Electromagnetics. Electronics & Communication Engineering. For

Electromagnetics

For

Electronics & Communication Engineering

By

www.thegateacademy.com

✆ 080-40611000

Syllabus

Syllabus for Electromagnetics

Electrostatics, Maxwell’s Equations, Differential and Integral forms and their Interpretation, Boundary Conditions, wave Equation, Pointing Vector, Plane Waves and Properties, Reflection and Refraction, Polarization, Phase and Group Velocity, Propagation Through Various Media, Skin Depth, Transmission Lines, Equations, Characteristic Impedance, Impedance Matching, Impedance Transformation, S-parameters, Smith Chart, Waveguides, Modes, Boundary Conditions, Cut-Off Frequencies, Dispersion Relations, Antennas, Antenna Types, Radiation Pattern, Gain and Directivity, Return Loss, Antenna Arrays, Basics of Radar, Light Propagation in Optical Fibers.

Previous Year GATE Papers and Analysis GATE Papers with answer key

Subject wise Weightage Analysis thegateacademy.com/gate-papers

thegateacademy.com/gate-syllabus

Contents

[email protected] ©Copyright reserved. Web:www.thegateacademy.com i

Co C on nt te en n ts t s

Chapters Page No.

#1. Electromagnetic Field 1 – 33



Introduction 1



Operators 2 – 7



Material and Physical Constants 7 – 8



Electromagnetic (EM Field) 8



Electric Field Intensity 9 – 11



Electric Dipole 11 – 17



Divergence of Current Density and Relaxation 17 –18



Boundary Conditions 18 – 20



The Magnetic Vector Potential 20 – 24



Faraday’s Law 24 – 26



Maxwell’s Equation’s 26 – 27



Solved Examples 27 – 33

#2. EM Wave Propagation 34 – 51



Introduction 34



General Wave Equations 34 – 35



Plane Wave in a Dielectric Medium 35 – 39



Loss Tangent 39 – 44



Wave Polarization 44 – 51

#3. Transmission Lines 52 –71



Introduction 52 – 53



Classification of Transmission Lines 53 – 57



Transients on Transmission Lines 57 – 58



Bounce Diagram 58 – 61



Transmission & Reflection of Waves on a Transmission Line 61 – 62



Impedance of a Transmission Line 63 – 68



The Smith Chart 68 – 70



Scattering Parameters 70 – 71

#4. Guided E.M Waves 72 – 93



Introduction 72



Wave Guide 72



Rectangular Wave Guide 73 – 75



Transverse Magnetic (TM) Mode: (H

= 0) 75 – 78

Contents



Transverse Electric (TE) Mode: (E

= 0) 78 – 83



Non-Existence of TEM Waves in Wave Guides 84



Circuter Wave Guide 84 – 85



Optical Fiber 85 – 87



Solved Examples 87 – 93

#5. Antennas 94 – 124



Inroduction 94



Hertzian Dipole 95 – 97



Field Regions 97



Antenna Characteristics 98 – 101



Radiation Pattern 101 – 102



Antenna Arrays 102 – 107



Basics of Radar 107 – 112



Solved Examples 112 – 124

Reference Books 125

[email protected] ©Copyright reserved. Web:www.thegateacademy.com 1 “Picture yourself vividly as winning and that alone will contribute immeasurably to success."

…Harry Fosdick

Electromagnetic Field

Learning Objectives

After reading this chapter, you will know:

1. Elements of Vector Calculus 2. Operators, Curl, Divergence

3. Electromagnetic Coulombs’ law, Electric Field Intensity, Electric Dipole, Electric Flux Density 4. Gauss's Law, Electric Potential

