# Quick High Pass Filter Question

#### jegues

Joined Sep 13, 2010
733
For a high pass filter of the form,

$$\frac{kj \omega}{1 + j\frac{\omega}{\omega_{o}}}$$

The point at which it levels off, in dB, is it,

$$20log(K \omega_{o})$$ or,

$$20log(K)$$

?

Thanks again!

#### tyblu

Joined Nov 29, 2010
199
$$S(j\omega) = k \frac{j\omega}{1+j\frac{\omega}{{\omega}_0}} \\ S(j\omega) = k \frac{j\omega}{1+j\frac{\omega}{{\omega}_0}} \cdot \frac{{\omega}_0/\omega}{{\omega}_0/\omega} = k \frac{j{\omega}_0}{{\omega}_0/\omega + j} \\ \lim_{\omega \to \infty} S(j\omega) = \lim_{\omega \to \infty} k \frac{j{\omega}_0}{{\omega}_0/\omega + j} = k \frac{j{\omega}_0}{j} = k {\omega}_0 \\ \\ 20 log | \lim_{\omega \to \infty} S(j\omega) | = 20 log | k {\omega}_0 |$$

$$S(j\omega) = k \frac{j\omega}{1+j\frac{\omega}{{\omega}_0}} \\ S(j\omega) = k \frac{j\omega}{1+j\frac{\omega}{{\omega}_0}} \cdot \frac{{\omega}_0/\omega}{{\omega}_0/\omega} = k \frac{j{\omega}_0}{{\omega}_0/\omega + j} \\ \lim_{\omega \to \infty} S(j\omega) = \lim_{\omega \to \infty} k \frac{j{\omega}_0}{{\omega}_0/\omega + j} = k \frac{j{\omega}_0}{j} = k {\omega}_0 \\ \\ 20 log | \lim_{\omega \to \infty} S(j\omega) | = 20 log | k {\omega}_0 |$$