# signal propagates in bending transmission line (unequal length between signal path and return path)

Suppose I have a square wave signal generated from a signal generator and its outputs are connected to the two terminals of a paired PCB traces (left side of the picture below). The other end of the paired PCB traces are connected to a oscilloscope. V+(I+) and V-(I-) should arrive the red dash line at the same time according to transmission line theory. However, the outer transmission line is longer than the inner transmission line because of the bend, thus, V- travels a longer distance than V+, so should I see a the two signals appears on the oscilloscope in different time with V+ arrives at oscilloscope earlier? I don't think this is right, but I cannot think of a reason why. My gut feelings tell me that the two signal should arrive at oscilloscope at the same time, but I cannot explain it in physics. Can anyone help me to explain it? Thanks

• it depends whether (1) your V+ and V- conductors are each a transmission line with ground with little coupling between them, or (2) the ground is remote and they are tightly coupled, or (3) some situation intermediate between the two. In case 1, both signals will arrive at different times, each with respect to ground. In case two, the signal, the differential signal, will arrive at the scope. Jun 4, 2019 at 18:45
• @Neil_UK, but even in situation (2) the differential signal would be distorted and/or coupled into a common-mode signal by the mismatch. Jun 4, 2019 at 18:52
• @Neil_UK, what if my V- is ground? Jun 4, 2019 at 19:02
• @Neil_UK, recently I've been thinking your case 3, can you please elaborate more what will happen in case 3? Would be perfect if you can shows some examples. Thank you! Jun 10, 2019 at 7:52
• The result will be intermediate between the two. Get a circuit simulator, LTSpice is good, free, widely supported, and have a play. Jun 10, 2019 at 9:05

However, the outer transmission line is longer than the inner transmission line because of the bend, thus, V- travels a longer distance than V+, so should I see a the two signals appears on the oscilloscope in different time with V+ arrives at oscilloscope earlier?

Yes, this can happen.

It also causes differential-to-common-mode conversion which can be bad for EMC.

But it's normally only a significant problem for signal frequencies above 2 GHz or so.

Reducing this effect is why you'll sometimes see a differential pair with a serpentine in one trace but not the other:

• thanks for your prompt response! I have a follow up question if you don't mind to answer. In your attached picture, the purpose of serpentine is to make the differential line has equal length, the downside is that it causes impedance mismatch since the space between two lines changes here and there because of it, am I thinking right? Jun 4, 2019 at 18:34
• @WenhaoChen, if you're being very fussy, you can make the track segments on both lines slightly wider where the tracks are separated, to maintain the Z0. (You'd basically use the track width for separate single-ended lines) I've not seen anyone bother to do that for frequencies below maybe 10 GHz, though. Jun 4, 2019 at 18:50
• "But it's normally only a significant problem for signal frequencies above 2 GHz or so". On typical bench-sized setups. Run the cables the length of a 747 and the "don't care" frequency range will be different. Run the cables from San Diego to Yreka California, and the frequency range will be different yet. Jun 4, 2019 at 19:30
• @TimWescott, I was thinking that very few people build circuit boards that reach from San Diego to Yreka. But now I see the question is tagged as [power-engineering], so maybe that is an important application to keep in mind. Jun 4, 2019 at 20:07
• @Tim, and in the 747 scenario, it doesn't really matter what's the total length of the line. A single bend, like presented in the question, will produce the same length mismatch regardless. Jun 4, 2019 at 20:09