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fix linebreaks

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Brendan Haines 2024-11-10 22:19:34 -07:00
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book/01_tx_lines.md
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@ -76,9 +76,11 @@ Note that if $Z_L = Z_0$ then $\Gamma = 0$ and there is no reflection. This is g
Also note that this is only valid at $z=0$ since the phase of $\hat{V}_f$ and $\hat{V}_r$ vary with opposite signs as we move along the transmission line so their relationship changes with position. Also note that this is only valid at $z=0$ since the phase of $\hat{V}_f$ and $\hat{V}_r$ vary with opposite signs as we move along the transmission line so their relationship changes with position.
A perfectly matched load results in no reflected wave. A perfectly matched load results in no reflected wave.
![Reflection with Z_L = Z0](../assets/tx_lines/reflection_matched.gif) ![Reflection with Z_L = Z0](../assets/tx_lines/reflection_matched.gif)
Open and short circuits both have locations where no voltage is seen. Note that both of these cases have an equal amplitude reflected wave, they just have different phase. This is actually true of all lossless loads (ideal capacitor and inductor as well). Open and short circuits both have locations where no voltage is seen. Note that both of these cases have an equal amplitude reflected wave, they just have different phase. This is actually true of all lossless loads (ideal capacitor and inductor as well).
![Reflection with Z_L = inf](../assets/tx_lines/reflection_open.gif) ![Reflection with Z_L = inf](../assets/tx_lines/reflection_open.gif)
![Reflection with Z_L = 0](../assets/tx_lines/reflection_short.gif) ![Reflection with Z_L = 0](../assets/tx_lines/reflection_short.gif)
@ -86,6 +88,7 @@ Open and short circuits both have locations where no voltage is seen. Note that
It is also worth noting that the peak voltage on the line is $2V$ when we only excited it with $1V$. How is this possible? Remember that there is another end of the transmission line on the left which could cause a re-reflection of $V_r$ to combine with $V_f$ however we are assuming no such reflection exists. This means that the impedance seen at the left end of the transmission line must be matched to $Z_0$, meaning that the driver has a series impedance equal to $Z_0$. Therefore $V_f = \frac{1}{2} V_{source}$ and the total voltage $V$ never exceeds $V_{source}$. It is also worth noting that the peak voltage on the line is $2V$ when we only excited it with $1V$. How is this possible? Remember that there is another end of the transmission line on the left which could cause a re-reflection of $V_r$ to combine with $V_f$ however we are assuming no such reflection exists. This means that the impedance seen at the left end of the transmission line must be matched to $Z_0$, meaning that the driver has a series impedance equal to $Z_0$. Therefore $V_f = \frac{1}{2} V_{source}$ and the total voltage $V$ never exceeds $V_{source}$.
An imperfectly matched but lossy ($R \ne 0$) load results in a reduced amplitude reflected wave An imperfectly matched but lossy ($R \ne 0$) load results in a reduced amplitude reflected wave
![Reflection with arbitrary real Z_L](../assets/tx_lines/reflection_real.gif) ![Reflection with arbitrary real Z_L](../assets/tx_lines/reflection_real.gif)
![Reflection with arbitrary complex Z_L](../assets/tx_lines/reflection_complex.gif) ![Reflection with arbitrary complex Z_L](../assets/tx_lines/reflection_complex.gif)