5. Divergence of Current Density and Relaxation 6. Boundary Conditions

7. Biot-Savart’s Law, Ampere Circuit Law, Continuity Equation

8. Magnetic Vector Potential, Energy Density of Electric & Magnetic Fields, Stored Energy in Inductance

9. Faraday’s Law, Motional EMF, Induced EMF Approach 10. Maxwell’s Equations

Introduction

Cartesian coordinates (x, y, z), −∞ < x < ∞, −∞ < y < ∞, −∞ < z < ∞ Cylindrical coordinates (ρ , ϕ, z), 0 ≤ ρ < ∞, 0 ≤ ϕ < 2π, −∞ < z < ∞ Spherical coordinates (r, θ, ϕ ) , 0 ≤ r < ∞, 0 ≤ θ ≤ π, 0 ≤ ϕ < 2π Other valid alternative range of θ and ϕ are---

(i) 0 ≤ θ < 2π, 0 ≤ ϕ ≤ π (ii) −π ≤ θ ≤ π, 0 ≤ ϕ ≤ π (iii) −^π

2≤ θ ≤ π 2⁄ , 0 ≤ ϕ < 2π (iv) 0 < θ ≤ π, −π ≤ ϕ < π Vector Calculus Formula

SL. No Cartesian Coordinates Cylindrical Coordinates Spherical Coordinates (a) Differential Displacement

dl = dx ax + dy ay + dz az

dl = dρa_ρ + ρdϕa_ϕ +dza_z dl = dra_r + rdθa_θ + r sin θdϕa_ϕ

(b) Differential Area dS = dy dz a_x = dx dz a_y = dx dy a_z

dS = ρ dϕ dz a_ρ = d ρ dz a_ϕ = ρdρd ϕ a_z

d_s = r²sin θ dθ dϕ a_r = r sin θ dr dϕ a_θ = r dr dθ a_ϕ (c) Differential Volume

dv = dx dy dz

dv = ρ dρ dϕ dz dv = r²sin θ dθ dϕ dr

C HAP TER

1

Electromagnetic Field

Operators

1) ∇ V – Gradient, of a Scalar V 2) ∇ V – Divergence, of a Vector V 3) ∇ × V – Curl, of a Vector V 4) ∇² V – Laplacian, of a Scalar V DEL Operator:

∇ = ∂

∂xa_x+ ∂

∂ya_y+ ∂

∂z a_z(Cartesian) = ∂

∂ρa_ρ+1 ρ

∂

∂ϕa_ϕ+ ∂

∂z a_z(Cylindrical) = ∂

∂ra_r+1 r

∂

∂θa_θ+ 1 rsi n θ

∂

∂ϕ a_ϕ(Spherical) Gradient of a Scalar field

V is a vector that represents both the magnitude and the direction of maximum space rate of increase of V.

∇V =∂V

∂xa_x+∂V

∂ya_y+∂V

∂z a_z For Cartisian Coordinates =∂V

∂ρa_ρ+1 ρ

∂V

∂ϕa_ϕ+∂V

∂z a_z For Spherical Coordinates =∂V

∂ra_r+1 r

∂V

∂θa_θ+ 1 rsi n θ

∂V

∂ϕ a_ϕ For Cylindrical Coordinates

The following are the fundamental properties of the gradient of a scalar field V 1. The magnitude of ∇V equals the maximum rate of change in V per unit distance.

2. ∇V points in the direction of the maximum rate of change in V.

3. ∇V at any point is perpendicular to the constant V surface that passes through that point.

4. If A = ∇V, V is said to be the scalar potential of A.

5. The projection of ∇V in the direction of a unit vector a is ∇V. a and is called the directional derivative of V along a. This is the rate of change of V in direction of a.

Example: Find the Gradient of the following scalar fields:

(a) V = e^−z sin 2x cosh y (b) U = ρ²z cos 2ϕ (c) W = 10r sin²θ cos ϕ Solution:

(a) ∇V =^∂V

∂xa_x+^∂V

∂ya_y+^∂V

∂za_z

= 2e^−z cos 2x cosh y a_x+ e^−z sin 2x sinh y a_y− e^−z sin 2x cosh y a_z (b) ∇U =^∂U

