ve216_l11.md

VE216 Lecture 11

CT Frequency Response and Bode Plots

Asymptotic Magnitude Behavior

Isolated Zero

Magnitude	Asymptotic

Isolated Pole

Magnitude	Asymptotic

Complicated Systems Asymptotic Behavior

$\begin{align} H(s_0) = K\frac {\prod^Q_{q=1} (s_0-z_q)}{\prod^P_{p=1} (s_0-p_p)} \end{align}$, then $\begin{align}|H(s_0)| = \Big|K\frac {\prod^Q_{q=1} (s_0-z_q)}{\prod^P_{p=1} (s_0-p_p)} \Big| = |K|\frac {\prod^Q_{q=1} |s_0-z_q|}{\prod^P_{p=1} |s_0-p_p|} \end{align}$

Thus $log |H(j\omega)| = log|K| + \sum^Q_{q=1} log|j\omega - z_q| - \sum^P_{p=1}log|j\omega - p_p|$

With proportion to the $log (\omega)$, we get the bode plot.

Bode Plot Angle

According to the previous lectures: $\begin{align}\angle H(s_0) = \angle(K \frac {\prod ^Q_{q=1}(s_0-z_q)} {\prod^P_{p=1}(s_0-p_p)}) \end{align} = \angle K + \sum^Q_{q=1} \angle(s_0-z_q) - \sum^P_{p=1}\angle(s_0-p_p)$

If we need more calculation, then we can add them together as a graph.

Bode Plot: dB

$|H(j\omega)|\text{[dB]} = 20log_{10}|H(j\omega)|$

For $H(j\omega) = \frac 1 {j\omega + 1}$, the approximation is here

$X$	$20 log_{10}X$
$1$	$0$ dB
$\sqrt 2$	$3$ dB
$2$	$6$ dB
$10$	$20$ dB
$100$	$40$ dB

Frequency Response of High-Q System

$H(s) = \frac 1 {1 + \frac 1 Q \frac s {\omega_0} + (\frac s {\omega_0})^2}$

Peak Magnitude Dependence

Assume $Q > 3$. Then we see on the $\frac s {\omega_0}$ plane:

So the peak magnitude increase with $Q$.

3dB Bandwidth and Phase Change

Change in phase approximately $\pi / 2$.

Frequency Response of a High-Q System

As $Q$ increases, the phase changes more abruptly with $\omega$.