# What is the output of a signal in time domain that passed through a High Pass Filter with simple transfer function

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.

What you need to do is

1. 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

1. multiply $$H(s)*U(s)$$

that will get you a function of s $$Y(s)$$

1. use the inverse laplace transform to Y(s) to get the $$y(t) = \mathcal{L}^{-1}(Y(s))$$

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 -> %]


• Thank you so much... – Tee Dec 3 '20 at 23:34