I want to simulate the power spectral densities of different modulations in MATLAB (without Simulink). Unfortunately I have only very basic understanding of signal processing, hence I struggle with this topic. Currently I have a very simple spectrum like this: Spectrum without modulations There are different objects (A and B in the picture above) sending signals on different frequencies. I am not simulating any bits or calculating a sine, I am just assuming that there is something going on on a certain frequency with a certain power level.

Now, I want to simulate different digital modulations, e.g. PSK, BPSK, GMSK, FSK. I understand that the spectrum should look something like this: Spectrum with digital modulations There is not a single center frequency, but also side bands. A and B shall illustrate different modulations. My question now is, how can I calculate these different modulations? In text books I may find figures displaying the 'power spectral density' of ASK, but often times there is no formula how to calculate this. I guess the spectral density depends on the actual bit sequence, but I just need the envelope.

In one text book ("Digital Modulation Techniques" by F. Xiong) I found this formula for ASK: $$\Phi_s(f) = {A^2T \over 4} \mathrm{sinc}^2\big[T(f - f_c)] + {A^2 \over 4} \delta(f - f_c)\big]$$ where $T$ is bit duration, $A$ is carrier amplitude, $f_c$ is carrier frequency. At first I found this formula promising, because it does not require a bit sequence, but I do not understand the last addend.

  • Are there any formulas like this for other modulation techniques?
  • If not, how could I calculate these graphs?
  • 1
    $\begingroup$ A number of graduate-level texts include the calculations you want: Sklar, Proakis, Lapidoth, Benedetto, Blahut.... $\endgroup$
    – MBaz
    Dec 12, 2017 at 22:46
  • $\begingroup$ The question of "how does the spectrum around the carrier frequency look like for a digital modulation" is identical to asking for the complex equivalent baseband spectral shape. As @MBaz pointed out, that's something that textbooks arrive at after an introduction to baseband signals and digital comms: After reading such a book, you'll know the answer to your question and a lot more (for example, why F. Xiong's formula can never be true for any bandwidth-limited (i.e. real world) system). $\endgroup$ Dec 13, 2017 at 10:18

1 Answer 1


At first I found this formula promising, because it does not require a bit sequence,

That's the point about PSDs: They are an expectation value of a stochastic process! That is, the randomness (in this case, the data you're transmitting) is being "averaged away".

Deriving the spectral shape of any modulation (including the pulse shaping, don't forget that – you don't want to end up with a result like the formula you cite, as that is hardly band-limited!) is actually not that hard: You find a formula for the autocorrelation of the signal, and fourier transform that, and then you find the expectation. The trick is in finding a suitable form of the autocorrelation formula!

but I do not understand the last addend.

That's a speciality of ASK: Most constellations are designed so that the mean of the information symbols is 0; that's not the case for ASK.

To make it short: The PSD $\Phi_{vv}$ of the baseband signal $v$, consisting of a sequence of information symbols $I_n$ with variance $\sigma_i^2$ and mean $\mu_i$, modulated at a symbol period $T$ with a pulse $g(t)$ (the pulse's spectrum being $G(f)$) is (from Proakis "Digital Communications", 3rd ed., eq. 4-4-17, on page 207):

$$\Phi_{ii}(f) = \frac{\sigma_i^2}{T} \left\lvert G(f) \right\rvert^2 + \frac{\mu_i^2}{T^2} \sum\limits_{m=-\infty}^{\infty} \left\lvert G\left(\frac mT\right) \right\rvert^2\delta\left(f-\frac mT\right)\tag{4-4-17}$$

where $\delta$ is the dirac delta distribution.

This applies to all linear modulations – PSK, QAM, ASK, OOK, but not to non-linear ones (FSK etc).

In the case of the example you cite, a rectangular pulse shaper $g(t)$ was chosen, which is $1$ for the period $-\frac T2, \frac T2$ around a "symbol time", and $0$ else. That one has a $\DeclareMathOperator{\sinc}{sinc}\sinc$ Fourier transform shape. That's where your $\sinc^2$ comes from!

The second term, the $\frac{A^2}4 \delta$ is what the sum in eq. $(4-4-17)$ collapses to, because the $\sinc$ is zero in every $mT$ but for $m=0$.

As said above, most constellations are designed to have zero mean ($\mu_i = 0$), so that the sum term just disappears, and your spectral shape is only defined by the spectrum of your pulse. (Only ASK and OOK differ, as far as commonly used modulations are concerned, as far as I can tell.)

Your formula uses $(f-f_c)$ instead of $f$, because it's already the bandpass signal's shape. Don't do that to you! Just consider the signal spectra in baseband, because we can mathematically shift those anywhere we like (especially by an $f_c$).

If you don't have access to good literature: Proakis 3rd ed can be had only for less than a fancy coffee at an American coffee house chain at online used book sellers. It's not my favourite edition of the book, as it cuts the math parts where the second edition really excelled, but it's still a very solid book, and as far as I can see, the cheapest edition of the "very established" digital comms books (being Proakis, Sklar, Tse/Viswanath, and, if you're German, Kammeyer). I like Lapidoth's book, too, and it can be had for free on his website. (update: I just found out that you can get Tse online, too, which is awesome! Might be a bit too advanced, though.)

  • $\begingroup$ Good answer so far, but it ends kind of abruptly ... :) $\endgroup$
    – Matt L.
    Dec 13, 2017 at 11:48
  • $\begingroup$ @MattL. I kind of feel the same, but I'm not sure what to add – the whole derivation to arrive at $(4-4-17)$? An intro to equivalent BB? Both seemed to be a bit "over the top", so, abusing you as (cheap) second opinion: what's missing? $\endgroup$ Dec 13, 2017 at 11:55
  • $\begingroup$ No, that's not what I mean. At least in my browser the last sentence is literally cut off: "It's not my favourite edition of the" the what? :) $\endgroup$
    – Matt L.
    Dec 13, 2017 at 11:57
  • $\begingroup$ oooooooooh! oh! fixing that! $\endgroup$ Dec 13, 2017 at 12:00
  • $\begingroup$ I wouldn't recommend Tse and Viswanath here -- it's a more advanced book which assumes you already know the basics inside and out. $\endgroup$
    – MBaz
    Dec 13, 2017 at 14:06

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