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Lecture 21 Transmission Lines

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Lecture 21 Transmission Lines Lecture 21 Transmission Lines: RF and Microwave Circuits In this lecture you will learn: • More about transmission lines • Impedance transformation in transmission lines • Transmission line circuits and systems ECE 303 – Fall 2007 – Farhan Rana – Cornell University Transmission Lines: A Review V− V+ Zo Some common examples of Voltage at any point on the line can be written as: transmission lines − j k z + j k z V ()z = V+ e + V− e Current at any point on the line can be written as: V V I()z = + e− j k z − − e+ j k z Zo Zo Co-axial line The characteristic impedance of a transmission line is: L Z = o C The dispersion relation for a transmission line is: Wire on a ground plane k = ω LC ECE 303 – Fall 2007 – Farhan Rana – Cornell University 1 Transmission Line Circuits Consider a transmission line connected as shown below: Zs Z Vs o ZL j ω t Vs ()t = Re [Vs e ] z = −l z = 0 z Zo = Transmission line impedance ZL = Load impedance Zs = Source impedance In general, voltage on a transmission line is a superposition of forward and backward going waves: − j k z + j k z V ()z = V+ e + V− e The corresponding current is also a superposition of forward and backward going waves: V V I()z = + e− j k z − − e+ j k z Zo Zo ECE 303 – Fall 2007 – Farhan Rana – Cornell University Load Boundary Condition Z o + Zs V+ Vs ZL V− j ω t - Vs ()t = Re [Vs e ] z = −l z = 0 z − j k z + j k z V+ − j k z V− + j k z V ()z = V+ e + V− e I()z = e − e Zo Zo Boundary condition: At z = 0 the ratio of the total voltage to the total current must equal the load impedance: V ()z = 0 V+ + V− = = ZL I()z = 0 V+ Zo − V− Zo V Z Z − 1 This gives us the backward going ⇒ − = L o wave amplitude in terms of the V+ ZL Zo + 1 forward going wave amplitude Define a load reflection coefficient ΓL as: V− ZL Zo − 1 ΓL = = V+ ZL Zo + 1 ECE 303 – Fall 2007 – Farhan Rana – Cornell University 2 Load Reflections Suppose ZL = 0 (short): Zs V+ Vs Zo V− z = −l z = 0 z V− ZL Zo − 1 ΓL = = = −1 ⇒ V− = −V+ V+ ZL Zo + 1 V+ V− V+ V ()z = 0 = V+ + V− = 0 and I()z = 0 = − = 2 Zo Zo Zo Suppose ZL = ∞ (open): Zs V+ Vs Zo V− z = −l z = 0 z V− ZL Zo − 1 ΓL = = = +1 ⇒ V− = V+ V+ ZL Zo + 1 V V V ()z = 0 = V+ + V− = 2V+ and I()z = 0 = + − − = 0 Zo Zo ECE 303 – Fall 2007 – Farhan Rana – Cornell University Impedance Transformation - I Zs Z Vs o ZL j ω t Vs ()t = Re [Vs e ] z = 0 z = −l z − j k z + j k z V+ − j k z + j k z V ()z = V+ (e + ΓL e ) I()z = (e − ΓL e ) Zo Question: What is the impedance Z(z) looking towards the load at the location z in the transmission line? resistance 2 j k z V ()z 1+ ΓL e Z()z = = Zo Z(z) = R(z) + j X (z) I()z 1− Γ e2 j k z L reactance Check: What is the impedance Z(z=0)? 1+ Γ e2 j k z 1+ Γ Z()z = 0 = Z L = Z L = Z o 2 j k z o 1− Γ L 1− ΓL e z=0 L ECE 303 – Fall 2007 – Farhan Rana – Cornell University 3 Impedance Transformation - II Zs Z Vs o ZL j ω t Vs ()t = Re [Vs e ] z = −l z = 0 − j k z + j k z V+ − j k z + j k z V ()z = V+ (e + ΓL e ) I()z = (e − ΓL e ) Zo Question: What is the impedance Z(z=-ℓ) looking towards the load at the location z=-ℓ in the transmission line 2 j k z −2 j k l V ()z = −l 1+ ΓL e 1+ ΓL e Z()z = − = = Zo = Zo l 2 j k z −2 j k l I()z = −l 1− ΓL e 1− ΓL e z=−l Knowing the impedance looking into the line at z=-ℓ we can use the following equivalent circuit: Z s The load impedance ZL has been transformed by the transmission line into the impedance Vs Z(z = −l) Z(z=- ℓ) at the other end ECE 303 – Fall 2007 – Farhan Rana – Cornell University Equivalent Circuit Zs Z Vs o ZL j ω t Vs ()t = Re [Vs e ] z = −l z = 0 Z V z = V e− j k z + Γ e+ j k z s + ( ) + ( L ) V Z(z = −l) s V+ − j k z + j k z - I()z = (e − ΓL e ) Zo The voltage across the impedance Z(z=-ℓ) in the above circuit is: Z(z = −l) Vs Zs + Z()z = −l So on