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 ω_{0} was positive; here we discuss the case if ω_{0} is negative. We start with

If you carefully examine the analysis (here) of the "Magnitude" plot you'll see that the only time ω_{0} 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 ω_{0}, only its absolute value, so we don't need to change anything to accomodate a negative value of ω_{0}.

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 ω_{0} is negative:

**Case 1)** ω<<ω_{0}. This is the low frequency case with
ω/ω_{0}→0, and doesn't depend on the sign of ω_{0}. 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>>ω_{0}. Here we will consider the cases of positive and negative ω_{0} side by side.

ω_{0} > 0 (the minimum phase case, discussed previously)

This is the high frequency case with
ω/ω_{0} → +∞. 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°.

ω_{0} < 0 (the non-minimum phase case)

This is the high frequency case with
ω/ω_{0} → -∞. 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=ω_{0}. Again consider the cases of positive and negative ω_{0} side by side.

ω_{0} > 0 (the minimum phase case, discussed previously)

$$\angle H\left( {j\omega } \right) = - \arctan \left( 1 \right) = - 45^\circ$$

ω_{0} < 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 ω_{0} 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) has a positive ω_{0} (ω_{0}=+10, so H_{1}(s)=1/(1+s/10)) and H_{2}(s) has a negative ω_{0} (ω_{0}=-10, so H_{2}(s)=1/(1-s/10)). The pole of H_{1}(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H_{2}(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero)

H_{1}(s) is plotted as a solid blue line, and H_{2}(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 ω_{0} (ω_{0}=+10) and one with a negative ω_{0} (ω_{0}=-10). The zero of H_{1}(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H_{2}(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero).

H_{1}(s) is plotted as a solid blue line, and H_{2}(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 H_{2}(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 H_{1}(s) are at s=-0.5±*j*9.987 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H_{2}(s) is at s=+0.5±*j*9.987 (a positive real part, the right half of the s-plane; a non-minimum phase pole). H_{1}(s) is plotted as a solid blue line, and H_{2}(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 H_{1}(s) are at s=-0.5±*j*9.987 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H_{2}(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). H_{1}(s) is plotted as a solid blue line, and H_{2}(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites.

If a 1^{st} 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

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 2^{nd} order (complex conjugate) poles, and for 1^{st} and 2^{nd} order zeros.

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