sine with increasing frequency

9 次查看(过去 30 天)
Max E.
Max E. 2015-9-5
in my opinion, the following code should produce a sine-wave that has a frequency of 2Hz at t=20. but when i count the periods between t=19 and t=20, i count more than 3 periods. what am i doing wrong?
clear all, close all, clc;
t=linspace(0,20,10000);
y=sin(2*pi*(1+(5/100)*t).*t);
plot(t,y);
hold on;
  2 个评论
Image Analyst
Image Analyst 2015-9-5
If you have the signal processing toolbox, you might also be interested in the chirp() function.
Max E.
Max E. 2015-9-5
i'm actually using this function in simulink, which genereally includes a chirp block. The problem is, that i can't make the chirpblock start at a specific time, but it always starts at the beginning of the simulation. But what is the processing toolbox and where can i get it?

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回答(1 个)

Star Strider
Star Strider 2015-9-5
You’re multiplying by t first, then element-wise by .*t. I also don’t understand what the 1 is for.
See if this does what you want:
t=linspace(0,20,10000);
y=sin(2*pi/20.*t.^2);
  3 个评论
Max E.
Max E. 2015-9-5
the 1 is the "initial-condition" for the frequency. if i use your function i get a sinus, that's starting with a frequency of 0Hz and then going up to a frequency of 2Hz. But since i got a solution to my problem on another board meanwhile, here's the answer: if you have function sin(x(t)) you can calculate the frequency as w=2*pi*f=dx(t)/dt. so what i had to do was integrating my frequency function over t. the result is this: y=sin(2*pi*t.*(1+1/40*t))
the orange one is your function, the blue one is mine. as you can see, mine starts at a frequency of 1Hz and goes up to 2Hz at the end. And just for your information: the first time you don't have to multiplicate element-wise since you don't have a vector yet, but a scalar. only the power function has to be an element-wise multiplication
Star Strider
Star Strider 2015-9-5
I still don’t understand what you want, but so long as you got your answer, we’ll consider this resolved.

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