Transmission lines fundamentals

Reference:
Slides of EE4C05, TUD
Pozar D M. Microwave engineering

What is a transmission line

A Tx-line is a two-port network connecting a generator circuit at the sending end to a load at the receiving end.

• Circuit VS Transmission Line

  Circuit: low frequency, wavelength is much larger than the circuit size, approximately same phases at different locations in the circuit

Transmission line: high frequency, wavelength is much smaller than circuit size, different phases at different locations in the circuit

Unlike in circuit theory, the length of a transmission line is very important in transmission line analysis

• EM waves in transmission lines

  Modes: vector functions that represent the distribution of the field in the plane transverse to the direction of propagation

In bounded problems, boundary conditions lead to field representations
as summation of modes
E ⃗ ( r ⃗ ) = ∑ m V m ( z ) e ⃗ m ( x , y ) , H ⃗ ( r ⃗ ) = ∑ m I m ( z ) h ⃗ m ( x , y ) \vec E(\vec r)=\sum_m V_m(z)\vec e_m(x,y),\qquad \vec H(\vec r)=\sum_m I_m(z)\vec h_m(x,y) E (r )=m∑​Vm​(z)e m​(x,y),H (r )=m∑​Im​(z)h m​(x,y)
Of all the modes only 1 is typically useful: fundamental mode
E ⃗ ( r ⃗ ) = V ( z ) e ⃗ 0 ( x , y ) , H ⃗ ( r ⃗ ) = I ( z ) h ⃗ 0 ( x , y ) \vec E(\vec r)= V(z)\vec e_0(x,y),\qquad \vec H(\vec r)=I(z)\vec h_0(x,y) E (r )=V(z)e 0​(x,y),H (r )=I(z)h 0​(x,y)
Voltage and currents: scalar amplitudes of the dominant mode

Solutions ( E ⃗ , H ⃗ \vec E,\vec H E ,H ) for TEM from Maxwell Equations

Maxwell’s Equations in absence of sources

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propagation constant and Helmholtz equations

From M.E., we have
∇ × E ⃗ = − j ω μ H ⃗ ∇ × H ⃗ = j ω ε E ⃗ \begin{aligned} \nabla \times \vec E=&-j\omega \mu \vec H\\ \nabla \times \vec H=&j\omega \varepsilon \vec E \end{aligned} ∇×E =∇×H =​−jωμH jωεE ​
Take curl of both sides of the first equation.
∇ × ∇ × E ⃗ = − j ω μ ∇ × H ⃗ = − j ω μ ( j ω ε E ⃗ ) = ω 2 ε μ E ⃗ \nabla \times \nabla \times \vec E=-j\omega\mu\nabla \times \vec H=-j\omega\mu(j\omega\varepsilon\vec E)=\omega^2\varepsilon\mu\vec E ∇×∇×E =−jωμ∇×H =−jωμ(jωεE )=ω2εμE
Note that ∇ × ∇ × E ⃗ = ∇ ( ∇ ⋅ E ⃗ ) − ∇ 2 E ⃗ = − ∇ 2 E ⃗ \nabla \times \nabla \times \vec E=\nabla(\nabla \cdot \vec E)-\nabla^2\vec E=-\nabla^2\vec E ∇×∇×E =∇(∇⋅E )−∇2E =−∇2E , we obtain
∇ 2 E ⃗ + ω 2 ε μ E ⃗ = 0 \nabla^2\vec E+\omega^2\varepsilon\mu\vec E=0 ∇2E +ω2εμE =0
Analogously, if we started from the other Maxwell equation,
∇ 2 H ⃗ + ω 2 ε μ H ⃗ = 0 \nabla^2\vec H+\omega^2\varepsilon\mu\vec H=0 ∇2H +ω2εμH =0
Defining the propagation constant k = ω ε μ k=\omega\sqrt{\varepsilon\mu} k=ωεμ ​, we obtain the Homogeneous wave equations (Helmholtz)
∇ 2 E ⃗ + k 2 E ⃗ = 0 ∇ 2 H ⃗ + k 2 H ⃗ = 0 \begin{aligned} \nabla^2 \vec E+k^2\vec E=&0\\ \nabla^2 \vec H+k^2\vec H=&0 \end{aligned} ∇2E +k2E =∇2H +k2H =​00​

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separation of variables

By assuming that the dependence from z z z is the one typical of a wave travelling along + z +z +z, the field can be written as the product of a function of only ( x , y ) (x,y) (x,y) and a function of only ( z ) (z) (z):
E ⃗ ( x , y , z ) = [ e x ( x , y ) x ^ + e y ( x , y ) y ^ + e z ( x , y ) z ^ ] e − j k z H ⃗ ( x , y , z ) = [ h x ( x , y ) x ^ + h y ( x , y ) y ^ + h z ( x , y ) z ^ ] e − j k z \vec E(x,y,z)=[e_x(x,y)\hat x+e_y(x,y)\hat y+e_z(x,y)\hat z]e^{-jkz}\\ \vec H(x,y,z)=[h_x(x,y)\hat x+h_y(x,y)\hat y+h_z(x,y)\hat z]e^{-jkz} E (x,y,z)=[ex​(x,y)x^+ey​(x,y)y^​+ez​(x,y)z^]e−jkzH (x,y,z)=[hx​(x,y)x^+hy​(x,y)y^​+hz​(x,y)z^]e−jkz
Define transverse part and longitudinal part:

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search for TEM solutions

Recall that for TEM, both E-and H-field orthogonal to direction of propagation.

