Examples

• Given $Z_1=Z_2=Z_0$ and $v_0(t)$, find $A$, $\eta_1$, $\eta_2$, $T_1$, $T_2$, $Z_{in}$, and $Z_{out}$,as well as $v_1(t)$ and $v_2(t)$.
  
  $$A=\frac{Z_0}{Z_0+Z_1}=\frac{1}{2},\;\;\;\;\;\; \eta_1=\frac{Z_1-Z_0}{Z_1+Z_0}=\eta_2=\frac{Z_2-Z_0}{Z_2+Z_0}=0$$
  
  

  
  

  $$T_1=\frac{A[1+\eta_2e^{-2j\omega T}]}{1-\eta_1\eta_2e^{-2j\om... ..._2e^{-2j\omega T}} =Ae^{-j\omega T}=\frac{1}{2}e^{-j\omega T}$$
  
  

  
  

  $$Z_{in}=\frac{1+\eta_2e^{-2j\omega T}}{1-\eta_2e^{-2j\omega T}... ...\frac{1-\eta_2e^{-2j\omega T}}{1+\eta_2e^{-2j\omega T}}Z_0=Z_0$$
  
  
  For an input $V_0={\cal F}[v_0(t)]$, we have
  
  $$V_1=\frac{1}{2}V_0,\;\;\;\;\;\;\;\;\;\;v_1(t)=\frac{1}{2}v_0(t)$$
  
  
  and
  
  $$V_2=T_2V_0=\frac{1}{2}e^{-j\omega T}V_0,\;\;\;\;\;\;\;\;\;\;\... ...left[\frac{1}{2}e^{-j\omega T}V_0\right] =\frac{1}{2}v_0(t-T)$$
  
  
  The load and the transmission line behave like a voltage divider with a pure delay of $T$. In particular, if the internal impedance of the source is zero $Z_1=0$, then $A=1$, $\eta_1=-1$, $\eta_2=0$, $T_1=1$, $T_2=e^{-j\omega T}$, $Z_{in}=Z_0$, and consequently,
  
  $$V_1=T_1V_0=V_0,\;\;\;\;\;\;\;v_1(t)=v_0(t),\;\;\;\;\;\;\; V_2=T_2V_0=V_0e^{-j\omega T},\;\;\;\;\;\; v_2(t)=v_1(t-T)=v_0(t-T)$$
  
  
  i.e., the input voltage $V_0$ is transmitted to the output (the load) with no distortion but only a delay of $T$.

• When $Z_1=Z_0$ but the load is an open circuit of $Z_2=\infty$, we have
  
  $$A=\frac{1}{2},\;\;\;\;\;\;\eta_1=0,\;\;\;\;\;\;\eta_2=1, \;\... ...=\frac{1}{2}(1+e^{-2j\omega T}),\;\;\;\;\;\;T_2=e^{-j\omega T}$$
  
  
  
  
  $$Z_{in}=\frac{1+e^{-2j\omega T}}{1-e^{-2j\omega T}}Z_0$$
  
  

  In particular, if $\vert\omega T\vert << 1$, we have (*)
  

  $$Z_{in}\approx \frac{Z_0}{j\omega T}=\frac{\sqrt{L/C}}{j\omega\;l\sqrt{LC}} =\frac{1}{j\omega\;lC}=\frac{1}{j\omega C_{line}}$$
  
  
  i.e., if $T$ is short compared to times of interest (pulse width or period of signal), the open circuit load behaves like a capacitor with a capacitance equal to that of the entire line $C_{line}=lC$.

  Give an input $V_0={\cal F}[v_0(t)]$, we have
  

  $$V_2=T_2V_0=V_0 e^{-j\omega T},\;\;\;\;\;\;\; v_2(t)={\cal F}^{-1}[V_2]={\cal F}[V_0 e^{-j\omega T}]=v_0(t-T)$$
  
  
  i.e., the output is just a delayed version of the input. But at the front, we have
  
  $$V_1=T_1V_0=\frac{1}{2}(1+e^{-2j\omega T}),\;\;\;\;\;\; v_1(t... ...}{2}(1+e^{-2j\omega T})\right] =\frac{1}{2}[v_0(t)+v_0(t-2T)]$$
  
  
  The second term $v(t-2T)$ is the reflection at the back end of the line with coefficient $\eta_2=1$, which arrives at the front after traveling along the line forward and backward in $2T$ time. But as $\eta_1=0$, it is no longer reflected at the front.

