# Is Transmission Line Reflection Frequency Dependent?

I'm currently sending high speed serial over a coax cable. I've got an oscilloscope hooked up (on the load end) trying to watch the clock pulse signal. I've got a 10 ft coax cable with a 12.5 Mhz clock signal. I'm unsure of the coax impedance and the load impedance.

The clock signal appears to be deformed. I'm curious if this deformation is due to the cable acting like a transmission line and specifically due to transmission line reflection.. However, I'm unsure on some of the properties of the reflection.

Is the reflection of a signal frequency dependent? Thus the superposition of all of the frequencies in the square wave would be phase adjusted and report a bad clock signal?

• The wavelength at of the 12.5 MHz clock signal is about 24 meters at the fundamental frequency, so some the harmonics will likely see reflections if the cable isn't loaded properly. The impedance of the cable is fixed (50 or 75 $\Omega$) so if you try loading the cable with either of those impedances you should see some improvement. However, if you want to look at the problem from a low frequency perspective the problem is mainly cable capacitance causing the distortion. – Captainj2001 Jul 13 '16 at 16:48
• You may want to show your signal so others can give hints at what may be the problem. – JimmyB Jul 13 '16 at 16:49
• The impedance of ordinary cable is constant with frequency, for frequencies like 12.5MHz and its harmonics. Show a circuit diagram, including source output impedance, scope input impedance, and any additional resistors, capcitors you have used. – Neil_UK Jul 13 '16 at 16:52
• @JimmyB The cable impedance is defined as $Z_0 = \sqrt{L/C}$ where L and C are the per-unit length capacitance and inductance of the cable. This is fixed (especially at low frequencies) until you get to areas of nonlinear behavior in the dielectric which usually won't occur until 10+ GHz, even in relatively cheap cables. – Captainj2001 Jul 13 '16 at 16:52
• You need to find out what is terminating the transmission line. Without that information, there's no answer possible. – The Photon Jul 13 '16 at 16:57

## 4 Answers

simulate this circuit – Schematic created using CircuitLab

If $Zc=\sqrt{\dfrac{L}{C}}=Z_{source}=Z_{load}$, then there is no frequency dependance, the signal is just delayed but not distorted. If any part of above equation isn't equal, then you get standing waves.

• So let's say my source impedance, transmission impedance, and load impedance are not equal and I get a standing wave. Will this standing wave produce a distorted signal on the oscilloscope if I measure at the load? – Izzo Jul 13 '16 at 19:42
• You only get standing waves if you put in a single frequency sine wave. If you put in a square wave and it bounces around, you'll get jagged edges, presuming the frequency is low wrt. the line length. – alex.forencich Jul 14 '16 at 4:05

Try reducing the reflections (if there are any).

By adding an attenuator. Something like 3 dB, 6 dB attenuation? Attach this in between the coax and the oscilloscope. Did the deformation go away? The rx signal will be much smaller, but the attenuator will make the load appear matched thus reducing reflections.

 If you don't have an attenuator, add more cable.

You can also try to see if you can change the oscilloscopes load impedance to 50Ω and see if the distortion goes away.

Assuming your clock source is producing (approximate) square waves with a fundamental frequency of 12.5MHz, an expected rise time might be about 4ns.

Assuming the signal propagates at the speed of light (in reality, it's less than this, about half or a third), then to travel 10ft it takes 10ns, which indicates that transmission line effects might be important.

I wrote this online transmission line simulator so you can visualize what is happening to the signal inside of the transmission line. The model assumes a constant characteristic impedance (which is pretty good in practice for most things).

Parameters tested:

• tRise = 4ns
• tFall = 4ns
• tOn = 36ns
• tOff = 36ns
• tDelay = 10ns
• Zeq = 50 ohms
• R1 = 0 ohms (source terminator)
• R2 = 1 megohms (sink terminator)
• other parameters at their default values

Here are a few snapshots of the results (green is ideally terminated, red is poorly terminated, and blue is the source voltage):

First reflection

Second reflection

During the first reflection, the poorly terminated signal was 40% higher than the desired line signal, and during the second reflection the signal was about 10% lower than the desired line signal.

Repeating by simulating a shorter cable (approx. 1ft cable, i.e. tDelay = 1ns):

First reflection

Second reflection

Notice that the overshoot is much smaller (about 10% for first reflection). This happens even if Zeq is constant!

tl;dr: You should consider properly terminating your cable when using a 10ft long cable and/or getting a short cable. Also, the measured signal is a product of wave interference, even if the cable's equivalent impedance is near constant. In all cases the properly terminated signal always propagates to the correct level without any distortion, and is only delayed because of the finite signal speed. Only with a properly terminated/short cable can you tell if your clock signal is really bad, or if you're just measuring reflections.

You're absolutely right. For every frequency source you'll have a different behaviour in your transmission line since its impedance varies with frequency. You didn't say but as you mentioned 12.5MHz clock I suppose this is a square wave, right? If so, you can expect different reflection magnitude for each frequency component of your clock, hence leading to a highly distortion signal in the end.

• The impedance of a transmission line is independent of frequency. – Chu Jul 13 '16 at 17:06
• That's only valid for ideal TL. You can never expect that in real life mostly due to dielectric behavior – PDuarte Jul 13 '16 at 17:10
• Unless you use some really crap cables, it should be pretty close to constant Z0 up to at least 500 MHz, which would be the 40th harmonic of OP's signal. – The Photon Jul 13 '16 at 17:13
• A capacitive load or parasitic at the termination is a much more likely explanation given OP says s/he doesn't know what the termination is. – The Photon Jul 13 '16 at 17:14
• @chu - not true except for frequencies above about 1 MHz. – Andy aka Jul 13 '16 at 18:48