Elsewhere we have discussed how to make Bode plots for a real pole. You should be familiar with that analysis. The discussion there assumed that the value of ω0 was positive; here we discuss the case if ω0 is negative. We start with
$$H(s)=\frac{1}{1+\frac{s}{\omega_0}}$$If you carefully examine the analysis (here) of the "Magnitude" plot you'll see that the only time ω0 is used, it is squared (e.g., $\left( \frac{\omega}{\omega_0} \right)^2$). Therefore, the magnitude plot does not depend on the sign of ω0, only its absolute value, so we don't need to change anything to accomodate a negative value of ω0.
The phase however does change. The phase of a single real pole is given by is given by
$$\angle H\left( {j\omega } \right) = \angle \left( {{1 \over {1 + j{\omega \over {{\omega _0}}}}}} \right) = - \angle \left( {1 + j{\omega \over {{\omega _0}}}} \right) = - \arctan \left( {{\omega \over {{\omega _0}}}} \right)$$
Let us again consider three cases for the value of the frequency, but we assume ω0 is negative:
Case 1) ω<<ω0. This is the low frequency case with ω/ω0→0, and doesn't depend on the sign of ω0. At these frequencies We can write an approximation for the phase of the transfer function
$$\angle H\left( {j\omega } \right) \approx -\arctan \left( 0 \right) = 0^\circ = 0\;rad$$
Case 2) ω>>ω0. Here we will consider the cases of positive and negative ω0 side by side.
ω0 > 0 (the minimum phase case, discussed previously)
This is the high frequency case with ω/ω0 → +∞. We can write an approximation for the phase of the transfer function
$$\angle H\left( {j\omega } \right) \approx - \arctan \left( \infty \right) = - 90^\circ $$
The high frequency approximation is at shown in green on the diagram below. It is a horizontal line at -90°.
ω0 < 0 (the non-minimum phase case)
This is the high frequency case with ω/ω0 → -∞. We can write an approximation for the phase of the transfer function
$$\angle H\left( {j\omega } \right) \approx - \arctan \left(- \infty \right) = + 90^\circ $$
The high frequency approximation is at shown in green on the diagram below. It is a horizontal line at +90°.
Case 3) ω=ω0. Again consider the cases of positive and negative ω0 side by side.
ω0 > 0 (the minimum phase case, discussed previously)
$$\angle H\left( {j\omega } \right) = - \arctan \left( 1 \right) = - 45^\circ$$
ω0 < 0 (the non-minimum phase case)
$$\angle H\left( {j\omega } \right) = - \arctan \left( -1 \right) =+ 45^\circ$$
From the above discussion you can see that the only effect of the pole having a negative value of ω0 is that the phase is inverted (it increases from 0 → +90° as ω increases from 0 → ∞. The images below show the Bode plots for
$$H_1(s)=\frac{1}{1+\frac{s}{10}}, \quad\quad H_2(s)=\frac{1}{1-\frac{s}{10}}$$H1(s) has a positive ω0 (ω0=+10, so H1(s)=1/(1+s/10)) and H2(s) has a negative ω0 (ω0=-10, so H2(s)=1/(1-s/10)). The pole of H1(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H2(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero)
H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites.
The same conclusion holds for first order poles and second order poles and zeros (see below).
The images below show the Bode plots for two functions, one with a positive ω0 (ω0=+10) and one with a negative ω0 (ω0=-10). The zero of H1(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H2(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero).
$$H_1(s)=1+\frac{s}{10}, \quad\quad H_2(s)=1-\frac{s}{10}$$H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites. Recall that 360° is equivalent to 0° so you can think of the plot for the angle of H2(s) as starting at 0° and dropping by 90° (though the plot shows it as starting at 360°).
The images below show the Bode plots for (note the sign of the middle term in the numerator is different)
$$H_1(s)=\frac{1}{1+0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2}, \quad\quad H_2(s)=\frac{1}{1-0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2}$$
The poles of H1(s) are at s=-0.5±j9.987 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H2(s) is at s=+0.5±j9.987 (a positive real part, the right half of the s-plane; a non-minimum phase pole). H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that, again, the magnitudes are identical, but the phases are opposites.
The images below show the Bode plots for (note the sign of the middle term in the numerator is different)
$$H_1(s)=1+0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2, \quad\quad H_2(s)=1-0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2$$The zeros of H1(s) are at s=-0.5±j9.987 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H2(s) is at s=+0.5±j 9.987(a positive real part, the right half of the s-plane; a non-minimum phase zero). H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites.
If a 1st order pole has a positive real part (i.e., a nonminimum phase system, pole is in right half of s-plane) at say s=+5, so
$$H(s)=\frac{1}{1 - \frac{s}{5}}$$the magnitude of the Bode plot is unchanged from the case of a corresponding pole with negative value, at s=-5 (pole is in left half of s-plane)
$$H(s)=\frac{1}{1 + \frac{s}{5}}$$but the phase of the plot is inverted.
The same rule holds Bode plots for 2nd order (complex conjugate) poles, and for 1st and 2nd order zeros.