Therefore, the longitudinal parts are e z = h z = 0 e_z=h_z=0 ez​=hz​=0. Then
E ⃗ ( r ⃗ ) = e ⃗ t ( x , y ) e − j k z \vec E(\vec r)=\vec e_t(x,y)e^{-jkz} E (r )=e t​(x,y)e−jkz
Plugging it into ∇ 2 E ⃗ + k 2 E ⃗ = 0 \nabla^2 \vec E+k^2\vec E=0 ∇2E +k2E =0:
∇ 2 e ⃗ t ( x , y ) e − j k z + k 2 e ⃗ t ( x , y ) e − j k z = 0 \nabla^2 \vec e_t(x,y)e^{-jkz}+k^2\vec e_t(x,y)e^{-jkz}=0 ∇2e t​(x,y)e−jkz+k2e t​(x,y)e−jkz=0
Expanding the Laplacian using
∇ 2 f = ∂ x 2 f + ∂ y 2 f + ∂ z 2 f ∇ 2 A ⃗ = ∇ 2 A x x ^ + ∇ 2 A y y ^ + ∇ 2 A z z ^ \begin{aligned} \nabla^2 f&=\partial _x^2 f+\partial _y^2 f+\partial _z^2 f\\ \nabla^2 \vec A&=\nabla^2 A_x \hat x+\nabla^2 A_y \hat y+\nabla^2 A_z\hat z \end{aligned} ∇2f∇2A ​=∂x2​f+∂y2​f+∂z2​f=∇2Ax​x^+∇2Ay​y^​+∇2Az​z^​
We obtain
∇ 2 e ⃗ t ( x , y ) e − j k z = ( ∇ t 2 + ∂ z 2 ) e ⃗ t ( x , y ) e − j k z = e − j k z ∇ t 2 e ⃗ t ( x , y ) − k 2 e ⃗ t ( x , y ) e − j k z \begin{aligned} \nabla^2 \vec e_t(x,y)e^{-jkz}&=(\nabla_t^2+\partial_z^2)\vec e_t(x,y)e^{-jkz}\\ &=e^{-jkz}\nabla_t^2 \vec e_t(x,y)-k^2\vec e_t(x,y)e^{-jkz} \end{aligned} ∇2e t​(x,y)e−jkz​=(∇t2​+∂z2​)e t​(x,y)e−jkz=e−jkz∇t2​e t​(x,y)−k2e t​(x,y)e−jkz​
Therefore,
e − j k z ∇ t 2 e ⃗ t ( x , y ) = 0 ⇒ ∇ t 2 e ⃗ t ( x , y ) = 0 e^{-jkz}\nabla_t^2 \vec e_t(x,y)=0\Rightarrow \nabla_t^2 \vec e_t(x,y)=0 e−jkz∇t2​e t​(x,y)=0⇒∇t2​e t​(x,y)=0
Similarly
∇ t 2 h ⃗ t ( x , y ) = 0 \nabla_t^2 \vec h_t(x,y)=0 ∇t2​h t​(x,y)=0
They are like static fields in absence of sources:
{ ∇ t ⋅ e ⃗ t = 0 ∇ t × e ⃗ t = 0 { ∇ t ⋅ h ⃗ t = 0 ∇ t × h ⃗ t = 0 \left\{\begin{array}{l} \nabla_{t} \cdot \vec{e}_{t}=0 \\ \nabla_{t} \times \vec{e}_{t}=0 \end{array}\qquad \left\{\begin{array}{l} \nabla_{t} \cdot \vec{h}_{t}=0 \\ \nabla_{t} \times \vec{h}_{t}=0 \end{array}\right.\right. {∇t​⋅e t​=0∇t​×e t​=0​{∇t​⋅h t​=0∇t​×h t​=0​
In electrostatics, the E-field can be expressed as gradient of a scalar potential Φ ( x , y ) : e ⃗ t ( x , y ) = − ∇ t Φ ( x , y ) \Phi(x, y): \quad \vec{e}_{t}(x, y)=-\nabla_{t} \Phi(x, y) Φ(x,y):e t​(x,y)=−∇t​Φ(x,y)
which also satisfies Laplace equation: ∇ t 2 Φ ( x , y ) = 0 \quad \nabla_{t}^{2} \Phi(x, y)=0 ∇t2​Φ(x,y)=0

Voltage and current ( V , I V,I V,I) in a transmission line

Like static fields in (x,y)
∇ t 2 e ⃗ t ( x , y ) = 0 ∇ t 2 h ⃗ t ( x , y ) = 0 \begin{aligned} \nabla_t^2\vec e_t(x,y)=&0\\ \nabla_t^2\vec h_t(x,y)=&0 \end{aligned} ∇t2​e t​(x,y)=∇t2​h t​(x,y)=​00​
Analogous to the statics, we can also define

1. the voltage between the two conductors:
   V = ϕ 1 − ϕ 2 = ∫ 1 2 E ⃗ ⋅ d l ⃗ V=\phi_1-\phi_2=\int_1^2\vec E\cdot d\vec l V=ϕ1​−ϕ2​=∫12​E ⋅dl

2. the current flowing on a conductor (Ampere’s law):
   I = ∮ C H ⃗ ⋅ d l ⃗ I=\oint_C\vec H\cdot d\vec l I=∮C​H ⋅dl
   ????: closed path surrounding the cross-section of the conductor

3. We define the characteristic impedance of the line as
   Z 0 = V / I Z_0=V/I Z0​=V/I

Like waves in (z)
V ( z ) = e − j k z I ( z ) = e − j k z \begin{aligned} V(z)=&e^{-jkz}\\ I(z)=&e^{-jkz} \end{aligned} V(z)=I(z)=​e−jkze−jkz​

The characteristic impedance is
Z 0 = V + I + = − V − I − Z_0=\frac{V_+}{I_+}=-\frac{V_-}{I_-} Z0​=I+​V+​​=−I−​V−​​
Why minus? The change of travelling direction does not change the direction of E ⃗ \vec E E , but change the direction of H ⃗ \vec H H

Summary

structure that guides EM waves in a chosen direction and confines it in the plane transverse to the propagation direction

more in 3.1 Pozar D M. Microwave engineering[M]. John wiley & sons, 2009.

Relation between E ⃗ \vec E E and H ⃗ \vec H H fields

Note that ∇ × E ⃗ = − j ω μ H ⃗ \nabla \times \vec E=-j\omega\mu \vec H ∇×E =−jωμH , E ⃗ ( r ⃗ ) = e ⃗ t ( x , y ) e − j k z \vec E(\vec r)=\vec e_t(x,y)e^{-jkz} E (r )=e t​(x,y)e−jkz,
H ⃗ = − 1 j ω μ ∇ × E ⃗ = − 1 j ω μ ∇ × ( e ⃗ t ( x , y ) e − j k z ) . \vec H=-\frac{1}{j\omega \mu}\nabla \times \vec E=-\frac{1}{j\omega \mu}\nabla \times (\vec e_t(x,y)e^{-jkz}). H =−jωμ1​∇×E =−jωμ1​∇×(e t​(x,y)e−jkz).
Using the vector identity: ∇ × ( f A ⃗ ) = ∇ f × A ⃗ + f ∇ × A ⃗ \nabla \times (f\vec A)=\nabla f\times \vec A+f\nabla\times \vec A ∇×(fA )=∇f×A +f∇×A ,
H ⃗ = − 1 j ω μ ( ∇ e − j k z × e ⃗ t ( x , y ) + e − j k z ∇ × e ⃗ t ( x , y ) ) = a − 1 j ω μ ∇ e − j k z × e ⃗ t ( x , y ) = k ω μ e − j k z z ^ × e ⃗ t ( x , y ) \begin{aligned} \vec H&=-\frac{1}{j\omega\mu}(\nabla e^{-jkz}\times \vec e_t(x,y)+e^{-jkz}\nabla\times \vec e_t(x,y))\\&\stackrel{a}{=}-\frac{1}{j\omega\mu}\nabla e^{-jkz}\times \vec e_t(x,y)=\frac{k}{\omega\mu}e^{-jkz}\hat z\times \vec e_t(x,y) \end{aligned} H ​=−jωμ1​(∇e−jkz×e t​(x,y)+e−jkz∇×e t​(x,y))=a−jωμ1​∇e−jkz×e t​(x,y)=ωμk​e−jkzz^×e t​(x,y)​
where = a \stackrel{a}{=} =a is because that e ^ t ( x , y ) \hat e_t(x,y) e^t​(x,y) is analogous to the static field.