the transmission line we must have: Z(z = − ) j k l − j kl l V ()z = −l = V+ (e + ΓL e ) = Vs Zs + Z()z = −l V+ can be found from the above equation ECE 303 – Fall 2007 – Farhan Rana – Cornell University 4 Example – Short Circuit Load - I Zs Z Z = 0 Vs o L j ω t Vs ()t = Re [Vs e ] z = −l z = 0 ZL = 0 implies: −2 j k l ZL Zo − 1 1+ ΓL e ΓL = = −1 and Z()z = − = Zo = Zo j tan()k Z Z + 1 l −2 j k l l L o 1− ΓL e 2π λ kl << 1 ⇒ l << 1 ⇒ l << Suppose kℓ << 1: λ 2π ω v kl << 1 ⇒ l << 1 ⇒ ω << v l L Then: Z()z = − = Z j tan ()k ≈ j Z k = j ω LC = j ω ()L l o l o l C l l • Impedance seen at the source end is inductive • The transmission line appears like one big inductor ECE 303 – Fall 2007 – Farhan Rana – Cornell University Example – Short Circuit Load - II Zs Z Z = 0 Vs o L j ω t Vs ()t = Re [Vs e ] z = −l z = 0 ZL = 0 implies: −2 j k l ZL Zo − 1 1+ ΓL e ΓL = = −1 and Z()z = − = Zo = Zo j tan()k Z Z + 1 l −2 j k l l L o 1− ΓL e X Z()z = −l = j Zo tan ()k l inductive = j X π π π 2π π 0 3 5 kl 2 2 2 capacitive ECE 303 – Fall 2007 – Farhan Rana – Cornell University 5 Example – Open Circuit Load - I Zs Z Z = ∞ Vs o L j ω t Vs ()t = Re [Vs e ] z = −l z = 0 ZL = ∞ implies: −2 j k l ZL Zo − 1 1+ ΓL e ΓL = = +1 and Z()z = − = Zo = −Zo j cot()k Z Z + 1 l −2 j k l l L o 1− ΓL e 2π λ kl << 1 ⇒ l << 1 ⇒ l << Suppose kℓ << 1: λ 2π ω v kl << 1 ⇒ l << 1 ⇒ ω << v l Zo L 1 1 Then: Z()z = −l = −Zo j cot ()k l ≈ − j = − j = kl C ω LC l j ω ()Cl • Impedance seen at the source end is capacitive • The transmission line appears like one big capacitor ECE 303 – Fall 2007 – Farhan Rana – Cornell University Example – Open Circuit Load - II Zs Z Z = ∞ Vs o L j ω t Vs ()t = Re [Vs e ] z = −l z = 0 ZL = ∞ implies: −2 j k l ZL Zo − 1 1+ ΓL e ΓL = = +1 and Z()z = − = Zo = −Zo j cot()k Z Z + 1 l −2 j k l l L o 1− ΓL e X Z()z = −l = − j Zo cot ()k l inductive = j X π π π 2π π 0 3 5 kl 2 2 2 capacitive ECE 303 – Fall 2007 – Farhan Rana – Cornell University 6 Example – Matched Load Zs V+ Vs Zo Zo j ω t Vs ()t = Re [Vs e ] z = −l z = 0 ZL = Zo implies: −2 j k l ZL Zo − 1 1+ ΓL e ΓL = = 0 and Z()z = − = Zo = Zo Z Z + 1 l −2 j k l L o 1− ΓL e For matched loads: • The load reflection coefficient ΓL is zero • There is no reflected wave generated at the load end ( i.e. V_ = 0 ) • The impedance seen at the source end is Zo irrespective of the length of the transmission line ECE 303 – Fall 2007 – Farhan Rana – Cornell University Periodicity of Impedance Transformation Zs Z Vs o ZL j ω t Vs ()t = Re [Vs e ] z = −l z z = 0 2 j k z ZL Zo − 1 1+ ΓL e Γ = and Z()z = Z L Z Z + 1 o 2 j k z L o 1− ΓL e For locations z such that 2kz = 2mπ (m is an integer) then: 1+ Γ e2 j k z 1+ Γ Impedance seen at Z z = Z L = Z L = Z () o 2 j k z o L locations z = mπ/k is 1− ΓL e 1− ΓL also ZL 2π λ Note that: 2k z = 2mπ but k = ⇒ z = m λ 2 Impedance seen at locations z that are integer multiples of half-wavelength from the load end is the load impedance ZL ⎛ λ ⎞ Impedance is periodic with More generally: Z⎜ z ± m ⎟ = Z()z ⎝ 2 ⎠ period equal to half-wavelength ECE 303 – Fall 2007 – Farhan Rana – Cornell University 7 Standing Wave Ratio Z s V Z + Vs o ZL V− j ω t Vs ()t = Re [Vs e ] z = −l V− ZL Zo − 1 z = 0 ΓL = = V+ ZL Zo + 1 V ()z = V e− j k z + V e+ j k z = V e− j k z + Γ e+ j k z jφ + − + ( L ) ΓL = ΓL e 2 2 2 ⇒ V ()z = V+ []1+ ΓL + 2 ΓL cos()2 k z + φ 2 ⇒ V ()z = V+ 1+ ΓL + 2 ΓL cos()2 k z + φ The interference of forward and backward going waves leads to standing-wave-like behavior V (z) 1+ Γ Standing Wave Ratio = SWR = max = L V ()z min 1− ΓL ECE 303 – Fall 2007 – Farhan Rana – Cornell University Thevenin Equivalents of the Source - I So far we have taken a transmission line circuit: Zs Z Vs o ZL j ω t Vs ()t = Re [Vs e ] z = −l z = 0 And reduced it to the equivalent circuit: Zs Vs Z(z = −l) Now we want to reduce it to the equivalent circuit: Zth Vth ZL i.e.
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