Question

obtain the transfer function h9(s) for a prototype butterworth filter of order 9 in the quadratic factored form.

Answer 1

The transfer function
`\(H_9(s)\)` for a prototype Butterworth filter of order 9 in the quadratic factored form is:

`\[ H_9(s) = \frac{1}{{(s + 1.8019)^2 \cdot (s^2 + 1.2469s + 1) \cdot (s^2 + 0.4450s + 1) \cdot (s^2 + 0.9644s + 1)}} \]`

Explanation:

The transfer function of a Butterworth filter of order (n) can be derived using established formulae and mathematical procedures. For a Butterworth filter of order 9, the quadratic factored form involves multiple second-order and first-order terms in the denominator of the transfer function. These terms are determined based on the Butterworth filter's specifications, particularly the cutoff frequency and the order of the filter.

The quadratic factored form for a Butterworth filter typically includes pairs of complex conjugate poles in the form
`\((s^2 + a_n s + 1)\`) and real poles in the form
`\((s + b_n)^2\).`Each term in the factored form contributes to the filter's behavior and characteristics. For a ninth-order Butterworth filter, the derived transfer function showcases the multiplication of these quadratic terms.

By calculating the roots and factoring the denominator polynomials according to the Butterworth filter's order, the final expression represents the transfer function in terms of its poles in the \(s\)-plane. The specified quadratic factored form in the final answer corresponds to the precise arrangement of poles that define the frequency response and behavior of the Butterworth filter of order 9.

This intricate form of the transfer function illustrates the distribution of poles across the complex plane, determining the filter's frequency response characteristics and attenuation properties.