Since k = ω ε μ k=\omega\sqrt{\varepsilon\mu} k=ωεμ ​,
k ω μ = ω ε μ ω μ = ε μ = 1 ζ \frac{k}{\omega \mu}=\frac{\omega\sqrt{\varepsilon \mu}}{\omega \mu}=\sqrt{\frac{\varepsilon}{\mu}}=\frac{1}{\zeta} ωμk​=ωμωεμ ​​=με​ ​=ζ1​
Define the medium impedance:
ζ = μ ε \zeta=\sqrt{\frac{\mu}{\varepsilon}} ζ=εμ​ ​
Thus,
H ⃗ = 1 ζ z ^ × e ⃗ t ( x , y ) e − j k z = 1 ζ z ^ × E ⃗ \vec H =\frac{1}{\zeta} \hat z\times \vec e_t(x,y)e^{-jkz}=\frac{1}{\zeta} \hat z\times \vec E H =ζ1​z^×e t​(x,y)e−jkz=ζ1​z^×E
and wave impedance
Z T E M ≜ ∣ E ∣ ∣ H ∣ = ζ Z_{TEM}\triangleq \frac{|E|}{|H|}=\zeta ZTEM​≜∣H∣∣E∣​=ζ
is equal to the intrinsic medium impedance.

Procedure for analyzing a TEM line

Parallel plate waveguide (PPW)

We assume that W ≫ d W \gg d W≫d. With this argument we can neglect the fringing of the field around the edges.

Step 1: Solving Laplace eq. for Φ ( x , y ) \Phi(x,y) Φ(x,y)

Laplace’s eq. for the electrostatic potential
∇ t 2 Φ ( x , y ) = 0 for  0 ≤ x ≤ W , 0 ≤ y ≤ d \nabla_t^2 \Phi (x,y)=0 \qquad \text{for } 0\le x\le W , 0\le y\le d ∇t2​Φ(x,y)=0for 0≤x≤W,0≤y≤d
Boundary conditions: we assume that
Φ ( x , 0 ) = 0 , Φ ( x , d ) = V 0 \Phi(x,0)=0,\quad \Phi(x,d)=V_0 Φ(x,0)=0,Φ(x,d)=V0​
Since the potential is not varying in x x x, the general solution is of the form
Φ ( x , y ) = A + B y \Phi(x,y)=A+By Φ(x,y)=A+By
From the boundary conditions we find
Φ ( x , y ) = V 0 d y \Phi(x,y)=\frac{V_0}{d}y Φ(x,y)=dV0​​y
Step 2: Transverse fields
e ⃗ t ( x , y ) = − ∇ t Φ ( x , y ) = − ∇ t V 0 d y = − ∂ x V 0 d y x ^ − ∂ y V 0 d y y ^ = − V 0 d y ^ E ⃗ ( r ⃗ ) = e ⃗ t ( x , y ) e − j k z = − V 0 d e − j k z y ^ h ⃗ t ( x , y ) = 1 ζ z ^ × e ⃗ t ( x , y ) = 1 ζ z ^ × ( − V 0 d y ^ ) = 1 ζ V 0 d x ^ H ⃗ ( r ⃗ ) = h ⃗ t ( x , y ) e − j k z = V 0 ζ d e − j k z x ^ \begin{aligned} &\vec{e}_{t}(x, y)=-\nabla_{t} \Phi(x, y)=-\nabla_{t} \frac{V_{0}}{d} y=-\partial_{x} \frac{V_{0}}{d} y \hat{x}-\partial_{y} \frac{V_{0}}{d} y \hat{y}=-\frac{V_{0}}{d} \hat{y}\\ &\vec{E}(\vec{r})=\vec{e}_{t}(x, y) e^{-j k z}=-\frac{V_{0}}{d} e^{-j k z} \hat{y}\\ &\vec{h}_{t}(x, y)=\frac{1}{\zeta} \hat{z} \times \vec{e}_{t}(x, y)=\frac{1}{\zeta} \hat{z} \times\left(-\frac{V_{0}}{d} \hat{y}\right)=\frac{1}{\zeta} \frac{V_{0}}{d} \hat{x}\\ &\vec{H}(\vec{r})=\vec{h}_{t}(x, y) e^{-j k z}=\frac{V_{0}}{\zeta d} e^{-j k z} \hat{x} \end{aligned} ​e t​(x,y)=−∇t​Φ(x,y)=−∇t​dV0​​y=−∂x​dV0​​yx^−∂y​dV0​​yy^​=−dV0​​y^​E (r )=e t​(x,y)e−jkz=−dV0​​e−jkzy^​h t​(x,y)=ζ1​z^×e t​(x,y)=ζ1​z^×(−dV0​​y^​)=ζ1​dV0​​x^H (r )=h t​(x,y)e−jkz=ζdV0​​e−jkzx^​

Step 3: Voltage and Current

The voltage can be calculated from the electric field as
V = − ∫ 0 d E ⃗ ( r ⃗ ) ⋅ y ^ d y = ∫ 0 d V 0 d e − j k z y ^ ⋅ y ^ d y = V 0 e − j k z = V ( z ) V=-\int_0^d \vec E(\vec r)\cdot \hat y dy=\int_0^d \frac{V_0}{d}e^{-jkz}\hat y\cdot \hat y dy=V_0e^{-jkz}=V(z) V=−∫0d​E (r )⋅y^​dy=∫0d​dV0​​e−jkzy^​⋅y^​dy=V0​e−jkz=V(z)
The current can be calculated from the surface current density J ⃗ s \vec J_s J s​

From the boundary conditions of perfect conductor, we have
J ⃗ s = n ^ × H ⃗ , n ^ = − y ^ . \vec J_s=\hat n \times \vec H,\quad \hat n=-\hat y. J s​=n^×H ,n^=−y^​.
Thus
I = ∫ 0 W J ⃗ s ⋅ z ^ d x = ∫ 0 W − y ^ × H ⃗ ( r ⃗ ) ⋅ z ^ d x = ∫ 0 W − y ^ × ( V 0 ζ d e − j k z x ^ ) ⋅ z ^ d x = W ζ d V 0 e − j k z = I ( z ) I=\int_{0}^{W} \vec{J}_{s} \cdot \hat{z} d x=\int_{0}^{W}-\hat{y} \times \vec{H}(\vec{r}) \cdot \hat{z} d x=\int_{0}^{W}-\hat{y} \times\left(\frac{V_{0}}{\zeta d} e^{-j k z} \hat{x}\right) \cdot \hat{z} d x=\frac{W}{\zeta d} V_{0} e^{-j k z}=I(z)\\ I=∫0W​J s​⋅z^dx=∫0W​−y^​×H (r )⋅z^dx=∫0W​−y^​×(ζdV0​​e−jkzx^)⋅z^dx=ζdW​V0​e−jkz=I(z)
Step 4: Transmission line parameters

