What does it mean that the bandwidth of PSK modulated signal without applying any filtering or shaping of the pulse, is equal to the sampling rate with "1st lobe" equal to two times of symbol rate? What is "1st lobe" here? And how it equals to two times of symbol rate?

I've also read, that with applying pulse shaping techniques, "the null-to-null bandwidth is approximately equal to two times of symbol rate", but I don't quite understand what the null-to-null bandwidth is here, and how it equals two times of symbol rate (although I read that null-to-null bandwidth for PSK is $ 2/T $ but I don't know what it comes from).

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    $\begingroup$ Hi! Bandwidth depends on both symbol encoding (PSK, ASK, FSK, ...) and pulse shape (rectangular, raised cosine, ...), as both will shape the Power Spectral Density (PSD). For PSK and a rectangular pulse (no filter applied), the PSD will have a sinc^2() shape, which looks like a bunch of hills with varying amplitude, or "lobes". The first lobe (the largest) is contained between the first nulls (zeros): $\pm1/T$ (baseband) or $f_c\pm1/T$ (RF). The sinc^2() technically has infinite bandwidth, so this is a practical definition of bandwidth as it contains a significant majority of the signal power. $\endgroup$ – cjferes Feb 8 at 7:39
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    $\begingroup$ @cjferes nope! For linear modulations (i.e. PSK, ASK, not for FSK), as long as the expectation over all transmitted data symbols is 0 (which it usually is, unless you're doing OOK, which sucks for that very reason), the spectrum doesn't depends on the constellation, but only on the pulse shaping filter. That's a very fundamental theorem! $\endgroup$ – Marcus Müller Feb 8 at 12:15
  • $\begingroup$ (NB: transmit symbols need to be uncorrelated, which is an additional constraint to E(·)=0, but if you're not having that, your source coding lacks) $\endgroup$ – Marcus Müller Feb 8 at 14:45
  • $\begingroup$ I am well aware, but I was going for a more general reasoning here, hoping to address the underlying question (to the best of my understanding). Nevertheless, thanks for the specification, as it helps others as well! $\endgroup$ – cjferes Feb 9 at 17:02

With no filtering or pulse shaping what you have is your signal with the default rectangular pulse shape $$ \Pi(t) = \begin{cases}1 & \text{for} & -\frac T2\leq t\leq \frac T2\\0 & \text{otherwise}\end{cases} $$ Where $T$ is the symbol duration. The Fourier transform of the such a pulse result in a sinc function of the form $$ H(f) = T\operatorname{sinc}\big((f-f_c)T\big) \equiv T\frac{\sin\big(\pi (f - f_c)T\big)}{\pi \left(f-f_c\right)T}\tag{1} $$ The zeros in Equation $(1)$ are at \begin{align} (f - f_c)T &= k\quad \big\vert \quad k\in \mathbb{Z}_{\neq 0}\tag{with $f\neq f_c$}\\ \implies 2(f - f_c) &= 2kR_s\tag{2} \end{align}

The sinc function in Equation $(1)$ dies slowly resulting in infinite frequency content. Hence the choice of other practical pulse shaping filters such as the SRRC for bandwidth conservation (and ISI mitigation). In brief, having rectangular pulses one has remnant lobes outside the bandwidth of interest whilst with practical ones such as the SRRC you get high attenuation outside the main bandwidth.

For bandpass modulated signals (e.g. PSK) the required double-sided bandwidth is defined as $$ W_{\rm DSB} = (1 + \alpha)R_s\quad,\qquad 0 \leq \alpha \leq 1 $$ Where $\alpha$ is the roll-off of the raised-cosine pulse, a measure of excess bandwidth, and $R_s$ the symbol rate.

For bandpass systems, the null-to-null bandwidth is defined as $$f_2 − f_1\tag{3}$$ where $f_2$ and $f_1$ are the first nulls of the magnitude spectrum $|H(f)|$ above and below $\arg\max_\limits f\left(|H(f)|\right)$ respectively. So, from Equation $(3)$ and $(2)$ you have that \begin{align} \arg\max_\limits f\left(|H(f)|\right) &= f_c\\ \left(f_2 - f_1\right) = 2(f - f_c) &\equiv 2R_s\tag{with $k = 1$}\\ \end{align}

  • $\begingroup$ Thank you so much! One more question. Isn't T the duration of the symbol and not the symbol rate? Isn't symbol rate 1/T? $\endgroup$ – MagicMan Feb 9 at 1:41
  • $\begingroup$ Because I don't fully understand conversion from $(f - f_c)T = k$ to $2(f - f_c) = 2kR$ $\endgroup$ – MagicMan Feb 9 at 2:17
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    $\begingroup$ @MagicMan, correct $T$ is the symbol duration, my bad, it was a typo. The symbol rate $R_s = 1/T$. For this reason $(f-f_c)T = k \implies (f-f_c) = k/T = kR_s$ $\endgroup$ – Gilles Feb 9 at 7:10

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