If a signal function $ U(t) = 25 – (5 – t)^2$ is passed through a high pass filter with transfer function $\frac{s}{s + ω}$. what is the output signal Y(t). I know that the transfer function $H(s) = \frac{Y(s)}{U(s)}$. If I understand the basic concept, I can solve my own problems. I will also appreciate it if someone can reference a textbook.
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$\begingroup$ Look for a basic signal processing book like this one amazon.com/Digital-Signal-Processing-John-Proakis/dp/0131873741 $\endgroup$– Mahendra GunawardenaCommented Dec 2, 2020 at 12:08
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2 Answers
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What you need to do is
- use Laplace transform to $U(t)$ so you would get $\mathcal{L}(U(t))=U(s)$
for example, if you are rusty on Laplace transforms (or their inverse), you can use wolfram alpha
- multiply $H(s)*U(s)$
that will get you a function of s $Y(s)$
- use the inverse laplace transform to Y(s) to get the $y(t) = \mathcal{L}^{-1}(Y(s))$
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Luckily in this case there is a symbolic result.
OutputResponse[TransferFunctionModel[s/(s + w), s], 25 - (5 - t)^2, t]
res1 = Simplify[%[[1]]]
$-\frac{2 e^{-t w} \left(e^{t w} ((t-5) w-1)+5 w+1\right)}{w^2}$
A procedure to get this (as mentioned) is
s/(s + w) LaplaceTransform[25 - (5 - t)^2, t, s]
res2 = InverseLaplaceTransform[%, s, t]
$-\frac{2 e^{-t w} \left(e^{t w} ((t-5) w-1)+5 w+1\right)}{w^2}$
The results for various values of $w$
Table[w, {w, 0.1, 1, 0.2}];
Plot[Evaluate@Table[res, {w, %}], {t, 0, 10}, PlotLegends -> %]
