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Why do we know that $\omega_1$ and $\omega_2$ are corner frequencies (for the RLC filter examples)? Why do we switch to $\omega_1$ and $\omega_2$ instead of $\omega RC$ and such?

When you write the denominator of all the $RLC$ examples as $(1-\frac{\omega^2}{\omega_1 \omega_2} + j
\frac{\omega}{\omega_1}$, you can see that $\omega_1$ and $\omega_2$ divide different regimes by eyeballing the expression for different values of $\omega$.

We assumed $\omega_2»\omega_1$. Then the $\frac{\omega^2}{\omega_1 \omega_2}$ term will be guaranteed to be very large, and $\vert\hat{H}\vert\propto \omega^{-2}$, for $\omega»\omega_2$. (Since $\omega_1$ is smaller than $\omega_2$, then if $\omega»\omega_2$, then $\omega^2»\omega_1 \omega_2$). When does that $\propto \omega^2$ behavior in the denominator take over? When $\frac{\omega^2}{\omega_1 \omega_2} \sim \frac{\omega}{\omega_1}$, so at $\omega=\omega_2$. So that's one corner frequency.

Now what if $\omega$ is very small? Then the $\omega^2$ term can be ignored. Then it looks like a single-pole filter. The behavior transition in the denominator from 1 to $\propto \omega$ is when $1=\frac{\omega}{\omega_1}$, so when $\omega = \omega_1$.

What if $\omega_2=\omega_1$? Then both breakpoints are in the same place.

But what if $\omega_1»\omega_2$? Then it gets a bit less obvious, and only extreme behaviors are clear.

As for the second question: the formulation in terms of $\omega_1$ and $\omega_2$ is just a convenient one for understanding the behavior in different frequency regimes, as these quantities correspond to corner frequencies.


next up previous
Next: What is the physical Up: Content Questions Previous: If we have poles,
Kate Scholberg 2017-02-16