The characteristic impedance of a line is the ratio between voltage and current:
Z 0 = V ( z ) I ( z ) = d W ζ , Z_0=\frac{V(z)}{I(z)}=\frac{d}{W}\zeta, Z0​=I(z)V(z)​=Wd​ζ,
which is a constant depends on geometry and the material parameters. (while wave impedance Z T E M = ∣ E ∣ ∣ H ∣ = ζ = μ ε Z_{TEM}=\frac{|E|}{|H|}=\zeta=\sqrt{\frac{\mu}{\varepsilon}} ZTEM​=∣H∣∣E∣​=ζ=εμ​ ​ only depends on material)

The propagation constant
k = ω ε μ k=\omega\sqrt{\varepsilon\mu} k=ωεμ ​
The phase velocity
c = ω k = 1 ε μ c=\frac{\omega}{k}=\frac{1}{\sqrt{\varepsilon \mu}} c=kω​=εμ ​1​

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Power transported in PPW

An electromagnetic wave carries power, expressed as
s ⃗ ( r ⃗ , t ) = e ⃗ ( r ⃗ , t ) × h ⃗ ( r ⃗ , t ) ( W / m 2 ) \vec s(\vec r,t)=\vec e(\vec r,t)\times \vec h(\vec r,t)\quad (W/m^2) s (r ,t)=e (r ,t)×h (r ,t)(W/m2)
directed along the propagation direction ( z ^ ) (\hat z) (z^)

Phasor domain: S ⃗ a v = 1 2 ℜ { E ⃗ × H ⃗ ∗ } ( W / m 2 ) \vec S_{av}=\frac{1}{2}\Re\{\vec E \times \vec H^*\}\quad (W/m^2) S av​=21​ℜ{E ×H ∗}(W/m2)

Then
S ⃗ a v ( r ⃗ ) = 1 2 ℜ { − V + d e − j k z y ^ × ( I + W e − j k z x ^ ) ∗ } = 1 2 ℜ { − V + d e − j k z y ^ × I + ∗ W e j k z x ^ } = 1 2 1 d W ℜ { V + I + ∗ } z ^ P a v ( z ) = ∫ A S ⃗ a v ( z ) ⋅ n ^ d A = ∫ A 1 2 1 d W ℜ { V + I + ∗ } z ^ ⋅ z ^ d A = 1 2 ℜ { V + I + ∗ } = 1 2 Z 0 ∣ I + ∣ 2 = 1 2 ∣ V + ∣ 2 Z 0 \begin{aligned} \vec{S}_{a v}(\vec{r})&=\frac{1}{2} \Re\left\{-\frac{V^{+}}{d} e^{-j k z} \hat{y} \times\left(\frac{I^{+}}{W} e^{-j k z} \hat{x}\right)^{*}\right\}\\ &=\frac{1}{2} \Re\left\{-\frac{V^{+}}{d} e^{-j k z} \hat{y} \times \frac{I^{+*}}{W} e^{j k z} \hat{x}\right\}=\frac{1}{2} \frac{1}{d W} \Re\left\{V^{+} I^{+*}\right\} \hat{z} \\ P_{a v}(z)&=\int_{A} \vec{S}_{a v}(z) \cdot \hat{n} d A=\int_{A} \frac{1}{2} \frac{1}{d W} \Re\left\{V^{+} I^{+*}\right\} \hat{z} \cdot \hat{z} d A\\ &=\frac{1}{2} \Re\left\{V^{+} I^{+*}\right\} =\frac{1}{2}Z_0|I^+|^2=\frac{1}{2}\frac{|V^+|^2}{Z_0} \end{aligned} S av​(r )Pav​(z)​=21​ℜ{−dV+​e−jkzy^​×(WI+​e−jkzx^)∗}=21​ℜ{−dV+​e−jkzy^​×WI+∗​ejkzx^}=21​dW1​ℜ{V+I+∗}z^=∫A​S av​(z)⋅n^dA=∫A​21​dW1​ℜ{V+I+∗}z^⋅z^dA=21​ℜ{V+I+∗}=21​Z0​∣I+∣2=21​Z0​∣V+∣2​​
In TEM transmission lines, we can calculate power flow from V , I V,I V,I

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Transmission line circuit representation

More typical derivation using the lumped-element circuit model

From Pozar. 2.1

A transmission line is a distributed parameter network, where voltages and currents can vary in magnitude and phase over its length, while ordinary circuit analysis deals with lumped elements, where voltage and current do not vary appreciably over the physical dimension of the elements.

The piece of line of infinitesimal length Δ z \Delta z Δz can be modeled as a lumped-element circuit, where

R = R= R= series resistance per unit length, for both conductors, in Ω / m \Omega / \mathrm{m} Ω/m
L = L= L= series inductance per unit length, for both conductors, in H / m . \mathrm{H} / \mathrm{m} . H/m.
G = G= G= shunt conductance per unit length, in S / m \mathrm{S} / \mathrm{m} S/m
C = C= C= shunt capacitance per unit length, in F / m . \mathrm{F} / \mathrm{m} . F/m.

The series inductance L L L represents the total self-inductance of the two conductors, and the shunt capacitance C C C is due to the close proximity of the two conductors. The series resistance R R R represents the resistance due to the finite conductivity of the individual conductors, and the shunt conductance G G G is due to dielectric loss in the material between the conductors. R R R and G , G, G, therefore, represent loss.

