Transmission Line¶
Transmission-Line Equations¶
First define - \(ℭ\) : inductance per unit length - \(𝔏\) : inductance per unit length
Lossless Transmission Lines¶
\[
\left\{
\begin{align}
\frac{\partial V}{\partial z}&=-𝔏\frac{\partial I}{\partial t}
\\\\
\frac{\partial I}{\partial z}&= -ℭ\frac{\partial V}{\partial t}
\end{align}\right.
\]
tip
These equation can be easily derived from the definition of capacitance and capacitance in electric circuit theorem
\[
\begin{gather}
v = L\frac{di}{dt}
\\\\
i= C\frac{dv}{dt}
\end{gather}
\]
Similar to the derivation in EM wave equation, we can obtain the wave equation form
\[
\left\{
\begin{align}
\frac{\partial^{2} V}{\partial z^{2}} &= 𝔏ℭ\frac{\partial^{2} V}{\partial t^{2}}
\\\\
\frac{\partial^{2}I}{\partial z^{2}} &= 𝔏ℭ\frac{\partial^{2} I}{\partial t^{2}}
\end{align}\right.
\]
Thus we have the wave velocity
\[
\begin{gather}
u = \frac{1}{\sqrt{𝔏ℭ}}
\end{gather}
\]
Comparing to the velocity of EM wave, it tell us the important relation that
\[
\begin{gather}
\boxed{
𝔏ℭ = \mu\varepsilon
}
\end{gather}
\]
Furthermore, considering the phasor form of transmission-line equation, we have
\[
\begin{align}
\frac{d^{2}V_s}{dz^{2}} = (j\beta)^{2}\,V_s
\\\\
\frac{d^{2}I_s}{dz^{2}} = (j\beta)^{2}\,I_s
\end{align}
\]
in which
\[
\begin{gather}
(j\beta)^{2} = (j\omega)^{2}𝔏ℭ
\\\\
\implies
\boxed{
\beta = \omega\sqrt{𝔏ℭ} = \frac{\omega}{u}
}
\end{gather}
\]
General Solutions¶
\[
\begin{align}
V(z, t) &= V^{+}(z-ut)+V^{-}(z+ut)
\\\\
I(z, t) &= I^{+}(z-ut)+I^{-}(z+ut)
\end{align}
\]
Similarly, we have the characteristic impedance \(Z_0\)
\[
\begin{gather}
Z_0 = \sqrt\frac{𝔏}{ℭ}=\frac{V^{+}}{I^{+}}=-\frac{V^{-}}{I^{-}}
\end{gather}
\]
General Transmission LInes¶
Revise the transmission-line equation in the lossless case into
\[
\left\{
\begin{align}
-\frac{\partial V}{\partial z}&=RI+L\frac{\partial I}{\partial t}
\\\\
-\frac{\partial I}{\partial z}&= GV+C\frac{\partial V}{\partial t}
\end{align}\right.
\qquad\overset{𝔉ourier}{\iff}\qquad
\left\{
\begin{aligned}
\frac{d V_s}{d z}&=-(R+ j\omega L)I_s
\\\\
\frac{d \,I_s}{d z}&= -(G+j\omega C)V_s
\end{aligned}\right.
\]
Also, in the form of wave equations
\[
\begin{gather}
\frac{d^{2}V_s}{dz^{2}} = \gamma^{2}V_s
\\\\
\frac{d^{2}I_s}{dz^{2}} = \gamma^{2}I_s
\end{gather}
\]
in which \(\gamma\) is also called as propagation constant.
