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ELEC 20006 Antenna and Wave Propagation Assignment - Middle East College, Oman

Assignment Outcomes - The assignment will help you in assessing your understanding on the transmission line circuits and their solutions, in applying smith chart in analyzing and solving transmission line problems, impedance matching and also differentiate between various important antenna parameters. Specifically:

1. Analyze transmission line circuits based on the voltage and current distributions on them;

2. Apply Smith chart for analyzing and solving transmission-line problems and for impedance matching;

3. Discuss and analyze the important terminologies related to antennas.

Question 1 - a. How much power is delivered to the antenna?

Solution 1) a) Z_L = 8+j40 ?

Z_O = 50 ?

P_inc = 30 W

Γ = Z_L - Z_O / Z_L+ Z_O

= 8+j40-50/8+j40+50

= -42+j40/58+j40

= (58/_136)/(70.46/_34.6)

= 0.823/_101.4

| Γ | = 0.823

P_load = P_inc - P_ref = P_inc (1-|Γ|²)

= 30 (1- (0.823)²)

= 9.68 W

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b. Determine: i. The load reflection and ii. Load impedance.

Solution 1) b) Transmission Line (Lossless)

Z_L = ?

Z_O = 50 ?

Vmax = √2+1 V = Vinc + Vref

Vmin = √2 - 1 V = Vinc - Vref

Hence, Vinc = √2 V

Vref = 1 V

L= 15λ/16

(i) Load Reflection = ?

= ρ = Vref/Vinc e ^-j15∏/4

= 1/√2 e ^-j15∏/4

= 0.707 e ^-j15∏/4

(ii) ZL= ?

ρ = Z_L - Z_O / Z_L+ Z_O

1/√2 = Z_L - 50 / Z_L+50

Z_L + 50 = √2 (Z_L - 50) = √2 Z_L - 50 X √2

120.7 = 0.414 Z_L

Z_L = 291.55 ?

Question 2 - a. Discuss the practical importance of the antenna beam width.

Solution 2) a) Practical importance of antenna beam width:

A given antenna has a radiation pattern in a particular co-ordinate system. We can describe it in the form of angles. The radiation pattern defines how much power is radiated by an antenna in a given direction. In a given space, by observing the radiation pattern we can see how the distribution of power has been done. In some direction, it may be more. In other, it may be less. There may be a case where uniform power is being radiated in all directions. But generally, an antenna radiates more amount of power in a given angle range. Let the maximum power radiated by an antenna be Pmax in a direction. Then Half power beam Width defines the range where at least half of the maximum power Pmax is being radiated. So, it is very useful in practical situations to determine in which region antenna is mainly radiating power. We can position the antenna properly using this information.

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b. Find: i. Antenna's gain in dB, and ii. Radiation resistance.

Solution 2) b) Omnidirectional Antenna:

HPBW (vertical ) = 30 degree = 51λ/b

HPBW (horizontal) = 20 degree = 70λ/a

Rloss = 25?

Efficiency = η = 0.8

(i) Antenna Gain (in dB) = ? = 10A/λ2 = 10ab/λ2 = 10 X 51/30 X 70/20 = 10 log 59.5 = 17.74 dB

(ii) Rrad = ?

We know, η = Rrad/(Rrad+Rloss)

0.8 = Rrad/(Rrad+25)

0.8 x(Rrad+25)= Rrad

0.2 Rrad = 25 x 0.8

Rrad = 25 x 4 = 100 ?

Question 3 - a. find: i. Reflection coefficient; ii. Voltage standing voltage ratio;

Solution 3) a) Lossless Transmission Line

Z_L = 100+ j150 ?

Z_O = 75 ?

(i) Reflection coefficient = ρ = Z_L - Z_O / Z_L+ Z_O

= 100+ j150 -75 / 100+ j150 ? + 75

= 25+j150/175+j150

= (152,80.53)/(230.49,40.6)

= 0.66 /_39.93

(ii) Voltage SWR = 1+| ρ |/1-| ρ |

= 1+0.66/1-0.66

= 1.66/0.34

= 4.88

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b. Determine: i. Gain of the antenna (in dB). ii. Maximum radiation intensity of the antenna (in dB). iii. If the power radiated from the antenna is doubled, discuss how it affects the directivity of the antenna using necessary equations.

Solution 3) b) f= 500 MHz

Wavelength= λ = c/f= 300/500 m = 0.6m

Prad = 3.25 kW

Aeff = 3.5 m2

Total efficiency = e_o = 42%

Radiation efficiency = e_cd = 65%

Tilt angle = ψ = 35 degree

(i) Gain (dB) = 10 log10 (4∏ηA/λ²) = 10 log10 (4∏X0.42X3.5/0.6X0.6) = 17.1 dB

(ii) Maximum radiation intensity (dB) = Bo

Gain = 4∏ Radiation Intensity/Total Input power

51.29 = 4∏ Bo cos ψ/3250

Bo = 51.29 X 3250/4∏ cos 35 = 16193.54 = 42 dB

(iii) Prad' = 2 X Prad

Dmax = Umax/Uo = 4∏Umax/Prad

Dmax' = 4∏Umax/Prad' = 4∏Umax/ 2 X Prad = Dmax/2

Hence, directivity becomes half.

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Question 4 - a. Find the appropriate Z₀.

Solution 4) a) 15m , 300?

3m, 150?

ZL = 150?

VSWR on 300? line = ?

ρ = Z_L - Z_O / Z_L+ Z_O = 150-300/150+300 = -150/450 = -1/3 = -0.33

SWR = 1+| ρ |/1-| ρ | = 1+0.33/1-0.33 = 1.98

Zo (to match 2 sections - quarter λ line) = ?

Zo = √Z1 x Z2 = √300X150 = 150X1.414 = 212 ?

F= 50 MHz

b. i. Distance of 1st Vmax and 1st Vmin from load; ii. Number of voltage maximas and voltage minimas on this line; iii. Distance and length of an open series stub to match the antenna with the transmission line.

Solution 4) b) ZL = 150+j200 ? (Input Impedance)

Zchar = 400?

L = 37 cm

λ = 15 cm

Normalized Impedance = (150+j200)/400 = 0.38+j0.5

Smith Chart:

(i) Distance between first Vmax and first Vmin = ? = λ/4 = 15/4 = 3.75 cm

(ii) Number of Vmax and Vmin = ? = 37/3.75= 9

(iii) Open Series stub to match antenna with transmission line

Incase of single stub matching with series open stub, first we match the real part, ie, 0.38. Then we plot ZL. Then we rotate TWG (d) as per the matching circle.

Distance = ? = d =(0.325-0.21)λ = 0.115X15 cm = 1.725 cm.

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