From the circuit of the figure above, Kirchhoff’s voltage law can be applied to give
v ( z , t ) − R Δ z i ( z , t ) − L Δ z ∂ i ( z , t ) ∂ t − v ( z + Δ z , t ) = 0 (CM.1a) v(z, t)-R \Delta z i(z, t)-L \Delta z \frac{\partial i(z, t)}{\partial t}-v(z+\Delta z, t)=0 \tag{CM.1a} v(z,t)−RΔzi(z,t)−LΔz∂t∂i(z,t)​−v(z+Δz,t)=0(CM.1a)
and Kirchhoff’s current law leads to
i ( z , t ) − G Δ z v ( z + Δ z , t ) − C Δ z ∂ v ( z + Δ z , t ) ∂ t − i ( z + Δ z , t ) = 0 (CM.1b) i(z, t)-G \Delta z v(z+\Delta z, t)-C \Delta z \frac{\partial v(z+\Delta z, t)}{\partial t}-i(z+\Delta z, t)=0 \tag{CM.1b} i(z,t)−GΔzv(z+Δz,t)−CΔz∂t∂v(z+Δz,t)​−i(z+Δz,t)=0(CM.1b)
Dividing both equations by Δ z \Delta z Δz and taking the limit as Δ z → 0 \Delta z \rightarrow 0 Δz→0 gives the following differential equations:
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These are the time domain form of the transmission line equations, also known as the telegrapher equations. For the sinusoidal steady-state condition, with cosine-based phasors they can be simplified to
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The two equations can be solved simultaneously to give wave equations for V ( z ) V(z) V(z) and I ( z ) I(z) I(z): (Helmholtz equation)
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where
γ = α + j β = ( R + j ω L ) ( G + j ω C ) (CM.5) \gamma=\alpha+j \beta=\sqrt{(R+j \omega L)(G+j \omega C)} \tag{CM.5} γ=α+jβ=(R+jωL)(G+jωC) ​(CM.5)
is the complex propagation constant, which is a function of frequency. Traveling wave solutions can be found as
V ( z ) = V o + e − γ z + V o − e γ z I ( z ) = I o + e − γ z + I o − e γ z (CM.6) \begin{aligned} V(z) &=V_{o}^{+} e^{-\gamma z}+V_{o}^{-} e^{\gamma z}\\ I(z) &=I_{o}^{+} e^{-\gamma z}+I_{o}^{-} e^{\gamma z} \tag{CM.6} \end{aligned} V(z)I(z)​=Vo+​e−γz+Vo−​eγz=Io+​e−γz+Io−​eγz​(CM.6)
where the e − γ z e^{-\gamma z} e−γz term represents wave propagation in the + z +z +z direction, and the e γ z e^{\gamma z} eγz term represents wave propagation in the − z -z −z direction. Applying ( C M . 3 a ) (\mathrm{CM}.3 a) (CM.3a) to the voltage of ( C M . 6 a ) (\mathrm{CM}.6 a) (CM.6a) gives the current on the line:
I ( z ) = γ R + j ω L ( V o + e − γ z − V o − e γ z ) I(z)=\frac{\gamma}{R+j \omega L}\left(V_{o}^{+} e^{-\gamma z}-V_{o}^{-} e^{\gamma z}\right) I(z)=R+jωLγ​(Vo+​e−γz−Vo−​eγz)
Comparison with ( C M . 6 b ) (\mathrm{CM}.6 b) (CM.6b) shows that a characteristic impedance, Z 0 , Z_{0}, Z0​, can be defined as
Z 0 = R + j ω L γ = R + j ω L G + j ω C (CM.7) Z_{0}=\frac{R+j \omega L}{\gamma}=\sqrt{\frac{R+j \omega L}{G+j \omega C}}\tag{CM.7} Z0​=γR+jωL​=G+jωCR+jωL​ ​(CM.7)
to relate the voltage and current on the line as follows:
V o + I o + = Z 0 = − V o − I o − \frac{V_{o}^{+}}{I_{o}^{+}}=Z_{0}=\frac{-V_{o}^{-}}{I_{o}^{-}} Io+​Vo+​​=Z0​=Io−​−Vo−​​
Then ( 2.6 b ) (2.6 \mathrm{b}) (2.6b) can be rewritten in the following form:
I ( z ) = V o + Z 0 e − γ z − V o − Z 0 e γ z (CM.8) I(z)=\frac{V_{o}^{+}}{Z_{0}} e^{-\gamma z}-\frac{V_{o}^{-}}{Z_{0}} e^{\gamma z} \tag{CM.8} I(z)=Z0​Vo+​​e−γz−Z0​Vo−​​eγz(CM.8)
Converting back to the time domain, we can express the voltage waveform as
v ( z , t ) = ∣ V o + ∣ cos ⁡ ( ω t − β z + ϕ + ) e − α z + ∣ V o − ∣ cos ⁡ ( ω t + β z + ϕ − ) e α z (CM.9) \begin{aligned} v(z, t)=&\left|V_{o}^{+}\right| \cos \left(\omega t-\beta z+\phi^{+}\right) e^{-\alpha z} \\ &+\left|V_{o}^{-}\right| \cos \left(\omega t+\beta z+\phi^{-}\right) e^{\alpha z}\tag{CM.9} \end{aligned} v(z,t)=​∣∣​Vo+​∣∣​cos(ωt−βz+ϕ+)e−αz+∣∣​Vo−​∣∣​cos(ωt+βz+ϕ−)eαz​(CM.9)
where ϕ ± \phi^{\pm} ϕ± is the phase angle of the complex voltage V o ± V_{o}^{\pm} Vo±​. Using arguments similar to those in Section 1.4 , 1.4, 1.4, we find that the wavelength on the line is
λ = 2 π β (CM.10) \lambda=\frac{2 \pi}{\beta} \tag{CM.10} λ=β2π​(CM.10)
and the phase velocity is
v p = ω β = λ f . (CM.11) v_p=\frac{\omega}{\beta}=\lambda f.\tag{CM.11} vp​=βω​=λf.(CM.11)
The lossless line

The above solution is for a general transmission line, including loss effects, and it was seen that the propagation constant and characteristic impedance were complex. In many practical cases, however, the loss of the line is very small and so can be neglected, resulting in a simplification of the results. Setting R = G = 0 R=G=0 R=G=0 gives the propagation constant as
γ = α + j β = j ω L C \gamma=\alpha+j \beta=j \omega \sqrt{L C} γ=α+jβ=jωLC ​
or
β = ω L C α = 0 (CM.12) \begin{array}{l} \beta=\omega \sqrt{L C} \\\tag{CM.12} \alpha=0 \end{array} β=ωLC ​α=0​(CM.12)
As expected for a lossless line, the attenuation constant α \alpha α is zero. The characteristic impedance reduces to
Z 0 = L C (CM.13) Z_{0}=\sqrt{\frac{L}{C}} \tag{CM.13} Z0​=CL​ ​(CM.13)
which is now a real number. The general solutions for voltage and current on a lossless transmission line can then be written as
V ( z ) = V o + e − j β z + V o − e j β z I ( z ) = V o + Z 0 e − j β z − V o − Z 0 e j β z (CM.14) \begin{array}{l} V(z)=V_{o}^{+} e^{-j \beta z}+V_{o}^{-} e^{j \beta z} \\ I(z)=\frac{V_{o}^{+}}{Z_{0}} e^{-j \beta z}-\frac{V_{o}^{-}}{Z_{0}} e^{j \beta z} \tag{CM.14} \end{array} V(z)=Vo+​e−jβz+Vo−​ejβzI(z)=Z0​Vo+​​e−jβz−Z0​Vo−​​ejβz​(CM.14)
The wavelength is
λ = 2 π β = 2 π ω L C (CM.15) \lambda=\frac{2 \pi}{\beta}=\frac{2 \pi}{\omega \sqrt{L C}}\tag{CM.15} λ=β2π​=ωLC ​2π​(CM.15)
and the phase velocity is
v p = ω β = 1 L C (CM.16) v_{p}=\frac{\omega}{\beta}=\frac{1}{\sqrt{L C}}\tag{CM.16} vp​=βω​=LC ​1​(CM.16)

Summary: Transmission line parameters

Terminated lossless line

Voltage reflection coefficient

An incident wave is generated from a source:
V + e − j k 0 z V^+e^{-jk_0z} V+e−jk0​z
A reflected wave will be generated at the edge:
V − e j k 0 z V^-e^{jk_0z} V−ejk0​z
( V + V^+ V+ and V − V^- V− can be complex number)

