Laplace transforms: Simple Electrical Network: Difference between revisions

Revision as of 18:24, 1 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

$I_{1} (s) = \frac{25 s + 12500}{s (s^{2} + 500 s + 2500)} = \frac{5}{8 s} + \frac{- 5 s + 175}{s^{2} + 125 s + 20000}$

Taking the Inverse Laplace transform yields

$ℒ^{- 1} {I_{1} (s)} = \frac{5}{8} + \frac{39 \sqrt{103}}{824} s i n * (\frac{25}{2} \sqrt{103} * t)$

$ℒ^{- 1} {I_{2} (s)} =$

@@ Line 7: / Line 7: @@
 ==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}+80i_2=50</math>
+<math>4\frac{di_1}{dt}+20i_2=50</math>
-<math>80(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))+80\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>0.008(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)+80I_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>-125I_1(s)+[s+125]I_2(s)=0</math>
+<math>\Rightarrow-500I_1(s)+[s+500]I_2(s)=0</math>
 Solving for <math>I_2(s)</math>
-<math>I_2(s)= \frac{12500}{s(s^2+125s+20000)}</math>
+<math>I_2(s)= \frac{6250}{s(s^2+500s+2500)}</math>
 We find the partial decomposition
-Let <math>I_2(s)= \frac{12500}{s(s^2+125s+20000)}=\frac{A}{s}+\frac{Bs+C}{s^2+125s+20000}</math>
+Let <math>I_2(s)= \frac{6250}{s(s^2+500s+2500)}=\frac{A}{s}+\frac{Bs+C}{s^2+500s+2500}</math>
-<math>\Rightarrow12500=A(s^2+125s+20000)+(Bs+C)s</math>
+<math>\Rightarrow6250=A(s^2+500s+2500)+(Bs+C)s</math>
-<math>\Rightarrow12500=As^2+125As+20000A+Bs^2+Cs</math>
+<math>\Rightarrow62500=As^2+500As+2500A+Bs^2+Cs</math>
 Comparing the coefficients we get
-<math>A=\frac{5}{8},B=-5,C=-625</math>
+<math>A=\frac{5}{2},B=-5,C=-1250</math>
 Thus
-<math>I_2(s)=\frac{5}{8s}-\frac{5s+625}{s^2+125s+20000}</math>
+<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{100s+12500}{s(s^2+125s+20000)}=\frac{5}{8s}+\frac{-5s+175}{s^2+125s+20000}</math>
+<math>I_1(s)= \frac{25s+12500}{s(s^2+500s+2500)}=\frac{5}{8s}+\frac{-5s+175}{s^2+125s+20000}</math>
+Taking the Inverse Laplace transform yields
+<math>\mathcal{L}^{-1}\left\{I_1(s)\right\}=\frac{5}{8}+\frac{39\sqrt{103}}{824}sin*(\frac{25}{2}\sqrt{103}*t)</math>
+<math>\mathcal{L}^{-1}\left\{I_2(s)\right\}=</math>

Laplace transforms: Simple Electrical Network: Difference between revisions

Revision as of 18:24, 1 December 2009

Problem Statement

Solution

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