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Can you do this? This is part of my final year EE work. I need to solve this in order to figure out how my sensor is behaving.

$$y'' + 3y' + 2y = f(t), \; y(0) = y'(0) = 0$$ where $f(t)$ is a square wave.

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closed as off-topic by dustin, graydad, Rory Daulton, voldemort, Sujaan Kunalan Jan 18 '15 at 19:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – dustin, graydad, Rory Daulton, voldemort, Sujaan Kunalan
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You should probably mention things you've tried, or else people here may think you're just looking for easy answers for homework problems. – james Jan 20 '13 at 22:45
The things I've tried are far too long to put on this page. Literally pages and pages of workings. But I keep on getting to the wrong answer... This is part of my final year project in EE. I need to solve this in order to use a sensor – user59055 Jan 20 '13 at 23:18

Since this is a EE project, I will not solve everything for you, but this should help you move forward.

We have:

$$\tag 1 y'' + 3 y' + 2y = f(t), ~y(0) = y'(0) = 0, ~ \text{where f(t) is a square wave}$$

Taking the Laplace Transform of $(1)$, yields:

$\tag 2 \mathcal L(y'') +3 \mathcal L(y') + 2\mathcal L(y) = \mathcal L(f(t))$


$\tag 3 \mathcal L(y'') = s^{2}y(s) - s(0) -(0) = s^{2}y(s)$

$\tag 4 3 \mathcal L(y') = 3(s y(s) -(0)) = 3 s y(s)$

$\tag 5 2 \mathcal L(y) = 2 y(s)$

For periodic functions, $f(x + \omega) = -f(x)$, the Laplace Transform is given by (you can also calculate it by steps - it is not too bad):

$$\mathcal L(f(x)) = \frac{\int_0^\omega e^{-s x} f(x) dx}{1 + e^{- \omega s}}$$ Now, we have no idea what sort of square wave you have, so I assume it is $f(x+1) = -f(x)$, but you will have to modify the result below if your square wave varies from that (in amplitude and frequency).

Hence, with $\omega = 1$, we obtain:

$\tag 6 \mathcal L(f(t)) = \Large \frac{\int_0^\omega e^{-s t} f(t) dt}{1 + e^{- \omega s}} = \frac{\int_0^1 e^{-s t} (1) dt}{1 + e^{- s}} = \frac{(\frac{1}{s})(1 - e^{-s})}{1+e^{-s}} = \frac{1}{s} \tanh \frac{s}{2}$

Now, substituting $(3)$ through $(6)$ into $(2)$, yields:

$$(s^{2} + 3s +2)y(s) = \frac{1}{s} \tanh \frac{s}{2}$$

Solving for $y(s)$, yields:

$$y(s) = \frac{\tanh \frac{s}{2}}{(s)(s^2+ 3s + 2)} = \frac{\tanh \frac{s}{2}}{(s)(s+1)(s+2)}$$

Since this is a project, I think this is where I will stop, since you should now be able to move it forward given that you are in this class.

Against my better judgement, I will also use WolframAlpha, to solve and plot this DEQ. This is a further hint to help you to resolve the point where I stopped above.


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All this work, and good work, at that, needs recognition! +1 – amWhy May 7 '13 at 2:01
Yes, I understand. I think, for some askers, they are so absorbed in the urgency of solving a problem, they forget to return and acknowledge help...especially when the asker's use of the site is minimal and/or sporadic. – amWhy May 7 '13 at 2:07

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