(a) A lossless transmission line with Z, line is terminated with a load impedance, Zį = 100 + j100 N. Given the phase velocity 0.6c on the line, where c is the speed of light in the vacuum field. Using the Smith Chart method, evaluate the following: 50 N is 30 m long and operates at 2 MHz. The %3D (i) The reflection coefficient (ii) The standing wave ratio (iii) The load admittance (iv) The input impedance

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Question 3
A lossless transmission line with Z.
(a)
line is terminated with a load impedance, Z, = 100 + j100 N. Given the phase velocity
0.6c on the line, where c is the speed of light in the vacuum field. Using the Smith Chart
method, evaluate the following:
50 N is 30 m long and operates at 2 MHz. The
(i) The reflection coefficient
(ii) The standing wave ratio
(iii) The load admittance
(iv) The input impedance
Transcribed Image Text:Question 3 A lossless transmission line with Z. (a) line is terminated with a load impedance, Z, = 100 + j100 N. Given the phase velocity 0.6c on the line, where c is the speed of light in the vacuum field. Using the Smith Chart method, evaluate the following: 50 N is 30 m long and operates at 2 MHz. The (i) The reflection coefficient (ii) The standing wave ratio (iii) The load admittance (iv) The input impedance
Stepl
a)
niven data
YL= (100 +j100
70 = 50 S
30 m
V= 0.6 C
Whete, c= s peed of light
C= 3X 10 m/s
f= 2MH2
Step2
b)
Nbw, The refle ction coetficient which is denoted by
is
calculated os
ZL- 20
ZL +20
(100 +j100) - 50
(100+ji00) + s0
1-j2
3+j2
Now,
= (21
J32422
J13
17/=
D. 6201 - Magnitude.
Step3
c)
ii) The stonding wave 8atio con be given as
I+ 17)
SWR =
(2(-1
1+ 0.620)
1-0.620)
SWR = 4.2645
i) The 1oad odmittance con be colculated as
YL =
- (5 - si) x10"
100 tjibo
YL=7.071×1o3) L-45*
Step4
d)
According to the bartleby guidlines, i solved the first 3
subparts. Please repost the remaining one. Thank you.
Transcribed Image Text:Stepl a) niven data YL= (100 +j100 70 = 50 S 30 m V= 0.6 C Whete, c= s peed of light C= 3X 10 m/s f= 2MH2 Step2 b) Nbw, The refle ction coetficient which is denoted by is calculated os ZL- 20 ZL +20 (100 +j100) - 50 (100+ji00) + s0 1-j2 3+j2 Now, = (21 J32422 J13 17/= D. 6201 - Magnitude. Step3 c) ii) The stonding wave 8atio con be given as I+ 17) SWR = (2(-1 1+ 0.620) 1-0.620) SWR = 4.2645 i) The 1oad odmittance con be colculated as YL = - (5 - si) x10" 100 tjibo YL=7.071×1o3) L-45* Step4 d) According to the bartleby guidlines, i solved the first 3 subparts. Please repost the remaining one. Thank you.
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