Bode Plot

Bode Plot

This is the main graphical representation of a frequency response. It has magnitudes and phases. For a given frequency, shows that your frequency response is in a graphical way. The Bode plot contains a magnitude and phase. The x-axis is typically in a logarithmic scale, and the y-axis is in a decibel scale

You basically just, solve for the frequency response of a given system (transfer function where $s=j\omega$) and plug in different values of $\omega$ and see what the frequency response is

General Rules

General rules for approximate graphing

• Your magnitude break points are ${\omega }_n$ or the s roots in the numerator and denominator
• Your phase angle break points are 10x and 0.1x of each magnitude break point

1. Given a transfer function $G\left(s\right)$
2. Identify poles and zeroes
3. Transform into Bode forms (normalized forms) $G\left(s\right)=K_0{s}^p\frac{\left(\frac{s}{z_1}+1\right)\dots \left(\frac{s}{z_m}+1\right)}{\left(\frac{s}{p_1}+1\right)\dots \left({\left(\frac{s}{{\omega }_n}\right)}^2+2\zeta \frac{s}{{\omega }_n}+1\right)}$
   • These forms look like this:
     • You may need to complete the square?
     • 
     • 
4. List the breakpoints in ascending order
   1. For magnitude plots: $p_1<z_1<\dots <{\omega }_n$
   2. For phase plots: $0.1p_1,10p_1,0.1z_1,10z_1,0.1{\omega }_n,10{\omega }_n$ where you list them basically in the same order by also by the front magnitude
5. Continue plotting for subsequent breakpoints
   1. At each breakpoint, change the slope accordingly (for magnitudes)
      1. Ex. for a 1st order breakpoint, change the slope by 20, for a 2nd order, change the slope by 40
      2. The change is positive for zeros and negative for poles
      3. The last line needs a slope of -20
   2. At each point (0.1x and 10x), change the slope accordingly (for phases)
      1. For each 0.1x point, If it is for the 1st order term, change the slope by 45°, for a 2nd order, change the slope by 90°
      2. The change is positive for zeros and negative for poles
      3. For each 10x point, do it in negative way
      4. The final line must lie flat with a slope of -90°
6. If there is an $s^p$ in the TF, it changes the first line segment of the Bode plot
   1. Here, we draw the first line of the magnitude plot with a slope of 20p and its value $20{\log }_{10}{K}_0$ at $\omega =1$
   2. Draw the first line of the phase plot with a flat line with $90p$°
   3. If any break point coincides with the other one, apply the rules multiple times according to each rules

Minimum Phase System

• This is a system whose poles are all in the LHP

Sensitivity Function

• The ratio of the fractional change in $T\left(s\right)$ (closed-loop TF) to the fractional change in $G_p\left(s\right)$ (the plane TF)

Sensitivity function for the TF above is… $\frac{1}{1+G_P{G}_C}$

The complementary transfer function i defined as $C\left(s\right)=1-S\left(s\right)=\frac{G_C\left(s\right)G_P\left(s\right)}{1+G_C\left(s\right)G_P\left(s\right)}$

Gain Phase Relationship

• When the slope of $|G\left(j\omega \right)|$ versus $\omega$ on a log-log scale persists as a constant value for a decade of frequency, then $<G\left(j\omega \right)$ is related to $|G\left(j\omega \right)|$ as $<G\left(j\omega \right)=n\ast 90°$ where n is the slope of the magnitude in [decade/decade]
• This can be used as a guide to infer stability from $|G\left(j\omega \right)|$

Infinite Gain at $\omega$ = 0 for Tracking

• We want large amplitudes (gains) are low frequencies which means we need at least 1 s (we want a System Type of I or II)
• Open Loop FR: $\left|G\left(j\omega \right)\right|=\left|G_K\left(j\omega \right)G_P\left(j\omega \right)\right|$
• Closed Loop PR: $\left|T\left(j\omega \right)\right|=\left|\frac{G\left(j\omega \right)}{1+G\left(j\omega \right)}\right|$
• We want $\left|T\left(j\omega \right)\right|$ to be 0 for low frequencies…
• Consider ( |G(j\omega)| ) with its behaviour $\{\begin{array}{ll}|G(j\omega )|\gg 1,& \text{for }\omega \ll {\omega }_c\\ |G(j\omega )|\ll 1,& \text{for }\omega \gg {\omega }_c\end{array}$ where ${\omega }_C$ is the cross-over frequency
• Then, the closed-loop transfer function (TF) becomes $|T(j\omega )|\approx \{\begin{array}{ll}1,& \omega \ll {\omega }_c\\ \frac{1}{|G(j\omega )|},& \omega \gg {\omega }_c\end{array}$
• Where $\left|G\left(j\omega \right)\right|=\infty$ at $\omega =1$
• For our closed-loop ${\omega }_{BW}$
  • Range of frequencies that have good closed-loop tracking performance, or maximum freq. that can be tracked
  • A typical indicator of speed of response
  • The closed-loop bandwidth is closely related to crossover frequency of the open-loop frequency response as ${\omega }_c\le {\omega }_{BW}\le 2{\omega }_c$

Loop Shaping

We use our understanding of Bode Plots for Loop Shaping