\[
\begin{gather}
\gamma = \sqrt{(R+j\omega L)(G+j\omega C)} = \alpha+j\beta
\end{gather}
\]
General Solutions¶
solve the wave equations above
\[
\begin{align}
V_s &= V_0^{+}e^{-\gamma z}+V_0^{-}e^{\gamma z}
\\\\
I_s &= I_0^{+}e^{-\gamma z}+I_0^{-}e^{\gamma z}
\\\\
&= \frac{V_0^{+}}{Z_0}e^{-\gamma z}-\frac{V_0^{+}}{Z_0}e^{\gamma z}
\end{align}
\]
and the corresponding time domain form
\[
\begin{align}
V(z, t) &= |V_0^{+}|e^{-\alpha z}\cos(\omega t- \beta z + \phi^{+}) + |V_0^{-}|e^{\alpha z}\cos(\omega t+ \beta z + \phi^{-})
\\\\
I(z, t) &= |I_0^{+}|e^{-\alpha z}\cos(\omega t- \beta z + \varphi^{+}) + |I_0^{-}|e^{\alpha z}\cos(\omega t+ \beta z + \varphi^{-})
\end{align}
\]
- power attenuation
\[
\begin{align}
\text{attenuation const} &= \alpha \quad\text{(NP/m)}
\\\\
\text{power attenuation} &= 20 \log{e^{\alpha}}\quad\text{(dB/m)}
\end{align}
\]
thus,
\[
\begin{gather}
1\,\text{NP} = 20\log{e^{1}} = 8.69 \, \text{dB}
\end{gather}
\]
Power¶
- time-average power
\[
\begin{align}
P(z) &= \frac{1}{2}\text{Re}\bigg[VI^{*}\bigg]
\\\\
&= \frac{1}{2}\text{Re}\bigg[V_0^{+}(1+\Gamma)\left(\frac{V_0^{+}}{Z_0}\right)^{*}(1-\Gamma^{*})\bigg]
\\\\
&= \frac{1}{2}\text{Re}\bigg[\frac{{V_0^{+}}^{2}}{Z_0}(1+\Gamma)(1-\Gamma^{*})\bigg]
\\\\
&=\frac{1}{2}\frac{\left|V_0^{+}\right|^{2}}{Z_0}(1-|\Gamma|^{2})
\tag 6
\end{align}
\]
^equ-6
\[
\implies
\begin{dcases}
P^{+}=\frac{1}{2}\left|V_0^{+}\right||I_0^{+}|\,e^{+ 2\alpha z}
\\\\
P^{-}=\frac{1}{2}\left|V_0^{-}\right||I_0^{-}|\,e^{- 2\alpha z}
\end{dcases}
\]
- power loss
\[
\begin{gather}
\text{PL} =P(0)-P(z)
\end{gather}
\]
Distortionless¶
- conditions
\[
\begin{gather}
\frac{R}{L} = \frac{G}{C}
\end{gather}
\]
thus, we have the propagation constant \(\gamma\)
\[
\begin{align}
\gamma &= \sqrt{(R+j\omega L)(G+ j\omega C)}
\\\\
&= \sqrt{LC} \sqrt{\left(\frac{R}{L} +j\omega\right)\left(\frac{G}{C} +j\omega\right)}
\\\\
&= \sqrt{LC} \left(\frac{R}{L} +j\omega\right)
\\\\
&= \sqrt{RG} + j\omega\sqrt{LC}
\end{align}
\]
and the characteristic impedance \(Z_0\)
\[
\begin{align}
Z_0 &= \sqrt{\frac{R+j\omega L}{G+ j\omega C}}
\\\\
&= \sqrt {\frac{R(1+j\omega\frac{L}{R})}{G(1+j\omega(\frac{C}{G}))}}
\\\\
&= \sqrt{\frac{R}{G}}
= \sqrt{\frac{L}{C}}
\end{align}
\]
Standing Wave Ratio¶
- aka \(\text{SWR}\)
Reflection¶
- voltage reflection coefficient
\[
\begin{gather}
\Gamma(z) = \frac{V^{-}(z)}{V^{+}(z)}
