# Signals and system (difficult!)

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:

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

the solution for the equation is (according to the hint given by the homework):
$$y[t]=\int_{-\infty}^{t} e^{-(t-\lambda)/\tau}x'[\lambda] \,d\lambda$$ (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 $$x'[\lambda]$$ in (1) and replace by $$x[\lambda]$$.

the hint suggested us to use the integration by part to the solution $$y[t]$$.
I chose x'[t] for v' and the exponetial function as u.
$$\int_{a}^{b}u.v'd\lambda =[u.v] -\int_{a}^{b}u'. vd\lambda$$

Doing this, I got:
$$y[t]=x[t]+\lambda\int_{-\infty}^{t}x(\lambda)e^{-(t-\lambda)/\tau} \,d\lambda$$

My problem is that by intergrating, even though I took away what I wanted, I have introduced the variable t. 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 $$h[t]$$ so that the solution is has the form:
$$y[t]=\int_{-\infty}^{+\infty}h(t-\lambda) x(\lambda) \,d\lambda$$

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

$$\delta(t)$$ is the delta function.

please can I have help with this problem?
Thank you

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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:

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

the solution for the equation is (according to the hint given by the homework):
$$y[t]=\int_{-\infty}^{t} e^{-(t-\lambda)/\tau}x'[\lambda] \,d\lambda$$ (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 $$x'[\lambda]$$ in (1) and replace by $$x[\lambda]$$.

the hint suggested us to use the integration by part to the solution $$y[t]$$.
I chose x'[t] for v' and the exponetial function as u.
$$\int_{a}^{b}u.v'd\lambda =[u.v] -\int_{a}^{b}u'. vd\lambda$$

Doing this, I got:
$$y[t]=x[t]+\lambda\int_{-\infty}^{t}x(\lambda)e^{-(t-\lambda)/\tau} \,d\lambda$$

My problem is that by intergrating, even though I took away what I wanted, I have introduced the variable t. 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 $$h[t]$$ so that the solution is has the form:
$$y[t]=\int_{-\infty}^{+\infty}h(t-\lambda) x(\lambda) \,d\lambda$$

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

$$\delta(t)$$ is the delta function.

please can I have help with this problem?
Thank you
In the integral t is not a variable, it is a constant, since the integration variable is lambda. You can make $$e^{-(t-\lambda)} = e^{-t}e^\lambda$$ then take $$e^{-t}$$ outside of the integral.

Thank you CEL but I think that I need to keep the form so that I always get the $$t-\lambda$$ as argument.

Thank you CEL but I think that I need to keep the form so that I always get the $$t-\lambda$$ as argument.
Even if you want to keep the form $$t-\lambda$$, t is still a constant and not a variable. The only variable is still $$\lambda$$ .

Astronuc
Staff Emeritus
When one does

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

the integral involves the frequency domain, and lambda (frequency) is the variable, not t.

Frquency domain is the basis of Fourier and Laplace transforms and analyses, as opposed to the time-domain.

If x(t) = eat, then x'(t) = a eat = a x(t) assuming that x'(t) = d x(t) / dt

and the other part of that is d [x(t)]/da = t eat = t x(t)

Last edited:
Thank you Astronuc and CEL,

I 'll redo my computations. I will let you know if I still have an issue.