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Math Help - High Pass Filter Signal analysis! HELP HELP!!

  1. #1
    cwj
    Guest

    High Pass Filter Signal analysis! HELP HELP!!

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

    above is the given equation based on Kirchhoff law.
    And the question is:
    Find the solution of the ODE to obtain y[t] as some integral of x[t].

    I had search through google and found a solution here.
    y[t]=\int_{-\infty}^{t} e^{-(t-\lambda)/\tau}x'[\lambda] \,d\lambda
    But i do not know how to introduced the e^{-(t-\lambda)/\tau}!! Anyone help!! can any1 provide me some solution??
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  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    \frac{dx[t]}{dt}=\frac{y[t]}{RC}+ \frac{dy[t]}{dt}

    above is the given equation based on Kirchhoff law.
    And the question is:
    Find the solution of the ODE to obtain y[t] as some integral of x[t].

    I had search through google and found a solution here.
    y[t]=\int_{-\infty}^{t} e^{-(t-\lambda)/\tau}x'[\lambda] \,d\lambda
    But i do not know how to introduced the e^{-(t-\lambda)/\tau}!! Anyone help!! can any1 provide me some solution??
    Next time check that your post displays latex equations properly.

    The ODe of the electric circuit is:

    <br />
\frac{{dy}}<br />
{{dt}} + \frac{1}<br />
{{RC}}y = \frac{{dx}}<br />
{{dt}}<br />

    multiply the equation by: <br />
e^{\frac{t}<br />
{{RC}}} <br />

    <br />
\begin{gathered}<br />
  e^{\frac{t}<br />
{{RC}}} \frac{{dy}}<br />
{{dt}} + \frac{1}<br />
{{RC}}e^{\frac{t}<br />
{{RC}}} y = e^{\frac{t}<br />
{{RC}}} \frac{{dx}}<br />
{{dt}} \hfill \\<br />
  \frac{d}<br />
{{dt}}\left( {y(t)e^{\frac{t}<br />
{{RC}}} } \right) = e^{\frac{t}<br />
{{RC}}} \frac{{dx}}<br />
{{dt}} \hfill \\ <br />
\end{gathered} <br />

    as you might notice the left hand side became a perfect differential. Now we integrate both sides from { - \infty } to arbitrary time t:


    <br />
\int\limits_{ - \infty }^t {\frac{d}<br />
{{d\lambda }}\left( {y(\lambda )e^{\frac{\lambda }<br />
{{RC}}} } \right)d\lambda }  = \int\limits_{ - \infty }^t {e^{\frac{\lambda }<br />
{{RC}}} \frac{{dx}}<br />
{{d\lambda }}d\lambda } <br />

    <br />
\begin{gathered}<br />
  y(t)e^{\frac{t}<br />
{{RC}}} = \int\limits_{ - \infty }^t {e^{\frac{\lambda }<br />
{{RC}}} x'\left( \lambda  \right)d\lambda }  \hfill \\<br />
  y(t) = \int\limits_{ - \infty }^t {e^{ - \left( {\frac{{t - \lambda }}<br />
{{RC}}} \right)} x'\left( \lambda  \right)d\lambda }  \hfill \\ <br />
\end{gathered} <br />

    # RC is the time constant of the circuit which is usually denoted as \tau
    # This integral is called the convolution integral, and <br />
h(t) = e^{ - \frac{t}<br />
{{RC}}} <br />

    is the impulse response of the circuit, these are very important concepts which you'll probably meet in your future studies.
    Last edited by Peritus; February 25th 2008 at 05:19 AM.
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