In my opinion, the following code should produce a sine wave that has a frequency of $2\text{Hz}$ 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; My plot:
2 Answers
$\begingroup$If you look carefully at
y=sin(2*pi*(1+(5/100)*t).*t); you'll see that the variable $t$ appears to the second power. You've graphed a section of $$ y = \sin(C t^2) $$ rather than $$ y = \sin(C t). $$
Looking closely at your picture, you can see that the spacing of the "humps" changes from left to right, so what you've shown can't possibly be a true sinusoid.
General matlab debugging hint: it's so easy to do stuff -- especially big stuff, like building large matrices, or functions represented by 100,000 samples --- that it's easy to delude yourself into thinking you must have done it right. Always write your code so that you can look at corresponding small cases during debugging -- 20 samples of a function of low frequency rather than 20,000 samples of a high-freq function, or the 2 x 2 output matrix from a small example rather than the 200 x 200 output from a large one.
$\endgroup$4$\begingroup$The argument of sine function represents the phase of the wave. In general, to find the frequency of the wave at time 't', you have to differentiate it wrt 't'.
f = (1/2*pi) * d/dt (phase).
However for the phase that linearly varies with time, i.e., frequency being independent of time, you can just divide it by 't'.
f = (1/2*pi) * (1/t) * (phase).
$\endgroup$2
