# Bandwidth of a chirp signal

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Vinod am 27 Sep. 2018
Beantwortet: Hieu Nguyen am 22 Nov. 2019
I have generated three signals and their spectrum (with FFT). Two chirps and one pure cosine signal.
<https://pastebin.com/z2pzq0E7 Code for generating these signals is given here.>
Time variation -> -5 to +5 seconds
frequency variation -> -2.5 to +2.5 Hz (For the two chirps)
Sampling frequency = 25 Hz
Qn 1:
For the cosine chirp cos(2*pi*f.*t) [Line 12],
The spectrum obtained is spanning from -5 Hz to +5 Hz.
Even though the frequency variation (Line 10) is from -2.5 to +2.5 Hz,
why the spectrum is spanning from -5 Hz to +5 Hz?
Qn 2:
For the complex chirp exp(2j*pi*f.*t) [Line 31],
The spectrum obtained is spanning from -5 Hz to +5 Hz.
Even though the frequency variation (Line 10) is from -2.5 to +2.5 Hz,
why the spectrum is spanning from -5 Hz to +5 Hz?
Observation:
For the pure cosine tone cos(2*pi*(B/2)*t) [Line 44],
The spectrum obtained is centered at -2.5 Hz and +2.5 Hz.
This is expected.
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### Akzeptierte Antwort

Honglei Chen am 27 Sep. 2018
I think the two questions are the same, essentially your signal model is
cos(2*pi*B*t^2)
This means that when you look at the instantaneous frequency, it is given by 2*B*t. That's why you see it span from -5 to 5 Hz instead of from -2.5 to 2.5 Hz.
HTH
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Vinod am 28 Sep. 2018
May I know why cos(2*pi*B*t^2) and cos(2*pi*f.*t) are essentially the same.For the waveform cos(2*pi*f.*t), the instantaneous frequencies at the ends of pulse width are -B/2 and +B/2 respectively.For the waveform cos(2*pi*B*t^2), the instantaneous frequencies at the ends of pulse width are (-2*B*5 = -50) and (+2*B*5 = 50) respectively. I think this is a difference.
Honglei Chen am 28 Sep. 2018
Well your f in cos(2*pi*f*t) is just B/T*t, that's why I say it's essentially cos(2*pi*f*t) is essentially cos(2*pi*B/T*t^2). But your instantaneous frequency is not f, it's actually the derivative of B/T*t^2 over t, so it's actually 2*B/T*t, that's why at beginning it's -B and at the end it's B.
Hope this is clear.

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### Weitere Antworten (1)

Hieu Nguyen am 22 Nov. 2019
(There was a mistake he made in equation 8 - it should be fi(t) = 2a*t+f0 )
The answer is in equation 10 where he wrote the instantenous phase should be calculated as the integration of the angular frequency. If you follow that instruction, you will get the expression of x as: x=cos(2*pi*(k/2*t+f0).*t+phase)
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