Voltage reflection coefficient:
Γ V = V − V + \Gamma_V=\frac{V^-}{V^+} ΓV​=V+V−​
Total voltage and current at a generic point z z z:
V ( z ) = V + e − j k 0 z + V − e j k 0 z I ( z ) = 1 Z 0 ( V + e − j k 0 z − V − e j k 0 z ) = V + Z 0 ( e − j k z 0 − Γ V e j k z 0 ) \begin{aligned} V(z)&=V^+e^{-jk_0z}+V^-e^{jk_0z}\\ I(z)&=\frac{1}{Z_0}(V^+e^{-jk_0z}-V^-e^{jk_0z})=\frac{V^+}{Z_0}(e^{-jkz_0}-\Gamma_V e^{jkz_0}) \end{aligned} V(z)I(z)​=V+e−jk0​z+V−ejk0​z=Z0​1​(V+e−jk0​z−V−ejk0​z)=Z0​V+​(e−jkz0​−ΓV​ejkz0​)​
(because Z 0 = V + / I + = − V − / I − Z_0=V^+/I^+=-V^-/I^- Z0​=V+/I+=−V−/I−)

We assume V l = V ( 0 ) V_l=V(0) Vl​=V(0), the current and voltage at the load are
V ( 0 ) = V l = V + + V − I ( 0 ) = I l = 1 Z 0 ( V + − V − ) \begin{aligned} V(0)&=V_l=V^++V^-\\ I(0)&=I_l=\frac{1}{Z_0}(V^+-V^-) \end{aligned} V(0)I(0)​=Vl​=V++V−=Il​=Z0​1​(V+−V−)​
Since V l = Z l I l V_l=Z_lI_l Vl​=Zl​Il​, we have
V + + V − = Z l Z 0 ( V + − V − ) ⇒ Γ V = V − V + = Z l − Z 0 Z l + Z 0 V^++V^-=\frac{Z_l}{Z_0}(V^+-V^-)\Rightarrow \Gamma _V=\frac{V^-}{V^+}=\frac{Z_l-Z_0}{Z_l+Z_0} V++V−=Z0​Zl​​(V+−V−)⇒ΓV​=V+V−​=Zl​+Z0​Zl​−Z0​​
Similarly we can derive current reflection coefficient:
Γ I = I − I + = − V − V + = Z 0 − Z l Z l + Z 0 = − Γ V \Gamma_I=\frac{I^-}{I^+}=-\frac{V^-}{V^+}=\frac{Z_0-Z_l}{Z_l+Z_0}=-\Gamma_V ΓI​=I+I−​=−V+V−​=Zl​+Z0​Z0​−Zl​​=−ΓV​
They are equivalent.
From now we will use only voltage reflection coefficient, we indicate with Γ \Gamma Γ.

matching: Z l = Z 0 , Γ = 0 ⟹  No reflected wave, perfect matching Z_l=Z_0,\Gamma=0\Longrightarrow\text{ No reflected wave, perfect matching} Zl​=Z0​,Γ=0⟹ No reflected wave, perfect matching

standing waves: ∣ Γ ∣ = 1 |\Gamma|=1 ∣Γ∣=1

If they don’t have the same amplitude, oscillate between a max and a min

Voltage standing wave ratio (VSWR):
V S W R = V m a x V m i n = 1 + ∣ Γ ∣ 1 − ∣ Γ ∣ \mathrm{VSWR}=\frac{V_{max}}{V_{min}}=\frac{1+|\Gamma|}{1-|\Gamma|} VSWR=Vmin​Vmax​​=1−∣Γ∣1+∣Γ∣​

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Power transfer in a transmission line

Remember that the characteristic impedance Z 0 Z_0 Z0​ reduces to a real number for a lossless line, the instantaneous power carried by the incident wave is
P i ( t ) = v i ( t ) ⋅ i i ( t ) = R e { V + e j ω t } ⋅ R e { I + e j ω t } = R e { ∣ V + ∣ e j ϕ + e j ω t } ⋅ R e { ∣ V + ∣ Z 0 e j ϕ + e j ω t } = ∣ V + ∣ 2 Z 0 cos ⁡ 2 ( ω t + ϕ + ) \begin{aligned} P_{i}(t)&=v_{i}(t) \cdot i_{i}(t)=\mathcal{R} e\left\{V^{+} e^{j \omega t}\right\} \cdot \mathcal{R} e\left\{I^{+} e^{j \omega t}\right\}\\ &=\mathcal{R} e\left\{\left|V^{+}\right| e^{j \phi^{+}} e^{j \omega t}\right\} \cdot \mathcal{R} e\left\{\frac{\left|V^{+}\right|}{Z_{0}} e^{j \phi^{+}} e^{j \omega t}\right\} =\frac{\left|V^{+}\right|^{2}}{Z_{0}} \cos ^{2}\left(\omega t+\phi^{+}\right) \end{aligned} Pi​(t)​=vi​(t)⋅ii​(t)=Re{V+ejωt}⋅Re{I+ejωt}=Re{∣∣​V+∣∣​ejϕ+ejωt}⋅Re{Z0​∣V+∣​ejϕ+ejωt}=Z0​∣V+∣2​cos2(ωt+ϕ+)​
Taking time average, we have
P i , a v = 1 2 ∣ V + ∣ 2 Z 0 Incident power P r , a v = ∣ Γ ∣ 2 2 ∣ V + ∣ 2 Z 0 Reflected power P a v = P i , a v − P r , a v = ∣ V l ∣ 2 2 R e { 1 Z l } Power delivered to the load  \begin{aligned} P_{i, a v}=&\frac{1}{2} \frac{\left|V^{+}\right|^{2}}{Z_{0}} &&\text {Incident power} \\ P_{r, a v}=&\frac{|\Gamma|^{2}}{2} \frac{\left|V^{+}\right|^{2}}{Z_{0}} &&\text {Reflected power}\\ P_{av}=&P_{i, a v}-P_{r, a v}=\frac{\left|V_l\right|^{2}}{2} \mathcal{R} e\left\{\frac{1}{Z_l}\right\} &&\text {Power delivered to the load } \end{aligned} Pi,av​=Pr,av​=Pav​=​21​Z0​∣V+∣2​2∣Γ∣2​Z0​∣V+∣2​Pi,av​−Pr,av​=2∣Vl​∣2​Re{Zl​1​}​​Incident powerReflected powerPower delivered to the load ​

Often used in decibels: Γ d B = 10 log ⁡ 10 ∣ Γ ∣ 2 \Gamma_{\mathrm{dB}}=10\log_{10}|\Gamma|^2 ΓdB​=10log10​∣Γ∣2.

