I have two signals $x_1(t)$ and $x_2(t)$ both with a baseband bandwidth of $W$ Hz. $x_1(t)$ is transmitted over carrier frequency $f_1$ Hz, and $x_2(t)$ over carrier frequency $f_2=f_1 + \frac{W}{2}$ Hz. The received signal can be written as

$$ \begin{split} r(t)=&\Re\left\{x_1(t)e^{j2\pi f_1t}+x_2(t)e^{j2\pi f_2t}\right\}\\ =&\Re\left\{e^{j2\pi f_1t}\left[x_1(t)+x_2(t)e^{j2\pi (f_2-f_1)t}\right]\right\} \end{split} $$
The baseband signal $x_1(t)$ spans $[-W,W]$, while $x_2(t)e^{j2\pi(f_2-f_1)t}$ spans $[-\frac{W}{2},\frac{3W}{2}]$. I want to calculate the power of $x_2(t)e^{j2\pi (f_2-f_1)t}$ that interferes with $x_1(t)$ from the power spectral density (PSD) of the aforementioned signal.

I know to find the power of $x_1(t)$ from PSD, I need to find the area under the PSD in $[0, W]$, and multiply it with 2 to account for the negative frequencies. Where I am not sure is whether to calculate the power of $x_2(t)e^{j2\pi (f_2-f_1)t}$ from $[\frac{-W}{2}, W]$ or the power in $[0, W]$, and if I need to double any part of the frequencies?

In the bandpass it's clearer, since $x_2\cos\left(2\pi f_2t\right)$ interferes with $x_1\cos(2\pi f_1 t)$ in the frequency range $[f_1-\frac{W}{2}, f_1+W]$. However, to find the PSD in the bandpass, I need to use a sampling frequency greater than $2f_1+W$, which is very large in case $f_1$ is in GHz.

To sum-up:

  • Which is better to do the spectral analysis in baseband or bandpass
  • How to deal with spectral of $x_2(t)$ in baseband in the example above?
  • $\begingroup$ with or without filters? $\endgroup$ Nov 24, 2022 at 16:24
  • $\begingroup$ Without bandpass or baseband filters $\endgroup$ Nov 25, 2022 at 12:35

2 Answers 2


It should not matter whether you perform your PSD estimation in baseband or not. Basebanding a signal is carried out with a complex demodulation, which simply shifts the frequency of the measurement. For a bandlimited signal, appropriate low-pass filtering will practically prevent amplitude distortion. Just make sure your integration bounds are consistent with the frequency shift of the demodulation. For example, if you were finding the PSD of $x_1(t)$ and demodulating by $f_1$, your bounds would shift from $[f_1-W,f_1+W]$ to $[-W, W]$. PSD estimation typically only requires multiplying the Fourier transform by its conjugate - not scaling it by a factor of two. There are many PSD estimators out there, many of which have been discussed on DSP.SE.

Separating $x_1(t)$ from $x_2(t)$ spectrally is impossible unless you have prior information about those components, can make reasonable assumptions about the spectral structure, or can measure them individually at some point to aid you in either of those tasks. If you know nothing about the signals, how can you tell the difference between the signal you stated and a single signal without separable components and a bandwidth of $[-W, 3W/2]$?

  • $\begingroup$ Thanks for the response. For $x_1(t)$, if you find the area under PSD from $[-W,W]$, then you don't double the PSD. But, if you find the area under PSD from $[0,W]$ (the +ve frequency), then you do. For $x_2(t)$, I don't aim to separate its spectral from that of $x_1(t)$. I only want to calculate how much power from $x_2(t)$ interferes with $x_1(t)$. How can I find the area of PSD in negative frequencies $[-W,\frac{3W}{2}]$, since FFT in matlab, as far as I understand, starts with frequency 0 and upwards to $f_s/2$, where $f_s$ is the sampling rate? $\endgroup$ Oct 28, 2022 at 7:40

1.- There's an assumption in your question that is not correct, let me explain :

where you wrote

'.. while x2(t)exp(1j2pi*(f2-f1)) spans [-W/2 3/2W] ..'

you could have written

'.. while x2(t)*exp(1j*2*pi*(f2-f1)) may or may not span [-W/2 3/2*W] many modulations produce RF signals which are either symmetrical around f1 (the carrier or channel centre) or are Single Side Band spanning [f1 f1+W2] , as well as W2 may or may not be constant ..'

2.- Let be the signals x1 and x2 of your question as follows :

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Then whatever modulation on real signals produces symmetrical spectrum and SSB (Single Side Band) can be applied without losing signal.

But do x1 and x2 agree upon what side of f1 is SSB implemented?

x1(t) x2(t) real

enter image description here

enter image description here

3.- As a matter of fact all operators that make money out of the spectrum, more or less, but most of time in compliance, do filter and have to filter signals.

How well filtered may be up for debate, but they have to, otherwise :

3.1.- The spectrum would be wild-west, that it was in early wireless communications.

3.2.- Regulators that already filled up coffers with outrageous licence fees would and do readily apply fines when unfiltered signals start causing interference that ends up in loss of revenue.

4.- Even those operators that do not (don't need to) make money out of the spectrum (for instance military) also apply filtering to signals, both in transmission and reception.

Basically and just as bottom lines because :

4.1.- filtering transmission allow control of transmitted power, thus controlling for instance range, and signal spectrum shape, easying reception, and

4.2.- filtering reception 'cleans' signals allowing detection of lower power levels

5.- When you are about to design a modulation, what would be the point of leaving significant spectral tails (the W/2 mentioned in the question) below f1 if you are already halving the signal below f1 anyway?

So rewordking, most likely :

X2(f) has bandwidth [-W2+f1 f1+W2] or [f1 f1+W2] where W2 may or may not be same as W, and may or may not be constant.

6.- Bear in mind that for instance DVB-T and in general OFDM modulations stand a lot of in-channel interference, compared to analog and digital single carrier modulations.

I can attest for real time overlapped UHF PAL TV signal, your x1, and DVB-T 8k, your x2, right on same RF channel, the DVB-T 9dB(W) below the PAL signal, and both were doing wonderfully well ignoring each other.

7.- With CDMA (Code Division Multiple Access), SS (Spread Spectrum), and now IoT, in-channel robustness is even higher.

Because unless receive codes available, all signals can and in fact take the whole band, all at the same time, and as long as no signal 'too loud' such systems even have certain advantages compared to frequency-channel multiplexed schemes. If you don't believe me ask for instance Qualcomm.

Thanks for reading.


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