Laplace transforms: Simple Electrical Network: Difference between revisions

Latest revision as of 16:35, 13 December 2009

Problem Statement

Using the formulas

 $E (t) = L \frac{d i_{R}}{d t} + R i_{C}$ 
 $R C \frac{d i_{R}}{d t} + i_{R} - i_{C} = 0$

Solve the system when V0 = 50 V, L = 4 h, R = 20 Ω, C = 10^-4 f, and the currents are initially zero.

Solution

Solve the system when V0 = 50 V, L = 4 h, R = 20 Ω, C = 10^-4 f, and the currents are initially zero.

$4 \frac{d i_{1}}{d t} + 20 i_{2} = 50$

$20 (1 0^{- 4}) \frac{d i_{2}}{d t} + i_{2} - i_{1} = 0$

Applying the Laplace transform to each equation gives

$4 (s ℒ {i_{1}} - i_{1} (0)) + 20 ℒ {i_{2}} = 50$

$\Rightarrow 4 s I_{1} (s) + 20 I_{2} (s) = \frac{50}{s}$

$0.005 (s ℒ i_{2} - i_{2} (0)) + ℒ {i_{2}} - ℒ {i_{1}} = 0$

$\Rightarrow - 500 I_{1} (s) + [s + 500] I_{2} (s) = 0$

Solving for $I_{2} (s)$

$I_{2} (s) = \frac{6250}{s (s^{2} + 500 s + 2500)}$

We find the partial decomposition

Let $I_{2} (s) = \frac{6250}{s (s^{2} + 500 s + 2500)} = \frac{A}{s} + \frac{B s + C}{s^{2} + 500 s + 2500}$

$\Rightarrow 6250 = A (s^{2} + 500 s + 2500) + (B s + C) s$

$\Rightarrow 62500 = A s^{2} + 500 A s + 2500 A + B s^{2} + C s$

Comparing the coefficients we get

$A = \frac{5}{2}, B = - 5, C = - 1250$

Thus $I_{2} (s) = \frac{5}{2 s} - \frac{5 s + 1250}{s^{2} + 500 s + 2500}$

Now we do the same for $I_{1}$ where we solve the function in terms of $I_{1}$ and decomposing the partial fraction resulting in

$I_{1} (s) = \frac{25 s + 12500}{s (s^{2} + 500 s + 2500)} = \frac{5}{s} - \frac{5 s + 2475}{s^{2} + 500 s + 2500}$

In order to make it nicer on us we need to complete the square as follows

$s^{2} + 500 s + 2500 = 0$

$\Rightarrow s^{2} + 500 s = - 2500$

$\Rightarrow s^{2} + 500 s + {(\frac{500}{2})}^{2} = - 2500 + {(\frac{500}{2})}^{2}$

$\Rightarrow s^{2} + 500 s + 62500 = 6000$

$\Rightarrow (s + 250)^{2} - (100 \sqrt{6})^{2} = 0$

Thus

$I_{2} (s) = \frac{5}{2 s} - 5 \frac{s + 250}{(s + 250)^{2} - (100 \sqrt{6})^{2}} - \frac{5 \sqrt{6}}{12} \frac{100 \sqrt{6}}{(s + 250)^{2} - (100 \sqrt{6})^{2}}$

Taking the Inverse Laplace transform gives

$ℒ^{- 1} {I_{2} (s)} = i_{2} (t) = \frac{5}{2} - 5 e^{- 250 t} c o s h 100 \sqrt{6} t - \frac{5 \sqrt{6}}{12} 5 e^{- 250 t} s i n h 100 \sqrt{6} t$

Initial Value Theorem

$\lim_{s \to \infty} s I (s) = f (0^{+})$

$\lim_{s \to \infty} s \frac{25 s + 12500}{s (s^{2} + 500 s + 2500)} = i (0)$

$\Rightarrow i (0) = 0$

$\lim_{s \to \infty} s \frac{6250}{s (s^{2} + 500 s + 2500)} = i (0)$

$\Rightarrow i (0) = 0$

Final Value Theorem

$\lim_{s \to 0} s I (s) = f (\infty)$

$\lim_{s \to \infty} s \frac{25 s + 12500}{s (s^{2} + 500 s + 2500)} = i (\infty)$

