Impedance Matching?

Question

I am not entirely sure how it works. So, because of the difference of impedance b/t lets say a transmission line and a load creates a difference of voltage. To cross this voltage some of the energy of the signal is given up and the rest of the energy can't cross this distance and reflects back. I kinda don't get how the last part affects the signals. Is it just interference with the reflected waves when the incoming waves go through or does the energy loss across this potential effect the waveform.

Lets consider a 1,000 ohm source impedance, 50 ohm wire impedance, and 10,000,000 ohm oscilloscope impedance, which was used during class demonstration. The cable was connected where he put a resistor 50 ohm and a 50 ohm terminator to show reduction in the noise. Don't you apply the reflection coefficient to the source-cable junction and then scope-cable junction. I got for one (10M/50 -1 )/(10M/50-1) that would mean the cable is infinite and should not be much noise.

if I got it upside down I guess then it would be negative meaning almost total reflection. So, would it bounce back an forth until, it constructively interferes enough to make it through.

I see what you mean. It says 10M Ohm for the scope I would figure that is mega as in 10^6.

OH... so my Oscilloscope is not a load. But, it receives all the signal since its impedance is really high...?

Answer 1

You are talking about impedance mismatches in A/C transmission lines and the resulting standing waves.

First, be careful with the term "NOISE." It usually refers to a signal coming into the line from outside. In this case, you are not talking about signal from outside. You are talking about reflected signal, which is part of the original signal.

What exactly happens involve type of the source as well, but in theory, when a signal travels along a line, it will form certain voltage and current. Because of this ratio, you can arrive at a resistance value by ohms law. This is called a "characteristics impedance." When this signal meets a load that does NOT equal this impedance, not all energy is absorbed, and the difference goes BACK to the source.

At the source, the reflected signal meets the circuit, and often, it gets mixed with the original signal and in a way "distortes" the signal. With some oscilators, this reflected signal is absorbed.

You have a problem with your experiment because you have no load. The source is 1kohm, the cable is 50 ohm, but the input impedance of 10Mohm is not a load. To the source, it will look like an open end, which ALL the signal reflects back.

When you put 50 ohm load, depending on the frequency in use and the cable length, it could look like a perfect match on the load side.

As to the length of the cable, again, depending on the frequency in use, it is not that simple.... when there is a mismatch, the cable forms a transformer, and transforms the impedance, depending on their characteristic impedance and the length in relations to the wave length of the signal, and the dialectic factor. I won't even go into it....

Answer 2

A 50 ohm line will look like a 50 ohm resistor until the reflection comes back. 50 ohms is small enough to look like a short circuit to your 1 K source. The line will act much like a capacitor. The signal will bounce back and forth. Each reflection is a little bit smaller than the one before. Eventually, it will settle down to the value defined by the voltage divider made up of the 1 K source and 10 meg load.
If your driving signal is slow enough, the reflections will blur together and the line will behave almost exactly like a capacitor. this happens when the rise time is longer than the round trip delay.