Let's say data rate requirement for some application is 36Kbps.

For a binary Chirp spread spectrum system, SF=6 is taken so that M=64. Time * bandwidth = 64.

If binary 1 is represented as Upchirp and binary 0 is represented as Downchirp then what will be the bandwidth requirement for such a modem :

Since Ts× Bw = 64. Bw = 64× Rs = 64 × Rb = 64 * 36000 = 2.304 MHz.

So chirp bw will vary from -1.152 to + 1.152 and channel bw will be 2.304 MHz. Is this correct ?


1 Answer 1


Chirp Spread Spectrum is a "wideband" FM process where as a first estimate you can use Carson's Rule for the occupied bandwidth as the sum of the peak to peak frequency deviation and modulation rate. In this case the modulation rate is significantly smaller than the deviation so the occupied bandwidth will be close to the range of the chirp itself. This is consistent with the OP's conclusion.

According to Carson's Rule we would get the result of 2.304 MHz + 36 KHz = 2.34 MHz which represents the dominant occupied bandwidth (98% or so). This is useful for defining the passband required for modulation accuracy but is not useful for confirming out of band emissions since there is still significant energy in outer bands. The Fourier Transform of the base ramp up down pulse defines the Kernel spectral shape under which a random waveform will occupy and can be used for a more accurate representation of spectral occupancy to compare with emission limits.

In this case the base pulse is:

$$e^{j2\pi(f_c -f_c f_m| t |)t}, \text{ for } t \in [-1/f_m, 1/f_m]$$

Where $f_c = 2.304E6$ and $f_m= 36e3$.

Below shows the time domain of the base up-down chirp and the Fourier Transform as determined using a sampled pulse and FFT.



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.