6.

013 - Electromagnetics and Applications Fall 2005

Lecture 10 - Transmission Lines
Prof. Markus Zahn October 13, 2005

I. Transmission Line Equations

A. Parallel Plate Transmission Line

¯ must be perpendicular to the electrodes and H
E ¯ must be tangential, so

¯ = Ex (z, t)īx
E
¯ = Hy (z, t)īy
H

¯
¯ = −µ ∂H ⇒ ∂Ex = −µ ∂Hy
�×E
∂t ∂z ∂t
∂E¯ ∂H y ∂E x
�×H¯ =� ⇒ = −�
∂t ∂z ∂t

 � 2
v(z, t) = ¯ · d¯l = Ex (z, t)d
E
1
z = constant
i(z, t) = Kz (z, t)w = Hy (z, t)w

∂v ∂i µd
= −L L= henries / meter Inductance per unit length
∂z ∂t w
∂i ∂v �w
= −C C= farads / meter Capacitance per unit length
∂z ∂t d
µd� ��
w 1
LC = = µ� = 2
w d�
� c
B. Transmission Line Structures

C. Distributed Circuit Representation with Losses

∂v(z, t)
i(z, t) − i(z + Δz, t) = CΔz + GΔz v(z, t)
∂t
∂i(z + Δz, t)
v(z, t) − v(z + Δz, t) = LΔz + i(z + Δz, t)RΔz
∂t
i(z + Δz, t) − i(z, t) ∂i ∂v
lim = = −C − Gv
Δz →0 Δz ∂z ∂t
v(z + Δz, t) − v(z, t) ∂v ∂i
lim = = −L − iR
Δz →0 Δz ∂z ∂t

R is the series resistance per unit length, measured in ohms/meter, and G is the shunt
conductance per unit length, measured in siemens/meter.

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 If the line is lossless (R = G = 0), we have the Telegrapher’s equations:

∂i ∂v
= −C
∂z ∂t
∂v ∂i
= −L
∂z ∂t
Including loss, Poynting’s theorem for the circuit equivalent form is:
�
� ∂i ∂v

v·� = −C − Gv

�
� ∂z
∂t
�
� ∂v ∂i

i
·
� = −L − iR

�
�
∂z
∂t
� �

∂i ∂v
∂(vi)
∂
1
2 1
2
Add: v
+ i =
=− Cv + Li − Gv 2 − i2 R
∂z ∂z ∂z
∂t 2
2

D. Wave Equation (Lossless, R = 0, G = 0)

�
∂
�� ∂i ∂v
∂2i ∂ 2 v

= −C ⇒ = −C 2
∂t
�
∂z
∂t ∂z∂t ∂t
�
�
∂
�� ∂v ∂i
∂2i 1 ∂ 2 v

= −L =−
∂z
�
∂z
∂t ∂z∂t L ∂z 2
�

1 ∂2v ∂2v ∂2v ∂2v 1 ∂2v

−
= −C
⇒
= LC =

L
∂z 2 ∂t2 �∂z
2 ∂t
��
2 c2 ∂t2�
Wave equation

II. Sinusoidal Steady State

A. Complex Amplitude Notation

v(z, t) = Re
v̂(z)ejωt
�
�

� �

i(z, t) = Re î(z)ejωt

Substitute into the wave equation:

∂2v 1 ∂2v d2 v̂ ω2 ω
2
= 2 2
⇒ 2
= − 2
v̂(z), let k =
∂z c ∂t dz c c
d2 v̂
+ k 2 v̂ = 0 ⇒ v̂(z) = V̂+ e−jkz + V̂− e+jkz
dz 2
dv̂ 1
� �
= −Ljω î ⇒ î(z) = −
−j�k V̂+ e−jkz + j�k V̂− e+jkz
dz
Lj ω
√ �
√
�

k ω k LC C
= LC ⇒

�
= =
=
= Y0 is the Line Admittance
ω c�ω ωL
L L

1 L
Z0 = = is the Line Impedance
Y0 C
�
�

î(z) = Y0 V̂+ e−jkz − Vˆ− e+jkz

 v̂(z) = V̂+ e−jkz + V̂− e+jkz

� �
v(z, t) = Re V̂+ ej(ωt−kz) + V̂− ej(ωt+kz)
� �
j(ωt−kz) j(ωt+kz)
i(z, t) = Re Y0 V̂+ e − V̂− e
ω √ √
k = = ω LC = ω �µ
c

