ElectroMagnetic Theory Concepts & Solved Numericals

Q.4) A 70 Ω lossless line has s = 1.6 and θ Γ = 300 o . If the line is 0.6λ long, obtain (a) Γ, Z L , Z in (b) The distance of the first minimum voltage from the load Answer: a) Using the smith chart, locate S at s = 1.6. Draw a circle of radius OS. Locate P where θ Γ = 300 o . At P, | Γ | = OP / OQ = 2.1 cm / 9.2 cm = 0.228 Γ = 0.228 ∠300 o Also at P, Z L = 1.15 – j0.48, Z L = Z o  Z L = 70 (1.15 – j0.48) = 80.5 – j33.6 Ω l = 0.6 λ = 0.6 x 720 o = 432 o = 360 o + 72 o From P, move 432 o to R. At R, Z in = 0.68 – j0.25 Z in = Z o  Z in = 70 (0.68 – j0.25) = 47.6 – j17.5 Ω b) The maximum voltage (the only one) occurs at θ Γ = 180 o ; its distance from the load is      (180 – 60) λ / 720  =   λ / 6  =   0.1667 Ω Q.5) A lossless 60 Ω line is terminated by a 60 + j60 Ω load. (a) Find Γ and s. If Z in = 120 - j60 Ω, how far (in terms of wavelengths) is the load from the generator? Solve this without using the Smith chart

Q.1) A transmission line operating at 500 MHz has Z o = 80Ω, α = 0.04Np/m, β = 1.5 rad/m. Find the line parameters R, L, G and C. Answer: Since Z o is real & α ǂ 0, this is a distortionless line. R o = α Z o = 0.04 x 80 = 3.2 Ω / m G = α / Z o = 0.04 / 80 = 5 x 10-4 Ω / m L = β Z o / ω = 1.5 x 80 / (2 π x 5 x 108 ) = 38.2 nH / m C = L G / R = 38.2 x 10-9 x 5 x 10-4 / 3.2 = 5.97 pF / m Q.2) A telephone line R = 30 Ω /km, L = 100 mH/km, G = 0 , and C = 20 µF/km. At f = 1 kHz, obtain: a) The characteristics impedance of the line. b) The propagation constant. c) The phase velocity. Answer: Q.3) A 40 m long transmission line has V g = 15 ∠0 o V rms , Z o = 30 + j60 Ω, and V L = 5 ∠-48 o V rms . If the line is matched to the load, calculate: a) The input impedance Z in b) The sending end current I in and voltage V in c) The propagation constant γ Answer: a) Z g  =  Z 1  --> Z in  =  Z o  =  30 + j

19) At the far field, an antenna produces Calculate the directive gain and the directivity of the antenna? SOLUTION: 20) For a thin dipole λ/16 long, find the a) Directive gain b) Directivity c) Effective area d) Radiation resistance SOLUTION: a) On substituting we get, G φ = 1.5 sin 2 θ b) Directivity , D = G φ, max = 1.5 c) Effective Area , A e = (λ 2 / 4π) G φ = (1.5 λ 2 sin 2 θ) / 4π d) Radiation Resistance R rad

16) Sketch the normalized E-field and H-field patterns for a. A half-wave dipole b. A quarter-wave monopole. SOLUTION: a) b) The same as for λ/2 dipole except that the fields are zero for θ > π/2 as shown: 17) In free space, an antenna has a far zone field given by Determine the radiated power? SOLUTION: 18) At the far field, the electric field produced by antenna is Sketch the vertical pattern of the antenna. Your plot should include as many points as possible. SOLUTION:        f(θ) = | cosθ cosφ |             For the vertical pattern, φ = 0 which means,                        f(θ) = | cosθ | which is sketched below

13) A 1-m-long car radio antenna operates in the AM frequency of 1.5 MHz. How much current is required to transmit 4 W of power? SOLUTION: This is a monopole antenna λ = c / f = (3 x 108) / (1.5 x 106) = 200 m In this case, it’s pretty clear that l << λ 14) An antenna located on the surface of a flat earth transmits an average power of 200 kW. Assuming that all the power is radiated uniformly over the surface of a hemisphere with the antenna at the center, calculate (a) The time-average Poynting vector at 50 km, and (b) The maximum electric field at that location. SOLUTION: a)      P rad = ∫ P rad . ∂s = P ave . 2πr 2 (hemisphere)      P ave = P rad / 2πr 2 = (200 x 10 3 ) / (2π 50 x 10 6 ) 2             =12.73 µW / m 2                                              P ave = 12.73 a r µW / m 2 b)       P ave = (E max ) 2 / 2η       (E max ) 2 = 240π x 12.73 x 10 -6        E max = 0.098 V/m 15) A 100-turn loop antenna of ra

