Why mismatch causes reflected waves

reflected waves

Hi,

Why mismatch causes reflected waves ??

thanks

Can you elucidate your doubt please

why waves that are going trough a transmission line are reflected back,this happen when the impadance of the line is not equal to the the impadance of the generator or to the impadance of the load .

so, again, Why mismatch causes reflected waves

I'll take a stab at it. I think the focus is on the short at the resonant frequency. You know the length has to be so many wavelengths and the line appears as a short at the resonant frequency. This is not a short, but it is the characterstic impedance which is close to a short. This is what gives you a 50 ohm cable. Now, you must make it 50ohm cable in order to realize the property of cancelling inductance and capacitance and come up with a short. This 50ohms is the key. Remember that the line impedance is dependant upon load. Somewhere between an open and a short. This has to be 50 ohms in order to make it a 50 ohm cable. If the source impedance is too high, then the signal becomes dependent on the line resistance which is not good because of the reactance. You need a low source resistance so that the signal will only see this resistance.

Hi.

The terminating impedence roles as a boundary condition. Formally, you can sove the second-order wave equation with the terminating load as a boundary condition, then, you will find out two terms representing one wave incident to the terminating load and the other wave coming out of the terminating load.

I like to think of the reflected wave as a visual aid in determining the what the voltage does in the system. We all know that current or energy is returned with reactance in the system. You will remember that reactance alone does not cause standing waves. So, with a standing wave, you have a signal and DC offsets along the line.

you can assume a mirror.
when it can't absorb light then reflect it.

I know of two ways to explain this phenomenon. One uses lumped models, and the other uses quantum physics. I prefer the lumped model. I have drawn the lumped model case where the end of the transmission line is terminated in an open circuit. Note that the lumped model is a way of describing the transmission line with components such as R, L, and C.

1) The wave from Vs begins to propagate down the transmission line
2) The first capacitor is charged
3) The second capacitor is charged
4) The third capacitor is charged, but since current cannot flow through another branch (because of the open circuit), the inductor and its unstopping current charge the third capacitor beyond Vs.
5) Since the potential on the third capacitor is larger, now current flow will eventually reverse direction and flow back towards the source, charging the subsequent capacitors.

Now note that if the transmission line were terminated in a short, the current would flow through the short and back towards the source on the bottom of the circuit. This would cause each capacitor from right to left to discharge in turn, which causes 0 volts to be seen on the line. So, with both short and open explained, anything else is a case in between these two extremes.

to satisfy boundary conditions

Rudane's explanation is quite impressive.

Boundary condition can explain the stabilized solution not what happened at the first second.