GATE & UGC-NET: Digital Circuits & Electromagnetics (EC) Study Material

GATE-ESE-NET
NET & GATE : Digital Circuits & Electromagnetics (EC) Study Material

◈ Digital Circuits

◈ Electromagnetics



1.Digital Circuits:

What is a MUX?

MUX is abbreviation of 'Multiplexer'. Where numbers of data streams are multiplexed to be transmitted thru same channel as parallel data streams. In case of modulation techniques, for example in QPSK modulation scheme we see two signals are transmitted thru same channel and even at same time also are signals are with different phase. If we do not apply modulation for this particular situation then we need to send two signals separately one by one. For your knowledge, modulation enables multiplexing of data in a communication media.


What is K-map?

K-map is used to minimize the logic operations in a combining circuit. It is helps to improve power efficiency by reducing propagation time thru different gates by minimizing number of operation a well. Which directly impacts circuit's cost.


Question



In the above diagram, s1 and s0 are the select lines. I0 and I3 are input data line. 'EN' is the enable line. And output is F which is a function of P, Q, and R or F(P, Q, R) = ?


Answer

Here, in the above K-map,

F = PˉQˉR + PQˉR + PQ

Or, PˉQˉR + PQˉR + PQRˉ + PQR

Or, Σm (1, 5, 6, 7)


Explanation:

In the above question it was given that output is the function of P, Q, and R and I0 and I3 are input lines. So, in the K-map we've not counted as '1' the inputs like, I0, I3, and R=I0 and I2. In general I0 is PˉQˉR, I3 is PQˉRˉ, and I2 is PQˉR.


1's Complement and 2's Complement

1's Complement of 1100 is 0011

Alternatively,

1's Complement of 1100 is equals to (1111 - 1100) = 0011

1100 => 0011


2's Complement of 1100 is 0100

Alternatively, 2's Complement of 1100 is equal to (1's Complement + binary '1')

Or, 0011 + 1 = 0100


Signed magnitude of a binary number:

In case of signed binary number if 1 comes in prefix then the equivalent decimal number comes with negative sign; Otherwise it is treated as positive decimal.

Example, 1100 => -4

0100 => +4


Question

P, Q, and R are decimal integers. P is the signed magnitude of 4 bit binary number 1100. Q and R are the 1's Complement and 2's Complement of the same binary number. Now tell us 6 bit 2's Complement representation of (P+Q+R) is -------

Answer

Given, P = signed magnitude of 1100 = -4

Q = 1's Complement of 1100 = 0011 = 3

R = 2's Complement of 1100 = 0100 = 4

Now, (P+Q+R) = 3

6 bit representation of decimal 3 is 000011

2's Complement of 000011 is equals to (1's Complement of 000011 + binary '1') = 111100 + 1=111101


A/D or D/A Converter:

It is a converter which converts digital data to analog data. For example to digitalize a analog signal i.e., voice signal (as it is analog in nature) we digitalize it by quantize where analog signal is represented by different quantize levels. Before that it must go through sampling process. This is a example of analog to digital converter.


Question

There is a 10 bit digital to analog converter. It can be calibrated over a range of 0 volt to 10 volts. Now, question is if input of that digital to analog converter is 13A (It is a hexadecimal number). That what will be the output of the converter?

Answer

Firstly, convert the hexadecimal input into decimal value as convert ranges from 0 to 10 volts which are decimal numbers. So, when a system works with decimal value then for convenience we convert the hexadecimal input into decimal.

(13A) = 1X(16)^2 + 3X(16)^1 + 10X(16)^0 = 314 (As we know A in hexadecimal is equivalent to decimal 10)

As, given in question that it is 10 bit converter. So, number of levels in converter is equal to 2^10 = 1024

Now, resolution of that converter is (10 Volts / 1024) = 9.76 millivolt

If the input is 13A (equivalent decimal = 314) then output will be

Resolution X equivalent decimal input

= 3.067 volts


Number System (Definition, Types, Conversion & Examples):

The same decimal number is represented by its binary, octal, hexadecimal equivalent. In the following sections, we'll go over the specifics.

Because it employs ten digits from 0 to 9, the decimal number system has a base of ten as there are total ten digits in decimal number's set. In the same way, the binary number system has two digits - 0 and 1 (so, base - 2). The octal number system has eight digits ranging from 0 to 7, whereas the hexadecimal number system has sixteen digits ranging from 0 to 15. Without further ado, let's look at how to convert from one number system to another.

