Final answer:
The transfer function of the 2nd order Butterworth HPF with a cutoff frequency of 1000 Hz and a gain of 1 is H(s) = (4π²√10⁶s²) / (s² + 2√2π√1000s + 4π²√10⁶).
Step-by-step explanation:
Designing a 2nd Order Butterworth High-Pass Filter (HPF)
To design a 2nd order Butterworth HPF with a cutoff frequency (ƒc) of 1000 Hz and a gain of 1, we would first need to find the standard form of the 2nd order Butterworth filter transfer function. A 2nd order Butterworth filter has a squared magnitude response that is flat at low frequencies and falls off at -40 dB/decade beyond the cutoff frequency. The normalized transfer function H(s) for such a filter is given by:
H(s) = s^2 / (s^2 + √2*s + 1)
To convert this to a high-pass filter, we replace s with ƒc/s, where s is the complex frequency (jω). For our case, ƒc is 1000 Hz:
H(s) = (ƒc^2*s^2) / (s^2 + √2*ƒc*s + ƒc^2)
Substituting ƒc = 2π*1000 rad/s, the transfer function becomes:
H(s) = ((2π*1000)^2*s^2) / (s^2 + √2*2π*1000*s + (2π*1000)^2)
Therefore, the transfer function of the 2nd order Butterworth HPF with a cutoff frequency of 1000 Hz and a gain of 1 is:
H(s) = (4π^2*10^6*s^2) / (s^2 + 2√2π*1000*s + 4π^2*10^6)