Irregular Sinusoidal CurvesDate: 08/24/2004 at 23:39:08 From: David Subject: Irregular sinusoidal curves Is there a function to graph a sinusoidal curve where the periodicity remains constant but the distance between the maximum and minimum values is non-equidistant? For example, is there a function to plot a sinusoidal curve with a period of 24 hours but where the time span from the maximum value to the minimum value is 16 hours and the span from minimum value to maximum value is 8 hours? Date: 08/26/2004 at 16:01:54 From: Doctor Douglas Subject: Re: Irregular sinusoidal curves Hi, David. Thanks for writing to the Math Forum. Yes, you can construct a function that does this by varying parameters in the basic sinusoidal function to achieve what you want. There are many ways to do this, and I’ll demonstrate one method below. Strictly speaking, the resulting function is not sinusoidal, but it does have a fundamental period T. We start from a simple sinusoidal function of the form below, and let the phase phi be time-dependent so as to make the maximum and minimum occur when we want: y = A*sin[2*pi*t/T + phi(t)], period T = 24 hr. Clearly, if phi is a constant, all we do is shift the graph of y horizontally—-it is still a sine wave with a half-wavelength between maximum and minimum and between minimum and maximum. We know that the sine function takes on its (first) maximum when the argument of the sine is equal to pi/2 [using radians as the angular unit of measure]. Let’s force this to occur at t = tmax = +8 hr (i.e. t/T = 1/3) by demanding that the argument [**] of the sine above reach pi/2 when t = tmax: pi/2 = 2*pi*(1/3) + phi(T/3). We also enforce a similar condition on the argument at t = tmin = -8 hr, where we want the sine function to go thru its minimum: -pi/2 = 2*pi*(-1/3) + phi(-T/3). These two equations specify the values of phi at the maximum and minimum: phi(T/3) = -pi/6 phi(-T/3) = +pi/6 and while so far we have not required anything about the behavior of phi at other values of t, we might as well choose a simple function phi that passes through these two points, and is T-periodic: = +pi*(t/T + 1/2) -T/2 <= t < -T/3 phi(t) = -pi*t/(2*T) -T/3 <= t <= T/3 = -(pi/6) + pi*(t/T - 1/3) T/3 < t <= T/2 By using this expression for phi(t) in the equation for y, you will force the maximum to occur at t = tmax = 8 hr, and the minimum to occur at t = tmin = -8 hr, so that the durations between min-max and max-min are indeed imbalanced, as required. To reverse the intervals so that the max-min event is the longer of the two, we can simply invert the function by multiplying it by –1: y = -sin[2*pi*t/T + phi(t)]. I think that this function will serve your needs. Note that there are all sorts of functions phi(t) that will work, as long as they pass through the given points at t = T/3 and t = -T/3 and are T-periodic. The method above is an example of “phase modulation” (PM), and is a technique that is exploited in communication systems. You can also vary other parameters in the sinusoidal wave y = A(t) sin[2*pi*f(t)*t + phi(t)]. If you let A vary with t, it is called “amplitude modulation”. If you let f vary with t, it is called “frequency modulation”. You may have heard of these terms in the context of AM and FM radio. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
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