who can solve this simple problem????

[eng.Joseph]

hi everybody,
how can i found the maximum and the minimum frequencies that will be heard from this chirp matlab code

fs = 7000;
dt = 1/fs;
dur = 1.8;
t = 0:dt:dur;
psi = 2*pi*(100+200*t+500*t.*t);
x=real(7.7*exp(j*psi));
sound(x,fs);

thanks in advance for any help.

 

[zorro]

Use the fact that instantaneous omega is d(phase)/d(time), i.e., instantaneous frequency is 1/(2*pi) * delta(psi)/delta(t) .
Regards

Z

 

[Mityan]

exp(j*psi) = exp(j*2*pi*100) * exp(j*2*pi*t*(200+500*t))
First multiplier is constant and it is related to amplitude of signal together with 7.7
Second is chirp wave. Argument 2*pi*f*t = 2*pi*t*(200+500*t))
So f = 200 + 500*t
So min is 200 and max is 500*dur.

 

[eng.Joseph]

thank you zorro and thank you Mityan but how can i substitute the maximum f value in the script written above and to verify about minimum frequency.
regards