∂ρa_ρ+¹

ρ

∂U

∂ϕa_ϕ+^∂U

∂za_z

= 2ρz cos 2ϕ a_ρ− 2ρz sin 2ϕ a_ϕ+ ρ² cos 2 ϕ a_z (c) ∇W =^∂W

∂r a_r+¹

r

∂W

∂θa_θ+ ¹

r sin θ

∂W

∂ϕa_ϕ

= 10 sin²θ cos ϕ a_r+ 10 sin 2θ cos ϕ a_θ− 10 sin θ sinϕ aϕ

Electromagnetic Field

[email protected] ©Copyright reserved. Web:www.thegateacademy.com 3

Divergence of a Vector

Statement: Divergence of A at a given point P is the outward flux per unit volume as the volume shrinks about P.

Hence,

DivA = ∇. A = lim

∆v→0

∮ A . ds_S

∆v … … … (1)

Where, ∆v is the volume enclosed by the closed surface S in which P is located. Physically, we may regard the divergence of the vector field A at a given point as a measure of how much the field diverges or emanates from that point.

∇. A =∂A_x

∂x +∂A_y

∂y

∂A_z

∂z Cartisian System

=1 ρ

∂

∂ρ(ρAρ) +1 ρ

∂A_ϕ

∂ϕ + ∂A_z

∂z Cylindrical System

= 1 r²

∂

∂r(r²A_r) + 1 r sin θ

∂

∂θ(A_θsin θ) + 1 r sin θ

∂Aϕ

∂ϕ Sphearical System From equation (1),

∮ A . dS

S

= ∫ ∇ . A dv

V

This is called divergence theorem which states that the total outward flux of the vector field A through a closed surface S is same as the volume integral of the divergence of A.

Example: Determine the divergence of these vector field (a) P = x²yza_x+ xza_z

(b) Q = ρ sin ϕ a_ρ+ ρ²za_ϕ+ z cos ϕ a_z

(c) T =_r¹₂cos θ a_r+ r sin θ cos ϕ a_θ+ cos θ a_ϕ Solution:

(a) ∇. P = ^∂

∂xP_x+ ^∂

∂yP_y+ ^∂

∂zP_z = ∂

∂x(x²yz) + ∂

∂y(0) + ∂

∂z(xz) = 2xyz + x

(b) ∇ . Q =¹

ρ

∂

∂ρ(ρQ_ρ) +¹

ρ

∂

∂ϕQ_ϕ+ ^∂

∂zQ_z =1

ρ

∂

∂ρ(ρ²sin ϕ) +1 ρ

∂

∂ϕ(ρ²z) + ∂

∂z (z cos ϕ) = 2 sin ϕ + cos ϕ

(c) ∇. T = ¹

r²

∂

∂r(r²T_r) + ¹

r sin θ

∂

∂θ(T_θsin θ) + ¹

r sin θ

∂

∂ϕ (T_ϕ) = 1

r²

∂

∂r(cos θ) + 1 r sin θ

∂

∂θ(r sin ²θ cos ϕ) + 1 r sin θ

∂

∂ϕ(cos θ) = 0 + 1

r sin θ2r sin θ cos θ cos ϕ + 0 = 2 cos θ cos ϕ

Curl of a Vector

Curl of a Vector field provides the maximum value of the circulation of the field per unit area and indicates the direction along which this maximum value occurs.

P

Electromagnetic Field That is,

Curl A = ∇ × A = lim

ΔS→0(∮ A . dl_L

∆S )

max

a_n… … … . . (2)

∇ × A = ||

a_x a_y a_z

∂

∂x

∂

∂y

∂

∂z A_x A_y A_z

||

= 1 ρ ||

a_ρ ρa_ϕ a_z

∂

∂ρ

∂

∂ϕ

∂

∂z A_ρ ρA_ϕ A_z

||

= 1 r²sin θ ||

a_ρ ra_θ r sin θ a_ϕ

∂

∂r

∂

∂θ

∂

∂ϕ A_r rA_θ r sin θ A_ϕ

||

From equation (2) we may expect that

∮ 𝐀 𝐝𝐥 = ∫(𝛁 × 𝐀

𝐒

) . 𝐝𝐬

𝐋

This is called stoke’s theorem, which states that the circulation of a vector field A around a (closed) path L is equal to the surface integral of the curl of A over the open surface S bounded by L, Provided A and 𝚫 × 𝐀 are continuous no s.

