engineering recuitment GATE ECE 2023-24 Test Series Control Systems Frequency Response Analysis Bode Plot
For an LTI system, the Bode plot for its gain is as illustrated in the figure shown. The transfer function of the system is:
1
\(H\left( s \right) = \frac{{10^5\left( {1 + \frac{s}{{1000}}} \right){{\left( {1 + \frac{s}{{{{10}^4}}}} \right)}}}}{{\left( {1 + \frac{s}{{10}}} \right){{\left( {1 + \frac{s}{{100}}} \right)}^2}{{\left( {1 + \frac{s}{{{{10}^5}}}} \right)}^2}\left( {1 + \frac{s}{{{{10}^6}}}} \right)}}\)
2
\(H\left( s \right) = \frac{{10^5\left( {1 + \frac{s}{{1000}}} \right){{\left( {1 + \frac{s}{{{{10}^4}}}} \right)}^2}}}{{\left( {1 + \frac{s}{{10}}} \right){{\left( {1 + \frac{s}{{100}}} \right)}^2}{{\left( {1 + \frac{s}{{{{10}^5}}}} \right)}^2}\left( {1 + \frac{s}{{{{10}^6}}}} \right)}}\)
3
\(H\left( s \right) = \frac{{\left( {1 + \frac{s}{{1000}}} \right){{\left( {1 + \frac{s}{{{{10}^4}}}} \right)}^2}}}{{\left( {1 + \frac{s}{{10}}} \right){{\left( {1 + \frac{s}{{100}}} \right)}^2}{{\left( {1 + \frac{s}{{{{10}^5}}}} \right)}^2}\left( {1 + \frac{s}{{{{10}^6}}}} \right)^2}}\)
4
\(H\left( s \right) = \frac{{\left( {1 + \frac{s}{{1000}}} \right)^2{{\left( {1 + \frac{s}{{{{10}^4}}}} \right)}^2}}}{{\left( {1 + \frac{s}{{10}}} \right){{\left( {1 + \frac{s}{{100}}} \right)}^2}{{\left( {1 + \frac{s}{{{{10}^5}}}} \right)}^2}\left( {1 + \frac{s}{{{{10}^6}}}} \right)^2}}\)