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 December 6th, 2016, 07:46 AM #1 Newbie   Joined: Dec 2016 From: United States Posts: 5 Thanks: 0 Amplitude and frequency of forced square waves I'm working on two similar problems, both with a forced square wave. For both equations, f(t) is as follows $\displaystyle f(t) = u(t) + 2 * sum[ (-1)^k * u(t-k*pi) ]$ , k from 0 to inf The first asks for the amplitude of a DAMPED, FORCED square wave. The equation is $\displaystyle y'' + 0.1y' + y = f(t)$ I know that the solution to this ODE is $\displaystyle y = 1 - e^(-.05t)*cos(sqrt(.9975)t) - (.05/sqrt(.9975)t)*e^(-.05t)*sin(sqrt(.9975)t) + 2* sum[ (-1)^k * u(t-k*pi)*(1-e^(-.05(t-k*pi))*cos(sqrt(.9975)*(t-k*pi)) - (.05/sqrt(.9975)*e^(-.05(t-k*pi))*sin(sqrt(.9975)*(t-k*pi))$ I have been trying to use the equation to solve for amplitude but am unsure what to plug in for F0, or if this is even the correct approach. $\displaystyle R = F0 / (gamma*w0*sqrt(1-(gamma^2/4mk)))$, where gamma = 0.1, w0 = 1, m = 1, k = 1. The second question asks for the slow and fast frequency (there are beats) of an UNDAMPED, FORCED square wave. The equation is $\displaystyle y'' + y = f(t)$ The solution to this ODE is $\displaystyle y = 1 - cos(t) + 2 * sum( (-1)^k * u(t-11k/4) * (1-cos(t-11k/4)$ From the plot I have estimated the slow frequency to be 14 pi and the fast frequency to be 2 pi. I know that slow f = $\displaystyle abs(w-w0)/2$ and fast f = $\displaystyle abs(w+w0)/2$, but I am unsure what to plug in for each omega. Would it be correct that the forced frequency $\displaystyle w0$ would be 11/4? Last edited by lcollins19; December 6th, 2016 at 07:49 AM.

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