Filtering a QPSK signal

Question

Envelope variation due to filtering of a QPSK signal becomes more pronounced at greater phase steps.

Why is that? Why do greater phase steps (180 vs 90 degrees for example) lead to more pronounced envelope variations?

Answer 1

I think a picture of the QPSK constellation would be helpful to intuitively see what is occuring. Consider two possible transitions in the QPSK constellation as shown below, from "B" to "C" (180° transition) and from "A" to "C" (90° transition):

Note the difference between a 180° transition as when we move from "B" to "C" compared to a 90° transition in moving from "A" to "C". The instantaneous amplitude will be the vector magnitude at any point along the dashed lines during the transition. If the signal and has infinite bandwidth (not realistic, but if), the transition will be instant and therefore the vector magnitude will never change (will be constant envelope). However with any filtering there will be a finite time to complete the transition, and we will observe the magnitude change according to the vector length along the path taken. I placed stars where the magnitude is minimum in each case, and it is clear how in the 180° case where we pass through the origin, the magnitude approaches zero while in the 90° case the reduction in magnitude is much less.

Based on actual filtering and prior symbol history the paths will deviate somewhat from what I shown in the picture above, but the concept still holds. For example below shows a constellation of a root raised cosine signal where we can make out the same effect with the different transitions.

Minimizing the peak to average ratio is important in power sensitive applications such as Satcom and battery operated devices. This is the motivation for modified waveforms such that no trajectories go through the origin such as $\pi/4$-QPSK and waveforms that are constant envelope such as MSK and and FQPSK.

Answer 2

That is actually specific to the transmit filter, and whether your system tries to link symbol rate to RF carrier frequency.

In general, I don't see that statement being true, so the answer to your question really is

because the convolution of the transmit filter with two consecutive opposite symbols has a lower envelope

and that's about it – the filter was designed that way. Not all filters are.

A note on the figure of waveforms you show:

I'm a bit biased, as anyone is, by the way I've been educated, but really, explaining QPSK with passband signals feels very 1980's to me, and it would probably be easier to understand what happens if you looked at QPSK happening in baseband, and at the filter happening in baseband.

Because, a) that's easier to understand (in my humble opinion) and b) it's closer to how one would implement a QPSK transmitter – you don't filter the bandpass signal, you filter the baseband signal to get your pulse-shaped QPSK, and only then you modulate it, exactly as the upper block diagram shows.

Now, when you consider QPSK in baseband, then you realize that the pulse-shaping's job is to find a smooth line between the symbol instants, where the values must be the symbol values (in case of QPSK, that'd be {1, j, -1, -j}). The way you "connect" these points defines the envelope variation you get. If you're using a linear filter (which you really should for many reasons), then you've got limited freedom in how to choose these trajectories, but, for example, a half-$\cos$-shaped pulse shape is something that stays on exactly the same envelope, and hence doesn't have the envelope variations in your question. That's easy to see in the constellation diagram, but it's hard to see in the passband signal's shape!

Answer 3

Think of the signal in the IQ domain, ie the complex plane. The unfiltered signal can be thought of as jumping instantaneously between the individual constellation points (on the unit circle for QPSK), so that its always at unit amplitude.

Once filtered the signal can be thought of as moving around the complex plane smoothly in loops, crossing the the relevant phase points only on the sample times. This can involve the signal spending a lot more time nearer the origin when the neighbouring phases are opposite, ie the amplitude can be less than unity (and also it can be more as the signal loops outside the unit circle sometimes).