# Second order transfer function with second order numerator?

I was wondering if this is even possible.

For example, if the standard form of a transfer function is is it still possible to have: I've only ever seen this done in one place where they said this had to be the equation for a high pass filter, but every other place I've seen has said the standard form should have less of an order in the numerator.

Would this even be correct, and would it have to be a high pass filter or could it also possibly be a low pass filter?

This would be a high pass filter. You have two zeros at the origin which are manifested as differentiators. This means that in the frequency domain you start off with a slope of +40dB per decade. This implies that as frequency increase, magnitude increases. Inversely, as frequency decreases, magnitude decreases. You can verify this by setting s=0, the numerator is 0, denominator is non-zero, your magnitude response is 0 at DC.

Lets look at s-> inf now to see high frequency behaviour. At large values of s, s^2 >> s, so dropping lower order terms. This leaves you with $$\\frac{ks^2}{s^2}\$$, which is simply k, your high frequency gain. So zero response at DC, bounded flat response at high frequency.

There are many standard (canonical) forms. Among them are canonical 2nd order lowpass, highpass, and bandpass transfer functions. This is a 2nd order highpass. The highest degree of numerator is equal to highest degree of denominator, meaning this equation is proper. A strictly proper equation has the denominator being one degree higher.

https://en.wikipedia.org/wiki/Proper_transfer_function

• Okay this makes sense to me, I was thinking that the numerator determined the filter, but the zeros understanding makes sense, thanks! Dec 5, 2020 at 2:51

It is important to verify the homogeneity of transfer functions. The first one is not unitless unless $$\k\$$ has the dimension of a frequency. The way to write a transfer function is to respect a so-called low-entropy format, a term forged by Dr. Middlebrook some years ago. What it means is simply to express the formula so that you gain insight in what the expression does just by looking at it, for instance inferring from the expression if there are poles, zeroes, some gain etc. Also, you write a transfer function for a design purpose. In a bandpass filter, what matters is surely the resonant frequency but also the gain at the resonance. It is important to account for this goal when writing the transfer function. This is called design-oriented analysis or D-OA.

A transfer function should be expressed with a leading term carrying the unit or the dimension of what you want to express. If this is a gain that you write, then the leading term is a unitless term [V]/[V] or [A]/[A] for instance. If you write an impedance, then the leading term is resistance expressed in [$$\\Omega\$$]. This is extremely useful to check homogeneity of the formulas. In the case of a bandpass filter, a possible canonical expression is this one: In this expression $$\Q\$$ is unitless and $$\\omega_0\$$ is expressed in rad/s. $$\H_0\$$ represents the gain at the resonance.

This is truly the most compact form you can find where the gain appears as the leading term. I have built a quick Mathcad sheet showing the factorization steps in $$\H_1\$$, $$\H_2\$$ and $$\H_3\$$: You have here a good source of information on the bandpass filters.

It is possible to perform polynomial division so that

$$\\frac{ks^2}{s^2+\frac{\omega_0}{Q}s+\omega_0^2}\$$ takes the form

$$\A + \frac{Bs+C}{Ds^2+Es+F}\$$

• Since the two are equivalent, would the filter change because of the order or would it stay the same (assuming high pass filter) even in this scenario? Dec 5, 2020 at 2:48
• The filter is unchanged by how the transfer function is represented. The two transfer functions are equal, they just appear different. Dec 5, 2020 at 2:50