Bode Plot Example of First-Order System using Matlab
In this article, Bode Plot of Simple Phase-Lag Network (First Order System) is obtained using Matlab. In order to draw Bode Plot, we need transfer function from which we deduce the equations for Magnitude and Phase.
\[G(s)=\frac{1}{2s+1}\]
Function in the frequency domain can be written as:
\[G(s)=\frac{1}{2j\omega +1}\]
From above expression, we can deduce the corner frequency or break point as;
\[\omega =\frac{1}{2}\]
For Magnitude Plot:
When ω<<1 (very very small value), then
$G(s)\approx 1$
So, for very small value of ω, log magnitude of the transfer function would be;
\[|G(j\omega ){{|}_{dB}}=20\text{ }log|G(j\omega )|=20\text{ }log(1)=0\]
Hence, magnitude response would be constant below breakpoint.
When ω>>1 (very very large value), then
\[G(s)\approx \frac{1}{2j\omega }\]
So, for very large value of ω, log magnitude of the transfer function would be;
\[|G(j\omega ){{|}_{dB}}=20\text{ }log|G(j\omega )|=20\text{ }log\left| \frac{1}{|2j\omega |} \right|=20\text{ }log\left( \frac{1}{2\omega } \right)=20log\left| 1 \right|-20log\left| 2\omega \right|=-20log(2\omega )\]
So, above the break point, the magnitude plot would be a straight line with -20 dB/decade slope
Now, phase of the transfer function G(s) can be calculated as;
\[\angle G(j\omega )=0-{{\tan }^{-1}}(\omega T)={{\tan }^{-1}}(\omega T)\]
For Phase Plot:
When ω is very very small (ω≈0), then
\[\angle G(j\omega )H(j\omega )={{0}^{\centerdot }}\]
When ω is very very large (ω→∞), then
\[\angle G(j\omega )H(j\omega )=-{{90}^{\centerdot }}\]
Bode Plot Example Matlab Code
Here, we implemented the bode-plot for the comprehensive understanding of the readers.
% Bode Plot for Phase-Lag Network Example clc % Transfer function K = [1]; T = 2; num = [K]; den = [T 1]; H = tf(num, den) % Bode Plot grid on bode(H) grid %For Asymptotic Plot % num=[1]; % den=[2 1]; % bode_asymptotic(num,den);
Bode Plot for Phase-Lag Network Example
- You May Also Read: Bode Plot Detailed Overview with Matlab
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