What is the Optimal Resistance Matching for a Transmission Line?

[likephysics]

Homework Statement

I have source with some impedance (say 10 ohms-not shown in fig). since the tx line is 50 ohms, I added a series termination resistor of 40 ohms to match it with the tx line.
After the series resistor there is a parallel termination of 50 ohms, which is followed by a tx line of 50 ohms and another 50 ohm parallel termination resistor.
source voltage is 1.8v step
(see attached).

Homework Equations

The Attempt at a Solution

1. The source impedance and the series termination form 50 hms. This will form a voltage divider with the parallel termination resistor.

2. Now you have a voltage source of 0.9v with impedance of 50 ohms. This 50 ohms and tx line are matched, no nothing happens.

3. When the step reaches the 50 ohms parallel termination, its halved again to 0.45v?

 

[NascentOxygen]

Hi likephysics! I'd like it made clear what is given as part of the problem, and what is your addition. I'm having difficulty agreeing with some key features. You haven't stated exactly what the exercise is, either, but it seems to be matching a transmission line.

I have source with some impedance (say 10 ohms-not shown in fig). since the tx line is 50 ohms, I added a series termination resistor of 40 ohms to match it with the tx line.

So I think that end of the transmission line can now be considered matched. (Though the driver may not consider itself well matched: with its output Z being 10Ω, it might prefer to work into a load of 10Ω.)

After the series resistor there is a parallel termination of 50 ohms,

Why? I can't see it being necessary. In fact, it makes the impedance of the feed to the transmission line as 50Ω||(10+40)Ω

which is followed by a tx line of 50 ohms and another 50 ohm parallel termination resistor

Does that end's amplifier have an input Z of 10Ω, too?

 

[likephysics]

Nascentoxygen, this is not exactly a problem. I made it up myself.
DDR2 terminations are similar to this problem.
There are 2 recommended termination schemes.
One is parallel termination at both ends (DDR controller and Memory), the other is only at Memory.
Unlike typical RF amps, which have output impedance of 50 ohms, CMOS circuits do not have 50 ohm output impedance, but the PCB traces are 50 ohms. To avoid reflections, we add a series resistor between cmos IC and PCB trace.


I got a little confused, when the source is terminated parallely(we never did this in transmission line problems in class). I tried to make a tx line equivalent and solve it.
you have the source impedance(10 ohms), then the matching impedance(40 ohms) and the 50 ohm parallel termination, 50 ohm tx line and a 50 ohm load.
Please see attached for circuit equivalent and reduced ckt.
1. First the source voltage(1.8v) gets divided between the source impedance and matching impedance
2. The 1.44v gets divided between matching impedance and parallel combo of termination resistor and tx line impedance.
3. The load sees the same voltage as the tx line since they are matched.

 

[NascentOxygen]

I'd surmise that you incorporate the 40Ω to boost the output Z to 50Ω and so create a match for the cable. But I'm not convinced that you should also add that parallel 50Ω to ground.

I can see your thinking here, but looking at it purely as impedance matching, it doesn't seem right. However, with RAM I'm out of my depth, so I'll say no more and hope someone else can offer better advice.

Can you provide a web reference on this topic?

 

[skeptic2]

Here is a resistive matching calculator: http://chemandy.com/calculators/matching-pi-attenuator-calculator.htm

Enter 10 for the input impedance, 50 for the output impedance and 13 for the required attenuation and click calculate. The results are on the right and show that for a pi matching network you should us an 11.177 ohm shunt resistor, a 47.44 ohm series resistor and a 970.61 shunt resistor to match 10 to 50 ohms. This network will look like 10 ohms to the source and like 50 ohms to the transmission line.