Reflection and termination

The transmission line is typically used to connect a voltage source $V_0={\cal F}[v_0(t)]$ with output (internal) impedance $Z_1$ and a load of input impedance $Z_2$, as shown in the figure.

• At the front of the line ($x=0$), we have
  
  $$V_1=V_0-I_1Z_1$$
  
  
  But since
  
  $$V_1=V(x)\vert _{x=0}=\left[V_be^{j\omega Tx/l}+V_fe^{-j\omega Tx/l}\right]_{x=0}=V_b+V_f$$
  
  
  
  
  $$I_1=I(x)\vert _{x=0}=\left[I_be^{j\omega Tx/l}+I_fe^{-\omega Tx/l}\right]_{x=0}=I_b+I_f$$
  
  
  we have
  
  $$V_1=V_b+V_f=V_0-I_1Z_1=V_0-[I_b+I_f]Z_1 =V_0-[-V_b+V_f]\frac{Z_1}{Z_0}$$
  
  
  Solving for $V_f$, we get
  
  $$V_f=\frac{Z_0}{Z_0+Z_1}V_0+\frac{Z_1-Z_0}{Z_1+Z_0}V_b =AV_0+\eta_1V_b$$
  
  
  where
  
  $$A\stackrel{\triangle}{=}\frac{Z_0}{Z_0+Z_1},\;\;\;\;\;\;\;\;\; \eta_1\stackrel{\triangle}{=}\frac{Z_1-Z_0}{Z_1+Z_0}$$
  
  
  are respectively the voltage attenuation ratio that describes the transmission line with $Z_0$ and the output impedance $Z_1$ of the source as a voltage divider with input $V_0$, and the reflection coefficient at the front end of the line that represents how much of the backward voltage $V_b$ wave is reflected at the front end.

  In particular, we consider the following two special cases:

  • When the internal impedance of the source matches the characteristic impedance, i.e., $Z_1=Z_0$, then $\eta_1=0$, indicating the backward voltage $V_b$ is no longer reflected at the front end, and $V_f=A V_0$.
  • When the internal impedance of the source is zero, i.e., $Z_1=0$, then $\eta_1=-1$; indicating the backward voltage $V_b$ is totally reflected, and $V_f=AV_0-V_b$.

• At the back end of the transmission line ($x=l$), we have
  
  $$V_2=I_2Z_2$$
  
  
  But since
  
  $$V_2=V(x)\vert _{x=l}=[V_be^{sx/\mu}+V_fe^{-sx/\mu}]_{x=l} =V_be^{j\omega T}+V_fe^{-j\omega T}$$
  
  
  
  
  $$I_2=I(x)\vert _{x=l}=[I_be^{sx/\mu}+I_fe^{-sx/\mu}]_{x=l} =I_be^{j\omega T}+I_fe^{-j\omega T}$$
  
  
  we have
  
  $$V_2=V_be^{j\omega T}+V_fe^{-j\omega T}=I_2Z_2=[I_be^{j\omega ... ... T}]Z_2 =[-V_be^{j\omega T}+V_fe^{-j\omega T}]\frac{Z_2}{Z_0}$$
  
  
  i.e.,
  
  $$[Z_0+Z_2]V_be^{j\omega T}=[Z_2-Z_0]V_fe^{-j\omega T}$$
  
  
  From this we can define the reflection coefficient at the back end of the line as the ratio between the backward voltage and the forward voltage at the back end of the line:
  
  $$\eta_2\stackrel{\triangle}{=}\frac{V_B(l)}{V_F(l)} =\frac{V_be^{j\omega T}}{V_fe^{-j\omega T}}=\frac{Z_2-Z_0}{Z_2+Z_0}$$
  
  
  which represents how much of the forward voltage wave is reflected at the back end. Now we have
  
  $$V_B(l)=V_b e^{j\omega T}=\eta_2 V_F(l)=\eta_2 V_f e^{-j\omega T}$$
  
  
  Consider these three special cases:
  • When the load $Z_2=\infty$ is an open-circuit, $\eta_2=1$, $V_B(l)=V_F(l)$; i.e., the forward voltage is totally reflected.
  • When the load $Z_2=0$ is a short-circuit, $\eta_2=-1$, $V_B(l)=-V_F(l)$; i.e., the forward voltage is totally reflected.
  • When the load impedance matches the characteristic impedance, $Z_2=Z_0$, $\eta_2=0$, and $V_B(l)=\eta_2 V_F(l)=0$; i.e., the forward voltage is not reflected.

In the following we will analyze the transmission line in the frequency domain, in which all impedances such as $Z_0$, $Z_1$, $Z_2$, the voltage and current $V$ and $I$ (in phasor forms), and all coefficients $A$, $\eta_1$, $\eta_2$, are functions of $\omega$. We will therefore drop the argument $\omega$ and use capitalized letters without argument to represent these variables.