difficulty to use "sin" function

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parham kianian
parham kianian 2020-5-13
评论： Stephen23 2020-5-13
Consider following:
T = 0.02; %period (sec)
w = 2 * pi / T; %w = 314.1593, angqular frequency (rad/sec)
t = 0 : 0.01 : 20; %time vector (sec)
x = sin(w*t);
plot(t,x)
What is wrong with MATLAB? A sine plot does not look like above. When I set w = 314.1593, the above figure will be plotted. Sine function is a periodic function. But above plot is something completely different. Maximum value should be equal to 1.0 but it is about 10e-12. I cannot figure out where is the problem?
  1 个评论
Stephen23
Stephen23 2020-5-13
You are plotting integer multiples of pi:
>> w*t(1:8)
ans =
0.00000 3.14159 6.28319 9.42478 12.56637 15.70796 18.84956 21.99115
What is sine of any integer multiple of pi ?

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采纳的回答

Mehmed Saad
Mehmed Saad 2020-5-13
Either increase you sampling rate
T = 0.02; %period (sec)
w = 2 * pi / T; %w = 314.1593, angqular frequency (rad/sec)
t = 0 : 0.001 : 20; %time vector (sec)
x = sin(w*t);
plot(t,x)
or add a phase in sin
T = 0.02; %period (sec)
w = 2 * pi / T; %w = 314.1593, angqular frequency (rad/sec)
t = 0 : 0.01 : 20; %time vector (sec)
x = sin(w*t+pi/10);
plot(t,x)

更多回答（1 个）

Walter Roberson
Walter Roberson 2020-5-13
Your w is approximately 100*pi and your t values are all integer multiples of 1/100 . When you multiply the two together, you get an integer multiple of 1/100 * 100*pi which will give you an integer multiple of pi. sin() of an integer multiple of pi is 0 to within round-off error. SO what you are plotting is just round-off error.

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