Talk:Chirp

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Why was the linear chirp formula changed to:

x(t) = \sin(2 \pi \int_0^t f(t') t' dt') = \sin(2\pi (f_0 + \frac{k}{2} t) t)

? — Omegatron 03:01, 31 October 2005 (UTC)

Because that is the correct formula for a chirped sine wave if the chirp function is given by

f(t) = f0 + kt

For a function of the form

x(t) = exp(z(t))

the frequency is defined to be

\nu(t) = \frac{1}{2 \pi i}\frac{dz(t)}{dt}

Lets pick an arbitrary chirp function:

z(t) = 2πif(t)t

Then:

\nu(t)=\frac{1}{2 \pi i} \frac{dz(t)}{dt}=( \frac{df(t)}{dt}t + f(t) )

So for a linear chirp

f(t) = f0 + kt

the frequency is

ν(t) = kt + kt + f0 = 2kt + f0

the chirp rate is

\frac{d\nu(t)}{dt}=2k

In other words one of your equations is off by a factor of two. Either you need to define your linear chirp as

ν(t) = ν0 + 2kt

or your chirped sine wave as

x(t) = \sin(2 \pi \int_0^t \nu(t') t' dt') = \sin(2\pi (f_0 + \frac{k}{2} t) t)

I haven't looked to see whether this means all the other chirp functions described are similarly wrong....

Korpela 05:58, 31 October 2005 (UTC)

Yes, the exponential chirp is similarly broken. In this case for a function of the type

x(t) = sin(2πf0ktt)

the instantaneous frequency is

ν(t) = f0kt(ln(k)t + 1)

If you define exponential chirp as one where the frequency has the form

ν(t) = f0kt

then the functional form of the waveform is

x(t) = \sin(2 \pi \int_0^t \nu(t') t' dt') = \sin(\frac{2\pi f_0}{\ln(k)^2} \left[ k^t (\ln(k) t-1) +1\right]) =\sin(2\pi f_1 \left[e^{k_1 t} ( k_1 t - 1 ) +1\right])

Neither is as simple as what's on the page already. You can make things look better by redefining k and t in one of the definitions. What is there now will confuse people who assume that that "k" and "f" in the frequency definition is the same as "k" and "f" in the waveform definition.

Korpela 17:22, 31 October 2005 (UTC) revised 18:48, 31 October 2005 (UTC)

Sorry, but I think there is a fundamental error in the previous reasoning: When you have an (angular) frequency ω(t) then the corresponding x(t) is

x(t) = exp(i \int_0^t \omega(t') dt') = exp(2 \pi i \int_0^t \nu(t') dt')

and not

x(t) = exp(i \int_0^t \omega(t') t' dt') = exp(2 \pi i \int_0^t \nu(t') t' dt')

Simple example to help against knots in the mind: Assume for a moment we wouldn't calculate a chirp and take ν(t) = ν = const.

right answer:

x(t) = exp(2 \pi i \int_0^t \nu dt') = exp(2 \pi i \nu t)

wrong answer:

x(t) = exp(2 \pi i \nu \int_0^t t' dt') = exp(\pi i t^2)

Now let's see what we get for the chirps: Well, if we then start with the linear chirp and assume:

f(t) = f0 + kt

then we get

x(t) = \sin(2 \pi \int_0^t f(t') dt') = \sin(2 \pi \int_0^t (f_0 + kt') dt') = \sin\left(2\pi (f_0 + \frac{k}{2} t) t \right)

Checking the result:

\nu(t) = \frac{d}{dt}((f_0 + \frac{k}{2} t) t) = f_0 + kt

And for the exponential chirp with

f(t) = f0kt

we get

x(t) = \sin(2 \pi f_0 \int_0^t k^{t'} dt') = \sin(2 \pi f_0 \int_0^t exp(ln(k) t') dt') = \sin\left(\frac{2\pi f_0}{\ln(k)} ( k^t - 1)\right)

I have tested these results, they produce correct chirps. SiriusGrey 00:31, 24 March 2006 (UTC)

[edit] Graphs

I added the examples to the article, which I have now been told were wrong, and made the graphs with the same equations. I haven't put in the time to understand the differences regarding instantaneous frequency, but, in the meantime, can someone post some references that show these equations? (I don't see any references at all right now.) I want to know for certain that this is what people actually use to make Doppler-immune chirp radars and such. It's a little confusing. — Omegatron 14:37, 4 June 2006 (UTC)

[edit] Nature of chirp signal

How can we say that a chirp signal is bandlimited as well as time limited? does the term time-limited make sense? please explain. —The preceding unsigned comment was added by Krishna2531985 (talk • contribs) 05:13, 23 January 2007 (UTC).

[edit] Geometric same as exponential

Surely the geometric and exponential chirps are the same. If the frequency is k^t, say, then k^(t+const) = (k^const)*(k^t), ie the frequency at t2 = (t1 + c) is a fixed multiple of the frequency at t1. —Preceding unsigned comment added by 86.150.1.188 (talk) 18:01, 17 January 2008 (UTC)