![[Pasted image 20261013205800.png|434]] [Smith Charts (Veritasium)](https://www.youtube.com/watch?v=GK2pZ_oVU1o) # Transmission Lines A TEM line has voltage and current waves. The same math describes plane waves at layered interfaces (see the dictionary below and [[reflection]]). ## Telegrapher's Equations A differential section $dz$ has series $R\,dz$ and $L\,dz$, and shunt $G\,dz$ and $C\,dz$. KVL and KCL give: $ \frac{dV}{dz}=-(R+j\omega L)\,I,\qquad \frac{dI}{dz}=-(G+j\omega C)\,V $ Combined: $ \frac{d^2V}{dz^2}=\gamma^2V,\qquad \gamma=\alpha+j\beta=\sqrt{(R+j\omega L)(G+j\omega C)} $ Solution, characteristic impedance, and the lossless case ($R=G=0$): $ V(z)=V_+e^{-\gamma z}+V_-e^{+\gamma z},\qquad I(z)=\frac{V_+e^{-\gamma z}-V_-e^{+\gamma z}}{Z_0} $ $ Z_0=\sqrt{\frac{R+j\omega L}{G+j\omega C}},\qquad Z_0=\sqrt{\frac{L}{C}},\quad \beta=\omega\sqrt{LC},\quad v=\frac{1}{\sqrt{LC}} $ - **Distortionless** when $\dfrac{R}{L}=\dfrac{G}{C}$ (Heaviside): $\alpha$ and $v$ don't depend on $\omega$. Loading coils add $L$ to reach this. - For any TEM line, $LC=\mu\epsilon$. ## Coaxial Line Inner radius $a$, outer radius $b$, dielectric $\epsilon$. Solve Laplace in the cross-section ($\nabla_t^2\Phi=0$, so $\Phi\propto\ln r$), then multiply by $e^{-j\beta z}$: $ \vec E=\frac{V}{r\ln(b/a)}\,\hat r\,e^{-j\beta z},\qquad \vec H=\frac{I}{2\pi r}\,\hat\phi\,e^{-j\beta z},\qquad \beta=\omega\sqrt{\mu\epsilon} $ $ Z_0=\frac{V}{I}=\frac{\eta}{2\pi}\ln\frac{b}{a},\qquad L=\frac{\mu}{2\pi}\ln\frac{b}{a},\qquad C=\frac{2\pi\epsilon}{\ln(b/a)} $ Check: $\dfrac{E_r}{H_\phi}=\eta$ (TEM), and $\displaystyle\int\vec E\times\vec H^*\cdot\hat z\,dA=VI^*$. The field is largest at the inner conductor, which is where breakdown starts. ## Reflection Coefficient and Impedance Along the Line Put the load $Z_L$ at $z=0$, with the line at $z<0$, and let $d=-z$ be the distance back from the load. Lossless: $ \Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_0},\qquad \Gamma(d)=\Gamma_L\,e^{-2j\beta d},\qquad Z(d)=Z_0\,\frac{1+\Gamma(d)}{1-\Gamma(d)} $ $ Z_\text{in}(d)=Z_0\,\frac{Z_L+jZ_0\tan\beta d}{Z_0+jZ_L\tan\beta d} $ | Load / length | $Z_\text{in}$ | Note | | ---------------------- | -------------------- | ---------------------------------------------------------- | | short, any $d$ | $jZ_0\tan\beta d$ | inductive for $\beta d<\dfrac{\pi}{2}$ | | open, any $d$ | $-jZ_0\cot\beta d$ | **looks like a short** when $\beta d=\dfrac{\pi}{2}+m\pi$ | | $d=\dfrac{\lambda}{2}$ | $Z_L$ | invisible | | $d=\dfrac{\lambda}{4}$ | $\dfrac{Z_0^2}{Z_L}$ | **impedance inversion**: open ↔ short, small ↔ large | **Why an open looks like a short:** at the open end, $I=0$ and $V$ is maximum. A quarter wavelength back, the standing wave has $V=0$ and $I$ maximum. The source "sees" a node of $V$, which is a short. The quarter-wave line is a 90° phase rotation that swaps the $V$ and $I$ standing-wave patterns. ## Standing Waves and Power $ |V(d)|=|V_+|\,\big|1+\Gamma_Le^{-2j\beta d}\big|,\qquad \text{VSWR}=\frac{|V|_\text{max}}{|V|_\text{min}}=\frac{1+|\Gamma|}{1-|\Gamma|} $ - $V$ maxima are where $\Gamma(d)$ is real and positive. Minima are $\dfrac{\lambda}{4}$ away. Adjacent minima are $\dfrac{\lambda}{2}$ apart. - Power delivered (lossless line): $ P=\frac{|V_+|^2}{2Z_0}\left(1-|\Gamma|^2\right) $ - Real-time picture (pset 3 P1e): an open line with $\beta L=\dfrac{3\pi}{2}$ has a pure standing wave, $V\propto\cos\beta z\cos\omega t$ and $I\propto\sin\beta z\sin\omega t$. $V$ and $I$ are 90° out of phase in both space and time. ## Smith Chart ![[Pasted image 20261019110048.png|237]] The map below takes the passive half-plane $\mathrm{Re}\,z\ge0$ onto the unit disk $|\Gamma|\le1$: $ \Gamma=\frac{z-1}{z+1},\qquad z=\frac{Z}{Z_0} $ - Circles of constant $r=\mathrm{Re}\,z$ and arcs of constant $x=\mathrm{Im}\,z$ make up the chart. - Center: matched. Left edge: short. Right edge: open. Upper half: inductive. Lower half: capacitive. - Moving along a lossless line is a rotation about the center (toward the generator = clockwise). A full turn is $\dfrac{\lambda}{2}$. - Rotating by 180° ($\dfrac{\lambda}{4}$) maps $z$ to $\dfrac{1}{z}$, so the chart doubles as an admittance chart. - Changing $\lambda$ changes the rotation angle $2\beta d$ ## Matching - **Quarter-wave transformer**: match real $Z_L$ to $Z_1$ with a $\dfrac{\lambda}{4}$ section of $Z_m=\sqrt{Z_1Z_L}$. In optics this is the AR coating $n_c=\sqrt{n_1n_s}$. - Single stub: move to where $\mathrm{Re}\,y=1$, then cancel $\mathrm{Im}\,y$ with a shorted stub. ## Wave ↔ Line Dictionary (plane wave at angle $\theta$; Kong §6.4) | Plane wave (TE) | Plane wave (TM) | Line | |---|---|---| | $E_y$ | $E_x$ | $V$ | | $-H_x$ | $H_y$ | $I$ | | $k_z=k\cos\theta$ | $k_z=k\cos\theta$ | $\beta$ | | $Z_0=\dfrac{\eta}{\cos\theta}=\dfrac{\omega\mu}{k_z}$ | $Z_0=\eta\cos\theta=\dfrac{k_z}{\omega\epsilon}$ | $Z_0$ | | $\dfrac{1}{2}\mathrm{Re}\{\vec E\times\vec H^*\}\cdot\hat z$ | same | $\dfrac{1}{2}\mathrm{Re}\{VI^*\}$ | - Continuity of tangential $E$ and $H$ at an interface is continuity of $V$ and $I$ at a junction, so $Z$ is continuous across interfaces. - An evanescent layer (TIR) is a line with imaginary $k_z$ and imaginary $Z_0$. The same formulas give frustrated-TIR tunneling (Kong 6.4.1). - Multilayers are a cascade of line sections: use $Z_\text{in}$ repeatedly, or ABCD matrices ([[optical operators]]).