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Reflection Coefficient: Special cases

Γ = Z l − Z 0 Z l + Z 0 \Gamma=\frac{Z_l-Z_0}{Z_l+Z_0} Γ=Zl​+Z0​Zl​−Z0​​

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Short circuit line
Γ = Z l − Z 0 Z l + Z 0 = − 1 \Gamma=\frac{Z_l-Z_0}{Z_l+Z_0}=-1 Γ=Zl​+Z0​Zl​−Z0​​=−1

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Open circuit line
Γ = Z l − Z 0 Z l + Z 0 = 1 \Gamma=\frac{Z_l-Z_0}{Z_l+Z_0}=1 Γ=Zl​+Z0​Zl​−Z0​​=1

v ( z , t ) = 2 cos ⁡ ( k 0 z ) cos ⁡ ( ω t ) i ( z , t ) = 2 sin ⁡ ( k 0 z ) sin ⁡ ( ω t ) \begin{aligned} v(z,t)&=2\cos (k_0z)\cos (\omega t)\\ i(z,t)&=2\sin (k_0z)\sin (\omega t) \end{aligned} v(z,t)i(z,t)​=2cos(k0​z)cos(ωt)=2sin(k0​z)sin(ωt)​

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Matched line
Γ = Z l − Z 0 Z l + Z 0 = 0 \Gamma=\frac{Z_l-Z_0}{Z_l+Z_0}=0 Γ=Zl​+Z0​Zl​−Z0​​=0

Summary

Input Impedance

If we are interested in knowing the voltage at the generator:

Impedance transfer: Impedance seen looking toward the load
Z i n ( − l ) = V ( − l ) I ( − l ) = Z 0 Z l + j Z 0 tan ⁡ k 0 l Z 0 + j Z l tan ⁡ k 0 l Z_{in}(-l)=\frac{V(-l)}{I(-l)}=Z_0\frac{Z_l+jZ_0\tan k_0l}{Z_0+jZ_l\tan k_0l} Zin​(−l)=I(−l)V(−l)​=Z0​Z0​+jZl​tank0​lZl​+jZ0​tank0​l​
Proof:

Z ( − l ) = V ( − l ) I ( − l ) = V + ( e j k 0 l + Z l − Z 0 Z l + Z 0 e − j k 0 l ) V + Z 0 ( e j k 0 l − Z l − Z 0 Z l + Z 0 e − j k 0 l ) = Z 0 ( Z l + Z 0 ) e j k 0 l + ( Z l − Z 0 ) e − j k 0 l ( Z l + Z 0 ) e j k 0 l − ( Z l − Z 0 ) e − j k 0 l = Z 0 Z l ( e j k 0 l + e − j k 0 l ) + Z 0 ( e j k 0 l − e − j k 0 l ) Z 0 ( e j k 0 l + e − j k 0 l ) + Z l ( e j k 0 l − e − j k 0 l ) = Z 0 Z l cos ⁡ k 0 l + j Z 0 sin ⁡ k 0 l Z 0 cos ⁡ k 0 l + j Z l sin ⁡ k 0 l = Z 0 Z l + j Z 0 tan ⁡ k 0 l Z 0 + j Z l tan ⁡ k 0 l \begin{aligned} Z(-l)=& \frac{V(-l)}{I(-l)}=\frac{V^{+}\left(e^{j k_{0} l}+\frac{Z_{l}-Z_{0}}{Z_{l}+Z_{0}} e^{-j k_{0} l}\right)}{\frac{V^{+}}{Z_{0}}\left(e^{j k_{0} l}-\frac{Z_{l}-Z_{0}}{Z_{l}+Z_{0}} e^{-j k_{0} l}\right)}=Z_{0} \frac{\left(Z_{l}+Z_{0}\right) e^{j k_{0} l}+\left(Z_{l}-Z_{0}\right) e^{-j k_{0} l}}{\left(Z_{l}+Z_{0}\right) e^{j k_{0} l}-\left(Z_{l}-Z_{0}\right) e^{-j k_{0} l}} \\ =&Z_{0} \frac{Z_{l}\left(e^{j k_{0} l}+e^{-j k_{0} l}\right)+Z_{0}\left(e^{j k_{0} l}-e^{-j k_{0} l}\right)}{Z_{0}\left(e^{j k_{0} l}+e^{-j k_{0} l}\right)+Z_{l}\left(e^{j k_{0} l}-e^{-j k_{0} l}\right)}=Z_{0} \frac{Z_{l} \cos k_{0} l+j Z_{0} \sin k_{0} l}{Z_{0} \cos k_{0} l+j Z_{l} \sin k_{0} l} \\ =&Z_{0} \frac{Z_{l}+j Z_{0} \tan k_{0} l}{Z_{0}+j Z_{l} \tan k_{0} l} \end{aligned} Z(−l)===​I(−l)V(−l)​=Z0​V+​(ejk0​l−Zl​+Z0​Zl​−Z0​​e−jk0​l)V+(ejk0​l+Zl​+Z0​Zl​−Z0​​e−jk0​l)​=Z0​(Zl​+Z0​)ejk0​l−(Zl​−Z0​)e−jk0​l(Zl​+Z0​)ejk0​l+(Zl​−Z0​)e−jk0​l​Z0​Z0​(ejk0​l+e−jk0​l)+Zl​(ejk0​l−e−jk0​l)Zl​(ejk0​l+e−jk0​l)+Z0​(ejk0​l−e−jk0​l)​=Z0​Z0​cosk0​l+jZl​sink0​lZl​cosk0​l+jZ0​sink0​l​Z0​Z0​+jZl​tank0​lZl​+jZ0​tank0​l​​

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Input impedance: Special cases

Z i n ( − l ) = Z 0 Z l + j Z 0 tan ⁡ k 0 l Z 0 + j Z l tan ⁡ k 0 l Z_{in}(-l)=Z_0\frac{Z_l+jZ_0\tan k_0l}{Z_0+jZ_l\tan k_0l} Zin​(−l)=Z0​Z0​+jZl​tank0​lZl​+jZ0​tank0​l​

short circuit
Z l = 0 , Z i n = j Z 0 tan ⁡ k 0 l Z_l=0, Z_{in}=jZ_0\tan k_0l Zl​=0,Zin​=jZ0​tank0​l

open circuit
Z l = ∞ , Z i n = − j Z 0 cot ⁡ k 0 l Z_l=\infty,Z_{in}=-jZ_0\cot k_0l Zl​=∞,Zin​=−jZ0​cotk0​l

matched load: input impedance is constant and independent on the length
Z l = Z 0 = Z i n Z_l=Z_0=Z_{in} Zl​=Z0​=Zin​

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Different impedances

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Power delivered to the load

For lossless lines, power delivered to the load is
P l = ∣ I i n ∣ 2 2 Re ⁡ { Z i n } = ∣ V i n ∣ 2 2 Re ⁡ { 1 Z i n } P_{l}=\frac{\left|I_{i n}\right|^{2}}{2} \operatorname{Re}\left\{Z_{i n}\right\}=\frac{\left|V_{i n}\right|^{2}}{2} \operatorname{Re}\left\{\frac{1}{Z_{i n}}\right\} Pl​=2∣Iin​∣2​Re{Zin​}=2∣Vin​∣2​Re{Zin​1​}
This method is equivalent to the method in the previous section [Power transfer in a transmission line](# Power transfer in a transmission line).

