Locations of poles
The transfer function of Butterworth Low pass is given as
\[
|T_n(j\omega)|^2 = \frac{1}{1 + \omega^{2n}_0}
\]
Where, \(\omega_0\) is the cutoff frequency.
For \(\omega_0\) normalized to 1 rad/sec,
\[
|T_n(j\omega)|^2 = \frac{1}{1 + \omega^{2n}}
\]
In the \(s\)-domain, putting \(s = j\omega\) or \(s = j \omega\),
\[
|T_n(s)|^2 = \frac{1}{1 + (j\omega)^{2n}} = \frac{1}{1 + s^{2n} \cdot (j)^{2n}} = \frac{1}{1 + s^{2n} \cdot (j^2)^n}
\]
Since \(j^2 = -1\),
\[
|T_n(s)|^2 = \frac{1}{1 + s^{2n} \cdot (-1)^n}
\]
Alternatively, it can also be expressed as:
\[
|T_n(s)|^2 = \frac{1}{1 + (-1)^n \cdot s^{2n}}
\]
\[
\text{Properties of } |T_n(s)|
\]
\[
\text{The poles of } |T_n(s)| \cdot |T_n(-s)| \text{ are uniformly spaced around a unit circle whose angles are } \frac{360^\circ}{2n} \text{ or } \frac{180^\circ}{n}
\]
\[
\text{If } n \text{ is odd, two of the poles lie on the real axis.}
\]
\[
\text{If } n \text{ is even, all poles are in complex conjugate pairs.}
\]
\[
\text{To obtain } T_n(s), \text{ only poles in the left half of the } s\text{-plane are used.}
\]
\[
\text{Here, Two complex poles}
\]
\[
s = -\frac{1}{\sqrt{2}} \pm j \frac{1}{\sqrt{2}}
\]
In the \(s\)-domain,
\[
|T_n(s)|^2 = \frac{1}{1 + (-1)^n s^{2n}}
\]
\[
\text{Note:}
\]
\[
\text{If } s^n = 1, \text{ then } s = 1 \angle \frac{180^\circ + k \times 360^\circ}{n} \quad \text{for } k = 0, 1, \ldots, (n-1)
\]
\[
\text{If } s^n = -1, \text{ then } s = 1 \angle \frac{k \times 360^\circ}{n} \quad \text{for } k = 0, 1, \ldots, (n-1)
\]
---
\[
\text{When } n=1:
\]
\[
|T_1(s)|^2 = \frac{1}{1 + (-1)^1 s^{2}} = \frac{1}{1 - s^2}
\]
\[
\text{Poles are located at: } 1 - s^2 = 0 \implies s^2 = 1 \implies s = \pm 1
\]
\[
\text{Only take poles in the negative half-plane, so pole at } s = -1
\]
\[
B_1 = (s + 1)
\]
\[
\text{Transfer function: } \quad T_1(s) = \frac{1}{B_1} = \frac{1}{s + 1}
\]
---
\[
\text{When } n=2:
\]
\[
|T_2(s)|^2 = \frac{1}{1 + (-1)^2 s^{4}} = \frac{1}{1 + s^4}
\]
\[
\text{Poles are located at: } 1 + s^4 = 0 \implies s^4 = -1
\]
\[
s = 1 \angle \frac{180^\circ + k \times 360^\circ}{4}, \quad k = 0, 1, 2, 3
\]
For \(k=0\):
\[
s = 1 \angle 45^\circ
\]
For \(k=1\):
\[
s = 1 \angle 135^\circ
\]
For \(k=2\):
\[
s = 1 \angle 225^\circ
\]
For \(k=3\):
\[
s = 1 \angle 315^\circ
\]
Expressing in rectangular form:
\[
s = 1 \angle 45^\circ = \cos 45^\circ + j \sin 45^\circ = 0.707 + j\,0.707
\]
\[
s = 1 \angle 135^\circ = \cos 135^\circ + j \sin 135^\circ = -0.707 + j\,0.707
\]
\[
s = 1 \angle 225^\circ = \cos 225^\circ + j \sin 225^\circ = -0.707 - j\,0.707
\]
\[
s = 1 \angle 315^\circ = \cos 315^\circ + j \sin 315^\circ = 0.707 - j\,0.707
\]
\[
\text{Taking only poles in the left half-plane: } s = -0.707 \pm j 0.707 = s_1, s_2
\]
\[
B_2 = (s - s_1)(s - s_2)
\]
\[
= \left(s - (-0.707 + j 0.707)\right) \left(s - (-0.707 - j 0.707)\right)
\]
\[
= (s + 0.707 - j 0.707)(s + 0.707 + j 0.707)
\]
\[
= (s + 0.707)^2 - (j 0.707)^2
\]
\[
= s^2 + 2 \times 0.707 \times s + (0.707)^2 - (-1) \times (0.707)^2
\]
\[
= s^2 + \sqrt{2} s + 0.5 + 0.5 = s^2 + \sqrt{2} s + 1
\]
\[
\text{Transfer function: } \quad T_2(s) = \frac{1}{B_2} = \frac{1}{s^2 + \sqrt{2} s + 1}
\]
---
\[
\text{When } n=3:
\]
\[
|T_3(s)|^2 = \frac{1}{1 + (-1)^3 s^{6}} = \frac{1}{1 - s^6}
\]
\[
\text{Poles are located at: } 1 - s^6 = 0 \implies s^6 = 1
\]
\[
s = 1 \angle \frac{k \times 360^\circ}{6}, \quad k = 0,1, \ldots, 5
\]
Specifically:
\[
s = 1 \angle 0^\circ = 1
\]
\[
s = 1 \angle 60^\circ = \cos 60^\circ + j \sin 60^\circ = \frac{1}{2} + j \frac{\sqrt{3}}{2}
\]
\[
s = 1 \angle 120^\circ = \cos 120^\circ + j \sin 120^\circ = -\frac{1}{2} + j \frac{\sqrt{3}}{2}
\]
\[
s = 1 \angle 180^\circ = -1
\]
\[
s = 1 \angle 240^\circ = \cos 240^\circ + j \sin 240^\circ = -\frac{1}{2} - j \frac{\sqrt{3}}{2}
\]
\[
s = 1 \angle 300^\circ = \cos 300^\circ + j \sin 300^\circ = \frac{1}{2} - j \frac{\sqrt{3}}{2}
\]
Taking only poles in the left half-plane:
\[
s = -1, \quad s = -\frac{1}{2} \pm j \frac{\sqrt{3}}{2} = s_1, s_2, s_3
\]
\[
B_3 = (s - s_1)(s - s_2)(s - s_3)
\]
\[
= (s + 1)(s - (-\tfrac{1}{2} + j \tfrac{\sqrt{3}}{2}))(s - (-\tfrac{1}{2} - j \tfrac{\sqrt{3}}{2}))
\]
\[
= (s + 1)(s + \tfrac{1}{2} - j \tfrac{\sqrt{3}}{2})(s + \tfrac{1}{2} + j \tfrac{\sqrt{3}}{2})
\]
\[
= (s + 1) \left[(s + \tfrac{1}{2})^2 - (j \tfrac{\sqrt{3}}{2})^2 \right]
\]
\[
= (s + 1) \left[s^2 + s + \tfrac{1}{4} - (-1) \times \tfrac{3}{4}\right]
\]
\[
= (s + 1)(s^2 + s + 1)
\]
\[
\text{Transfer function: } \quad T_3(s) = \frac{1}{B_3} = \frac{1}{(s + 1)(s^2 + s + 1)}
\]