# Homework Help: Proper termination with 75Ohm cable

1. Jun 16, 2011

1. The problem statement, all variables and given/known data
A pulse generator is connected to an oscilloscope, as in the image. The pulse generator has an internal impedance of 50Ohm and the oscilloscope is set to 1MOhm. The cable between the devices has a characteristic impedance of 75Ohm. The pulse from the generator has an amplitude of 1V and it's long enough to interfer with the reflecting pulse (if any). What is the proper resistance of the termination at B?

2. Relevant equations
$V_r = V_0 \frac{Z-Z_0}{Z+Z_0}$

3. The attempt at a solution
At A there will be a reflection because of the higher impedance. The reflection amplitude will be
$V_r = 1\text{V} \frac{75-50\Omega}{75+50\Omega}=\frac{1}{5}\text{V}$

Now, would this reflection increase (decrease seems more reasonable...) the amplitude of the pulse arriving at B, with 1/5V?
Would then a proper calculation of the termination be using V0= 1.2V, Z0=75Ohm and Z=-0.2?

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2. Jun 16, 2011

### lewando

You are correct with "decrease is more reasonable".

No, "-0.2" looks more like the reflection coefficient from the mismatch at A and is not an impedance. The question asks "what is the proper resistance of the termination at B". For a 75-ohm cable, one should be thinking: 75-ohms. So, how do you make the oscilloscope that looks look like a 1M-ohm load look like a 75-ohm load? I'll give you a chance to think about that.

3. Jun 16, 2011

Yes, thats Vr, which is the amplitude of the reflected pulse at A. I'm thinking that the first step would be to calculate the amplitude of the incident pulse at B.
Well, I would say $\frac{1}{1\text{M}\Omega} + \frac{1}{Z}= \frac{1}{75\Omega} \rightarrow Z \approx 75\Omega$, but that assumes that the signal at B is 1V, which I don't think it is since the pulse generator is not matched to the cable...

4. Jun 16, 2011

### lewando

True, the pulse generator is not matched to the cable. That is a shame. As for the oscilloscope, its termination is independent of the voltage of the signal applied. It just needs to match the cable.

5. Jun 16, 2011

Ok, but putting a 25Ohm resistor in series between cable and generator would give a match at A aswell, right?

6. Jun 16, 2011

### Staff: Mentor

I agree with lewando. The question asks for the match at B, and that is just adding the 75 ohms in parallel with the 'scope.

You can add this "back termination" resistor at A, but it does not do any matching in the forward direction. It is only useful if there is a [STRIKE]mismatch at A[/STRIKE] mismatch at B (like just the 'scope), because you will not get a re-reflection at A when the reflected wave comes back and sees 50+25=75 Ohms looking back into the signal generator.

Last edited: Jun 17, 2011
7. Jun 17, 2011