Figure: Different types of number systems are converted into one another

Decimal to Binary conversion:

Let's say we take a decimal number, such as '123'.

We need to convert it to a binary equivalent.

We can convert '123' to another form as follows:

1*(10^2) + 2*(10^1) + 3*(10^0) = 123

The weights that apply here are ....., 100, 10, 1 (as per the position of a digit in a decimal number)

For binary number system applicable weights are

...., 2^5, 2^4, 2^3, 2^2, 2^1, 2^0

or, ...., 32, 16, 8, 4, 2, 1

Let me give an example, if a binary number is 1001, then equivalent decimal number is

1*(2^3) + 0*(2^2) + 0*(2^1) + 1*(2^0) = 8+0+0+1=9



Returning to the decimal to binary conversion of the decimal number '123,' we may use the binary weightage rule and use the procedure below.

1. The decimal number '123' is given.

As we all know, a binary number's weightage value is like this.

....., 256, 128, 64, 32, 16, 8, 4, 2, 1, 256, 128, 64, 32, 16, 8, 4, 2, 1

2. The number 128 is now greater than the decimal number 123.

3. So, start with 64 or 2^6, then add 32, making sure it doesn't exceed '123'. So, if you add the next (16), you get '112'. As a result, add next (8). The total addition is now '120'. When we add the next (4), the value of '123' is crossed. As a result, we reject it and add the following (2), followed by 1. The total addition is now '123'.

The binary equivalent of the decimal '123' is '1111011'.



Alternative Method:


In the alternate manner, you can see how we organize all of the reminders in a specific order. The MSB (most significant bits) is on the upper side of the arrow, while the LSB (least significant bits) is on the lower side.



Solve: The binary number 0.246 in binary is (up to 3 decimals) is ?

Solution:

0.246 * 2 = 0.492 =>0 (MSB)

0.492 * 2 = 0.984 =>0

0.984 * 2 = 1.968 =>1

0.968 * 2 = 1.936 =>1

....

So, answer is .001


Binary to Decimal:

Example: (111001)2 = (?)10

Solution:

(111001)2 = 1*(2^5) + 1*(2^4) + 1*(2^3) + 0*(2^2) + 0*(2^1) + 1*(2^0)

=32+16+8+0+0+1 = 57 (Ans.)



What if binary number continues after decimal point?

Example: (111001.11)2 = (?)10

(.11)2 = 1*(2^-1) + 1*(2^-2)

= 0.5 + 0.25

=.75

and, (111001)2 = 57 (as shown above)

So, (111001.11)2 = (57.75)10



Solve: The binary number 110011.11 is equivalent to which decimal number

A. 51.25

B. 49.75

C. 60

D. 50.75

Solution:

110011 (in binary) = 51 (in decimal)

Now, .11 (in binary) = 1*(2^-1)+1*(2^-2)= .75

So, 110011.11 (in binary) = 51.75 (in decimal)

Correct answer in option D.



Binary to octal:

Octal number set {0,1,2,3,4,5,6,7}

Example: (111001)2 = (?)8

Take three bits at a time and convert them to decimal equivalents. If there is no decimal point, always choose numbers from the left side.

(111001)2 = (71)8



Octal to Binary:

Similarly, if we want to convert an octal number to a binary number, we must convert each octal digit to its binary 3 bit equivalent.

For example

(72)8 = 111 010



Binary to hexadecimal:

Hexadecimal number set {0,1,2,3,4,5,6,7, 8, 9, A, B, C, D, E, F}

Example: (10011001)2 = (?)16

Take four bits at a time and convert them to decimal equivalents. If there is no decimal point, always choose numbers from the left side.

(10011001)2 = (99)16



Hexadecimal to Binary:

Similarly, if we want to convert a hexadecimal number to a binary number, we must convert each octal digit to its binary 4 bit equivalent.

For example

(99)16 = 1001 1001



Decimal to octal:

In this scenario, convert the decimal number to binary first, then convert binary to octal using the similar steps outlined above.

For example: (22)10 = (?)8

. (22)10 = (10110)2

Now, (10110)2 = (26)8



Octal to Decimal:

To convert an octal number to a binary number, first convert the octal number to binary, then convert the binary number to decimal.



