50 Ohm to 75 Ohm to 50 Ohm line?

[Helmholtzerton]

I'm trying to understand how I can mathematically describe a particular situation in order to calculate what I should expect to read on an oscilloscope.

Let's say I have a signal generator that is generating a 1MHz sinusoidal signal 1Vpp. The source impedance of the signal generator is 50ohm. This goes into a 50ohm BNC coax cable. This cable is connected to a 75ohm BNC coax cable. That cable is connected to a 50ohm BNC coax cable with the other end connected to an oscilloscope terminated at 50 ohms.

Now, without any losses, if I just went...
Signal generator -> 50ohm coax -> oscilloscope @ 50ohms, I should expect to see 1Vpp on the oscilloscope.

But by adding a 75ohm coaxial cable to the middle of that circuit, how would that effect the reading on the oscilloscope? Can I treat each mismatching point as both a source and a load along the transmission line?

 

[Charles Link]

What happens in the r-f case with the impedance mismatch in the configuration you described is precisely analogous to the Fabry-Perot case in Optics of light incident on a dielectric slab. In this case., $\frac{1}{Z}$ replaces the index $n$ in the expressions for the Fresnel reflection and transmission coefficients $\rho$ and $\tau$. There will be cable lengths where constructive interference occurs on transmission, and cable lengths where the signal is reduced because of the way the multiple reflections add to each other. Hecht and Zajac's Optics textbook treat the optics problem in detail. $\\$ In a google of the topic, I was unsuccessful at getting a "link" that showed the very well known results that are presented in Hecht and Zajac. See: https://www.amazon.com/dp/0201838877/?tag=pfamazon01-20

 

[Charles Link]

One additional item: With this kind of configuration, even if you terminate the line with a 50 ohm resistor, there will still necessarily be a wave that reflects off the 75 ohm "slab" that returns to the source. I would need to research this further to see how to send a 1 volt sinusoidal signal from the source and get a good reading on it that isn't affected by the reflected wave.

 

[Baluncore]

Because both ends of the circuit are matched to the 50 ohm cables, there will be a reflection at the 50 to 75 ohm step, there will be another reflection at the 75 to 50 ohm step. But the length of the 75 ohm cable resonator between the steps will confuse the situation. The only thing more you need to know to work out the response will be the transit time in the 75 ohm cable.

 

[tech99]

I'm trying to understand how I can mathematically describe a particular situation in order to calculate what I should expect to read on an oscilloscope.

The wavelength is 300m. An engineering solution is that, as the oscilloscope cables are short compared with the wavelength, they act as a single cable having an intermediate value of characteristic impedance. This is reasonably correct up to an eight of a wavelength. If the 50 and 75 Ohm sections are equal length, then Zo = SQRT (75 x 50). Now the question becomes that you have a 50 Ohm load on a 61 Ohm cable which is, say, 1 electrical degree in length. The generator will now see the 50 Ohm CRO in series with a miniscule inductance. It is possible to calculate this if wanted.
Incidentally, you do not clarify whether the signal generator provides 1 volt P-P pd or emf, that is terminated or open circuit. This could trap you!

 

[sophiecentaur]

When a short length of 'incorrect' cable is used (i.e. short in terms of wavelength), there is a simple approximate solution. Find the published / typical Capacitance per metre of the 75Ω cable and compare it with the capacitance of a 50Ω cable. Add a shunt capacitor across the 75Ω connector. For a narrow bandwidth and a low carrier frequency, a lumped component can often give a good compensation.
The actual lengths of cable involved are very relevant here. If you already have the equipment with you, you could 'dab' a small capacitor of the calculated amount and actually eyeball the effect on a scope. I suggest that you may see no difference at all in a non-critical experiment. Alternatively, buy some cable of the right impedance ( ? ).
I don't usually think in terms of Simulators but this could help the OP to give some idea of the possible effect.

 

[CWatters]

+1 to #5.

Just for info Google found a simple simulator. Looks like it might be worth having a play.

http://svn.clifford.at/tools/trunk/electrotools/tdtlsim.html

Maybe modify the example "Standing wave with broken cable" which already has two lengths of transmission line?

 

[sophiecentaur]

Best not to suggest digging out the Smith Chart - except to show how little difference rotating it by just a few degrees would make.

 

[tech99]

One additional item: With this kind of configuration, even if you terminate the line with a 50 ohm resistor, there will still necessarily be a wave that reflects off the 75 ohm "slab" that returns to the source. I would need to research this further to see how to send a 1 volt sinusoidal signal from the source and get a good reading on it that isn't affected by the reflected wave.

The wave is reflected with a standing wave ratio of 1.5 but very small time delay. So it is not significant in practical terms. If the cables had been significantly long in terms of wavelength, we could calculate the input impedance of each one when terminated by the last. We could also obtain the worst case condition by adding the reflection coefficients from each joint. Ref Coeff = (VSWR-1) /(VSWR+1). This will occur at certain frequencies.

 

[Charles Link]

The wave is reflected with a standing wave ratio of 1.5 but very small time delay. So it is not significant in practical terms. If the cables had been significantly long in terms of wavelength, we could calculate the input impedance of each one when terminated by the last. We could also obtain the worst case condition by adding the reflection coefficients from each joint. Ref Coeff = (VSWR-1) /(VSWR+1). This will occur at certain frequencies.

I didn't look closely at the frequency here to determine its effect on the cable length. The OP really has a problem with some interesting physics when the frequency is sufficiently high that the stretch of 75 ohm cable is more than 1/8 of a wavelength. $\\$ I first saw the calculations for this case in Optics, where a monochromatic wave is incident on a thin film, etc. About 15 years later, I saw the same calculations in a problem on transmission lines where we had impedance mismatches. It could be useful for the OP to study this in detail. The best source I know of for this problem is the Optics text by Hecht and Zajac.

 

[Baluncore]

Where is the OP in this discussion.