Example: Determine the curl of each of the vector fields.

(a) 𝐏 = 𝐱^𝟐𝐲𝐳 𝐚_𝐱+ 𝐱𝐳𝐚_𝐳

(b) 𝐐 = 𝛒 𝐬𝐢𝐧 𝛟𝐚_𝛒+ 𝛒^𝟐 𝐳𝐚_𝛟+ 𝐳 𝐜𝐨𝐬 𝛟𝐚_𝐳

(c) 𝐓 =_𝐫^𝟏_𝟐𝐜𝐨𝐬 𝛉 𝐚_𝐫+ 𝐫 𝐬𝐢𝐧𝛉 𝐜𝐨𝐬 𝛟𝐚_𝛉+ 𝐜𝐨𝐬 𝛟 𝐚_𝛟 Solution:

(𝐚) 𝛁 × 𝐏 = (𝛛𝐏_𝐳

𝛛𝐲 −𝛛𝐏_𝐲

𝛛𝐳) 𝐚𝐱+ (𝛛𝐏_𝐱

𝛛𝐳 −𝛛𝐏_𝐳

𝛛𝐱) 𝐚_𝐲+ (𝛛𝐏_𝐲

𝛛𝐱 −𝛛𝐏_𝐱

𝛛𝐲) 𝐚𝐳

= (𝟎 − 𝟎)𝐚_𝐱+ (𝐱^𝟐𝐲 − 𝐳)𝐚_𝐲+ (𝟎 − 𝐱^𝟐𝐳)𝐚_𝐳 = (𝐱^𝟐𝐲 − 𝐳)𝐚_𝐲− 𝐱^𝟐𝐳𝐚_𝐳

(𝐛) 𝛁 × 𝐐 = [𝟏 𝛒

𝛛𝐐_𝐳

𝛛𝛟 −𝛛𝐐_𝛟

𝛛𝐳 ] 𝐚_𝛒+ [𝛛𝐐_𝛒

𝛛𝐳 −𝛛𝐐_𝐳

𝛛𝛒] 𝐚_𝛟+𝟏 𝛒 [𝛛

𝛛𝛒(𝛒𝐐_𝛟) −𝛛𝐐_𝛒

𝛛𝛟] 𝐚_𝐳 = (−𝐳

𝛒 𝐬𝐢𝐧 𝛟 − 𝛒^𝟐) 𝐚_𝛒+ (𝟎 − 𝟎)𝐚_𝛟+𝟏

𝛒(𝟑𝛒^𝟐𝐳 − 𝛒 𝐜𝐨𝐬 𝛟)𝐚_𝐳 = −𝟏

𝛒(𝐳 𝐬𝐢𝐧 𝛟 + 𝛒^𝟑)𝐚_𝛒+ (𝟑𝛒𝐳 − 𝐜𝐨𝐬 𝛟)𝐚_𝐳

(𝐜) 𝛁 × 𝐓 = 𝟏 𝐫 𝐬𝐢𝐧 𝛉[𝛛

𝛛𝛉(𝐓_𝛟𝐬𝐢𝐧 𝛉) − 𝛛

𝛛𝛟𝐓_𝛉] 𝐚_𝐫 +𝟏

𝐫[ 𝟏 𝐬𝐢𝐧 𝛉

𝛛

𝛛𝛟𝐓_𝐫− 𝛛

𝛛𝐫(𝐫𝐓_𝛟)] 𝐚_𝛉 +𝟏

𝐫 [𝛛

𝛛𝐫(𝐫𝐓_𝛉) − 𝛛

𝛛𝛉𝐓_𝐫] 𝐚_𝛟

Electromagnetic Field

[email protected] ©Copyright reserved. Web:www.thegateacademy.com 5

