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Suppose I have a signal $x(t)$ with baseband bandwidth $W$ Hz and power $P$, and consists of $K$ samples, and I draw the two-sided power spectral density (PSD) of this [baseband] signal in MATLAB as

xFFT = fftshift(fft(x));
psd = (1/(fs*K))*abs(xFFT).^2; %fs is the sampling frequency
freqAxis = -fs/2+fs/K:fs/K:fs/2;
plot(freqAxis, 10*log10(psd));
xlabel('Frequency');
ylabel('PSD [dB/Hz]');

Obviously, most of the signal power is in $[-W,W]$, and in MATLAB, I can find an approximate of the signal power by finding the area under the PSD in that range.

Are things the same in practice? For example, at the output of the analog-to-digital converter (A/D), can I do the same operations I did above, and find the area under the PSD between $[-W,W]$ Hz, or negative frequencies don't exist in practice, and this is just a mathematical representation?

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6 Answers 6

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negative frequencies don't exist in practice, and this is just a mathematical representation?

I'd say it's exactly the the opposite. Signals with only positive frequencies do not exist in nature and they are indeed just a convenient mathematical representation.

All physical signals are real (as in "not complex"). We often use complex numbers to describe and model them but whenever we go back to the actual physical world (voltage, excursion, pressure, current, field strength, antenna signal, DAC output, etc.) everything becomes real again and every signal that's real has the same amount of energy at both negative and positive frequencies.

This is a direct consequence of the definition of the Fourier Transform and the associated definition of "frequency". There are other transforms that are not complex (Discrete Cosine Transform, for example) and one could argue that they use a somewhat different definition of "frequency".

But as long as you use Fourier Spectra: all physical signals are real and hence their Fourier spectra are conjugate symmetric with equal magnitudes and opposite phases at positive and negative frequencies .

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  • $\begingroup$ In the realm of radio we can use complex I/Q signals which are frequency-shifted radio passbands. Then a negative frequency is a frequency that was originally less than the shift frequency, and a positive frequency is one that was originally above. The same works in reverse if you shift upwards: negative frequencies go to below the shift and positive frequencies go to above it. $\endgroup$
    – user253751
    Nov 8, 2022 at 17:43
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Negative frequencies do/don't...

First

In practice, if you have a single-valued real signal, the spectrum is symmetric across 0Hz (specifically, the negative frequency component is the complex conjugate of the positive frequency component).

If I hand you a spectrum of a signal going from 0Hz up and I tell you the signal is real then reconstructing the spectrum for negative frequencies is trivial. So it's not uncommon that folks will consider negative frequencies to be meaningless.

Second

There are times, in practice, where it is very handy indeed to represent a signal as a complex number. Usually this is done because you're processing some RF signal and your first step is inphase/quadrature downconversion. Sometimes it's done because you have a sine and cosine signal from a rotating shaft.

In these cases, then the arithmetic on an inphase/quadrature pair is exactly the same as the arithmetic over the complex numbers -- for $x = \begin{bmatrix}x_i & x_q\end{bmatrix}$, you can treat $x$ as $x = x_i + j x_q$. When you do this, then taking the Fourier transform of $x$ results in a spectrum that is, in general, not symmetrical around 0Hz -- and suddenly, negative frequencies acquire a weight and depth of meaning that cannot be ignored.

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Yes these are the same in practice; you can do the same operations on the real ADC samples. The FFT of a set of samples from the ADC will be a complex result consisting of positive frequencies for the lower half of the result and negative frequencies for the upper half of the result. The real signal that the ADC in sampling have a Nyquist range in frequency from $-f_s/2$ to $+f_s/2$ (where $f_s$ is the sampling rate), which the FFT output is representing. For real signals, such a spectrum is complex conjugate symmetric (Hermitian symmetric), so the negative frequencies are redundant and therefore can be discarded. However, and beyond the detail of this post, we can instead have two real ADC's for which we can sample complex signals in which case the positive and negative frequencies can be completely independent. This exists in practice.

It's all equally abstractions: Real numbers, imaginary numbers, complex numbers, positive frequencies, negative frequencies. How can we say one "exists" and others don't? Because we use two real numbers (on paper) to describe a single complex number, does that mean the complex number is somehow less of a "reality" than a real number? It is the same thing in implementation. We implement complex data paths routinely in the construction of modern radio hardware. We just need two data paths to do such an implementation. In the same fashion, negative frequencies "exist", just as much as positive frequencies do. These are our tools to describe the physical world.

A common sinusoid, which we are first introduced to and therefore so comfortable with as being "reality" is composed of a positive and negative frequency given Euler's identity:

$$\cos (\omega t) = \frac{1}{2}e^{j \omega t} + \frac{1}{2}e^{-j \omega t}$$

Now from that, hopefully we have some clarity on what exactly a positive or negative frequency is (and that it "exists"!). A sinusoid itself does not represent a positive or negative frequency, but consists of both. A positive or negative frequency is specifically of the form of $e^{j \omega t}$ where the sign of the exponent indicates if this is a positive or negative frequency. Since $e^{j\phi}$ is simply a phasor with magnitude $1$ and angle $\phi$, a positive frequency is represented in the time domain as a complex phasor rotating counter-clockwise, and negative frequency is represented in the time domain as a complex phasor rotating clockwise. Any complex phasor rotating in the time domain at a constant rate with constant magnitude will be an impulse in the frequency domain.

After answering I realized this was asked and answered before, please see: Does negative frequency actually exist or it is just theoretical?

Please also see my favorite question here related to this (use of negative frequencies and specifically complex numbers to describe physical signals). So as long as we have two scope probes we can measure actual signals that are not real, but composed of pairs of real numbers (complex!), and so to the same extent a real signal "exists in practice", so too does a complex signal. Under the same definition of "existence", which is applicable to the question if negative frequencies exist in practice, I say yes they do!

