# Thread: High Pass Filter Signal analysis! HELP HELP!!

1. ## 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].

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??

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

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].

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

$
\frac{{dy}}
{{dt}} + \frac{1}
{{RC}}y = \frac{{dx}}
{{dt}}
$

multiply the equation by: $
e^{\frac{t}
{{RC}}}
$

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

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

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

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

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

is the impulse response of the circuit, these are very important concepts which you'll probably meet in your future studies.