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In a book on optical solitons, the author says that a Gaussian signal pulse has the form-

$u(t) = \exp((\eta + i\beta)t^2)$

where $\beta$ is the chirp parameter.

My question is: what does the $i$ signify here? As a chirped signal is still real, what does the complex nature of the signal represent?

robert bristow-johnson
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Paddy
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2 Answers2

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Depending on the context, the use of the complex form could be for mathematical convenience or for a no-kidding need for both real and imaginary parts.

When you factor the expression, you get

$$u(t) = e^{{\eta}t^2}e^{j{\beta}t^2}$$

Where the first exponential is a generic magnitude envelope, in this case Gaussian. The second exponential is the chirp itself and is where all the action is.

So to simplify things, lets assume that the envelope is ideal so that the signal is just the chirp

$$u(t) = e^{j{\beta}t^2}$$

To view the chirp, you can take either the real or imaginary part, and it looks something like this

enter image description here

In a system that does not use I/Q, the real part is what you would expect to transmit as your waveform. The type of system will determine whether using a real signal or some type of I/Q is best. I'm going to use a radar example here.

In frequency-modulated continuous wave (FMCW) radars, a real chirp like the one above is transmitted and has the form

$$x(t) = cos({{\beta}t^2})$$

Which is just the real part of the complex form. It is received after a delay and mixed with itself, and without going into the mixing process, produces a single frequency sinusoid that can be used to determine range. Here, using a real part only is practical. Using I/Q in FMCW is also beneficial (SNR improvement), but is not usually necessary and many systems do not use it.

Another type of radar, pulse-Doppler, benefits greatly from using the complex form. The same chirp is considered, except now the imaginary version is used. This is important because pulsed-Doppler radars usually operate on performing pulse compression, which is just correlating the transmitted waveform with the received one.

The autocorrelation of a complex chirp looks like

enter image description here

Using a complex waveform allows us to mix our signals to baseband, which give the classic autocorrelation responses we expect without additional mixing and filtering.

Envidia
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$i$ is the symbol for $\sqrt{-1}$

There is a very important formula called Euler's Equation.

$$ e^{i\theta}=\cos(\theta) + i \sin(\theta) = (e^i)^\theta$$

"$ e^i $" is a point on the unit circle one radian along the circumference. Any point on the unit circle raised to a power will stay on the unit circle and its distance along the circumference will be multiplied by the power.

$$ (e^{i\theta})^p=e^{ip \theta } $$

Simply factor it.

$$ u(t) = e^{\left(\eta t^2\right)} \cdot \left(e^i\right)^{\beta t^2} $$

The first factor is your real Gaussian (bell curve) acting as an envelope.

The second factor is a point spinning around the complex unit circle. At a steady pace you would get a steady tone. This one's pace isn't steady, but one of linearly increasing frequency (in absolute terms away from the center).

Your signal/function is complex.

$$ \begin{aligned} u(t) &= e^{\left(\eta t^2\right)} \cdot \left[ \cos\left(\beta t^2\right) + i \sin\left(\beta t^2\right)\right]\\ &= e^{\left(\eta t^2\right)} \cdot \left[ \cos\left([\beta t] t\right) + i \sin\left([\beta t]t\right)\right]\\ &= \left[ e^{\left(\eta t^2\right)} \cdot \cos\left([\beta t] t\right) \right] + i \left[ e^{\left(\eta t^2\right)} \sin\left([\beta t]t\right)\right]\\ \end{aligned} $$

