• Bode discovered that the phase can be uniquely derived from the slope of the magnitude for minimum-phase system.

Basic Factor	M. Slope(Low)	P. (Low)	M. Slope(High)	P. (High)
K	0	0	0	0
\(s^N\)	20N	90N	20N	90N
1st Order	0	0	-20	-90
2nd Order	0	0	-40	-180

• A transfer function \(G(s)\) is minimum phase if both \(G(s)\) and \(1/G(s)\) are causal and stable.
• Roughly speaking it means that the system does not have zeros or poles on the right-half plane. Moreover, it does not have delay.

Consider two transfer functions: \[G_1(s) = 1,\,G_2(s) = \frac{1-s}{1+s}\]

• \(G_1(s)\) is minimum phase since it does not have unstable zeros/poles.
• The magnitude of \(G_1(s)\) is 0dB, and the phase is \(0^\circ\).
• \(G_2(s)\) is non-minimum phase, since \(G_2(1) = 0\).

\[|G_2(j\omega)| = \frac{\sqrt{1+\omega^2}}{\sqrt{1+\omega^2}} = 1,\,\angle G_2(j\omega) = -2\tan^{-1}\omega.\]

• Time delay often exists in communication systems, chemical and manufacturing processes, etc.
• Time delay systems are of non-minimum phase behavior.
• The transfer function of the time delay factor is \(G_3(s) = \exp(-\tau s)\), where \(\tau\) is the delay.

\[G_3(j\omega) = \exp(-j\omega\tau) = 1\angle -\omega\tau\times\frac{180^\circ}{\pi}.\]

Minimum phase system has minimum phase change amongst all transfer functions that have the same magnitude plot.

• Non-minimum phase systems are much more difficult to control than minimum phase system

Bode plots of a transfer function can be obtained using Matlab. However, to interpret the Bode plots, we shall take a look at how basic factors of the transfer function affect the Bode plots.

\begin{align} G(s) = \frac{2000(s+0.5)}{s(s+10)(s+50)} = \frac{2(2s+1)}{s(0.1s+1)(0.02s+1)}. \end{align}

Factor	Corner Freq.
2
\(1/s\)
\(2s+1\)	0.5
\(1/(0.1s+1)\)	10
\(1/(0.02s+1)\)	50

By adding the Bode plots of the above factors together, we can get the Bode plot of \(G(s)\)

• For low frequency, only need to look at the constant \(K\) and \(s^N\) term.
• Draw the plot of \(s^N\) first
  • Magnitude: a straight line passes through \((1,0dB)\) with a slope of \(20NdB\).
  • Phase: a horizontal line at \(90N^\circ\).
• Adding the constant term \(K\)
  • Magnitude: add \(20\log|K|\)
  • Phase: If \(K > 0\), do nothing. If \(K < 0\), subtract \(180^\circ\).

Factor	\(2s+1\)	\((0.1s+1)^{-1}\)	\((0.02s+1)^{-1}\)
Corner Freq.	0.5	10	50

• Magnitude

For \((\tau s+1)^N\), change the slope at corner frequency \(1/\tau\) by \(20N\).

Frequency	Low	0.5	10	50
Slope Change		+20	-20	-20
Slope	-20	0	-20	-40

Factor	\(2s+1\)	\((0.1s+1)^{-1}\)	\((0.02s+1)^{-1}\)
Corner Freq.	0.5	10	50

• Phase: For \((\tau s+1)^N\)
• change the slope at \(0.1/\tau\) by \(45N\).
• change the slope at \(10/\tau\) by \(-45N\).

Frequency	Low	0.05	1	5	100	500
Slope Change		+45	-45	-45 -45	+45	+45
Slope	0	45	0	-90	-45	0

\begin{align} G(s) = \frac{10}{s(s^2+0.4s+4)} = \frac{2.5}{s((s/2)^2+2\times 0.1s/2+1)}. \end{align}

Factor	Corner Freq.
2.5
\(1/s\)
\((s/2)^2+0.2s/2+1\)	2

The low frequency factors are \(2.5\) and \(1/s\).

• Magnitude
  • Draw a straight line passing through \((1,0dB)\) with slope -20.
  • Shift the magnitude plot of \(1/s\) up by \(20\log(2.5)\).
• Phase
  • Draw a constant line at \(-90^\circ\).

Consider a second order term:

\[\left[\left(\frac{s}{\omega_n}\right)^2+2\zeta\frac{s}{\omega_n}+1\right]^{\pm 1}.\]

• Magnitude
  • change the slope at the corner frequency \(\omega_n\) by \(\pm 40dB\)
  • add \(\pm 20\log(2\zeta)\) correction at the corner frequency \(\omega_n\).
• Phase
  • draw the phase plot using calculators.

\begin{align} G(s) &= \frac{2500(s+10)}{s(s+2)(s^2+30+2500)}\\ &=\frac{5(0.1s+1)}{s(0.5s+1)\left[(s/50)^2+0.6s/50+1\right]}. \end{align}