High pass filter and signal

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September 13th 2007, 10:22 PM
braddy

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High pass filter and signal

Hi,

I have this problem about signal that I have a little trouble to finish:
I attached a small figure of the circuit.

I have a High pass filter, and the equation is:

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

the solution for the equation is:
$y[t]=\int_{-\infty}^{t} e^{-(t-\lambda)/\tau}x'[\lambda] \,d\lambda$

This is what I found and it matches with the answer from the professor.

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

Now I need to take away the $x'[\lambda]$ and replace by $x[\lambda]$ .
What i did is to use the integration by part to the solution $y[t]$ .

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.

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: $x[t]=\int_{-\infty}^{+\infty}\delta(t-\lambda) x(\lambda) \,d\lambda$

Here I am confused with the boundaries of the integration

Please can I have some help with the problem
Thank you
September 25th 2007, 12:27 PM
albi

First of all let's check if your solution is really a solution of differential equation:

$y'(t) + \frac{y(t)}{\tau} = u(t)$

Let: $y(t) = \int_{-\infty}^{t} e^{-\frac{t-\lambda}{\tau}} u(\lambda) d\lambda$

We can see that:
$y(t) = \int_{-\infty}^{+\infty} 1(t - \lambda)e^{-\frac{t-\lambda}{\tau}} u(\lambda) d\lambda$

where $1(t)$ denotes unit step function.

$\frac{y(t)}{\tau} = \frac{1}{\tau} \int_{-\infty}^{+\infty} 1(t - \lambda)e^{-\frac{t-\lambda}{\tau}} u(\lambda) d\lambda$

$y'(t) = \int_{-\infty}^{+\infty} \left(\delta(t-\lambda)e^{-\frac{t-\lambda}{\tau}} - \frac{1}{\tau} 1(t-\lambda)e^{-\frac{t-\lambda}{\tau}} \right) u(\lambda) d\lambda$

where $\delta(t)$ denotes Dirac delta function.

Hence:

$y'(t) + \frac{y(t)}{\tau} = \int_{-\infty}^{+\infty} \delta(t-\lambda) u(\lambda) d\lambda = u(t)$

So I can see that this solution is valid...

You can express it by $x(t)$ if you use "integration by parts" formula:

$y(t) = \int_{-\infty}^{t} e^{-\frac{t-\lambda}{\tau}} x'(\lambda) d\lambda = \left. e^{-\frac{t-\lambda}{\tau}} x(\lambda) \right|_{-\infty}^{t} - \frac{1}{\tau} \int_{-\infty}^{t} e^{-\frac{t-\lambda}{\tau}} x(\lambda) d\lambda$

So:
$y(t) = x(t) - \frac{1}{\tau} \int_{-\infty}^{t} e^{-\frac{t-\lambda}{\tau}} x(\lambda) d\lambda$

I did check the validity of your solution because it was unusual. Usually we assume that signals are casual, which means that the circuit response (with zero initial conditions) is equal:

$y(t) = \int_{0}^{t} e^{-\frac{t-\lambda}{\tau}} x'(\lambda) d\lambda = x(t) - \frac{1}{\tau} \int_{0}^{t} e^{-\frac{t-\lambda}{\tau}} x(\lambda) d\lambda$

$= \int_{0}^{t} \left(\delta(t-\lambda) - \frac{1}{\tau} e^{-\frac{t-\lambda}{\tau}}\right) x(\lambda) d\lambda$

Noting that circuit response due to excitation x(t) is convolution of x(t) and impulse response h(t) we can see that:

$h(t) = \delta(t) - \frac{1}{\tau} e^{-\frac{t}{\tau}}$
September 25th 2007, 01:17 PM
albi

Laplace transform

I can understand that you do it for educational reasons, but usually we use Laplace transform instead of directly solving differential equations...

In this case capacitor and resistor constitutes "voltage divider". Using well known formula we can obtain transfer function for the circuit:

$H(s) = \frac{R}{R + \frac{1}{sC}} = \frac{\tau s}{\tau s + 1} = 1 - \frac{1}{\tau} \frac{1}{s + \frac{1}{\tau}}$

We know that inverse Laplace transform of the function $\frac{1}{s+a}$ is equal $e^{-a t}$

This way we can find impulse response (as an inverse transform of the transfer function)