My understanding of a naive PSK scheme is that you have some frequency(s), f(n), at baseband, and you modulate information by shifting the phase (what "point in time") the signal is at. Isn't the phase shift instantaneous? In this case, isn't the frequency constant? Why do PSK modes look vaguely like MFSK in a waterfall?


3 Answers 3


Binary PSK with instantaneous phase shifts would be equivalent to multiplying a sine wave (the carrier) with a square wave with values at 1 or -1.

When two signals are multiplied, this forms a frequency mixer. A mixer with inputs at frequencies $f_1$ and $f_2$ creates outputs at $f_1 + f_2$ and $|f_1 - f_2|$.

A sine wave is just one frequency, let's call that $f_c$ for the carrier frequency. And the square wave will be at the symbol rate, which for PSK31 is 31.25 symbols per second. A square wave is a series of odd harmonics. More specifically, a square wave at frequency $f$ is equivalent to the infinite sum:

$$ \sin(2\pi f) + {1\over 3} \sin(3\pi f) +{1 \over 5} \sin(5\pi f) + \dots$$

This means a square wave at 31.25 Hz has frequency components at:

  • 31.25 Hz
  • 93.75 Hz (31.25 * 3)
  • 156.25 Hz (31.25 * 5)
  • 187.5 Hz (31.25 * 7)
  • ...

So say you're transmitting PSK at 14.075 MHz at a symbol rate of 31.25 per second. This means you'll be emitting power on frequencies:

  • $14.075\:\mathrm{MHz} \pm 31.25\:\mathrm{Hz} $
  • $14.075\:\mathrm{MHz} \pm 93.75\:\mathrm{Hz} $
  • $14.075\:\mathrm{MHz} \pm 156.25\:\mathrm{Hz} $
  • $14.075\:\mathrm{MHz} \pm 187.5\:\mathrm{Hz} $
  • $\dots$

As you can see, the bandwidth extends out to infinity. The power diminishes as you get away from the carrier frequency, but not very rapidly, and it never reaches zero. If you're transmitting with 1 kW then you'll be spewing significant harmonics over the entire band, and even outside it.

Consequently, except for very low power, cheap radios you might find in part 15 devices, the phase shifts are not instantaneous but gradual. For example PSK31 uses a cosine envelope, meaning in the case of alternating between phases it multiplies the carrier not by a square wave, but rather by a cosine. Since a cosine consists of just one frequency component, this generates not an infinite series of frequency components in the output of the mixer, but just two: the carrier frequency, plus and minus 31.25 Hz.

Things get a little worse when the phase isn't strictly alternating between states, because the first derivative of phase is discontinuous. This does generate an infinite series of harmonics (I have a graph in another answer) but one which decreases much more rapidly than the square wave case before. It should be noted the technical design of PSK31 isn't especially good, and professionally design PSK implementations often use a root-raised cosine pulse shaping filter which is better in this regard.

In general, the only thing which occupies just one frequency is a sinusoid with no start and no end which isn't modulated at all. Changing the amplitude or phase in any way will cause the signal to occupy more bandwidth. It's rather easy to demonstrate why this must be true intuitively: were it possible to transmit information with just one frequency, signals could be crammed infinitely close together, so an infinite number of users could be crammed in a finite amount of bandwidth. There'd be no need to license or sell spectrum because there'd always be room to add more users. Also, we could fit infinite information bandwidth in any slice of spectrum, so we wouldn't need more bandwidth signals for higher data rates.

The more gradually amplitude or phase change, the less bandwidth will be occupied. Ideally the derivative of amplitude and phase are continuous functions, as well as the second, third, and so on derivatives. A gaussian function higher-order derivatives are all continuous, which is why you see gaussian functions come up in modulations like GMSK.


Isn't the phase shift instantaneous?

Ideally yes, in practice it is not.

In this case, isn't the frequency constant?

No, phase and frequency are related. A shift in phase is equivalent to a shift in frequency. People found that looking for a phase shift instead of a frequency shift can take less RF bandwidth for the same data throughput.

Why do PSK modes look vaguely like MFSK in a waterfall?

Because, in a manner of speaking, PSK is a lot like MFSK. Nyquist-Shannon says that data transmission takes bandwidth. There is a minimum bandwidth required for any data to move at a given rate. The more noise on the data path the greater the bandwidth is needed to overcome this. Because noise is, roughly speaking, correlated to the bandwidth of the channel it helps to minimize the bandwidth to minimize noise.

With a noiseless wire that has no resistance or capacitance the bandwidth needed for infinite data throughput is zero. Since we don't live in an ideal world data takes bandwidth.

I don't know if I'm helping here since my vocabulary may have some nuanced differences to yours. Looking up the theory on Shannon and Nyquist bandwidth will help. As will the relationships between phase, frequency, and amplitude.

  • 1
    $\begingroup$ I would say an instantaneous phase shift is very far from ideal. Probably also illegal, as it would generate spurious emissions on many out-of-band harmonics. $\endgroup$ Commented Jan 6, 2021 at 18:24
  • $\begingroup$ " I would say an instantaneous phase shift is very far from ideal. Probably also illegal, as it would generate spurious emissions on many out-of-band harmonics. " Ideally anything outside of the intended channel would be filtered, or ideally nobody would care about out of band harmonics. $\endgroup$
    – MacGuffin
    Commented Jan 6, 2021 at 19:00
  • 1
    $\begingroup$ If you filter the out of band harmonics, you aren't changing phase instantaneously anymore. And if you are imagining a world where no one cares about out of band harmonics so there's no need to filter, that isn't an idealized world, just a simplified one. $\endgroup$ Commented Jan 6, 2021 at 19:06
  • $\begingroup$ "the bandwidth needed for infinite data throughput is zero" The Shannon-Hartley theorem is $C=B \log_2(1+{S/N})$. Even if $S/N$ approaches infinity, the capacity is zero if the bandwidth is zero. Transmission of information with zero bandwidth is impossible, even in the absence of noise. $\endgroup$ Commented Jan 6, 2021 at 20:02
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    $\begingroup$ @Expectator probably the relevant thing is not the Nyquist-Shannon sampling theorem but rather the Shannon-Hartley theorem. A PSK signal is an analog thing in the electromagnetic field -- sampling there is not directly relevant to your question. $\endgroup$ Commented Jan 6, 2021 at 20:36

For a mathematical sinusoid, instantaneous frequency is the first derivative of the phase of that sinusoid with respect to time. So if the phase isn't changing at a constant rate with respect to time, the first derivative will change, and thus so will the instantaneous frequency.

Also, in the real world, there can be no instantaneous discontinuous phase changes, as all capacitors (including all the parasitic ones and in the wires) require finite time to charge up or down in order to change signal levels. Any band-limiting filters reduce the rate of change even further.


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