hello i don't want to use chirp inbuilt function and i have written a code for chirp. could any one tell me is the code is correct ? i have simulate it and getting plot but at instantaneous time t1, I am not getting frequency change.
i have written a code according to equation of chirp signal generation.
please tell me this is correct ?
*****************************************
f1 =10;
f2 = 50 ;
t =0:0.001:20;
t1 = 10;
alpha = (f2-f1)/t1;
f= (alpha)*t + f1;
xx =1*cos(f);
plot (t,xx);
**********************************************

1 Kommentar

Neal Bambha
Neal Bambha am 16 Dez. 2019
This is not the correct expression for a linear chirped signal. See the wikipedia page for "Chirp" There is a t-squared term in the sine. Otherwise you will not have the correct spectrum
chirp_slope = (f2 - f1)/t(end);
chirp_signal = sin(2*pi*(0.5*chirp_slope*t.^2 + f1*t));

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

Youssef  Khmou
Youssef Khmou am 5 Feb. 2013
Bearbeitet: Youssef Khmou am 5 Feb. 2013

0 Stimmen

hi, try this :
Fs=1000; % sample rate
tf=2; % 2 seconds
t=0:1/Fs:tf-1/Fs;
f1=100;
f2=400; % start @ 100 Hz, go up to 400Hz
semi_t=0:1/Fs:(tf/2-1/Fs);
sl=2*(f2-f1/2);
f1=f1*semi_t+(sl.*semi_t/2);
f2=f1(end)+f2*semi_t-sl.*semi_t/2;
f=[f1 f2];
y=1.33*cos(2*pi*f.*t);
plot(t,y)

6 Kommentare

Aniket
Aniket am 7 Feb. 2013
Bearbeitet: Aniket am 7 Feb. 2013
Hello Youssef, i have used your code and got some results. but i am not getting what is going in the following step
f1=f1*semi_t+(sl.*semi_t/2);
f2=f1(end)+f2*semi_t-sl.*semi_t/2;
could you please explain
hi, consider the slope as :
K=f2-f1/2 ;
then :
f1=f1*semi_t+(K*semi_t);
f2=f1(end)+f2*semi_t-(K*semi_t);
We added "f(end)" to have continuity in the frequency , to test do this :
f1=f1*semi_t+(K*semi_t);
f2=f2*semi_t-(K*semi_t);
f=[f1 f2]; plot(t,f)
vamsi
vamsi am 23 Aug. 2013
hi youssef i execuded the code but the output is not comming for fs=500e6,f1=160e6 ,f2=170e6 ,t=5e-6 could u please explain that...
hi vamsi,
Your parameters do not fit the last example because of scaling problem; the frequency continues to increase or cant reach the final value f2 . To overcome this issue, consider the frequency as function F=at+constant that when time ends F=f2 and when time starts F=f1 :
Fs=500e6;
f=5e-6;
t=0:1/Fs:tf-1/Fs;
f1=160e6 ;
f2=170e6;
SLOPE=(f2-f1)./t(end);
F=f1+SLOPE*t;
y=1.33*cos(2*pi*F.*t);
plot(t,y)
fy=fft(y);
Freq=(0:length(t)-1)*Fs/length(t);
figure, plot(Freq(1:end/2),abs(fy(1:end/2)))
Have you ever used the function chirp(t,f0,t1,f1) ?
Atilla Golan
Atilla Golan am 21 Jan. 2018
Bearbeitet: Atilla Golan am 21 Jan. 2018
Hi Youssef, Can we try up-going and down going chirp together? (without using chirp function on Matlab). In a time range of [1.2s 2.7s] (for example) my function should go up 0.6s (since 1.2s to 1.8s) and go down for the rest (since 1.8s to the end:2.7s). I mean in a constant frequency range, function goes up and down during time like quadratic chirp function but without using chirp.
Dhananjay Singh
Dhananjay Singh am 25 Feb. 2019
hi youssef,
the code you last sent doesn't work when i set the values of f1, f2 in Khz. can you help?

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (3)

Honglei Chen
Honglei Chen am 5 Feb. 2013

0 Stimmen

It should be
cos(2*pi*f.*t)
instead of
cos(f)
REEM ALI
REEM ALI am 30 Jan. 2014

0 Stimmen

please i want to generate triangular modulation sweep of fmcw of rx and tx
farouk behar
farouk behar am 11 Aug. 2015

0 Stimmen

f = 150; a =5; w =2*pi*f; t =0:0.1:20; theta = pi/2; X = a*cos((w*t)+theta); y0 = X + 2*rand(size(t));
plot(t,y0)

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by