Example:

The power delivered to the load can be calculated both using
P l = ∣ I i n ∣ 2 2 Re ⁡ { Z i n } = ∣ V i n ∣ 2 2 Re ⁡ { 1 Z i n } P_{l}=\frac{\left|I_{i n}\right|^{2}}{2} \operatorname{Re}\left\{Z_{i n}\right\}=\frac{\left|V_{i n}\right|^{2}}{2} \operatorname{Re}\left\{\frac{1}{Z_{i n}}\right\} Pl​=2∣Iin​∣2​Re{Zin​}=2∣Vin​∣2​Re{Zin​1​}
or
P l = ∣ I l ∣ 2 2 Re ⁡ { Z l } = ∣ V l ∣ 2 2 Re ⁡ { 1 Z l } P_{l}=\frac{\left|I_{l}\right|^{2}}{2} \operatorname{Re}\left\{Z_{l}\right\}=\frac{\left|V_{l}\right|^{2}}{2} \operatorname{Re}\left\{\frac{1}{Z_{l}}\right\} Pl​=2∣Il​∣2​Re{Zl​}=2∣Vl​∣2​Re{Zl​1​}

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We can first compute Z i n Z_{in} Zin​ by
k 0 l = 2 π λ λ λ 8 = π 4 tan ⁡ ( k 0 l ) = tan ⁡ ( π 4 ) = 1 Z i n ( − l ) = Z 0 Z l + j Z 0 tan ⁡ k 0 l Z 0 + j Z l tan ⁡ k 0 l = 40 − 30 j   Ω \begin{aligned} k_{0} l=\frac{2 \pi \lambda}{\lambda} \frac{\lambda}{8}=\frac{\pi}{4} \\ \tan \left(k_{0} l\right)=\tan \left(\frac{\pi}{4}\right)=1 \end{aligned}\\ Z_{in}(-l)=Z_0\frac{Z_l+jZ_0\tan k_0l}{Z_0+jZ_l\tan k_0l}=40-30j~\Omega k0​l=λ2πλ​8λ​=4π​tan(k0​l)=tan(4π​)=1​Zin​(−l)=Z0​Z0​+jZl​tank0​lZl​+jZ0​tank0​l​=40−30j Ω

Therefore,
V i n = V g Z i n Z g + Z i n = 7.33 − 1.33 j   Ω P l = ∣ V i n ∣ 2 2 Re ⁡ { 1 Z i n } = 0.444   W V_{in}=V_g\frac{Z_{in}}{Z_g+Z_{in}}=7.33-1.33j~\Omega\\ P_{l}=\frac{\left|V_{i n}\right|^{2}}{2} \operatorname{Re}\left\{\frac{1}{Z_{i n}}\right\}=0.444~W Vin​=Vg​Zg​+Zin​Zin​​=7.33−1.33j ΩPl​=2∣Vin​∣2​Re{Zin​1​}=0.444 W

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Or we can first compute Γ \Gamma Γ:
Γ = Z l − Z 0 Z l + Z 0 = 1 3 \Gamma=\frac{Z_l-Z_0}{Z_l+Z_0}=\frac{1}{3} Γ=Zl​+Z0​Zl​−Z0​​=31​
Then according to the definition,
V ( z ) = V + ( e − j k 0 z + 1 3 e j k o z ) V(z)=V^+(e^{-jk_0z}+\frac{1}{3}e^{jk_oz}) V(z)=V+(e−jk0​z+31​ejko​z)
By imposing the boundary conditions V ( z = − l ) = V i n V(z=-l)=V_{in} V(z=−l)=Vin​, we obtain V + V^+ V+:
V + = V i n e − j k 0 z + 1 3 e j k o z = 5.6555 − 4.2384 j   V V^+=\frac{V_{in}}{e^{-jk_0z}+\frac{1}{3}e^{jk_oz}}=5.6555-4.2384j~V V+=e−jk0​z+31​ejko​zVin​​=5.6555−4.2384j V
Thus
V l = V ( z = 0 ) = 7.5407 − 5.6513 j   V P l = ∣ V l ∣ 2 2 Re ⁡ { 1 Z l } = 0.444   W V_l=V(z=0)=7.5407-5.6513j~V\\ P_{l}=\frac{\left|V_{l}\right|^{2}}{2} \operatorname{Re}\left\{\frac{1}{Z_{l}}\right\}=0.444~W Vl​=V(z=0)=7.5407−5.6513j VPl​=2∣Vl​∣2​Re{Zl​1​}=0.444 W

Therefore,
V i n = V g Z i n Z g + Z i n = 7.33 − 1.33 j   Ω P l = ∣ V i n ∣ 2 2 Re ⁡ { 1 Z i n } = 0.444   W V_{in}=V_g\frac{Z_{in}}{Z_g+Z_{in}}=7.33-1.33j~\Omega\\ P_{l}=\frac{\left|V_{i n}\right|^{2}}{2} \operatorname{Re}\left\{\frac{1}{Z_{i n}}\right\}=0.444~W Vin​=Vg​Zg​+Zin​Zin​​=7.33−1.33j ΩPl​=2∣Vin​∣2​Re{Zin​1​}=0.444 W

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Or we can first compute Γ \Gamma Γ:
Γ = Z l − Z 0 Z l + Z 0 = 1 3 \Gamma=\frac{Z_l-Z_0}{Z_l+Z_0}=\frac{1}{3} Γ=Zl​+Z0​Zl​−Z0​​=31​
Then according to the definition,
V ( z ) = V + ( e − j k 0 z + 1 3 e j k o z ) V(z)=V^+(e^{-jk_0z}+\frac{1}{3}e^{jk_oz}) V(z)=V+(e−jk0​z+31​ejko​z)
By imposing the boundary conditions V ( z = − l ) = V i n V(z=-l)=V_{in} V(z=−l)=Vin​, we obtain V + V^+ V+:
V + = V i n e − j k 0 z + 1 3 e j k o z = 5.6555 − 4.2384 j   V V^+=\frac{V_{in}}{e^{-jk_0z}+\frac{1}{3}e^{jk_oz}}=5.6555-4.2384j~V V+=e−jk0​z+31​ejko​zVin​​=5.6555−4.2384j V
Thus
V l = V ( z = 0 ) = 7.5407 − 5.6513 j   V P l = ∣ V l ∣ 2 2 Re ⁡ { 1 Z l } = 0.444   W V_l=V(z=0)=7.5407-5.6513j~V\\ P_{l}=\frac{\left|V_{l}\right|^{2}}{2} \operatorname{Re}\left\{\frac{1}{Z_{l}}\right\}=0.444~W Vl​=V(z=0)=7.5407−5.6513j VPl​=2∣Vl​∣2​Re{Zl​1​}=0.444 W