Final answer:
To design a 2nd order Butterworth High Pass Filter (HPF) with a cutoff frequency of 1000 Hz and a gain of 1, the transfer function is H(s) = (2000π)² / (s² + (2000π/Q)s + (2000π)²). the Bode plots for the magnitude and phase of the frequency response function can be plotted by substituting the complex frequency variable s with jω.
Step-by-step explanation:
In order to design a 2nd order Butterworth High Pass Filter (HPF) with a cutoff frequency of 1000 Hz and a gain of 1, we can start by finding the transfer function. the transfer function of a 2nd order Butterworth HPF is given by:
H(s) = ω² / (s² + ω/Qs + ω²)
Where ω is the cutoff frequency in radians per second, s is the complex frequency variable, and Q is the quality factor.
In this case, ω = 2πf = 2π(1000) = 2000π rad/s. Since the gain is 1, Q can be calculated as Q = 1 / √2.
Substituting the values into the transfer function equation, we get:
H(s) = (2000π)² / (s² + (2000π/Q)s + (2000π)²)
To plot the Bode plots for the magnitude and phase of the frequency response function, we can substitute the complex frequency variable s with jω. The magnitude plot will show us the gain of the filter at different frequencies, and the phase plot will show us the phase shift introduced by the filter at different frequencies.