enter image description here

My Try:

enter image description here

\$Z_{in}=Z_0*\frac{Z_L+jZ_0tan\beta l}{Z_0+jZ_L tan\beta l}\$


\$\beta l=\frac{2\pi\times37.5\times10^6\times10}{3\times10^8}=\frac{5\pi}{2}\$

so its becomes

\$Z_{in}=\frac{Z_0^2}{Z_L} \$


So for part \$(a)\$

\$I_{i}=\frac{200}{600}=\frac{1}{3}A\$ also \$V_{i}=\frac{400\times200}{600}=\frac{400}{3}\$

Now for part \$(b)\$ from the generator we know the equation for lossless line is \$ I(z)=\frac{-j}{Z_0}sin\beta z V_i+Cos\beta z I_{i}\$

so current at the load, that means at \$\lambda/4\$ distance from the generator will be \$I=\frac{-j}{200}sin(\frac{5\pi}{2})\times\frac{400}{3}+0\times\frac{1}{3}=\frac{2}{3}\angle-90^{\circ}\$

but for part\$(b)\$ the actual answer is \$\frac{1}{3}\angle-90^{\circ}\$

What is the mistake i am doing can anyone help please?


2 Answers 2


In part (a) you calculated the current as 1/3 so why should it suddenly be any different to the load current in part (b). As for phase angle, you have shown the line to be a equivalent to a quarter wavelength long at 37.5 MHz so, given the load is resistive, the phase angle of the load current has to be 90 degrees lagging as you have also shown.

  • \$\begingroup\$ Thanks for the answer sir, but why its not coming by TL equations? \$\endgroup\$
    – Rohit
    Commented Sep 13, 2017 at 6:56
  • \$\begingroup\$ @Rohit I have no idea. One method produces X and another method produces Y. Both should produce the same OR there is a math error or formula error. I have only tried to show that your method for calculating answer (b) is incorrect based on the fact that Iin = Iout (RMS values) \$\endgroup\$
    – Andy aka
    Commented Sep 13, 2017 at 10:36

For these problems, remember that the \$z=0\$ location (highlighted in the image below) is defined at the terminals of the load. So you are looking for \$I(0)\$ in part b.

For the input impedance, you do use the length because it is some distance \$l\$ away from the load and towards the generator. This is \$z=-l\$.

enter image description here

Now, I don't know whether you're using the right equation (can't remember off the top of my head all the TL equations) but hopefully this points you in the right direction.


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