=\frac{V_0^{-}e^{+\gamma z}}{V_0^{+}e^{-\gamma z}}
=\frac{V_0^{-}}{V_0^{+}}e^{+2\gamma z}
\end{gather}
\]
- current reflection coefficient
\[
\begin{gather}
\Gamma_I(z)=\frac{I^{-}}{I^{+}}=-\frac{V^{-}}{V^{+}}=-\Gamma(z)
\end{gather}
\]
- \(\Gamma_L\)
Let's define the impedance from load \(Z_L=V_L/I_L\)
\[
\begin{gather}
V_0^{+}e^{-\gamma ℓ}+V_0^{-}e^{+\gammaℓ}=\frac{Z_L}{Z_0}(V_0^{+}e^{-\gammaℓ}-V_0^{-}e^{+\gamma ℓ})
\\\\
\implies
(Z_0-Z_L)V_0^{+}e^{-\gamma ℓ}=-(Z_L+Z_0)V_0^{-}e^{+\gamma ℓ}
\\\\
\implies
\boxed{
\Gamma_L=\frac{V_0^{-}e^{+\gamma ℓ }}{V_0^{+}e^{-\gamma ℓ}}=\frac{V_0^{-}}{V_0^{+}}e^{+2\gamma ℓ }=\frac{Z_L-Z_0}{Z_L+Z_0}
}
\tag{2}
\end{gather}
\]
^equ-2
We can further write the reflection coefficient from load at \(d\) as
\[
\begin{align}
\Gamma(d)= \Gamma_{L}\,e^{-2\gamma d}
\tag 3
\end{align}
\]
^equ-3
tips
from \(\text{equation (3-1)}\),
\[
\begin{gather}
Z(z) = Z_0\frac{1+\Gamma}{1-\Gamma}
\\\\
\implies
Z(z) - Z(z)\,\Gamma = Z_0+Z_0\,\Gamma
\\\\
\implies
\boxed{
\Gamma(z) = \frac{Z(z) - Z_0}{Z(z) + Z_0}
}
\end{gather}
\]
we can also derive \(\text{equation (2)}\) by virtue of this equation
\[
\begin{gather}
\Gamma_L = \Gamma(z=ℓ) = \Gamma(d=0) = \frac{Z(z=ℓ)-Z_0}{Z(z=ℓ)+Z_0}
\end{gather}
\]
Furthermore, at the generator-end (i.e. \(z=0\)), we have
\[
\begin{gather}
\Gamma(z=0)=\Gamma(d=ℓ)=\frac{Z_{in}-Z_0}{Z_{in} + Z_0}
\end{gather}
\]
impedance¶
- Line impedance at \(z\), \(Z(z)\)
\[
\begin{align}
Z(z) &= \frac{V_s}{I_s}
\\\\
&=
Z_0\frac{V_0^{+}e^{-\gamma z}+V_0^{-}e^{+\gamma z}}{V_0^{+}e^{-\gamma z}-V_0^{-}e^{+\gamma z}}
= Z_0\frac{1+\displaystyle \frac{V_0^{-}}{V_0^{+}}e^{+2\gamma z}}{1-\displaystyle \frac{V_0^{-}}{V_0^{+}}e^{+2\gamma z}}
\\\\
&=
Z_0\frac{1+\Gamma(z)}{1-\Gamma(z)} \tag{3-1}
\end{align}
\]
^equ-3-1
- impedance at load \(Z_L\)
\[
\begin{align}
Z_L &= Z(z=ℓ) = Z_0\frac{1+\Gamma(z=ℓ)}{1-\Gamma(z=ℓ)}
\\\\
&= Z_0\frac{1+\Gamma_L}{1-\Gamma_L}
\end{align}
\]
- input impedance \(Z_{in}\)
\[
\begin{align}
Z_{in} &= Z(z=0) = Z_0\frac{1+\Gamma(z=0)}{1-\Gamma(z=0)}
\\\\
&= Z_0\,\frac{1+ \displaystyle \frac{V_0^{-}}{V_0^{+}}}{1- \displaystyle \frac{V_0^{-}}{V_0^{+}}}
\\\\
&= Z_0\,\frac{1+ \Gamma_L\,e^{-2\gamma ℓ}}{1- \Gamma_L\,e^{-2\gamma ℓ}}
\end{align}
\]
from \(\text{equation (2)}\) we have
\[
\begin{align}
Z_{in}
&= Z_0\,\frac{1+ \Gamma_L\,e^{-2\gamma ℓ}}{1- \Gamma_L\,e^{-2\gamma ℓ}}
\\\\
