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

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].

$\displaystyle y[t]=\int_{-\infty}^{t} e^{-(t-\lambda)/\tau}x'[\lambda] \,d\lambda$
But i do not know how to introduced the $\displaystyle e^{-(t-\lambda)/\tau}$!! 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:

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

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

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

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

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

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

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

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