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The negative frequency must always be taken into account. I suppose you are intrigued because the frequency interpreted as the inverse of the period should not be negative as the period is always positive.

However, FFT decomposes the signal as the superposition of complex harmonics. Complex harmonics are functions describing rotating complex phasors, they can rotate in clockwise or counter-clokwise direction, this is where the sign of the frequency comes from.

A real signal require the negative frequency components to cancel the imaginary part of the positive frequent components, e.g. $cos(\omega t) = (e^{j \omega t} + e^{-j \omega t})/2$.

As such, real signals will have exactly half of the energy in the positive frequency and the other half in the negative spectrum. So you could calculate the power by integrating the positive frequencies and multiplying by two.

Edit

By your comment It seems like you are interested in knowing if it would be possible to recover the positive frequency of a signal that has interference in the negative frequencies.

You have to remember that the received signal also has negative frequencies.

Consider two signals where one low frequencies of the s1 (the square), interfere with the high frequencies of s2.

illustration

One could think, great, if we move the signal to base-band the positive side will be intact, and we could reconstruct the signal. That is not true, because the negative side of the spectrum (in red) will also be moved to base-band, and on that side the frequencies above the center frequency are corrupted.

enter image description here

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  • $\begingroup$ In my system, I partition the bandpass spectrum into subbands, and when I shift them to the baseband, some of these subbands fall in the negative frequencies. For examples, if I have two subbands each of bandwidth $2W$ in passband centered at $f_1$ and $f_2=f_1+2W$, respectively, and I down-converted the spectrum by $f_2$, then the 1st subband would fall in the negative frequencies. Even half of the 2nd suuband would also. Is this OK in practice? Can I manipulate subband 1 normally, like finding the power under its PSD? $\endgroup$ Nov 7, 2022 at 10:54
  • $\begingroup$ In that case it is OK, if the passing band of the filter the first subband uses $f_1 - W < f < f_1 + W$, the second subband uses $f_2 - W = f_1 + W < f < f_2 + W = f_1 + 3W$ $\endgroup$
    – Bob
    Nov 8, 2022 at 8:53
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Since you asked for "in practice" and the other answers are a bit more theoretical:

A real-valued signal - such as a direct measurement - always has a symmetrical FFT. If it works for your purposes, you can simply discard one half and process the other.

However, it's also possible to have a complex-valued signal. This occurs frequently in radio, where we take some block of frequencies (such as 99-101MHz) and down-shift it by some frequency (such as 100MHz). Glossing over some details, this gives us a complex signal (also known as I/Q) from -1MHz up to 1MHz, which is not symmetrical. The frequency -0.8MHz after downshifting corresponds to 99.2MHz in the original signal, and +0.8Mhz corresponds to 100.8MHz. Completely different and unrelated frequencies.

It also works the other way: you can generate a complex signal and then up-shift it, so a -0.8MHz frequency without +0.8MHz, upshifted by 100MHz produces 99.2MHz, and +0.8MHz produces 100.8MHz.

If you simply produce an 0.8MHz sine wave with the sin or cos function and up-shift it, your 0.8MHz sine wave is actually -0.8MHz and +0.8MHz together. To produce only one or the other, you need a complex sine wave consisting of sin and imaginary cos components.

In a real-valued sine wave, the positive and negative frequencies can be seen to have the same real parts and opposite imaginary parts, with the imaginary parts cancelling out. If they don't cancel out, it means the imaginary part isn't zero and the FFT isn't symmetrical.

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  • $\begingroup$ Actually, I have a similar situation. I have two signals $x_1(t)$ and $x_2(t)$ both of baseband bandwidth $W$. The carrier frequency of $x_1(t)$ is $f_1$, while that for $x_2(t)$ is $f_2=f_1+2W$. I want to shift the spectrum $[f_L, f_U]$ that contains both signals by $f_2$, and then do some spectral analysis on the shifted spectrum. Obviously, the spectrum of $x_1(t)$ would be in the negative frequencies after down-converting the spectrum. Thus my question: can I do spectral analysis in negative frequencies? $\endgroup$ Nov 9, 2022 at 9:17
  • $\begingroup$ Maybe this is just a question of semantics, but in my opinion physical signals are always real. In your example of an IQ signal you simply generate two (real) voltage signals by downmixing with cosine and a sine. You can call one signal "real part" and the "imaginary part" but that's just a mathematical trick and doesn't change the fact that each individual signal is real and has both positive and negative frequencies. You cannot generate a physical signal that has no negative frequencies. $\endgroup$
    – Hilmar
    Nov 9, 2022 at 16:30
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    $\begingroup$ @Math_novice Yes, absolutely. Negative frequencies are just frequencies that were less than $f_2$ before the downshift. $\endgroup$
    – user253751
    Nov 9, 2022 at 17:53
  • $\begingroup$ @Hilmar The two parts of a complex number are always called "real" and "imaginary". The physical implementation of a complex number signal is two real signals but that's just the physical implementation and not the theory. $\endgroup$
    – user253751
    Nov 9, 2022 at 17:53
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An FFT returns a complex number result. That's just the way the most common mathematical formulation and software implementations work.

But strictly real number data has no imaginary components, or it has imaginary components of value zero.

So how do you get complex results to represent something (data) that is strictly real? By producing two results whose imaginary components are equal and opposite, and thus cancel out. Thus you need something other than just the positive frequency result (which can be a complex number with a non-zero imaginary component). You need the negative frequency complex conjugate. Add the two FFT result bins and you return to the real world.

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