Transfer functions of continuous-time systems Problem 1 : Consider the following system:
figure 1 : assignment-1-circuit
Let R 1 = 40 Ω , R 2 = 20 Ω , L = 10 m H , C = 1 μ F R_1 = 40\Omega, R_2 = 20\Omega, L = 10mH, C= 1\mu F R 1 = 40Ω , R 2 = 20Ω , L = 10 m H , C = 1 μ F . The input is v i n v_{in} v in the output is v o u t v_{out} v o u t . Give both transfer function and state space representation for the system.
Solution
Given circuit is a second-order linear system due to presence of one inductor (L) and one capacitor (C).
Given transfer function H ( s ) H(s) H ( s ) is given by the ratio over Laplace domain:
H ( s ) = V o u t ( s ) V i n ( s ) H(s) = \frac{V_{out}(s)}{V_{in}(s)} H ( s ) = V in ( s ) V o u t ( s )
Given that the impedance of the inductor Z l = s L Z_l = sL Z l = s L and the impedance of the capacitor Z c = 1 s C Z_c = \frac{1}{sC} Z c = s C 1 , the total impedance of the circuit is given by:
Z total = 1 1 s L + s C Z_{\text{total}} = \frac{1}{\frac{1}{sL} + sC} Z total = s L 1 + s C 1
Using voltage divider rule, the transfer function is given by:
H ( s ) = V o u t ( s ) V i n ( s ) = 1 s C 1 s L + 1 s C H(s) = \frac{V_{out}(s)}{V_{in}(s)} = \frac{\frac{1}{sC}}{\frac{1}{sL} + \frac{1}{sC}} H ( s ) = V in ( s ) V o u t ( s ) = s L 1 + s C 1 s C 1