Difference between revisions of "Second Order Filters"
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\mathbb{H}_B(j\omega)=\frac{K_B(2\zeta \omega_n(j\omega))}{(j\omega)^2+2\zeta \omega_n(j\omega)+\omega_n^2} | \mathbb{H}_B(j\omega)=\frac{K_B(2\zeta \omega_n(j\omega))}{(j\omega)^2+2\zeta \omega_n(j\omega)+\omega_n^2} | ||
$$</center> which is the form we will use here. | $$</center> which is the form we will use here. | ||
− | + | === Alternate Representation === | |
To analyze this transfer function more easily, we can divide through by the $$2\zeta\omega_n(j \omega)$$ term to get: | To analyze this transfer function more easily, we can divide through by the $$2\zeta\omega_n(j \omega)$$ term to get: | ||
<center>$$ | <center>$$ | ||
Line 30: | Line 30: | ||
\end{align*} | \end{align*} | ||
$$</center> | $$</center> | ||
+ | |||
+ | === Magnitude and Phase === | ||
+ | This alternate arrangement makes it easier to determine how the magnitude and phase change as the frequency changes: | ||
+ | ==== Magnitude ==== | ||
+ | To find the magnitude of $$\mathbb{H}_B(j\omega)$$, find the magnitude of the numerator and divide it by the magnitude of the denominator: | ||
+ | <center> | ||
+ | $$\begin{align*} | ||
+ | |\mathbb{H}_B(j\omega)|&=\frac{|K_B|}{\sqrt{1^2+Q^2\left(\frac{\omega}{\omega_n}-\frac{\omega_n}{\omega} \right)^2}} | ||
+ | \end{align*}$$ | ||
+ | </center> | ||
+ | From this, we can see that the numerator has a constant magnitude. Furthermore, the denominator has a constant real part. | ||
+ | This means that the magnitude of the denominator is going to be at its ''smallest'' when $$\omega=\omega_n$$; any deviation from this will create a non-zero imaginary part and thus increase the size of the denomintor. From this we can assert the following: | ||
+ | * '''The largest magnitude of this band-pass filter is $$|K_B|$$ and it occurs when $$\omega=\omega_n$$.''' |
Revision as of 13:55, 20 March 2021
This page is currently a sandbox for things related to second-order filters.
Contents
General Form
For the cases below, we will be looking at specific examples of second-order filters, and in each case we will turn on specific components of a general second-order filter by setting $$K_H$$, $$K_B$$, and $$K_L$$ to a non-zero value in:
Band-Pass
For the band-pass filter, with $$K_B$$ set to some non-zero value and $$K_H$$ and $$K_L$$ both set to zero, the transfer function becomes:
or, as a Fourier transform,
which is the form we will use here.
Alternate Representation
To analyze this transfer function more easily, we can divide through by the $$2\zeta\omega_n(j \omega)$$ term to get:
At this point, we introduce a new quantity, the quality factor of the filter $$Q$$, where $$Q=\frac{1}{2\zeta}$$, such that:
Magnitude and Phase
This alternate arrangement makes it easier to determine how the magnitude and phase change as the frequency changes:
Magnitude
To find the magnitude of $$\mathbb{H}_B(j\omega)$$, find the magnitude of the numerator and divide it by the magnitude of the denominator:
$$\begin{align*} |\mathbb{H}_B(j\omega)|&=\frac{|K_B|}{\sqrt{1^2+Q^2\left(\frac{\omega}{\omega_n}-\frac{\omega_n}{\omega} \right)^2}} \end{align*}$$
From this, we can see that the numerator has a constant magnitude. Furthermore, the denominator has a constant real part. This means that the magnitude of the denominator is going to be at its smallest when $$\omega=\omega_n$$; any deviation from this will create a non-zero imaginary part and thus increase the size of the denomintor. From this we can assert the following:
- The largest magnitude of this band-pass filter is $$|K_B|$$ and it occurs when $$\omega=\omega_n$$.