&= Z_0\,\frac{1+ \displaystyle \frac{Z_L-Z_0}{Z_L+Z_0}\,e^{-2\gamma ℓ}}{1- \displaystyle \frac{Z_L-Z_0}{Z_L+Z_0}\,e^{-2\gamma ℓ}}
\\\\
&= Z_0\,\frac{Z_L+Z_0+ (Z_L-Z_0)\,e^{-2\gamma ℓ}}{Z_L+Z_0- (Z_L-Z_0)\,e^{-2\gamma ℓ}}
\\\\
&= Z_0\,\frac{Z_L(1+e^{-2\gamma ℓ})+Z_0(1-e^{-2\gamma ℓ})}{Z_L(1-e^{-2\gamma ℓ})+Z_0(1+e^{-2\gamma ℓ})}
\\\\
&= Z_0\,\frac{Z_L+Z_0\,\tanh{\gammaℓ}}{Z_L\,\tanh{\gammaℓ}+Z_0}
\\\\
&=
\boxed{
Z_0\,\frac{Z_L+Z_0\,\tanh{\gammaℓ}}{Z_0+Z_L\,\tanh{\gammaℓ}}
}
\\\\
&\xlongequal{\alpha = 0}
Z_0\,\frac{Z_L+jZ_0\,\tan{\betaℓ}}{Z_0+jZ_L\,\tan{\betaℓ}}
\end{align}
\]
Complex form¶
As usual, we can represent \(\Gamma\) with complex coordinate.
\[
\begin{gather}
\Gamma = |\Gamma |\,\angle{\Gamma}
\end{gather}
\]
Consider \(\text{equation (3)}\), we further have
\[
\begin{align}
\boxed{
\Gamma=\Gamma_L\,e^{-2\alpha d}\,e^{-j2\beta d}
}
\end{align}
\]
Besides, we can rewrite \(V_s, I_s\) as
\[
\begin{align}
|V_s| &= \left|V^{+}\right||1+\Gamma|
\\\\
|I_s| &= \left|I^{+}\right||1-\Gamma|
\end{align}
\]
Standing Wave Ratio for Lossless¶
\[
\begin{align}
\text{VSWR} &\equiv \frac{V_{max}}{V_{min}} = \frac{|V_0^{+}||1+\Gamma|_{max}}{|V_0^{+}||1+\Gamma|_{min}}
\\\\
&= \frac{1+ |\Gamma|}{1- |\Gamma|}
\xlongequal{\alpha\,=\,0}\frac{1+ |\Gamma_L|}{1- |\Gamma_L|}
\end{align}
\]
and similarly for \(\text{ISWR}\),
\[
\begin{gather}
\frac{I_{max}}{I_{min}}=\frac{V_{max}}{V_{min}}=\text{SWR}
\end{gather}
\]
according to \(\text{equation (5)}\),
\[
\begin{align}
z=\frac{1+\Gamma}{1-\Gamma}=r+jx
\end{align}
\]
we can obtain \(\text{SWR}\) by picking the point that \(\angle\Gamma = 0\), then
\[
\begin{align}
\text{SWR} = \frac{1+|\Gamma|}{1-|\Gamma|}= r+j0=r
\end{align}
\]
Smith Chart¶
Smith chart help us quickly obtain the normalized impedance \(Z\) w.r.t \(Z_0\).
\[
\begin{align}
Z= z\times Z_0 = (r+jx)\times Z_0
\end{align}
\]
then, we can have the normalized impedance \(z\)
\[
\begin{align}
z=r+jx= \frac{Z}{Z_0}=\frac{1+\Gamma}{1-\Gamma}
\tag 5
\end{align}
\]
^equ-5
-
\(r\)-circles
-
\(x\)-circles
Admittance Chart¶
for the admittance of transmission line, we can easily obtain the relationship of admittance and impedance.
\[
\begin{gather}
y=\frac{Y}{Y_0}=\frac{1}{z}=\frac{1-\Gamma}{1+\Gamma}
\\\\
\implies
y=z\,e^{-j\pi} \tag 4
\end{gather}
\]
^equ-4
the \(\text{equation (4)}\) tell us that we can obtain the normalized admittance \(y\) with Smith chart by simply rotate the corresponding \(z\) with \(180\degree\).