B. Short Circuited Line (v(z = 0, t) = 0, v(z = −l, t) = V0 cos(ωt))

v̂(z) = V̂+ e−jkz + V̂− e+jkz ⇒ v̂(z = 0) = 0 = V̂+ + V̂− ⇒ V̂+ = −V̂−
� �
v̂(z = −l) = V0 = V̂+ e+jkl + V̂− e−jkl = V̂+ ejkl − e−jkl

 = 2j sin(kl)V̂+
V0
Vˆ+ = −V̂− =
2j sin(kl)

V0 � � V (−2j) sin(kz)
0
v̂(z) = e−jkz − e+jkz =
2j sin(kl) 2j sin(kl)
V0 sin(kz)
=−
sin(kl)
� � Y0 V0
� �
î(z) = Y0 V̂+ e−jkz − V̂− ejkz = e−jkz + e+jkz
2j sin(kl)
2�Y0 V0 cos(kz)
=
2�j sin(kl)
jY0 V0 cos(kz)
=−
sin(kl)
� �
V0 sin(kz) jωt V0 sin(kz) cos(ωt)
v(z, t) = Re v̂(z)ejωt = Re −
� �
e =−
sin(kl) sin(kl)
� �
� � jY0 V0 cos(kz) jωt Y0 V0 cos(kz) sin(ωt)
i(z, t) = Re î(z)ejωt = Re − e =
sin(kl) sin(kl)
ωl nπc
We have resonance when sin(kl) = 0 ⇒ kl = nπ = c ⇒ ω = ωn ≡ l ,n = 1, 2, 3, . . .
v̂(z)
Complex impedance: Z(z) = î(z)
= −jZ0 tan(kz)

Z(z = −l) = +jZ0 tan(kl)

In the following, take n = 1, 2, 3, . . .:

kl = nπ Z(z = −l) = 0 short circuit

π
kl = (2n − 1) Z(z = −l) = ∞ open circuit
2
π
(n − 1)π < kl < (2n − 1) Z(z = −l) = +jX, X > 0 (positive reactance, inductive)
2
1
(n − )π < kl < nπ Z(z = −l) = −jX, X > 0 (negative reactance, capacitive)
2
kl � 1 ⇒ Z(z) = −jZ0 k
�
L �
= −j ω L�
C
C
�
= −jLZ
Z(z = −l) = j(Ll) inductive
V0 z
|kz| � 1 v(z, t) = − cos(ωt) ⇒ v(z = −l, t) = V0 cos(ωt)
l
di
= (Ll) (z = −l, t)
dt
V0 Y0 V0 sin(ωt)
i(z, t) = sin(ωt) i(z = −l, t) =
kl (Ll)ω

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C. Open Circuited Line (i(z = 0, t) = 0)

v(z = −l, t) = V0 sin(ωt)

� � � �
î(z) = Y0 V̂+ e−jkz − V̂− e+jkz ⇒ î(z = 0) = 0 = Y0 V̂+ − V̂− ⇒ V̂+ = V̂−
� �
v̂(z = −l) = −jV0 = V̂+ e+jkl + V̂− e−jkl = V̂+ ejkl + e−jkl = 2V̂+ cos(kl)
jV0
V̂+ = V̂− = −
2 cos(kl)
jV0 � −jkz �
v̂(z) = − e + e+jkz
2 cos(kl)
jV0 · 2� cos(kz)
=−
2� cos(kl)
jV0 cos(kz)
=−
cos(kl)
jY0 V0 � −jkz �
î(z) = − e − e+jkz
2 cos(kl)
(−jY0 V0 )(−2j) sin(kz)
=
2 cos(kl)
Y0 V0 sin(kz)
=−
cos(kl)

� V0 cos(kz)
v(z, t) = Re v̂(z)ejωt =
�
sin(ωt)
cos(kl)
� � −V0 Y0
i(z, t) = Re î(z)ejωt = sin(kz) cos(ωt)
cos(kl)
π
Resonance: cos(kl) = 0 ⇒ (kl) = (2n − 1) , n = 1, 2, 3, . . .
2
(2n − 1) π2
ωn =
2l
Complex Impedance

v̂(z)
Z(z) = = Z0 j cot(kz)
î(z)
Z(z = −l) = −jZ0 cot(kl)
kl � 1 ⇒ v(z, t) = V0 sin(ωt)
i(z, t) = −V0 Y0 kz cos(ωt)
dv
i(z = −l, t) = (Cl)ωV0 cos(ωt) = (Cl) (z = −l, t)
dt

 Open circuited line

Impedance for short and open circuited wires