Question

Am (t) → volt) Figure Q1 R = 10, L = H, C = F Q1(a) [2 marks] The transfer function for the circuit in Figure V.(s)__ LC Vin(5) sz+s+ + Form the corresponding differential equation that governs the behaviour of the system Q1(b) [4 marks] Determine a state-space representation for the circuit in Figure Q1 with the states selected as x;(6) v.) and x2(t) = 1,t). Show that it is equal to the following 408 J18+) y(t) = [1 o [:] Q1(c) [4 marks] Confirm that the eigenvalues of the state matrix in the state-space representation are equal to the poles of the transfer function for the circuit in Figure Q1. Q1(d) [3 marks] The state-transition matrix for the state-space representation given in part Q1(b) is $(t) = 2e* + aeft e-+ +ye-2 1-3e-+ + 3e-2t -e-t + 2e-2t) What are a, ß and y?

Answer 1

Q1(a)

................................by KVL ................(1)

Vin(t) = Ri(t) + 2 * +1/0 | idt

From the given transfer function

Vo(S) Vin(S) 52+

...................multiplying numerator and denominator by 1/LC

Vo(s) Vin(S) SPLC + RCS + 1

$Vo(S)=\frac{Vin(S)}{(S^2LC+{RCS}+1)}$

$Vo(S)=\frac{1}{CS}\frac{Vin(S)}{(SL+{R}+\frac{1}{SC})}$

(b)

$x_{1}=V_{0}$

$x_{2}=i_{L}$

$V_{o}=\frac{1}{c}\int i_{L}dt$

that is

$\dot{x_{1}}=\frac{1}{C}i_{L}(t)$...........(2)

$\dot{x_{1}} =\frac{1}{C}x_{2}$......................(3)

on re arranging equation 1

$\dot{i_{L}}=-\frac{V_{O}}{L}-\frac{i_{2}}{L}+\frac{V_{in}}{L}$.......................(4)

$\dot{x_{2}}=-\frac{x_{1}}{L}-\frac{x_{2}}{L}+\frac{V_{in}}{L}$....................(5)

substituting the values of L ,C R

$\dot{x_{1}}=\frac{2}{3}x_{2}$........(6)

$\dot{x_{2}}=-3{x_{1}-3{x_{2}}+3{V_{in}}$............(7)

from equation 6 and 7

$\begin{bmatrix}\dot{x _{1}} \\\dot{x _{2}} \end{bmatrix}=\begin{bmatrix} 0 & \frac{2}{3}\\ -3 &-3 \end{bmatrix} \begin{bmatrix}x_{1} \\ x_{2} \end{bmatrix} +\begin{bmatrix} 0\\ 3 \end{bmatrix}V_{in}$

and output voltage Vo is voltage across capacitor

$\begin{bmatrix} y \end{bmatrix}=\begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}$ hence part b is verified

(c) substituting the values of L and c in transfer function we get

$\frac{V_{o}}{V_{in}}=\frac{\frac{1}{2}}{S^2+3S+2}$

$\frac{V_{o}}{V_{in}}=\frac{\frac{1}{2}}{S+2)(S+1)}$

therefore poles are at S=-2,-3

from the state space representation

system matrix A is given by

$\begin{bmatrix}0 &\frac{2}{3} \\ -3 & -3 \end{bmatrix}$

eigen values are found by

determinant of $\begin{bmatrix}S & 0\\ 0 & S \end{bmatrix}-\begin{bmatrix} 0 &\frac{2}{3} \\ -3 & -3 \end{bmatrix}=0$

on solving

S(S+3)+(2/3)*3=0

s^2+3S+2=0

therefore S=-2,-1 are the eigen values from thesytem matrix which matches with the poles of the tranfer function

(d)

state transmission matrix is given by

L^-1{[SI-A]^-1}

[SI-A]^-1=(1/(S+2)(S+3))[ S+3 2/3

-3 S ]

taking the laplace inverse

State transmission matrix=[ S+3/(S+1)*(S+2) (2/3)/(S+1)*(S+2)

-3/(S+1)*(S+2) S/(S+1)*(S+2) ]

=[2e^-t -0.5e^-2t -(2/3)e^-t+(2/3)e^-2t

-3e^-t+3e^-2t e^-t+2e^-2t ]

there fore alpha=-0.5 , beta=-2 and gama=2/3