Alternative Method:

Octal to decimal number:

(26)8 = (?)10

(26)8 = 2*(8^1) + 6*(8^0) =22



Decimal to hexadecimal:

In this scenario, convert the decimal number to binary first, then convert binary to hexadecimal using the similar steps outlined above.



Hexadecimal to Decimal:

Similarly, to convert a hexadecimal number to a binary number, convert the hexadecimal number to a binary number first, then the binary number to a decimal number.



Alternative Method:

hexadecimal to decimal number:

(26)16 = (?)10

(26)16 = 2*(16^1) + 6*(16^0) =32+6=38



Why are 1's and 2's complements Important in number system?

Consider the situation when we need to store a binary number in an 8-bit register.

Now we'll look at how to represent the decimal numbers +10 and -10 in binary form in an 8-bit register.

+10 will be stored in the register as 0 0 0 0 1 0 1 0; here, MSB bit '0' is used as the sign bit, where '0' represents a positive binary number and '1' represents a negative binary number.

If we need to store a negative number in the register, we will do so as follows:

1 1 1 1 0 1 0 1;

where we take the binary equivalent of +10's 1st complement. To get 1's complement, we just reverse the bits. To represent '-10,' we utilize bit '1' at the MSB bit location.

Simply add '1' to the LSB bit of 1's complement to get 2's complement of a binary number.

The 1's complement of the binary number 1100, for example, is 0011. (0011) + 1 = 0100 is now the 2's complement of the same binary value.

We can see from the preceding sections that we don't need to do anything with positive numbers, but when the number is negative, we use '1' as the MSB and then 1's complement of the equivalent binary integer.
Digital Filters IIR vs FIR (Click here)



Electromagnetics:

What is a transmission line's characteristic impedance?

In the case of a transmission line, the EM current and EM volts are constant over the length of the line. Coaxial cable is a common example of a transmission line. We'll assume EM current and EM volt are I and E, respectively, for the sake of simplicity, so characteristic impedance,

Z0=E/I



Question

A transmission line with a characteristic impedance of 50 ohm and a length of 3λ/4 is terminated with a load of 400 ohm. At the input, the impedance is ——-? ———

Answer

Given,

Characteristic impedance, Z0 =50 ohm

Load impedance, ZL = 400 ohm

So, input impedance, Z­in = (Z0)2/ZL

Or, Z­in = 502/400=6.25 ohm



Question

The electric field E0 in the far field proportions to (e-jkr /r)sin(θ) for an infinitesimally small dipole in free space, where k=2*π/λ. A vertically infinitesimally small dipole (δl << λ) is positioned above an infinite ideal conducting plane at a distance h (h>0). At theta = 60 degrees, what is the minimal value of h for which one of the far field radiation pattern maxima occurs? (Options: 0.75λ, λ, 0.25λ, and 0.5λ)

Answer

The correct response is λ.

An antenna, in general, radiates vertically from its axis for electromagnetic wave propagation. As the distance between the transmitting antenna and the receiving antenna increases, the power of the radiation weakens. The radiation region is divided into two portions as follows:

1. Near field

2. Far field

The amplitude of an EM wave in the far field is proportional to 1/r. It is 1/r3 for the near field reactive zone and 1/r2 for the near field radioactive region.

The far field region distance from the antenna ranges from 2λ to infinity. Where the wavelength of the operating frequency is equal to λ. Similarly, the area between the antenna and the length λ is referred to as the near field zone. And the area between λ and 2λ is known as the transition region.

For example, the length of a dipole antenna should be λ/2.



Question

H(x,y,z,t) = (ax + 2ay + baz) * cos(wt +3x –y –z) is the magnetic field of a uniform plane wave in vacuum.

Determine the value of b.

Answer

We know, in general, for electromagnetic wave propagation electric and magnetic fields are perpendicular in nature. The direction of an EM wave, on the other hand, is perpendicular to both the electric and magnetic field vectors.

The magnetic field vector in the given question is made up of two parts:

H = H0 * cosine part

Where H0 = (ax + 2ay + baz)

And direction vector, k=3ax –ay -az

In vector equations, if two vectors are perpendicular then dot product of them equals to 0.

So, H0.k=0

Or, (ax + 2ay + baz) . (3ax –ay -az­) = 0

Or, 3-2-b =0

Or, b=1
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