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As we move from an imaginary load towards generator through a transmission line, we can see that real and imaginary part of that impedance changes. In fact, an ideal transmission line is comprised of inductors and capacitors. It can alter only the imaginary part of a load; how does the real part of the load change?

Once the transmission line becomes lossy, the constant gamma circle on Smith-chart gradually shrinks down to the center and meets the characteristic impedance. Can anyone explain how this happens?

Thanks in Advance!

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If you assume the load is purely reactive, its real part=0 or infinite and the reflection factor is 100%. The real part stays=0 when you rotate the line length in Smith's chart. You will never meet other real part circles with that reflection factor

If you talk about general loads, not only pure reactances, you have a big false assumption "ideal transmission line is made of instructors and capacitors. It can alter only the imaginary part of a load"

Take a resistor, say 100 Ohms. Put it in parallel with an inductor whose impedance in the operating frequency is j100 Ohms. Calculate the total impedance phasor. It's about (50 + j50) Ohms; the real part has decreased substantially.

=> A resistor seen through a reactive network generally doesn't keep its real part.

I think you didn't see that the LC-model of ideal transmission line puts parts as well in parallel as in series with the load altough the visible metallic wire is only in series with the load.

The other question: Obviously nothing reflects when everything is dissipated in the line.

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  • \$\begingroup\$ Thanks for your Answer. Though i visualized transmission line as inductors and capacitors, i did not see that LC combination puts it in parallel too with the load. Yes , that can change the real part of the load. \$\endgroup\$ – Bth Sep 4 '18 at 16:58

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