Forward and backward traveling signals

Summarizing the above, we have two simultaneous equations of two unknowns $V_b=V_B(0)$ and $V_f=V_F(0)$ for the forward and backward voltages at the front end ($x=0$):

$$\left\{ \begin{array}{ll} V_f=AV_0+\eta_1V_b & \mbox{(at fron... ...{-j\omega T} & \mbox{(at back end $x=l$)} \end{array} \right.$$

Solving these simultaneous equations we get

$$V_f=\frac{AV_0}{1-\eta_1\eta_2e^{-2j\omega T}},\;\;\;\;\;\;\;... ...T}}{1-\eta_1\eta_2e^{-2j\omega T}} =\eta_2e^{-2j\omega T} V_f$$

The numerators represent the sources of the signals. The forward signal $V_f=V_F(0)$ is caused by the input $AV_0$ first entering the line at $t=0$. The backward signal $V_b=V_B(0)$ is due to the forward signal reflected at the back end of the line ($\eta_2$), delayed by the round-trip traveling time $2T$ ( $e^{-j2\omega T}$) for the signal to travel forward and then backward along the length of the line. The common denominator represents the subsequent reflections at both ends. As in general $\vert\eta_1\vert<1$ and $\vert\eta_2\vert<1$, we have $\vert\eta_1\eta_2e^{-2j\omega T}\vert<1$ and the denominator can be expanded to become:

$$\frac{1}{1-\eta_1\eta_2e^{-2j\omega T}}=1+\eta_1\eta_2e^{-2j\... ...ta_2e^{-2j\omega T})^2+(\eta_1\eta_2e^{-2j\omega T})^3+\cdots$$

Here the general term $(\eta_1\eta_2)^ke^{-2kj\omega T}$ represents the signal arriving at the front of the line after traveling forward and backward along the line and being reflected at both ends $k$ times ( $k=0,1,2,\cdots$). When $\eta_1\ne 0$ and $\eta_2\ne 0$, there will be infinite reflections between the two ends.

Given $V_f$ and $V_b$, the voltages $V(0)=V_1$ and $V(l)=V_2$ at the two ends. At the front, we have

$$V_1=V_b+V_f=\frac{A(1+\eta_2e^{-2j\omega T})}{1-\eta_1\eta_2e^{-2j\omega T}}V_0 = T_1V_0$$

where $T_1$ is the voltage transfer function at the front defined as

$$T_1\stackrel{\triangle}{=}\frac{A(1+\eta_2e^{-2j\omega T})}{1-\eta_1\eta_2e^{-2j\omega T}}$$

The two terms of the numerator correspond to the in-coming voltage and the reflection from the back end, respectively. The input current is

$$I_1=I_b+I_f=\frac{-V_b+V_f}{Z_0} =\frac{A(1-\eta_2e^{-2j\omega T})}{1-\eta_1\eta_2e^{-2j\omega T}} \frac{V_0}{Z_0}$$

The input impedance of the transmission line can be obtained as:

$$Z_{in}\stackrel{\triangle}{=}\frac{V_1}{I_1}=\frac{1+\eta_2e^{-2j\omega T}}{1-\eta_2e^{-2j\omega T}} Z_0$$

At the back end, we have

$$V_2=V_be^{j\omega T}+V_fe^{-j\omega T} =\frac{A(1+\eta_2)e^{-j\omega T}}{1-\eta_1\eta_2e^{-2j\omega T}}V_0 = T_2V_0$$

where $T_2$ is the voltage transfer function at the back end defined as

$$T_2\stackrel{\triangle}{=}\frac{A(1+\eta_2)e^{-j\omega T}}{1-\eta_1\eta_2e^{-2j\omega T}}$$

