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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(
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Answer #1

Q1(a)

Vin(t)=Ri(t)+L\frac{\mathrm{di} }{\mathrm{d} t}+1/C\int idt................................by KVL ................(1)

From the given transfer function

\frac{Vo(S)}{Vin(S)}=\frac{\frac{1}{LC}}{S^2+\frac{RS}{\frac{}{}L}+\frac{1}{LC}}

Vo(s) Vin(S) SPLC + RCS + 1...................multiplying numerator and denominator by 1/LC

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

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