robert bristow-johnson
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Cedron Dawg
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    $\eta$ better the hell be negative. – robert bristow-johnson Aug 12 '20 at 14:14
  • Thanks, but I get that. I am having trouble understanding how I must "look" at the signal. If the magnitude is in Volts, at any given time, should I just look at the envelope? If not, how does the presence of the imaginary component come into what I actually measure? – Paddy Aug 12 '20 at 14:42
  • @Paddy Kind of out of my area. I would venture that your measurement device strips the imaginary part, a graph of what it shows would confirm that. Other systems deal in I/Q signals like this, so the imaginary part is relevant. If your graph is showing the envelope then it is showing $|u(t)|$. – Cedron Dawg Aug 12 '20 at 15:02
  • @CedronDawg, I am only running a simulation on a computer (so I have no means to make actual measurements). I was hoping I could get an answer here and I can plot the appropriate signal. I'll look into this more deeply before accepting the answer – Paddy Aug 13 '20 at 04:17
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    @Paddy Whatever your platform, that is the math of the situation. RB-J is actually quite a chirp expert, especially in the continuous domain. I happen to be studying Discrete Gaussian Eigenvectors of the DFT right now. No hurry on the check, you may get a better answer yet. – Cedron Dawg Aug 13 '20 at 04:24
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    @robertbristow-johnson Where's your sense of adventure? Who doesn't want to see something spiral out of control spectacularly? – Cedron Dawg Aug 13 '20 at 04:25
  • Ced, your $$\left[ e^{\left(\eta t^2\right)} \cdot \cos\left([\beta t] t\right) \right] + i \left[ e^{\left(\eta t^2\right)} \sin\left([\beta t]t\right)\right]$$ expression is a little, uhm, misleading. The $[\beta t]$ factor is *not* the instantaneous frequency. What multiplies "$t$" is not in general the frequency but what is the derivative w.r.t. $t$ of the time-variant phase is the instantaneous frequency. $2\beta t$ is the instantaneous frequency. – robert bristow-johnson Aug 15 '20 at 15:29
  • @robertbristow-johnson You're quibbling about a time scaling issue? The purpose of the block is to show that the signal is complex, not real as the OP asserted. The separation of the $\beta t^2$ into $[\beta t]t$ was to put real sinusoids in standard form so it can be plainly seen that the frequency value is not constant but described as "linearly increasing frequency (in absolute terms away from the center)". What's misleading about that? You can see the graphs for $t>0$ in Envidia's post. He think $\eta=0$ is "ideal", you say it "better the hell be negative". Me, I try to stay neutral. – Cedron Dawg Aug 15 '20 at 15:42
  • @robertbristow-johnson To me, the concept of instantaneous frequency has always been very troublesome. I only consider it well defined for a steady tone. Any thing else should reference it as "the derivative of...", "phase value" would follow, only in a pure tone definition. – Cedron Dawg Aug 15 '20 at 15:47
  • Instantaneous frequency is quite well defined. If you have a sinusoid with time-varying amplitude and frequency

    $$ x(t) = A(t) \cos( \theta(t) ) $$

    then the instantaneous frequency is the time rate of change of the angle of the sinusoidal function $\theta(t)$

    $$ \omega(t) \triangleq \frac{\mathrm{d}\theta}{\mathrm{d}t} $$.

    – robert bristow-johnson Aug 15 '20 at 16:25
  • That means that to express this sinusoid explicitly in terms of its possibly time variant amplitude and frequency (and initial phase, which is a constant), it looks like

    $$ x(t) = \cos\left( 2 \pi \int_0^t f(u) \ \mathrm{d}u \ + \ \theta(0) \right) $$

    – robert bristow-johnson Aug 15 '20 at 16:42
  • @robertbristow-johnson Like I said, "only in a pure tone definition" which is what you just did. A varying envelope has instantaneous frequency implications. You can pick "a" definition of instantaneous frequency, but you can't pick "the" definition, unless is is a steady pure tone. What if your $A(t)$ is $ \sin( \theta(t) ) $ ? – Cedron Dawg Aug 15 '20 at 17:29
  • sorry, $A(t) \ge 0$. I should have said that. Probably the guys that do this stuff with the Hilbert Transform would say that the bandwidth of $A(t)$ is much smaller than that of the instantaneous frequency at any time $t$. – robert bristow-johnson Aug 15 '20 at 17:55
  • if $\hat{x}(t)$ is the Hilbert Transform of $x(t)$, then $$A(t) = \sqrt{x^2(t) + \hat{x}^2(t)}$$ and $$ \omega(t) \triangleq 2\pi f(t) = \frac{\mathrm{d}\theta}{\mathrm{d}t} $$ where $$ \theta(t) = \arg {x(t) + i \hat{x}(t) }$$ I think that is always true. – robert bristow-johnson Aug 15 '20 at 17:59
  • Look at you doing back handsprings trying to come up with a definition, no where close to be able to claim that it is the definition. As far as I am concerned, unless your signal looks like a single tone with a relatively slowly wavering envelope, trying to say "instantaneous frequency" is a matter of having your favorite definition with the properties you like. Toss in discrete vs continuous and it gets even more complicated. (The different use of $x$ is a little, uhm, misleading.) Last comment from me, just got the warning. – Cedron Dawg Aug 15 '20 at 18:43