- #1
brad sue
- 281
- 0
Hi,
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 variable t. I don't 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
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 variable t. I don't 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