Summary of steps to sketch asymptotic Bode plots¶
To construct an asymptotic Bode magnitude and phase plot given a known transfer function $P(s)$, follow the steps below.
Asymptotic magnitude plot¶
- Substitute $s=j\omega$ to obtain the steady-state frequency response transfer function $P(j\omega)$.
- Compute the magnitude response equation $\frac{Y}{U} = \left|P(j\omega)\right|$.
- Compute the steady-state gain for the transfer function by setting $\omega=0$. If the transfer function has a pole (or set of repeated poles) at $\omega=0$, you will not be able to find this (you'll get a division by zero). Find the gain at the lowest frequency you wish to see on your asymptotic Bode magnitude plot.
- The magnitude plot on a decibel-log-frequency scale will be flat until the first (slowest) corner frequency (pole or zero). At a frequency equal to each zero, the slope of the magnitude plot changes by $+20\frac{dB}{decade}$ per order of the zero. At a frequency equal to each pole, the slope of the magnitude plot changes by $-20\frac{dB}{decade}$ per order of the pole.
Asymptotic phase plot¶
- Substitute $s=j\omega$ to obtain the steady-state frequency response transfer function $P(j\omega)$.
- Compute the phase shift equation $\phi = \angle(P(j\omega))$
- Draw an asymptotic phase plot for each pole and each zero (including multiple plots for repeated and underdamped 2nd order poles and zeros):
- For each pole, the asymptotic phase plot for this pole is flat at $0$ until $\omega=\frac{a}{5}$ where $a$ is the pole, unless the pole is at $\omega=0$, in which case the phase plot is flat at low frequencies beginning at $\phi=\frac{\pi}{2}p(-90 ^\circ)$, where $p$ is the order of the pole at zero. Then, between $\omega=\frac{a}{5}$ and $\omega={5a}$, the plot slopes down with a slope of $-\frac{1}{2}\log_e(10)$ and then is flat at $\phi=\frac{\pi}{2}(-90 ^\circ)$ from $\omega=5a$ until $\omega=\infty$.
- For each zero, the asymptotic phase plot for this pole is flat at $0(0 ^\circ)$ until $\omega=\frac{a}{5}$ where $a$ is the pole, unless the zero is at $\omega=0$, in which case the phase plot is flat at low frequencies beginning at $\phi=90p^\circ$, where $p$ is the order of the zero at $\omega=0$. Then, between $\omega=\frac{a}{5}$ and $\omega={5a}$, the plot slopes up with a slope of $\frac{1}{2}\log_e(10)$ and then is flat at $\phi=\frac{\pi}{2}(90 ^\circ)$ from $\omega=5a$ until $\omega=\infty$.
- The total asymptotic phase plot can be constructed by summing the component "first order" phase plots.
Remember that you can always check your work using MATLAB's "bode" command.