical positioning of oscilloscope display.

Invert: this control inverts the input signal,it multiplies by -1 .Then +ve inputvoltage become –ve.

*10: This control makes the gain of vertical amplifier 10 times as great as normal

. Variable Sensitivity:  Provides a continuous range of sensitivities between the calibrated steps. Normally the sensitivity is calibrated only when the variable knob is in the fully clockwise position.

Vertical coupling:AC-DC-GND:  Selects desired coupling (ac or dc) for incoming signal applied to vertical amplifier, or grounds the amplifier input. Selecting dc couples the input directly to the amplifier; selecting ac send the signal through a capacitor before going to the amplifier thus blocking any constant component.

Vertical mode control:1: only signal at input 1 is displayed, 2: only signal at input 2 is displayed, 1+2: sum of inputs 1 and2 is displayed, 1-2: difference of inputs 1 and 2 is displayed:,alternate: input 1 is displayed first and then 2 is displayed,then input 1 again.So the relation between two can be studied This mode is useful for high frequency.. chop: input 1is displayed for fraction of a microsecond then input 2.inthiswayplot of boththe inputs can be drawn at the same time.This mode is useful for low frequency.

HORIZONTAL-SWEEP SECTION

Sweep time/cm:  Selects desired sweep rate from calibrated steps or admits external signal to horizontal amplifier.

Sweep time/cm Variable:  Provides continuously variable sweep rates. Calibrated position is fully clockwise.

Position:  Controls horizontal position of trace on screen.

Synchronization: Synchronisation is a must to obtain the stationary pattern on the screen.The types of signals selected by synchronous selector switch are:

Internal :The trigger is obtained from signal being measured through the vertical amplifier.

Line: The input to the trigger circuit is from a.c mains supply.

External: The input to the trigger circuit is from the external trigger circuit.

Z-AXIS INTENSITY CONTROL

It is useful for brightening the display. Periodic positive pulse are applied to the grid and alternatively negative pulse are applied to cathode,to brighten the beamduring its sweep

period.This control is obtained by inserting a signal between the ground and the control grid or ground and the cathode.

IV. Lissajous Figures:  When sine-wave signals of different frequencies are input to the horizontal and vertical amplifiers a stationary pattern is formed on the CRT when the ratio of the two frequencies is an intergral fraction such as 1/2, 2/3, 4/3, 1/5, etc. These stationary patterns are known as Lissajous figures and can be used for comparison measurement of frequencies.

Lissajous pattern method is the quickest method of measuring the frequency.The patterns obtained by applying simultaneously two different sinewave horizontal and vertical deflection plates are called Lissajous figures.The shape of these figures depend on Amplitudes,phase dfference and ratio of frequencies between two waves.

If 2 signals are applied ,having same amplitude and frequency having phasedifference between them e1=Em sinώt ,e2=Em sin(ώt+Ф)

The shapes of Lissajous figures vary for various values of Ф.

Phase angle can be obtained as, Ф=sinֿ1(y1/y2)

= sinֿ1(x1/x2)

Ex:Lissajous figure obtained on the CRO is shown in fig.find the phase difference between two waves applied.