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:
where the - sign in the last part comes from the fact that $$1/j=-j$$. 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.
Half-Power Frequencies
Next we will look at the cutoff, or half-power, frequencies for this filter. Recall that the half-power frequencies $$\omega_{hp}$$ are defined as the frequencies where:
or, since it will be much easier to calculate without the square roots involved in finding magnitudes,
Since we know the formula for the magnitude as well as the maximum magnitude, we can substitute that formula and that constant in for this band-pass filter to get:
Since the numerators are the same, the denominators need to be the same, which gives:
To solve for $$\omega_{hp}$$ requires...some...algebra:
Now we can solve the quadratic equation for $$\omega_{hp}$$:
This would seem to indicate that there are four half-power frequencies. There are two negative and two positive values, so we will only consider the positive ones. Which ones are those? We know that $$\omega_n\sqrt{1+4Q^2}\geq\omega_n$$ for real values of $$Q$$ so to get the positive $$\omega_{hp}$$, we will center the solutions on $$\omega_n\sqrt{1+4Q^2}$$:
which can be re-arranged and presented in the following two ways:
From this, you can see that there are two positive half-power frequencies centered on $$\omega_n\sqrt{1+\zeta^2}$$ with a total distance of $$\omega_n/Q$$ or $$2\zeta\omega_n$$ between them. That gives rise to the following definitions:
- The linear center (algebraic average) frequency is the average of the cutoff frequencies:
$$\begin{align*} \omega_{cen, lin}&=\frac{\omega_{hp,low}+\omega_{hp,high}}{2}\\ ~&=\frac{\omega_n\sqrt{1+\zeta^2}-\zeta\omega_n+\omega_n\sqrt{1+\zeta^2}+\zeta\omega_n}{2}\\ ~&=\omega_n\sqrt{1+\zeta^2}~\mbox{ or }~\omega_n\sqrt{1+\frac{1}{4Q^2}} \end{align*}$$ - The bandwidth is the difference between the cutoff frequencies:
$$\begin{align*} BW&=\omega_{hp,high}-\omega_{hp,low}\\ ~&=\omega_n\sqrt{1+\zeta^2}+\zeta\omega_n-\omega_n\sqrt{1+\zeta^2}+\zeta\omega_n\\ ~&=2\zeta\omega_n~\mbox{ or }~\frac{\omega_n}{Q} \end{align*}$$
We can also look at the product of the high and low half-power frequencies (recall that $$(a-b)(a+b)=a^2-b^2$$):
This means we can make the following assertion:
- The logarithmic center (geometric average) frequency is the square root of the product of the low and high half-power frequencies and is equal to the natural frequency: $$\omega_{cen, log}=\omega_n=\sqrt{\omega_{hp,low}\omega_{hp,high}}$$