9) A Hertzian dipole at the origin in free space has dl = 20 cm and I = 10 cos2π10 7 t A, find |E θs | at the distant point (100, 0, 0). SOLUTION: λ = c / f = ( 3 x 108 ) / (107) = 30 m At (100, 0, 0), r = 100 m & θ = π / 2 10) A 2-A source operating at 300 MHz feeds a Hertzian dipole of length 5 mm situated at the origin. Find E s and H s at (10, 30°, 90°). SOLUTION:                    λ = c / f = (3 x 108) / (3 x 108) = 1m                    β = 2π / λ = 2π                   At r = 10, θ = 30 o , φ = 90 o                    λ = 120π = 377 m                    E θs = η H φs = 94.25 mV/m 11) An antenna can be modelled as an electric dipole of length 5 m at 3 MHz. Find the radiation resistance of the antenna assuming a uniform current over its length. SOLUTION:           dl = 5 m           λ = c / f = (3 x 10 8 ) / (3 x 10 6 ) = 100 m           dl / λ = 5 / 100 = 1/20          Now since dl / λ = 1/20 < 1/10           R r

6) An antenna in air radiates a total power of 100 kW so that maximum radiated electric field strength of 12 mV/m is measured 20 km from the antenna.  Find its: (a) Directivity in dB (b) Maximum power gain if η r = 98%. SOLUTION: a) G = 10 log 10 G d = 10 log 10 (2.16) = 3.34 dB b) G = η r G d = 0.98 x 2.16 = 2.117 7) A C-band radar with an antenna 1.8 m in radius transmits 60 kW at a frequency of 6000 MHz. If the minimum detectable power is 0.26 mW, for a target cross section of 5 m 2 , calculate the maximum range in nautical miles and the signal power density at half this range. Assume unity efficiency and that the effective area of the antenna is 70% of the actual area. SOLUTION: 8) The magnetic vector potential at point P(r, θ, φ) due to a small antenna located at the origin is given by where r 2 = x 2 + y 2 + z 2 . Find E(r, θ, φ, t) and H(r, θ, φ, t) at the far field. SOLUTION: Using Vector transformation,          A rs = A

3) A certain antenna with an efficiency of 95% has maximum radiation intensity of 0.5 W/sr. Calculate its directivity when (a) The input power is 0.4 W (b) The radiated power is 0.3 W SOLUTION: a) P rad = η r P in = (0.95) (0.4) = 0.38 4) Evaluate the directivity of an antenna with normalized radiation intensity SOLUTION: 5) Determine the maximum effective area of a Hertzian dipole of length 10 cm operating at 10 MHz. If the antenna receives 3 µW of power, what is the power density of the incident wave? SOLUTION: For the hertzian dipole, G d = 1.5 sin 2 θ By definition, P r = A e P ave

1) An electric field strength of 10 µV/m is to be measured at an observation point θ = π/2, 500 km from a λ/4 monopole antenna operating in air at 50 MHz. (a) What is the length of the dipole? (b) Calculate the current that must be fed to the antenna. (c) Find the average power radiated by the antenna. (d) If a transmission line with Z o = 75 Ω is connected to the antenna, determine the standing wave ratio. SOLUTION: a) The wavelength λ = c / f =( 3 x 108) / (50 x 106) = 6 cm.                              Length Of The Dipole (l) = λ/4 = 6 / 4 = 1.5 cm. b) c) R rad = 73 Ω (For λ/2 (Half Wave) Dipole Antenna) R rad = 36.56 Ω (For λ/4 Monopole Antenna) d) Standing Wave ratio (S)      Z L = 73 + j42.5 (For Half wave dipole Antenna)      Z L = 36.5 + j21.25 (For λ/4 Monopole Antenna) 2) Calculate the directivity of a) The Hertzian monopole b) The quarter-wave monopole SOLUTION: a) For the hertzian monopole b)  

Polarization is defined as the orientation of electric field as a function of direction. The polarization of the radio wave can be defined by direction in which the electric vector E is aligned during the passage of at least one full cycle. Also polarization can also be defined the physical orientation of the radiated electromagnetic waves in space. The polarization are of three types. They are: Elliptical polarization. Circular polarization. Linear polarization. Linear Polarisation: A linearly polarized wave is one in which the electric field remains in only one direction.For a linearly polarized wave,the axial ratio(AR) is infinity. Elliptical Polarization: The electric field vector rotates and form a ellipse called polarization ellipse. The ratio of the major to the minor axes of the polarization ellipse is called the Axial Ratio (AR). AR is greater than 1 . Circular Polarization: The electric filed vector rotates and form a circle and this wave is called