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!

 

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

 

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

Thank you.