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Transmission lines II 1
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Power transmission Poynting vector:
the flow of power density in [W/m2] both E/M fields are in real expressions Or time averaged Poynting vector: both E/M fields are in complex expressions Power: Or time averaged power: Source: picture shows what happened physically in a (DC) transmission line
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Load and wave reflection
why place the picture in the negative part? ZL Z0 z z<0 z=0 From the transmission line theory, right at z=0, we have: From the load point of view, we still have: We then find: reflection coefficient transmission coefficient
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Load and wave reflection
Be careful with: think about why? Power reflection: Power transmission:
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Voltage standing wave ratio
By launching a forward going voltage wave V0 along a lossless transmission line with a reflection (from a load or another unmatched transmission line), we have: In time-domain complex form: In time-domain real form: travelling wave standing wave Also from: VSWR provides us with an easy measure of reflection
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Voltage standing wave ratio
By reading the waveform (ripples) along the transmission line, we can easily read VSWR (s). Therefore, we obtain: By reading the wave period span in length, we find the wavelength. By reading the 1st minimum or maximum from the load, we find the phase (ϕ) of the reflection coefficient (Γ). We then get the full information about the reflection coefficient (Γ). We can further resolve the load: Particularly, if ϕ=0 if ϕ=π resistive load higher than Z0 resistive load lower than Z0
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Voltage standing wave ratio
z z=0 |V(z)| position of the load 2|Γ|V0 V0 (1-|Γ|)V0 (1+|Γ|)V0 λ/2 -ϕ/(2β) -(ϕ+π)/(2β) -(ϕ+2π)/(2β) d Measure |V(z)|~z on transmission line: starting from the load, trace back to the direction of the source for a few periods One period of the length is half of the wavelength (λ/2) β is obtained by 2β(λ/2)=2π → β=2π/λ This is always valid regardless, but λ is the wavelength of the transmission line, not necessarily the vacuum wavelength defined by c/f, unless n=(εrµr)1/2=1. Actually, we have λ=(c/f)/n, or β=2π/λ=2πnf/c. ϕ/(2β) can be obtained as the distance (d) from the load to the 1st maximum If d=0, the load is at the voltage maximum, ϕ=0 and we have a pure resistive load > Z0. If d=π/(2β), the load is the voltage minimum, ϕ=π and we have a pure resistive load < Z0. Phase of the reflection coefficient is obtained as ϕ=2βd=4πd/λ Amplitude of the reflection coefficient (|Γ|) is obtained as (ripple peak-to-peak voltage)/ [2(averaged voltage at the centre line)], or can be taken by s=(maximum voltage)/(minimum voltage), then |Γ|=(s-1)/(s+1)
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Transmission line impedance
For a lossless transmission line, the wave impedance at any position is: If the load is at z=0, trace back along the line for an arbitrary distance L, the wave impedance at z=-L is the input impedance of the transmission line with a length of L and is given as: If L=mλ/2, m=0, 1, 2, … If L=mλ/2+λ/4, m=0, 1, 2, … 4 special scenarios Short circuit load, ZL=0 Open circuit load, ZL=∞
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Transmission line impedance
transmission line: introduce a time delay only, no other effect If L=mλ/2, m=0, 1, 2, … If L=mλ/2+λ/4, m=0, 1, 2, … often used for impedance matching to avoid reflection, same idea used for anti-reflection coating for glasses, due to the simple fact: TEM wave (guided by TL) = cross-sectionally localized plane wave (in free-space) Short circuit load, ZL=0 purely inductive load conversion Open circuit load, ZL=∞ purely capacitive load conversion
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Impedance calculation/matching: graphical method – the Smith chart
With a given transmission line (Z0) and a known load (ZL), use the Smith chart for: find Γ (both amplitude |Γ| and phase ϕ) find the input impedance (Zin) of the transmission line in any given length (L) find VSWR (s) design a matching network through reflection cancellation (through, e.g., a short-circuited stub) But this tool was powerful only in pre-PC age. Read textbook from page 341 (example 10.10) to page 346 for examples on how to exploit the Smith chart.
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Impedance matching – a general approach
Z0 Yin1 Yin ZL=Zr+jZi Yin2 Z0 L2 Strategy: 1. For any given ZL, at L1, the admittance is given as: 2. Adjust L1, so that: 3. Adjust L2, so that: 4. Therefore, we finally have: 5. From L1 towards left (the source side), there is no reflection as Γ=0. 11
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Impedance matching – a general approach
Solution technique: If zr=1, but zi≠0 If zr=1, but zi=0: the system is impedance matched already! If the result is negative, add mλ/2, where m is a proper integer, such that L1 will be a smallest positive value We find: 12
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Impedance matching – a general approach
with x given by the formula shown in the previous slide If choose SC TL stub we have: If choose OC TL stub we have: What to do if yx=0? 13
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Transient analysis t=0 V0 Rs ZL z=0 z=L V+ + - I+ V- I- as t→∞ where
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Transient analysis From we find:
For ideal source, VL=V1+=V0. Load matched or not only affects the line voltage transient behaviour: “charge” through single or multiple stairs. For non-ideal source, the load voltage can only take a fraction of the source voltage. For matched source impedance, the line voltage will be charged once (for matched load) or at most twice (for unmatched load). Ideal source, matched load: Ideal source, unmatched load: we still have: Non-ideal source, matched load: Non-ideal but matched source:
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Transient analysis Voltage reflection diagram By following we obtain:
z z=0 z=L L/v 3L/v 2L/v V1+ V1-=ΓLV1+ V2+=ΓSΓLV1+ d/v d V(L-d) t V0ZL/(Zs+ZL) (L-d)/v (L+d)/v (3L-d)/v (3L+d)/v V1++V1- V1++V1-+V2+ By following we obtain: I(L-d) t V0/(Zs+ZL) (L-d)/v (L+d)/v (3L-d)/v (3L+d)/v I1+ I1++I1- I1++I1-+I2+
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