Second Order Filters
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 $$\omega=\omega_n$$.
Phase
To find the phase of $$\mathbb{H}_B(j\omega)$$, find the phase of the numerator and subtract the phase of the denominator from it. For a complex number $$\mathbb{n}=n_r+jn_i$$, the angle is given by $$\arctan(n_i/n_r)$$; for the denominator of $$\mathbb{H}_N(j\omega)$$, the real part is simply 1 so the angle of the denominator is the arctangent of the imaginary part. The angle of the numerator depends on the sign of $$K_B$$:
$$\begin{align*} \angle \mathbb{H}_B(j\omega)&=\angle K_B - \arctan\left(Q\left(\frac{\omega}{\omega_n}-\frac{\omega_n}{\omega} \right) \right) \end{align*}$$
Once again, $$\omega=\omega_n$$ is an interesting point to consider - the denominator will contribute $$0^o$$ to the phase of the transfer function here. As $$\omega$$ gets larger, the phase angle of the transfer function will shift by $$-90^o$$ degrees while smaller values of $$\omega$$ will shift the transfer function by up to $$+90^o$$. There is therefore a total $$180^o$$ phase shift from the smallest frequencies to the highest frequencies - this is to be expected from a transfer function with two corners in the denominator. From this we can assert the following:
- The phase of this band-pass filter when $$\omega=\omega_n$$ is $$0^o$$ if $$K_B$$ is positive and $$180^o$$ if $$K_B$$ is negative.