• When $Z_1=Z_0$ but the load is a short circuit $Z_2=0$, we have
  
  $$A=\frac{1}{2},\;\;\;\;\;\;\eta_1=0,\;\;\;\;\;\eta_2=-1, \;\;\;\;\;\; T_1=\frac{1}{2}(1-e^{-2j\omega T}),\;\;\;\;\;\;T_2=0$$
  
  
  
  
  $$Z_{in}=\frac{1-e^{-2j\omega T}}{1+e^{-2j\omega T}}Z_0$$
  
  
  In particular, when $\omega T << 1$, we have (*)
  
  $$Z_{in}\approx Z_0 \;j\omega T=\sqrt{L/C}j\omega\;l\sqrt{LC} =j\omega\;lL=j\omega L_{line}$$
  
  
  i.e., if $T$ is short compared to times of interest, the short circuit load behaves like an inductor of inductance equal to that of the entire line $L_{line}=lL$.

  Given an input $V_0={\cal F}[v_0(t)]$, we have
  

  $$V_2=T_2V_0=0$$
  
  
  i.e., the output is zero due to the short circuit load. But at the front
  
  $$V_1=T_1V_0=\frac{1}{2}(1-e^{-2j\omega T}),\;\;\;\;\;\; v_1(t)={\cal F}^{-1}[V_1]=\frac{1}{2}[v_0(t)-v_0(t-2T)]$$
  
  
  The second term $-v_0(t-2T)$ is the reflection at the back end of the line with coefficient $\eta_2=-1$, which arrives at the front after traveling along the line forward and backward in $2T$ time. In particular, when $v_0(t)=u(t)$, the reflection $-u(t-2T)$ cancels the input $u(t)$ after $2T$ time, the step input becomes a pulse of width $2T$ at the front of the line.

• When $Z_1=Z_0$ but the load is purely inductive $Z_2=j\omega L_2$, we have
  
  $$A=\frac{1}{2},\;\;\;\;\;\;\eta_1=0,\;\;\;\;\;\; \eta_2=\frac... ..._2-Z_0}{j\omega L_2+Z_0}=\frac{j\omega-1/\tau}{j\omega+1/\tau}$$
  
  
  where $\tau\stackrel{\triangle}{=}\omega L_2/Z_0$ is the time constant of the circuit. Now we have
  
  $$T_1=A(1+\eta_2e^{-2j\omega T}) =\frac{1}{2}\left[1+\frac{j\omega-1/\tau}{\omega+1/\tau} e^{-2j\omega T}\right]$$
  
  
  
  
  $$T_2=A(1+\eta_2)e^{-j\omega T}=\frac{j\omega}{j\omega+1/\tau}e^{-j\omega T}$$
  
  

  Consider the response to a unit step $V_0={\cal F}[u(t)]=1/j\omega$. At the front of the line we have
  

  $$V_1 = T_1V_0 =\frac{1}{2}\left[1+\frac{j\omega-1/\tau}{j\omega+1/\tau}e^{-2j\omega T}\right]\frac{1}{j\omega}$$

  $$= \frac{1}{2j\omega}\left[1+\left(\frac{2j\omega}{j\omega+1/\tau}-1... ...2j\omega}\left(1-e^{-2j\omega T}\right)+\frac{1}{j\omega+1/\tau}e^{-2j\omega T}$$

  
  
  or in time domain
  
  $$v_1(t)={\cal F}^{-1}[V_1]=\frac{1}{2}[u(t)-u(t-2T)]+e^{-(t-2T)/\tau} u(t-2T)$$
  
  
  The response at the back end of the line is
  
  $$V_2=T_2V_0=\frac{j\omega}{j\omega+\tau}e^{-j\omega T} \frac{1}{j\omega} =\frac{1}{j\omega+\tau}e^{-j\omega T}$$
  
  
  or in time domain
  
  $$v_2(t)={\cal F}^{-1}[V_2]=u(t-T) e^{-(t-T)/\tau}$$
  
  
  We see that this is a delayed differentiator with delay time $T$.

