Pagel4 20 marks uestion 4: A 500 MHz generator with ,(1)-Scos ar (V) and interna
ID: 2266888 • Letter: P
Question
Pagel4 20 marks uestion 4: A 500 MHz generator with ,(1)-Scos ar (V) and internal resistance 4-50 is connected to a 0 characteristic lossless air transmission line (TL) as shown on Figure Q4. The TL parameters are 5 impedance, resistance per unit length R'=0, capacitance per unit length C-4 pF/m and ge was measured to conductance per unit length G'=0. The peak value of the forward incident volta be pat-6 . The TL length is 2 and is terminated by load ZL = (75-j50) . Determine the following: I4 Marks] [4 Marks] [2 Marks] [4 Marks] 3 Marks (i) The transmission line inductance per meter, L' . (ii) The standing-wave ratio on the line. (ii) The phase constant p (iv) The input impedance of the line. (v) The position of the voltage maximum nearest to the load. (vi) The voltage at an arbitrary location on the line. [3 Marks] 2 =50 Vg(t) Zi Figure Q4Explanation / Answer
Answer:-a) Speed of wave = 1/(LC)0.5 Since air is the medium hence we can write-
=> 3 x 108 = 1/(4 x 10-12 x L)0.5
=> L = 2.77 micro-H/m
b) ZL = 75 - j50 ohm, Z0 = 50 ohm hence reflection coefficient-
=> |gamma| = |(ZL - Z0)/(ZL + Z0)| = |(25 - j50)/(100 + j50)| = 0.632
Hence SWR = (1 + |gamma|)/ (1 - |gamma|) = 4.43
c) ohmega = 2 x pi x f = 2 x 3.14 x 500 x 106 rad/s
= 3.14 x 109 rad/s
phase constant, beta = ohmega x (LC)0.5 = 18.12 m-1
d) The length of transmission line is 2*lambda, and for every transmission line the value of impedance is same after 0.5*lambda.
Thus in this case we have input impedance equal to load impedance of 75 - j50 ohm. Since length of line is 4 times of 0.5*lambda.
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