1. Transmission Lines Neurons in the hippocampus Carbon nanotube
(Image courtesy Slice of Life Project)
Carbon nanotube

2. Summary l Worry about distributed circuits when l > 0.01l
Transmission Line
LOAD l
Worry about distributed circuits when l > 0.01l

3. Distributed model for TL
L’Dz
R’Dz
G’Dz
C’Dz
L’Dz
R’Dz
G’Dz
C’Dz
COPPER WIRE
INSULATION
COPPER MESH

4. AC Circuit model Z’Dz ejwt Y’Dz Z’ = R’ + jwL’ (Impedances in series)
Y’ = G’ + jwC’ (Admittances in series)
ejwt

5. Telegrapher’s Equations
dV/dz = -Z’I
dI/dz = -Y’V
d2V/dz2 = g2V
V = V0+(e-gz + Gegz)
I = V0+(e-gz - Gegz)/Z0
= Z’Y’ = (R’+jwL’)(G’+jwC’) = a + jb : Decay constant
Z0 = Z’/Y’ = (R’+jwL’)/(G’+jwC’): Characteristic Impedance

6. Lossless Line R’ = G’ = 0 g = jb = jwL’C’
Z0 = L’/C’ = 377 W in free space
(impedance of free space)
a=0 (lossless)
v = w/b: indep. of frequency (dispersionless)
Dispersionless: undistorted signal
Genl. Condition: R’/G’ = L’/C’

7. Reflection at a load Z0 ZL G = (ZL-Z0)/(ZL+Z0) = |G|ejqr
Transmission Line
LOAD Z0 ZL
V(z) = V+0(e-jbz + Gejbz)
I(z) = V+0(e-jbz - Gejbz)/Z0
V(0)/I(0) = ZL 
G = (ZL-Z0)/(ZL+Z0) = |G|ejqr
ZL=Z0 [(1+G)/(1-G)]

8. Work backwards to input impedance
Zin Z0 ZL
Transmission Line
LOAD
V(z) = V+0(e-jbz + Gejbz)
I(z) = V+0(e-jbz - Gejbz)/Z0
V(-l)/I(-l) = Zin
Zin=Z0 [(1+Gejbl)/(1-Gejbl)]
ie, to get Z(d), G  Ge-jbd in ZL formula

9. Nailing down V0+ Iin ZS Zin Z0 ZL VS Vin V(z) = V+0(e-jbz + Gejbz)
Set Vin = IinZin = VSZin/(ZS+Zin)
= V(-l) = V0+(ejbl + Ge-jbl)  gives V0+
Vin
Zin Z0
Transmission Line
LOAD ZL VS
V(z) = V+0(e-jbz + Gejbz)
Why is V(-l)/I(-l) ≠ Zs?
Because we haven’t included the source current Iin
Iin = VS/(ZS + Zin)

10. Special Cases: Impedance match. (ZL = Z0)
G = 0
No reflection

11. Special Cases: Short ckt. (ZL = 0)
G = -1
Refln. at a hard wall (phase p)

12. Special Cases: Open Ckt. (ZL = ∞)
G = 1
Refln. at a soft wall (phase 0)

13. Special Cases: Reactive load (ZL: imaginary)
= ejf
|G| = 1
Fully reflected, but with intermediate phase
(reactive component picked up)