Hi,(adsbygoogle = window.adsbygoogle || []).push({});

I have this problem about signals that I have a little trouble to finish:

I attached a small figure of the circuit.

I cannot figure out the answer and I struggled for 2 days.

It is a High pass filter, and the ODE of the circuit is:

[tex]\frac{dx[t]}{dt}=\frac{y[t]}{RC}+ \frac{dy[t]}{dt}[/tex]

the solution for the equation is (according to the hint given by the homework):

[tex]y[t]=\int_{-\infty}^{t} e^{-(t-\lambda)/\tau}x'[\lambda] \,d\lambda [/tex] (1)

But the question said:

Find the solution of the ODE to obtain y[t] as some integral of x[t].

This implies taking away the [tex]x'[\lambda][/tex] in (1) and replace by [tex]x[\lambda][/tex].

the hint suggested us to use the integration by part to the solution [tex]y[t][/tex].

I chose x'[t] for v' and the exponetial function as u.

[tex]\int_{a}^{b}u.v'd\lambda =[u.v] -\int_{a}^{b}u'. vd\lambda[/tex]

Doing this, I got:

[tex]y[t]=x[t]+\lambda\int_{-\infty}^{t}x(\lambda)e^{-(t-\lambda)/\tau} \,d\lambda [/tex]

My problem is that by intergrating, even though I took away what I wanted, I have introduced the variablet. I dont know if it is a problem. I am not sure of the answer.

My second question is the following:

Find the impulse response function [tex]h[t][/tex] so that the solution is has the form:

[tex]y[t]=\int_{-\infty}^{+\infty}h(t-\lambda) x(\lambda) \,d\lambda [/tex]

Hint: if you find a lonely x(t), remenber that: [tex]x[t]=\int_{-\infty}^{+\infty}\delta(t-\lambda) x(\lambda) \,d\lambda [/tex]

[tex]\delta(t)[/tex] is the delta function.

please can I have help with this problem?

Thank you

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# Signals and system (difficult!)

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