1
$\begingroup$

I've seen countless definitions of bandwidth that disregard negative frequencies.

Is that done due to the symmetrical nature of a certain signal eg. a cosine or due to the inherent lack of meaning that negative frequencies have?

The main source of my confusion is that based on how you interpret the bandwidth you also interpret the range of the signal. Sometimes I've seen bandwidth $W$ expanded as frequencies ranging from $-W$ to $+W$ and some other times (mainly on filters) I've seen $W$ expanded as $-\frac{W}{2}$ to $+\frac{W}{2}$.

Which of the above methods should I use?

$\endgroup$
1
  • 1
    $\begingroup$ It's an issue of convention. Real signals (that have no imaginary component) have symmetrical spectra about DC (or 0 Hz). Usually, for bandpass signals (like AM or FM or something modulated to a radio frequency) the bandwidth is of just one of the images (like one at the positive frequencies). And usually, even for lowpass (or baseband) signals, the bandwidth is from 0 Hz up to the bandedge, so that is also one-sided bandwidth. $\endgroup$ Commented Sep 24, 2023 at 0:00

1 Answer 1

1
$\begingroup$

Single-sided and double-sided bandwidth needs to be specified explicitly to be clear. The double-sided bandwidth of a modulated signal at baseband (including the negative frequencies) corresponds directly to the occupied bandwidth at RF (passband) in either the positive or negative frequencies: for example a baseband signal that extends from -1 MHz to +1 MHz in spectral occupancy when frequency translated to a real carrier at 100 MHz, will occupy the spectrum from 99 MHz to 101 MHz as well as the spectrum from -101 MHz to -99 MHz.

See the graphic below showing the related spectrums for complex baseband and real passband modulated signals. An IQ Mixer translates the baseband signal to passband.

IQ to RF

It is when the baseband signal is completely real that the positive and negative frequency depiction becomes redundant since the the spectrum will be complex conjugate symmetric (the positive and negative frequencies have the same magnitude and opposite phase). For that reason we can describe all the information using a single-sided bandwidth (positive frequencies only). It only makes sense to use single-sided bandwidth for real signals. However the relationship between this real spectrum at baseband considering both positive and negative frequencies and the spectrum in the passband is the same as described above. That said, if a real baseband spectrum has a single-sided bandwidth of 1 MHz (for example), then it's double-sided bandwidth at baseband including both the positive and symmetric negative frequencies will be 2 MHz, and the passband bandwidth will be 2 MHz as well.

For comparison to the prior graphic, below shows the case for spectrums for real baseband and real passband modulated signals. A single real multiplier (mixer) and local oscillator can be used in this case to translate the baseband signal to passband. In this graphic, the double-side bandwidth at baseband is $B$ and the single-sided bandwidth would be $B/2$.

Real to RF

When we use “negative frequencies” we are referring to frequency components of the complex form $e^{j\omega t}$ in contrast to real sinusoidal components such as $\cos(\omega t)$. These other existing posts provide more details on the frequency translation process and how the baseband and passband signals are related as well as intuition for "positive" and "negative" frequencies:

Frequency shifting of a quadrature mixed signal

About the process to convert basedband signal into passband

$\endgroup$
6
  • $\begingroup$ Single-sideband modulation is common enough that I'm not sure you can safely assert that the double-sided bandwidth at baseband corresponds directly to occupied bandwidth at RF. You at least need a qualifier in there that there are complicated modulation schemes (i.e., SSB) that don't follow that rule. $\endgroup$
    – TimWescott
    Commented Sep 23, 2023 at 17:42
  • $\begingroup$ @TimWescott I'm not sure I follow that--- Wouldn't the SSB at real passband then match the DSB bandwidth at complex baseband (since the occupancy will only be one side, the SSB and DSB at baseband would have the same spectral occupancy---so to me it still holds. Basically whatever bandwidth we see at baseband (including all spectrum positive and negative) IS the bandwidth you see in the real passband in just the positive frequencies as well as just the negative frequencies. It may not be symmetrical at baseband, or even occupy both positive and negative- but I didn't suggest it had to be. $\endgroup$ Commented Sep 23, 2023 at 18:33
  • $\begingroup$ I'm not sure I'm parsing what you're saying -- but a 300-3000Hz audio signal occupies 6kHz for AM or narrow-band PM, and 2.7-ish kHz for SSB. That's why people go to the extra expense and bother of designing, manufacturing, and using SSB equipment. $\endgroup$
    – TimWescott
    Commented Sep 23, 2023 at 22:26
  • $\begingroup$ @TimWescott I am referring to the modulated signal and it’s bandwidth: For both of those cases the complex baseband signal (as modulated) will match the occupied bandwidth of the real passband signal when you include the positive and negative frequencies. For SSB the baseband signal will only have spectrum in either the positive or the negative frequencies with the bandwidth you gave. $\endgroup$ Commented Sep 23, 2023 at 22:30
  • $\begingroup$ OK -- this is a point of view, technical English dialect thing. You're going with strict signal processing technical English. However, if I look in my 1953 ARRL Hand book and build the circuit called a "SSB Modulatior", then I'll talk into the microphone (300Hz-3kHz input, "single ended") and I'll get a 3kHz-ish bandwidth out. If, instead, I use the big tubes and build an AM modulator, I'll get a 6kHz-ish bandwidth out. $\endgroup$
    – TimWescott
    Commented Sep 28, 2023 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.