Question

3. Find the transfer function E, (s)/E, (s) of the circuit below using impedance methods Find expressions for the natural frequency a, and the damping ratio ζ . ei(t) For L-0.2H, 2 mF RI 10 ohms, R2 20 ohms, use MATLAB to calculatew, and ζ and then also plot the unit step response of the system. Apply the final value theorem and initial value theorem to the solution in the Laplace domain and show that the results agree with your plot.

I need help with this Dynamics II electrical system analysis.
Please provide detailed explanation and show work.

Thank you.

Answer 1

The given circuit can be redrawn in the Laplace domain as follows.

12. Sl SC

Let $Z_1$ be the impedance of the parallel combination of the inductor and series RC branch and $E_1$ be the voltage across it.

$Z_1=\frac{sL\times(R_2+\frac{1}{sC})}{sL+(R_2+\frac{1}{sC})}=\frac{s^2R_2CL+sL}{s^2LC+sR_2C+1}$

Now the total impedance of the circuit as referred to the source is

$Z_{eq}=R_1+Z_1=R_1+\frac{s^2R_2CL+sL}{s^2LC+sR_2C+1}=\frac{R_1(s^2LC+sR_2C+1)+s^2R_2CL+sL}{s^2LC+sR_2C+1}$

$\Rightarrow Z_{eq}=\frac{s^2LC(R_1+R_2)+s(L+R_1R_2C)+R_1}{s^2LC+sR_2C+1}$

According to the voltage division principle, the voltage across the impedance $Z_1$ is g iven by

$E_1(s)=\frac{Z_1}{Z_{eq}}E_i(s)=\frac{\frac{s^2R_2CL+sL}{s^2LC+sR_2C+1}}{\frac{s^2LC(R_1+R_2)+s(L+R_1R_2C)+R_1}{s^2LC+sR_2C+1}}E_i(s)=\frac{s^2R_2CL+sL}{s^2LC(R_1+R_2)+s(L+R_1R_2C)+R_1}E_i(s)$ ...............Eq.1

Now the output voltage $E_o$ according the voltage principle si given by

$E_o=\frac{\frac{1}{Cs}}{R_2+\frac{1}{Cs}}E_1(s)=\frac{1}{1+sCR_2}E_1(s)$ ........Eq.2

From Eq.1 and Eq.2

$E_o=\frac{1}{1+sCR_2}\cdot\frac{s^2R_2CL+sL}{s^2LC(R_1+R_2)+s(L+R_1R_2C)+R_1}E_i(s)$

$\Rightarrow E_o=\frac{1}{1+sCR_2}\cdot\frac{sL(sR_2C+1)}{s^2LC(R_1+R_2)+s(L+R_1R_2C)+R_1}E_i(s)$

$\Rightarrow E_o=\frac{sL}{s^2LC(R_1+R_2)+s(L+R_1R_2C)+R_1}E_i(s)$

From the above equation, the transfer function $\frac{E_o(s)}{E_i(s)}$ is given by

$\frac{E_o(s)}{E_i(s)}=\frac{sL}{s^2LC(R_1+R_2)+s(L+R_1R_2C)+R_1}$ ...........Eq.3

The above equation can be rearranged as

$\frac{E_o(s)}{E_i(s)}=\frac{sL}{LC(R_1+R_2)\left(s^2+s\frac{(L+R_1R_2C)}{LC(R_1+R_2)}+\frac{R_1}{LC(R_1+R_2)}\right)}$

$\frac{E_o(s)}{E_i(s)}=\frac{s\frac{1}{C(R_1+R_2)}}{\left(s^2+s\frac{(L+R_1R_2C)}{LC(R_1+R_2)}+\frac{R_1}{LC(R_1+R_2)}\right)}$ .............Eq.4

From the above equation, the characteristic equation is

$s^2+s\frac{(L+R_1R_2C)}{LC(R_1+R_2)}+\frac{R_1}{LC(R_1+R_2)}=0$ ...........Eq.5

Comparing Eq.5 with the standard characteristic equation of a second order system $s^2+2\zeta\omega_ns+\omega_n^2=0$ we have

$\omega_n=\sqrt{\frac{R_1}{LC(R_1+R_2)}}$

$2\zeta\omega_n=\frac{L+R_1R_2C}{LC(R_1+R_2)}$

$\Rightarrow \zeta=\frac{1}{2\omega_n}\frac{L+R_1R_2C}{LC(R_1+R_2)}$

$\Rightarrow \zeta=\frac{1}{2\sqrt{\frac{R_1}{LC(R_1+R_2)}}}\frac{L+R_1R_2C}{LC(R_1+R_2)}=\frac{L+R_1R_2C}{2\sqrt{R_1LC(R_1+R_2)}}$

So, the expressions for the natural frequency and the damping ratio are

$\omega_n=\sqrt{\frac{R_1}{LC(R_1+R_2)}}$

$\zeta=\frac{L+R_1R_2C}{2\sqrt{R_1LC(R_1+R_2)}}$

Given the values of the circuit elements are

$\\L=0.2H \\C=2mF \\R_1=10\Omega \\R_2=20\Omega$

Substituting the above values in Eq.3

$\frac{E_o(s)}{E_i(s)}=\frac{s(0.2)}{s^2(0.2)(0.002)(10+20)+s(0.2+(10)(20)(0.002))+10}$

$\frac{E_o(s)}{E_i(s)}=\frac{0.2s}{0.012s^2+0.6s+10}$

The output voltage to a step response in the Laplace domain can be given as

$E_o(s)=\frac{0.2s}{0.012s^2+0.6s+10}\cdot\frac{1}{s}$

According to Initial value theorem, the initial value is given

$e_o(t=0)=\lim_{s\rightarrow\infty}sE(s)=\lim_{s\rightarrow\infty}s\frac{0.2s}{0.012s^2+0.6s+10}\cdot\frac{1}{s}$

$e_o(t=0)=\lim_{s\rightarrow\infty}\frac{0.2s}{0.012s^2+0.6s+10}=0$

According to the final value theorem, the final value is given by

$e_o(t=\infty)=\lim_{s\rightarrow0}sE(s)=\lim_{s\rightarrow0}s\frac{0.2s}{0.012s^2+0.6s+10}\cdot\frac{1}{s}$

$e_o(t=\infty)=\lim_{s\rightarrow0}\frac{0.2s}{0.012s^2+0.6s+10}=0$

The MATLAB code to find the natural frequency and the damping ratio and to plot the step response is given below.

%values of elements
R1=10;
R2=20;
L=0.2;
C=0.002;
%coefficients of numerator
num=[L,0];
%coefficients of numerator
den=[L*C*(R1+R2), L+R1*R2*C R1];
%transfer function
sys1=tf(num,den);
%obtaining the damping ratio and the natural frequency
[wn,zeta]=damp(sys1);
%displaying the above results on command window
fprintf('Natural Frequency\t:%1.4f\n',wn(1))
fprintf('Damping Ratio\t\t:%1.4f\n',zeta(1))
%plotting the step response
step(sys1)

The Output on the command window is

Natural Frequency   :28.8675
Damping Ratio       :0.8660

The step response plot is

Step Response 0.25 0.2 0.15 0.1 0.05 0.05 0.05 0.15 0.2 0.25 0.3 Time (seconds)

We got zero as the initial and final value of the system from the step response plot. The same is verified from the initial and final value theorems.