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sine wave problem.

Started by s_blue Apr 27, 2000 at 5:04 PM 3 replies 600+ views

Original Post

Apr 27, 2000 at 5:04 PM

I really can''t think of a way to solve this problem: I need to create a sine wave that has a specific length(n), an applitude on the positive(a1) an amplitude on the minus(a2) The applitudes on the positive and minus differ. E.g This is the positive of the sine( a1 ) \ \ / \+ / + +/ --+---+---+--- Between the +''s is the length(n) /+ + / + / +¬ / The negative of the sine( a2 ) / /Midpoint( the cross between a1 & a2 ) The problem is, how do I create half the sine, at the specfific aplitude, maintaning the exactly half the lenght(n). If anyone could help me with this, I would appreciate. I wouldn''t mind some C/C++ source for something like this, cheers.

s_blue

Author

s_blue

Apr 27, 2000 at 5:23 PM

Darn it the diagram went weird. Basically, I want to create a half sine wave, with a specific length, and height. Any help appreciated, maybe some code aswell. Cheers.

NickGA

Apr 27, 2000 at 5:36 PM

quote:
Original post by s_blue

Darn it the diagram went weird. Basically, I want to create a half sine wave, with a specific length, and height. Any help appreciated, maybe some code aswell. Cheers.

Ok... lets see. Half a sine wave means 0->pi on the x-axis.
Lets say A is the amplitude and n is the number of samples of the sine wave.

double mysizehalf[n]; // insert a number n here
for (i=0;i{
mysinehalf = A*sin((pi/n)*i);
}

does that work? Not sure if I get your question.

"... you act as if stupidity were a virtue."
-- Flight of the Phoenix

NickGA

Apr 27, 2000 at 5:38 PM

oops what happenned?

ok here is the code again

double mysinehalf[n]; // insert number n

for (i=0;i{
mysinehalf = A*sin((pi/n)*i);
}

"... you act as if stupidity were a virtue."
-- Flight of the Phoenix

NickGA

Apr 27, 2000 at 5:40 PM

ok, this is strange......
for(i = 0; i < n; i++)
mysinehalf = A*sin((pi/n)*i);

if this doesn''t work I quit grrrrr darn html

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