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In a transmission line consisting of 2 parallel wires, one can show that the voltage and current exist as a wave propagating along the wire.

When calculating, for example, the reflection coefficient due to an unmatched load, one uses the waveforms $$V_0e^{i(\omega t - kz)}$$ $$I_0e^{i(\omega t - kz)}$$ to derive the result. However, why do we not use $$E_0e^{i(\omega t - kz)}$$ $$H_0e^{i(\omega t - kz)}$$ (representing the propagating E and H fields) for this derivation?

Essentially what I am unsure of is the difference between the propagating voltage and current waves with the propagating EM waves in a transmission line. Are they one of the same thing?

Any help will be much appreciated.

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The E field and the voltage are the same thing. The current is its own thing, it depends on the impedance of the line and the voltage or E. The magnetic field is its own thing but depends on the current.

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  • $\begingroup$ Thank you for the response, but then why is it that when we calculate the reflection coefficient that we use the voltage and current wave instead? $\endgroup$
    – Student 1
    Nov 18 '18 at 12:39
  • $\begingroup$ Usually when calculating reflection coefficient we are concerned with efficiency, or how much energy or power is transferred or reflected, so V and I are convenient because impedance is considered to get I and P=VI. $\endgroup$ Nov 18 '18 at 12:45
  • $\begingroup$ I see, and I guess even if we were to use H instead of I then we just multiply everything by a constant so they would cancel anyway? $\endgroup$
    – Student 1
    Nov 18 '18 at 12:49
  • $\begingroup$ Impedance can be complex, so getting I from V is the key. I don't remember all the details, but H is proportional to rate of change of current. $\endgroup$ Nov 18 '18 at 12:52

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