# Baseband representation of a LFM chirp signal

I need to generate a chirp signal between $$f_{start} = -f_{limit}$$ and $$f_{stop} = f_{limit}$$, to compare it to a FrFT generated signal.

The FrFT signal of order $$\alpha = 0$$ is PAM modulated data signal, so is in baseband. Therefore, the FrFT of that same signal, with $$\alpha \neq 0$$, sweeps between $$f_{start} = -f_{limit}$$ and $$f_{stop} = f_{limit}$$.

I was wondering, how do I generate a chirp between those two frequencies, $$-f_{limit}$$ and $$f_{limit}$$? And, is there any documentation for this?

• you might want to explain what you mean by a baseband signal in your problem, some would say a real signal with symmetric spectra around DC is a baseband. what is your notion
– user28715
Jun 12 '19 at 12:17
• I understand that a baseband signal is a signal around DC, which spectrum could not be symmetrical. As a matter of fact I think that the baseband representation of a QAM modulated data signal is not real, and therefore the spectrum is not symmetrical. Jun 12 '19 at 12:31
• then what’s your difficulty? since you know what a baseband signal is
– user28715
Jun 12 '19 at 12:45
• The difficulty comes when creating a LFM chirp in baseband that sweeps between a negative frequency and a positive frequency, does it make sense? How do I create it? And more importantly, is there any kind of article or paper on this? Jun 12 '19 at 13:06
• DaGuy, the difference between negative frequency and positive frequency is really only operational with complex-valued signals. in David's answer below $p(t)$ is a complex-valued signal. if your using this chirp to drive a system and then measure the system response, to be mathematically complete, you may need to make two sweeps, one of the real part and another sweep using the imaginary part and assemble the two results as the sweep of the superposition of the real and the imaginary parts. Jun 13 '19 at 5:48

The one element missing from your problem is the time duration of the pulse - I'll assume it is $$T$$. The general form of a chirp is $$p(t)=\exp\left(j2\pi(f_0t+\tfrac{1}{2}\alpha t^2)\right),$$ where $$f_0$$ is the frequency at the start of the chirp and $$\alpha$$ is the chirp rate. For you case $$f_0 =-f_{limit}$$. To find $$\alpha$$ we set the end frequency to be $$f_{limit}$$ and solve $$f_0+\alpha T =f_{limit}$$ which gives: $$\alpha = \frac{2f_{limit}}{T}$$
Thus the pulse is given by: $$p(t) =\exp \left(j2\pi\left(-f_{limit}t+\frac{f_{limit}}{T}t^2\right)\right)$$
• i wonder if the initial phase is an issue. at $t=0$ you are stuck with a phase of zero. just something i'm curious about. Jun 13 '19 at 5:37
• maybe i'm more worried about the phase at time $t= \frac{T}{2}$. Jun 13 '19 at 5:43
• David, i like your approach and your mathematical expression. the only thing i would maybe change, since your doing this in continuous time, is your pulse of duration $T$ should go from $t=-\frac{T}{2}$ to $t=+\frac{T}{2}$. maybe also, if this turns into code for a discrete-time chirp, we should reserve the symbol "$T$" for the sampling period; $T = \frac{1}{f_\mathrm{s}}$. Jun 13 '19 at 5:51
• @robertbristow-johnson Robert, From my background in Radar and Sonar signal processing the above expression is quite common. It is quite easy to see that what the initial frequency is at time $t=0$. It is also easier to apply to Principle of Stationary Phase to obtain an approximate Fourier Transform. It is easier to add in the initial phase and time delay(linear phase) in afterwards. Yes, the adjustments you suggest could be made, but I think they're easier to do after you've figured out the chirp rate. Jun 13 '19 at 11:58