Transmission Line¶

Transmission-Line Equations¶

First define - $ℭ$ : inductance per unit length - $𝔏$ : inductance per unit length

Lossless Transmission Lines¶

{\begin{aligned} \frac{\partial V}{\partial z} & = - 𝔏 \frac{\partial I}{\partial t} \\ \frac{\partial I}{\partial z} & = - ℭ \frac{\partial V}{\partial t} \end{aligned}

tip

These equation can be easily derived from the definition of capacitance and capacitance in electric circuit theorem

\begin{matrix} v = L \frac{d i}{d t} \\ i = C \frac{d v}{d t} \end{matrix}

Similar to the derivation in EM wave equation, we can obtain the wave equation form

{\begin{aligned} \frac{\partial^{2} V}{\partial z^{2}} & = 𝔏 ℭ \frac{\partial^{2} V}{\partial t^{2}} \\ \frac{\partial^{2} I}{\partial z^{2}} & = 𝔏 ℭ \frac{\partial^{2} I}{\partial t^{2}} \end{aligned}

Thus we have the wave velocity

\begin{matrix} u = \frac{1}{\sqrt{𝔏 ℭ}} \end{matrix}

Comparing to the velocity of EM wave, it tell us the important relation that

\begin{matrix} 𝔏 ℭ = μ ε \end{matrix}

Furthermore, considering the phasor form of transmission-line equation, we have

\begin{array}{r} \frac{d^{2} V_{s}}{d z^{2}} = (j β)^{2} V_{s} \\ \frac{d^{2} I_{s}}{d z^{2}} = (j β)^{2} I_{s} \end{array}

in which

\begin{matrix} (j β)^{2} = (j ω)^{2} 𝔏 ℭ \\ ⟹ β = ω \sqrt{𝔏 ℭ} = \frac{ω}{u} \end{matrix}

General Solutions¶

\begin{aligned} V (z, t) & = V^{+} (z - u t) + V^{-} (z + u t) \\ I (z, t) & = I^{+} (z - u t) + I^{-} (z + u t) \end{aligned}

Similarly, we have the characteristic impedance $Z_{0}$

\begin{matrix} Z_{0} = \sqrt{\frac{𝔏}{ℭ}} = \frac{V^{+}}{I^{+}} = - \frac{V^{-}}{I^{-}} \end{matrix}

General Transmission LInes¶

Revise the transmission-line equation in the lossless case into

{\begin{aligned} - \frac{\partial V}{\partial z} & = R I + L \frac{\partial I}{\partial t} \\ - \frac{\partial I}{\partial z} & = G V + C \frac{\partial V}{\partial t} \end{aligned} \overset{𝔉 o u r i e r}{⟺} {\begin{aligned} \frac{d V_{s}}{d z} & = - (R + j ω L) I_{s} \\ \frac{d I_{s}}{d z} & = - (G + j ω C) V_{s} \end{aligned}

Also, in the form of wave equations

\begin{matrix} \frac{d^{2} V_{s}}{d z^{2}} = γ^{2} V_{s} \\ \frac{d^{2} I_{s}}{d z^{2}} = γ^{2} I_{s} \end{matrix}

in which $γ$ is also called as propagation constant.

\begin{matrix} γ = \sqrt{(R + j ω L) (G + j ω C)} = α + j β \end{matrix}

General Solutions¶

solve the wave equations above

\begin{aligned} V_{s} & = V_{0}^{+} e^{- γ z} + V_{0}^{-} e^{γ z} \\ I_{s} & = I_{0}^{+} e^{- γ z} + I_{0}^{-} e^{γ z} \\ = \frac{V_{0}^{+}}{Z_{0}} e^{- γ z} - \frac{V_{0}^{+}}{Z_{0}} e^{γ z} \end{aligned}

and the corresponding time domain form

\begin{aligned} V (z, t) & = | V_{0}^{+} | e^{- α z} \cos (ω t - β z + ϕ^{+}) + | V_{0}^{-} | e^{α z} \cos (ω t + β z + ϕ^{-}) \\ I (z, t) & = | I_{0}^{+} | e^{- α z} \cos (ω t - β z + φ^{+}) + | I_{0}^{-} | e^{α z} \cos (ω t + β z + φ^{-}) \end{aligned}

power attenuation

\begin{aligned} attenuation const & = α (NP/m) \\ power attenuation & = 20 \log e^{α} (dB/m) \end{aligned}

thus,

\begin{matrix} 1 NP = 20 \log e^{1} = 8.69 dB \end{matrix}

Power¶

time-average power

\begin{aligned} P (z) & = \frac{1}{2} Re [V I^{*}] \\ = \frac{1}{2} Re [V_{0}^{+} (1 + Γ) {(\frac{V_{0}^{+}}{Z_{0}})}^{*} (1 - Γ^{*})] \\ = \frac{1}{2} Re [\frac{{V_{0}^{+}}^{2}}{Z_{0}} (1 + Γ) (1 - Γ^{*})] \\ (6) & = \frac{1}{2} \frac{{| V_{0}^{+} |}^{2}}{Z_{0}} (1 - | Γ |^{2}) \end{aligned}