$\Rightarrow i (\infty) = 0$

$\lim_{s \to 0} s \frac{6250}{s (s^{2} + 500 s + 2500)} = i (\infty)$

$\Rightarrow i (\infty) = 0$

Bode Plots

The following are bode plots for the transfer functions

Error creating thumbnail: File missing

H (s)_{1} = \frac{25 s + 12500}{s (s^{2} + 500 s + 2500)}

Error creating thumbnail: File missing

H (s)_{2} = \frac{6250}{s (s^{2} + 500 s + 2500)}

@@ Line 4: / Line 4: @@
   <math>RC\dfrac{di_R}{dt}+i_R-i_C=0</math>
-Solve the system when V0 = 50 V, L = 0.5 h, R = 80 Ω, C = 10<sup>-4</sup> f, and the currents are initially zero.
+Solve the system when V0 = 50 V, L = 4 h, R = 20 Ω, C = 10<sup>-4</sup> f, and the currents are initially zero.
 ==Solution==
-Solve the system when V0 = 50 V, L = 0.5 h, R = 60 Ω, C = 10^{-4} f, and the currents are initially zero.
+Solve the system when V0 = 50 V, L = 4 h, R = 20 Ω, C = 10<sup>-4</sup> f, and the currents are initially zero.
-Substituting numbers into the equations, we have
-<math>0.5\frac{di_1}{dt}+60i_2=50</math>
+<math>4\frac{di_1}{dt}+20i_2=50</math>
-<math>60(10^{-4})\frac{di_2}{dt}+i_2-i_1=0</math>
+<math>20(10^{-4})\frac{di_2}{dt}+i_2-i_1=0</math>
 Applying the Laplace transform to each equation gives
-<math>\frac{1}{2}(s\mathcal{L}\left\{i_1\right\}-i_1(0))+60\mathcal{L}\left\{i_2\right\}=50</math>
+<math>4(s\mathcal{L}\left\{i_1\right\}-i_1(0))+20\mathcal{L}\left\{i_2\right\}=50</math>
-<math>0.006(s\mathcal{L}{i_2}-i_2(0))+\mathcal{L}\left\{i_2\right\}-\mathcal{L}\left\{i_1\right\}=0</math>
+<math>\Rightarrow4sI_1(s)+20I_2(s)=\frac{50}{s}</math>
-<math>\Rightarrow\frac{1}{2}sI_1(s)+60I_2(s)=\frac{50}{s}</math>
+<math>0.005(s\mathcal{L}{i_2}-i_2(0))+\mathcal{L}\left\{i_2\right\}-\mathcal{L}\left\{i_1\right\}=0</math>
-<math>-200I_1(s)+[s+200]I_2(s)=0</math>
+<math>\Rightarrow-500I_1(s)+[s+500]I_2(s)=0</math>
-Solving for <math>\mathcal{L}\left\{i_1\right\}</math> gives
+Solving for <math>I_2(s)</math>
+<math>I_2(s)= \frac{6250}{s(s^2+500s+2500)}</math>
+We find the partial decomposition
+Let <math>I_2(s)= \frac{6250}{s(s^2+500s+2500)}=\frac{A}{s}+\frac{Bs+C}{s^2+500s+2500}</math>
+<math>\Rightarrow6250=A(s^2+500s+2500)+(Bs+C)s</math>
+<math>\Rightarrow62500=As^2+500As+2500A+Bs^2+Cs</math>
+Comparing the coefficients we get
+<math>A=\frac{5}{2},B=-5,C=-1250</math>
+Thus
+<math>I_2(s)=\frac{5}{2s}-\frac{5s+1250}{s^2+500s+2500}</math>
+Now we do the same for <math>I_1</math> where we solve the function in terms of <math>I_1</math> and decomposing the partial fraction resulting in
+<math>I_1(s)= \frac{25s+12500}{s(s^2+500s+2500)}=\frac{5}{s}-\frac{5s+2475}{s^2+500s+2500}</math>
+In order to make it nicer on us we need to complete the square as follows
+<math> s^2+500s+2500=0</math>
+<math>\Rightarrow s^2+500s=-2500</math>
+<math>\Rightarrow s^2+500s+\left(\frac{500}{2}\right)^2=-2500+\left(\frac{500}{2}\right)^2</math>
+<math>\Rightarrow s^2+500s+62500=6000</math>
+<math>\Rightarrow (s+250)^2-(100\sqrt{6})^2=0</math>
+Thus
+<math>I_2(s)=\frac{5}{2s}-5\frac{s+250}{(s+250)^2-(100\sqrt{6})^2}-\frac{5\sqrt{6}}{12}\frac{100\sqrt{6}}{(s+250)^2-(100\sqrt{6})^2}</math>
+Taking the Inverse Laplace transform gives
+<math>\mathcal{L}^{-1}\left\{I_2(s)\right\}= i_2(t) =\frac{5}{2}-5e^{-250t}cosh100\sqrt{6}t-\frac{5\sqrt{6}}{12}5e^{-250t}sinh100\sqrt{6}t</math>
+==Initial Value Theorem==
+<math>\lim_{s \to \infty}sI(s)=f(0^+)</math>
+<math>\lim_{s \to \infty} s\frac{25s+12500}{s(s^2+500s+2500)}=i(0)</math>
+<math> \Rightarrow i(0)=0</math>
+<math>\lim_{s \to \infty}s\frac{6250}{s(s^2+500s+2500)}=i(0)</math>
+<math> \Rightarrow i(0)=0</math>
+==Final Value Theorem==
+<math>\lim_{s \to 0}sI(s)=f(\infty)</math>
+<math>\lim_{s \to \infty} s\frac{25s+12500}{s(s^2+500s+2500)}=i(\infty)</math>
+<math>\Rightarrow i(\infty)=0</math>
+<math>\lim_{s \to 0}s\frac{6250}{s(s^2+500s+2500)}=i(\infty)</math>
+<math>\Rightarrow i(\infty)=0</math>
+==Bode Plots==
+The following are bode plots for the transfer functions
+[[Image:I1_.jpg|400px|thumb|left|<math>H(s)_1=\frac{25s+12500}{s(s^2+500s+2500)}</math>]]
+[[Image:I2_.jpg|400px|thumb|left|<math>H(s)_2=\frac{6250}{s(s^2+500s+2500)}</math>]]

Laplace transforms: Simple Electrical Network: Difference between revisions

Latest revision as of 16:35, 13 December 2009

Contents

Problem Statement

Solution

Initial Value Theorem

Final Value Theorem

Bode Plots

Navigation menu

Laplace transforms: Simple Electrical Network: Difference between revisions

Latest revision as of 16:35, 13 December 2009

Problem Statement

Solution

Initial Value Theorem

Final Value Theorem

Bode Plots

Navigation menu

Search