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e157: radio frequency circuit design

Lecture 4: propagation coefficient and standing wave

Profile picture of Samson ZhangSamson Zhang
Sep 11, 20232 min read

analytic form

Given input

V_{in} = V_{zp} \cos(\omega t) Vin=Vzpcos⁡(ωt) V_{in} = V_{zp} \cos(\omega t)
we can say each point sees
V(x,t) = V_{zp} \cos(\omega t - kx) V(x,t)=Vzpcos⁡(ωt−kx) V(x,t) = V_{zp} \cos(\omega t - kx)
, where
k = \frac{\omega}{v} = \frac{2\pi}{\lambda} k=ωv=2πλ k = \frac{\omega}{v} = \frac{2\pi}{\lambda}


These can be put into analytic form:

V_{in}(t) = V_{zp}e^{j\omega t}, V(x,t) = V_{zp}e^{j(\omega t - kx)} Vin(t)=Vzpejωt,V(x,t)=Vzpej(ωt−kx) V_{in}(t) = V_{zp}e^{j\omega t}, V(x,t) = V_{zp}e^{j(\omega t - kx)}


propagation coefficient

now if we go back to the rlcg divider and solve for the transfer function:



We get an exponential

V(x,t) = V(0,t)e^{-\sqrt{zy}x} V(x,t)=V(0,t)e−zyx V(x,t) = V(0,t)e^{-\sqrt{zy}x}


We define the propagation coefficient

\gamma = \sqrt{zy} = \alpha + jk γ=zy=α+jk \gamma = \sqrt{zy} = \alpha + jk


So

V(x,t) = V(0,t)e^{-\gamma x} = V(0,t)e^{-\alpha x}e^{-jkx} V(x,t)=V(0,t)e−γx=V(0,t)e−αxe−jkx V(x,t) = V(0,t)e^{-\gamma x} = V(0,t)e^{-\alpha x}e^{-jkx}




\alpha α \alpha
causes attenuation, while
k k k
adds phase with
x x x
(what does that mean?)



\alpha, k α,k \alpha, k
can be expressed as follows



Special conditions result in no dispersion:

  • lossless,
    r = g = 0 r=g=0 r = g = 0
    :
    \alpha = 0, k = \omega \sqrt{lc} α=0,k=ωlc \alpha = 0, k = \omega \sqrt{lc}
  • low loss,
    j\omega l << r, j\omega c << g jωl<<r,jωc<<g j\omega l << r, j\omega c << g
    :
    \alpha = \frac{r}{2Z_0} + \frac{gZ_0}{2}, k = \omega\sqrt{lc} α=r2Z0+gZ02,k=ωlc \alpha = \frac{r}{2Z_0} + \frac{gZ_0}{2}, k = \omega\sqrt{lc}
    
  • heaviside,
    rc = gl rc=gl rc = gl
    :
    \alpha = \sqrt{rg}, k = \omega \sqrt{lc} α=rg,k=ωlc \alpha = \sqrt{rg}, k = \omega \sqrt{lc}
    

No dispersion bc

\frac{d(kx)}{d\omega} = \sqrt{lc} = \text{const} d(kx)dω=lc=const \frac{d(kx)}{d\omega} = \sqrt{lc} = \text{const}
so sine waves in fourier transform all scale the same



k = \frac{\omega}{v}, v = \frac{1}{\sqrt{lc}} k=ωv,v=1lc k = \frac{\omega}{v}, v = \frac{1}{\sqrt{lc}}
consistent with earlier



\alpha α \alpha
typically has units of dB/m

standing wave

when sine wave reflects from end of transmission line we get a standing wave with twice the wavenumber



also with magnitude swing from

\text{amp} \cdot (1 + |\Gamma|) amp⋅(1+∣Γ∣) \text{amp} \cdot (1 + |\Gamma|)
to
\text{amp} \cdot (1 - |\Gamma|) amp⋅(1−∣Γ∣) \text{amp} \cdot (1 - |\Gamma|)


practically, voltage standing wave ratio (VSWR)

= \frac{1 + |\Gamma|}{1 - |\Gamma|} > 5 =1+∣Γ∣1−∣Γ∣>5 = \frac{1 + |\Gamma|}{1 - |\Gamma|} > 5
is difficult to work with

also useful to know

|\Gamma| = \frac{\text{VSWR} - 1}{\text{VSWR} + 1} ∣Γ∣=VSWR−1VSWR+1 |\Gamma| = \frac{\text{VSWR} - 1}{\text{VSWR} + 1}


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e157: radio frequency circuit design

fall 2023 class w prof. spencer