• So far we have only considered some special cases where either $\eta_1=0$ or $\eta_2=0$ (reflection at either or both ends is zero). As the result, the denominator of $T_1$ and $T_2$ is always 1. Now we consider the general case where $\eta_1\ne 0$ and $\eta_2\ne 0$. Assume $Z_1=Z_0/2$ and $Z_2=2Z_0$, then we have
  
  $$A=\frac{Z_0}{Z_1+Z_0}=\frac{2Z_1}{Z_1+2Z_1}=\frac{2}{3}$$
  
  
  
  
  $$\eta_1=\frac{Z_1-Z_0}{Z_1+Z_0}=\frac{Z_1-2Z_1}{Z_1+2Z_1}=-\frac{1}{3}$$
  
  
  
  
  $$\eta_2=\frac{Z_2-Z_0}{Z_2+Z_0}=\frac{2Z_0-Z_0}{2Z_0+Z_0}=\frac{1}{3}$$
  
  

  If the input is a step $v_0(t)=u(t)$, find $v_2(t)$:
  

  $$V_2 = V_0A(1+\eta_2)e^{-j\omega T}\left[1+\eta_1\eta_2 e^{-2j\omega T} +(\eta_1\eta_2)^2 e^{-4j\omega T}+\cdots \right]$$

  $$= \frac{8}{9}V_0 e^{-j\omega T}\left[1-\frac{1}{9} e^{-2j\omega T} +\left(\frac{1}{9}\right)^2 e^{-4j\omega T}-\cdots \right]$$

  $$= V_0\left[ \frac{8}{9} e^{-j\omega T}-\frac{8}{81}e^{-3j\omega T} +\frac{8}{729}e^{-5j\omega T}-\cdots \right]$$

  
  
  Find the value of the output for each of the following time durations.
  • $t<T:$
    
    $$v_2(t)=0$$
    
    

  • $T<t<3T:$
    
    $$v_2(t)=\frac{8}{9}$$
    
    

  • $3T<t<5T:$
    
    $$v_2(t)=\frac{8}{9}-\frac{8}{81}=\frac{64}{81}$$
    
    

  • $5T<t<7T:$
    
    $$v_2(t)=\frac{8}{9}-\frac{8}{81}+\frac{8}{729}=\frac{576}{729}$$
    
    

  • 
    
    $$\cdots \cdots \cdots \cdots$$
    
    

  The steady state value $v_2(\infty)$ can be found to be
  
  $$v_2(\infty)=\frac{8}{9}\sum_{k=0}^\infty \left(\frac{1}{9}\right)^k =\frac{8}{9}\frac{1}{1-(-1/9)}=\frac{4}{5}=0.8$$
  
  
  Plot the output $v_2(t)$ for $0<t<7T$.

  Next we repeat the above for an impulse input $v_0(t)=u(t)-u(T/3)$. As the signal only has a finite duration of $T/3$, at any time $t=2kT$ only the term $(\eta_1\eta_2)^ke^{-2kj\omega T}$ representing the wave arriving at the back end most recently is non-zero. Therefore we have:

  • $t=T:$
    
    $$v_2(t)=\frac{8}{9}$$
    
    

  • $T=3T:$
    
    $$v_2(t)=\frac{8}{9}\left(-\frac{1}{9}\right)=-\frac{8}{81}$$
    
    

  • $t=5T:$
    
    $$v_2=\frac{8}{9}\left(\frac{1}{81}\right)=\frac{8}{729}$$
    
    

  • 
    
    $$\cdots \cdots \cdots \cdots$$
    
    

  Obviously the steady state value $v_2(\infty)=0$.