^equ-6

⟹ {\begin{cases} P^{+} = \frac{1}{2} | V_{0}^{+} | | I_{0}^{+} | e^{+ 2 α z} \\ P^{-} = \frac{1}{2} | V_{0}^{-} | | I_{0}^{-} | e^{- 2 α z} \end{cases}

power loss

\begin{matrix} PL = P (0) - P (z) \end{matrix}

Distortionless¶

conditions

\begin{matrix} \frac{R}{L} = \frac{G}{C} \end{matrix}

thus, we have the propagation constant $γ$

\begin{aligned} γ & = \sqrt{(R + j ω L) (G + j ω C)} \\ = \sqrt{L C} \sqrt{(\frac{R}{L} + j ω) (\frac{G}{C} + j ω)} \\ = \sqrt{L C} (\frac{R}{L} + j ω) \\ = \sqrt{R G} + j ω \sqrt{L C} \end{aligned}

and the characteristic impedance $Z_{0}$

\begin{aligned} Z_{0} & = \sqrt{\frac{R + j ω L}{G + j ω C}} \\ = \sqrt{\frac{R (1 + j ω \frac{L}{R})}{G (1 + j ω (\frac{C}{G}))}} \\ = \sqrt{\frac{R}{G}} = \sqrt{\frac{L}{C}} \end{aligned}

Standing Wave Ratio¶

aka $SWR$

Reflection¶

voltage reflection coefficient

\begin{matrix} Γ (z) = \frac{V^{-} (z)}{V^{+} (z)} = \frac{V_{0}^{-} e^{+ γ z}}{V_{0}^{+} e^{- γ z}} = \frac{V_{0}^{-}}{V_{0}^{+}} e^{+ 2 γ z} \end{matrix}

current reflection coefficient

\begin{matrix} Γ_{I} (z) = \frac{I^{-}}{I^{+}} = - \frac{V^{-}}{V^{+}} = - Γ (z) \end{matrix}

$Γ_{L}$
Let's define the impedance from load $Z_{L} = V_{L} / I_{L}$

\begin{matrix} V_{0}^{+} e^{- γ ℓ} + V_{0}^{-} e^{+ γ ℓ} = \frac{Z_{L}}{Z_{0}} (V_{0}^{+} e^{- γ ℓ} - V_{0}^{-} e^{+ γ ℓ}) \\ ⟹ (Z_{0} - Z_{L}) V_{0}^{+} e^{- γ ℓ} = - (Z_{L} + Z_{0}) V_{0}^{-} e^{+ γ ℓ} \\ (2) & ⟹ Γ_{L} = \frac{V_{0}^{-} e^{+ γ ℓ}}{V_{0}^{+} e^{- γ ℓ}} = \frac{V_{0}^{-}}{V_{0}^{+}} e^{+ 2 γ ℓ} = \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}} \end{matrix}

^equ-2

We can further write the reflection coefficient from load at $d$ as

\begin{array}{r} (3) & Γ (d) = Γ_{L} e^{- 2 γ d} \end{array}

^equ-3

tips

from $equation (3-1)$ ,

\begin{matrix} Z (z) = Z_{0} \frac{1 + Γ}{1 - Γ} \\ ⟹ Z (z) - Z (z) Γ = Z_{0} + Z_{0} Γ \\ ⟹ Γ (z) = \frac{Z (z) - Z_{0}}{Z (z) + Z_{0}} \end{matrix}

we can also derive $equation (2)$ by virtue of this equation

\begin{matrix} Γ_{L} = Γ (z = ℓ) = Γ (d = 0) = \frac{Z (z = ℓ) - Z_{0}}{Z (z = ℓ) + Z_{0}} \end{matrix}

Furthermore, at the generator-end (i.e. $z = 0$ ), we have

\begin{matrix} Γ (z = 0) = Γ (d = ℓ) = \frac{Z_{i n} - Z_{0}}{Z_{i n} + Z_{0}} \end{matrix}

impedance¶

Line impedance at $z$ , $Z (z)$

\begin{aligned} Z (z) & = \frac{V_{s}}{I_{s}} \\ = Z_{0} \frac{V_{0}^{+} e^{- γ z} + V_{0}^{-} e^{+ γ z}}{V_{0}^{+} e^{- γ z} - V_{0}^{-} e^{+ γ z}} = Z_{0} \frac{1 + \frac{V_{0}^{-}}{V_{0}^{+}} e^{+ 2 γ z}}{1 - \frac{V_{0}^{-}}{V_{0}^{+}} e^{+ 2 γ z}} \\ (3-1) & = Z_{0} \frac{1 + Γ (z)}{1 - Γ (z)} \end{aligned}

