A question regarding transmission line

  • Thread starter yykcw
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  • #1
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When wave is reflected back means power is reflected back and not all power is consumed at the load. But I am thinking if I placed the load at the position of Vmax (of the stationary wave formed by the incident and reflected wave). Isn't that all the power will be dissipated at the load, so there should be no wave reflected back? Why is there a contradiction?
 

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  • #2
Svein
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If the load impedance is higher than the characteristic impedance, the load already experiences Vmax.
 
  • #3
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One way of looking at EM wave reflection is that either the voltage is at max and the current is zero (for an open) or the current is at max and the voltage is zero (for a short). The key is to get the voltage and current to work together. We do this by matching impedances.

There are other, more complicated solutions which rely on frequency and geometry, but those are usually reserved for weird applications. (I saw an air gap transmission line used as a lightning protector for a narrow frequency signal once; clever.)

It is poor design practice to use such clever tricks unless they are needed though. One ding and the system might literally burn.
 
  • #4
Baluncore
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When wave is reflected back means power is reflected back and not all power is consumed at the load.
To be more precise, it is energy that is reflected. Power is the rate of flow of energy.

There are two independent signals, one travelling in each direction. The impedance of a transmission line sets the ratio of voltage to current in each of those signals. The sum of those signals generates the standing wave pattern.

In order to get a voltage node there must be an impedance mismatch to reflect the energy in the first place. If you parallel another load somewhere on the line it will change the original line length and so move the voltage node somewhere else.
 

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