= 𝟏

𝐫 𝐬𝐢𝐧 𝛉[𝛛

𝛛𝛉(𝐜𝐨𝐬 𝛉 𝐬𝐢𝐧 𝛉) − 𝛛

𝛛𝛟(𝐫 𝐬𝐢𝐧 𝛉 𝐜𝐨𝐬 𝛟)] 𝐚_𝐫 +𝟏

𝐫 [ 𝟏 𝐬𝐢𝐧 𝛉

𝛛

𝛛𝛟

(𝐜𝐨𝐬 𝛉) 𝐫^𝟐 − 𝛛

𝛛𝐫(𝐫 𝐜𝐨𝐬 𝛉)] 𝐚_𝛉 +𝟏

𝐫[𝛛

𝛛𝐫(𝐫^𝟐𝐬𝐢𝐧 𝛉 𝐜𝐨𝐬 𝛟) − 𝛛

𝛛𝛉

(𝐜𝐨𝐬 𝛉) 𝐫^𝟐 ] 𝐚𝛟

= 𝟏

𝐫 𝐬𝐢𝐧 𝛉(𝐜𝐨𝐬 𝟐𝛉 + 𝐫 𝐬𝐢𝐧 𝛉 𝐬𝐢𝐧 𝛟)𝐚_𝐫+𝟏

𝐫(𝟎 − 𝐜𝐨𝐬 𝛉)𝐚_𝛉 +𝟏

𝐫(𝟐𝐫 𝐬𝐢𝐧 𝛉 𝐜𝐨𝐬 𝛟 +𝐬𝐢𝐧 𝛉 𝐫^𝟐 ) 𝐚_𝛟

= (𝐜𝐨𝐬 𝟐𝛉

𝐫 𝐬𝐢𝐧 𝛉+ 𝐬𝐢𝐧 𝛟) 𝐚_𝐫−𝐜𝐨𝐬 𝛉

𝐫 𝐚_𝛉+ (𝟐 𝐜𝐨𝐬 𝛟 + 𝟏

𝐫^𝟑) 𝐬𝐢𝐧 𝛉 𝐚_𝛟 Laplacian

(a) Laplacian of a scalar field V, is the divergence of the gradient of V and is written as 𝛁^𝟐𝐕.

𝛁^𝟐𝐕 =𝛛^𝟐𝐕

𝛛𝐱^𝟐+𝛛^𝟐𝐕

𝛛𝐲^𝟐+𝛛^𝟐𝐕

𝛛𝐳^𝟐 → 𝐅𝐨𝐫 𝐂𝐚𝐫𝐭𝐢𝐬𝐢𝐚𝐧 𝐂𝐨𝐨𝐫𝐝𝐢𝐧𝐚𝐭𝐞𝐬

𝛁^𝟐𝐕 =𝟏 𝛒

𝛛

𝛛𝛒(𝛒𝛛𝐕

𝛛𝛒) + 𝟏 𝛒^𝟐

𝛛^𝟐𝐕

𝛛𝛟^𝟐+𝛛^𝟐𝐕

𝛛𝐳^𝟐 → 𝐅𝐨𝐫 𝐂𝐲𝐥𝐢𝐧𝐝𝐫𝐢𝐜𝐚𝐥 𝐂𝐨𝐨𝐫𝐝𝐢𝐧𝐚𝐭𝐞𝐬 = 𝟏

𝐫^𝟐

𝛛

𝛛𝐫(𝐫^𝟐𝛛𝐕

𝛛𝐫) + 𝟏 𝐫^𝟐𝐬𝐢𝐧 𝛉

𝛛

𝛛𝛉(𝐬𝐢𝐧 𝛉𝛛𝐕

𝛛𝛉) + 𝟏 𝐫^𝟐𝐬𝐢𝐧 𝛉

𝛛^𝟐𝐕

𝛛𝛟^𝟐→ 𝐅𝐨𝐫 𝐒𝐩𝐡𝐞𝐫𝐢𝐜𝐚𝐥 𝐂𝐨𝐨𝐫𝐝𝐢𝐧𝐚𝐭𝐞𝐬 If 𝛁^𝟐𝐕 = 0, V is said to be harmonic in the region.