^equ-3-1

impedance at load $Z_{L}$

\begin{aligned} Z_{L} & = Z (z = ℓ) = Z_{0} \frac{1 + Γ (z = ℓ)}{1 - Γ (z = ℓ)} \\ = Z_{0} \frac{1 + Γ_{L}}{1 - Γ_{L}} \end{aligned}

input impedance $Z_{i n}$

\begin{aligned} Z_{i n} & = Z (z = 0) = Z_{0} \frac{1 + Γ (z = 0)}{1 - Γ (z = 0)} \\ = Z_{0} \frac{1 + \frac{V_{0}^{-}}{V_{0}^{+}}}{1 - \frac{V_{0}^{-}}{V_{0}^{+}}} \\ = Z_{0} \frac{1 + Γ_{L} e^{- 2 γ ℓ}}{1 - Γ_{L} e^{- 2 γ ℓ}} \end{aligned}

from $equation (2)$ we have

\begin{aligned} Z_{i n} & = Z_{0} \frac{1 + Γ_{L} e^{- 2 γ ℓ}}{1 - Γ_{L} e^{- 2 γ ℓ}} \\ = Z_{0} \frac{1 + \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}} e^{- 2 γ ℓ}}{1 - \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}} e^{- 2 γ ℓ}} \\ = Z_{0} \frac{Z_{L} + Z_{0} + (Z_{L} - Z_{0}) e^{- 2 γ ℓ}}{Z_{L} + Z_{0} - (Z_{L} - Z_{0}) e^{- 2 γ ℓ}} \\ = Z_{0} \frac{Z_{L} (1 + e^{- 2 γ ℓ}) + Z_{0} (1 - e^{- 2 γ ℓ})}{Z_{L} (1 - e^{- 2 γ ℓ}) + Z_{0} (1 + e^{- 2 γ ℓ})} \\ = Z_{0} \frac{Z_{L} + Z_{0} \tanh γ ℓ}{Z_{L} \tanh γ ℓ + Z_{0}} \\ = Z_{0} \frac{Z_{L} + Z_{0} \tanh γ ℓ}{Z_{0} + Z_{L} \tanh γ ℓ} \\ \overset{α = 0}{=} Z_{0} \frac{Z_{L} + j Z_{0} \tan β ℓ}{Z_{0} + j Z_{L} \tan β ℓ} \end{aligned}

Complex form¶

As usual, we can represent $Γ$ with complex coordinate.

\begin{matrix} Γ = | Γ | ∠ Γ \end{matrix}

Consider $equation (3)$ , we further have

\begin{array}{r} Γ = Γ_{L} e^{- 2 α d} e^{- j 2 β d} \end{array}

Besides, we can rewrite $V_{s}, I_{s}$ as

\begin{aligned} | V_{s} | & = | V^{+} | | 1 + Γ | \\ | I_{s} | & = | I^{+} | | 1 - Γ | \end{aligned}

Standing Wave Ratio for Lossless¶

\begin{aligned} VSWR & \equiv \frac{V_{m a x}}{V_{m i n}} = \frac{| V_{0}^{+} | | 1 + Γ |_{m a x}}{| V_{0}^{+} | | 1 + Γ |_{m i n}} \\ = \frac{1 + | Γ |}{1 - | Γ |} \overset{α = 0}{=} \frac{1 + | Γ_{L} |}{1 - | Γ_{L} |} \end{aligned}

and similarly for $ISWR$ ,

\begin{matrix} \frac{I_{m a x}}{I_{m i n}} = \frac{V_{m a x}}{V_{m i n}} = SWR \end{matrix}

according to $equation (5)$ ,

\begin{array}{r} z = \frac{1 + Γ}{1 - Γ} = r + j x \end{array}

we can obtain $SWR$ by picking the point that $∠ Γ = 0$ , then

\begin{array}{r} SWR = \frac{1 + | Γ |}{1 - | Γ |} = r + j 0 = r \end{array}

Smith Chart¶

Smith chart help us quickly obtain the normalized impedance $Z$ w.r.t $Z_{0}$ .

\begin{array}{r} Z = z \times Z_{0} = (r + j x) \times Z_{0} \end{array}

then, we can have the normalized impedance $z$

\begin{array}{r} (5) & z = r + j x = \frac{Z}{Z_{0}} = \frac{1 + Γ}{1 - Γ} \end{array}

^equ-5

$r$ -circles
$x$ -circles

Admittance Chart¶

for the admittance of transmission line, we can easily obtain the relationship of admittance and impedance.

\begin{matrix} y = \frac{Y}{Y_{0}} = \frac{1}{z} = \frac{1 - Γ}{1 + Γ} \\ (4) & ⟹ y = z e^{- j π} \end{matrix}

^equ-4

the $equation (4)$ tell us that we can obtain the normalized admittance $y$ with Smith chart by simply rotate the corresponding $z$ with $180 \degree$ .