The two terms of the numerator correspond to the in-coming voltage arriving at the back end after a time dealy of $T$ and the immediate reflection at the back end. The output current is

$$I_2=I_be^{j\omega T}+I_fe^{-j\omega T}=\frac{-V_be^{j\omega T... ...e^{-j\omega T}}{1-\eta_1\eta_2e^{-2j\omega T}} \frac{V_0}{Z_0}$$

We can easily verify that indeed $V_2/I_2=Z_2$:

$$\frac{V_2}{I_2}=\frac{1+\eta_2}{1-\eta_2}Z_0=Z_2$$

The output impedance of the transmission line can be defined as the ratio of the open-circuit voltage $V_2$ (when $Z_2=\infty$, $\eta_2=1$) and the short-circuit current $I_2$ (when $Z_2=0$, $\eta_2=-1$). But as

$$V_{2oc}=V_2\big\vert _{Z_2=\infty}=\frac{2Ae^{-j\omega T}}{1-... ...frac{2Ae^{-j\omega T}}{1+\eta_1e^{-2j\omega T}}\frac{V_0}{Z_0}$$

we get

$$Z_{out}\stackrel{\triangle}{=}\frac{V_{2oc}}{I_{2sc}} =\frac{1-\eta_2e^{-2j\omega T}}{1+\eta_2e^{-2j\omega T}} Z_0$$

We consider some special cases.

• When the length of the transmission line is integer multiple of half of the wavelength $\lambda=\mu/f$; i.e.,
  
  $$l=n\frac{\lambda}{2}=n\frac{\mu}{2f}=n\frac{\mu\pi}{\omega}$$
  
  
  we have
  
  $$e^{-j2\omega T}=e^{-j2\omega l/\mu}=e^{-j2n\pi}=1$$
  
  
  and
  
  $$Z_{in}=\frac{1+\eta_2}{1-\eta_2}Z_0=Z_2,\;\;\;\;\; Z_{out}=\frac{1-\eta_2}{1+\eta_2}Z_0=\frac{Z_0^2}{Z_2}$$
  
  

• If the transmission line matches the load, $Z_2=Z_0$ and $\eta_2=0$, we have $Z_{in}=Z_{out}=Z_0$. Moreover, if the transmission line also matches the internal impedance of the source, $Z_1=Z_0$ and $\eta_1=0$, we have $V_1=AV_0$, $V_{2oc}=V_2\big\vert _{Z_2=\infty}=2AV_0 e^{-j\omega T}$, and $V_2=AV_0e^{-j\omega T}$.

In summary, we have:

• Characteristic impedance and traveling velocity:
  
  $$Z_0=\sqrt{L/C},\;\;\;\;\;\;\;\;\;\;\mu=\frac{1}{\sqrt{LC}}$$
  
  

• Attenuation ratio and reflection coefficients:
  
  $$A=\frac{Z_0}{Z_0+Z_1},\;\;\;\;\;\;\;\; \eta_1=\frac{Z_1-Z_0}{Z_1+Z_0}, \;\;\;\;\;\;\;\;\eta_2=\frac{Z_2-Z_0}{Z_2+Z_0}$$
  
  

• Voltage transfer functions:
  
  $$T_1=\frac{A(1+\eta_2e^{-2j\omega T})}{1-\eta_1\eta_2e^{-2j\om... ...frac{A(1+\eta_2)e^{-j\omega T}}{1-\eta_1\eta_2e^{-2j\omega T}}$$
  
  

• Input and output impedances:
  
  $$Z_{in}=\frac{1+\eta_2e^{-2j\omega T}}{1-\eta_2e^{-2j\omega T}... ...ut}=\frac{1-\eta_2e^{-2j\omega T}}{1+\eta_2e^{-2j\omega T}}Z_0$$
  
  

• $L$ is of the order $10^{-6}\;\;[henry]/[meter]$, $C$ is of the order $10^{-11}\;\;[farad]/[meter]$, the speed is $\mu=1/\sqrt{LC}$ is of the order $3\times 10^8$ $[meter]/[second]$.