A vector field is solenoid if ∇.A = 0; it is irrotational or conservative if ∇ × A = 0

𝛁. (𝛁 × 𝐀) = 𝟎

𝛁 × (𝛁𝐕) = 𝟎 (b) Laplacian of Vector A̅

∇²A⃗⃗ = ⋯ is always a vector quantity

∇²A⃗⃗ = (∇²A_x)âx + (∇²A_y)ây + (∇²A_z)âz

∇²A_x→ Scalar quantity

∇²A_y→ Scalar quantity

∇²A_z→ Scalar quantity

∇²V =^−p

ϵ...Poission’s E.q.

∇²V = 0 ...Laplace E.q.

∇²E = μσ ∂E⃗⃗

∂t+ μE∂²E

∂t². . . wave E. q.

Example: The potential (scalar) distribution in free space is given as V = 10y⁴+ 20x³. If ε₀: permittivity of free space what is the charge density ρ at the point (2,0)?

𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Poission’s Equation ∇² V = −ρ ε (∂²

∂x²+ ∂²

∂y²+ ∂²

∂z²) (10 y⁴+ 20x³) =−ρ ε₀

∵ ε = ε_r ε₀[ε = ε₀ as ε_r= 1]

20 × 3 × 2x + 10 × 4 × 3y² =^−ρ

ε₀

Electromagnetic Field At pt(2, 10) ⇒ 20 × 3 × 2 × 2 =−ρ

ε₀ ρ = −240ε₀ Example: Find the Laplacian of the following scalar fields

(a) V = e^−z sin 2x cosh y (b) U = ρ²z cos 2ϕ (c) W = 10r sin² θ cos ϕ

Solution: The Laplacian in the Cartesian system can be found by taking the first derivative and later the second derivative.

(a) ∇²V =∂²V

∂x²+∂²V

∂y²+∂²V

∂z² = ∂

∂x(2e^−zcos 2x cosh y) + ∂

∂y(e^−zsin 2x sinh y) + ∂

∂z(−e^−zsin 2x cosh y) = −4e^−zsin 2x cosh y + e^−zsin 2x cosh y + e^−zsin 2x cosh y

= −2e^−zsin 2x cosh y (b) ∇²U =1

ρ

∂

∂ρ(ρ∂U

∂ρ) + 1 ρ²

∂²U

∂ϕ²+∂²U

∂z² =1

ρ

∂

∂ρ(2ρ²z cos 2ϕ) − 1

ρ²4ρ²z cos 2ϕ + 0 = 4z cos 2ϕ − 4z cos 2ϕ

= 0 (c) ∇²W = 1 r²

∂

∂r (r²∂W

∂r) + 1 r²sin θ

∂

∂θ(sin θ∂W

∂θ) + 1 r² sin²θ

∂²W

∂ϕ² = 1

r²

∂

∂r(10 r² sin²θ cos ϕ) + 1 r² sinθ

∂

∂θ(10r sin 2θ sin θ cos ϕ) −10r sin²θ cos ϕ r² sin²θ =20 sin² θ cos ϕ

r +20r cos 2θ sin θ cos ϕ

r²sin θ +10r sin 2θ cos θ cos ϕ

r²sin θ −10 cos ϕ r =10 cos ϕ

r (2 sin² θ + 2 cos 2θ + 2 cos² θ − 1) =10 cos ϕ

r (1 + 2 cos 2θ) Stoke’s Theorem

Statement: Closed line integral of any vector A⃗⃗ integrated over any closed curve C is always equal to the surface integral of curl of vector A⃗⃗ integrated over the surface area ‘s’ which is enclosed by the closed curve ‘c’.

∮ A⃗⃗ . dL⃗ = ∫ ∫(∇ × A⃗⃗ )

S

dS⃗

The theorem is valid irrespective of (i) Shape of closed curve ‘C’

S

C