Why are voltage pulses reflected back over long transmission lines or in circuits having high frequency? Could someone explain how this wave nature doesn't go against the basic principles of electronics(Kirchoff's law, etc.) ?

Thanks

 

Why are voltage pulses reflected back over long transmission lines or in circuits having high frequency? Could someone explain how this wave nature doesn't go against the basic principles of electronics(Kirchoff's law, etc.) ?

They are reflected only if there is a mismatch between the line's characteristic impedance and the termination impedance.

One way to see this is to look at the energy in the signal. The energy is equal to V\(^{2}\) / R where R is the characteristic impedance of the line. When this energy reaches the line end, the energy is dissipated in the termination impedance. If the impedance equals the line impedance all the energy is absorbed in the impedance with no change in voltage. If the impedance is different. than the voltage will have to change to dissipate the energy, but due to the mismatch not all the energy is dissipated. The difference in voltage will look like a new signal which travels down the line as a reflection back to the source. If the termination impedance is too high, then the reflected voltage will be a positive voltage. If the termination impedance is too low, then the reflected voltage is a negative voltage.

Whether this reflection is a problem or not depends upon the electrical length of the line (propagation time) as compared to the rise-time or frequency of the signal.

So there's no conflict between the reflection and any basic principles of electronics (it's actually in perfect agreement).

Make sense?

 

They are reflected only if there is a mismatch between the line's characteristic impedance and the termination impedance.

One way to see this is to look at the energy in the signal. The energy is equal to V\(^{2}\) / R where R is the characteristic impedance of the line. When this energy reaches the line end, the energy is dissipated in the termination impedance. If the impedance equals the line impedance all the energy is absorbed in the impedance with no change in voltage. If the impedance is different. than the voltage will have to change to dissipate the energy, but due to the mismatch not all the energy is dissipated. The difference in voltage will look like a new signal which travels down the line as a reflection back to the source. If the termination impedance is too high, then the reflected voltage will be a positive voltage. If the termination impedance is too low, then the reflected voltage is a negative voltage.

Whether this reflection is a problem or not depends upon the electrical length of the line (propagation time) as compared to the rise-time or frequency of the signal.

So there's no conflict between the reflection and any basic principles of electronics (it's actually in perfect agreement).

Make sense?

This was quite use full but still there is little doubt. Normally in open circuits whatever is the applied voltage is what we see across whole wire, but according to the reflection principle I should actually see twice the voltage that what is applied [ incident + reflected ] ???

Also the direction of current would be determined by the magnitude or direction of voltage ??

 

This was quite use full but still there is little doubt. Normally in open circuits whatever is the applied voltage is what we see across whole wire, but according to the reflection principle I should actually see twice the voltage that what is applied [ incident + reflected ] ???

Also the direction of current would be determined by the magnitude or direction of voltage ??

Normally all lines are transmission lines. It's just that you don't see the effects for a short piece of wire that carries DC or low frequency AC, where the wavelength is much longer than the electrical propagation time of the wire. In that case the voltage change is so slow at the source end that, for the time it takes the signal to travel to the wire end and back, only a minute change in the input occurs. However in theory, if you amplify the signal sufficiently you could see the reflection. In other words there has to be a significant difference in the source voltage in the time it takes to get to the end of the line and back for transmission line effects to be noticeable. This can occur for just a few inches of wire at microwave frequencies.

So yes, for an ideal lossless transmission line that is looking at an open circuit, the return pulse could double the voltage at the source, if the source impedance is equal to the line characteristic impedance.

The magnitude or polarity of the return voltage is determined by the degree of mismatch and whether it is higher or lower than the line characteristic impedance.