Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am reading a book about the telegraph equations on the transmission line, which is illustrated as follow

enter image description here

where the input impedance is $Z_0$ and the impedance of the load is $Z_L$, we setup the positive x direction pointing to the right and the load is at $x=0$, the length of the line is L so the input end is at $x=-L$. From the book, when the line is lossless, the general voltage and current can be written as

$$ V(x) = V_0^+e^{-\gamma x} + V_0^-e^{\gamma x}, \qquad I(x) = I_0^+e^{-\gamma x} + I_0^-e^{\gamma x} $$ where $\gamma$ is the propagation constant. $V_0^+$ and $I_0^+$ stand for the amplitude of the positive transmitted voltage and current while $V_0^-$ and $I_0^-$ stand for the amplitude of the reflected voltage and current. The book also gives the input impedance to be

$$ Z_0 = \frac{V_0^+}{I_0^+} = -\frac{V_0^-}{I_0^-} $$

In another section where the book trying to derive the impedance of the load, it then writes the current as

$$ I(x) = I_0^+e^{-\gamma x} - I_0^-e^{\gamma x} $$

I don't know why in previous section there is a PLUS in front of $I_0^-$ but change to MINUS here while they should refer to the same thing. I check another book but they do the same way. The only reason that I can think of is from the first voltage expression, we can divide both side with $Z_0$ and make use of

$$ Z_0 = \frac{V_0^+}{I_0^+} = -\frac{V_0^-}{I_0^-} $$


$$ I(x) = \frac{V(x)}{Z_0} = \frac{V_0^+}{Z_0}e^{-\gamma x} - \frac{V_0^-}{Z_0}e^{\gamma x} $$

so the current passing the load is just $$ \frac{V_L}{Z_0 } = \frac{V(x=0)}{Z_0 } = I_0^+ - I_0^- $$

Is that the correct way to derive the current (with minus sign) through the load? What confusing me is we divide the general voltage by the input impedance and figure out the load current by setting $x=0$, but I think if we know the load voltage, the load current should be related to the load impedance but not the input impedance!

share|improve this question
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.