Use Laplace transform to find H(s),
1. The highest degree of the expression is the order,
2. What is on the molecule can pass!
for example:
H(s)=as/(bs+c), there is a "high order" on the molecule, so it is a high pass.
The "higher degree" here means this: the denominator has the 0th and 1st degree of s, and the numerator is the 1st degree of s, so it is the higher one, referred to as "higher degree".
H(s)=a/(bs+c), there is a "low order" on the molecule, so it is a low pass.
H(s)=as^2/(bs^2+cs+d), there is a "high order" on the molecule, so it is a high pass.
H(s)=a/(bs^2+cs+d), there is a "low order" on the molecule, so it is a low pass.
H(s)=as/(bs^2+cs+d), there is a "middle order" on the molecule, so it is a bandpass.
H(s)=as^2+d/(bs^2+cs+d), there are "highest order and lowest order" on the molecule, so it is band resistance.
The method (2) has not yet found a theoretical basis. If the numerator and denominator are divided by the "high order", it can be understood when the frequency of judgment changes from small to infinity.