Math265 Course Project - Part C -Final

Math265 Project Part C Name: Taylor Johnson Please study the Module 7 Lesson and Module 7 Electronic Application first before attempting this project. Follow the steps demonstrated in the Module 7 Electronics Application and the MATLAB Tutorial video in Media Gallery. I. Solving the Differential Equation In the series RLC circuit shown in Figure 1, the switch SW is first placed at position P1 until the capacitor is fully charged. Then, the SW is moved to position P2 to form the discharging circuit. Figure 1: The Serial RLC Circuit for Discharging Process The differential equation for the above circuit, as derived in the Module 7 Electronics Application , can be expressed as the following. v C ' ' ( t ) + R L v C ' ( t ) + 1 LC v C ( t ) = 0 ( 1 ) Please show your detailed mathematical work for Steps 1 and 2. 1 | P a g e

Step 1: The component values are illustrated in Figure 1. R = 8 Ω, L = 1 H, and C = 1/15 F. Substitute the values to derive a general solution for the homogeneous second order differential equation. (10 points) V c (t)+ {R} over {L} {V} rsub {c} '(t)+(C) {V} rsub {c} (t)= V c (t)+8 {V} rsub {c} '(t)+15= Y 2 + 8 Y + 15 = 0 ( Y + 5 ) ( Y + 3 ) = 0 Y =− 3 , − 5 V c ( t ) = c 1 e − 5 t + c 2 e − 3 t Step 2: The initial conditions are given as: v C ( 0 ) = E = 20 V ∧ v C ' ( 0 ) = − E RC = − 75 2 V / s Use the initial conditions to solve for C 1 ∧ C 2 and then write the final solution. Highlight your final solution. (10 points) 20 = c 1 e − 3 ( 0 ) + c 2 e − 5 ( 0 ) ,c 1 = 20 − c 2 x = ( 20 − c 2 ) e − 3 t + c 2 e − 5 t x =− 3 e − 3 t ( 20 − c 2 ) − 5 c 2 e − 5 t − 75 2 =− 3 e − 3 ( 0 ) ( 20 − c 2 ) − 5 c 2 e − 5 ( 0 ) − 75 2 =− 3 ( 20 − c 2 ) − 5 c 2 − 75 2 =− 60 − 2 c 2, c 2 = − 45 4 V c ( t ) = 125 4 e − 3 t − 45 4 e − 5 t II. Solving the Second Order Differential Equation Using MATLAB You will use MATLAB to solve the second order differential equation. If you are not familiar with MATLAB programming, please watch the MATLAB Tutorial video in Media Gallery. 2 | P a g e

III. Using MATLAB to Plot the System Response Step 5: Calculate the time constant τ = 2L/R for the series RLC circuit. (5 points) τ = 2L/R = ______1/4_____ sec Step 6: Write a MATLAB program to plot the graph of the solution. Assign proper linspace() to display the x domain based on the τ = 2L/R value. Paste the program below. (10 points) Step 7: Label the X-axis and the Y-axis on the graph. Insert a meaningful title for the graph. Also, add a Data Tip to show the time constant and its corresponding voltage value. Paste the completed figure below. Is the v C ( τ ) close to the theoretical value? (10 points) 4 | P a g e

IV. Summary Write a two paragraph summary to conclude your findings from the above three sections. Research on the characteristics of an over-damped RLC circuit. What is meant by over-damped and what is the general behavior? Explain the discharging curve you obtained from the MATLAB plot. (15 points) First, I had to solve the second-order homogeneous equation for the series RLC circuit. The circuit was an overdamped case because the two exponentials are both driving the the current to zero. This means that the system response will lag. Because it is overdamped the system will respond slowly. The characteristic equation is vC(t) = C1e^(tγ1) + C2e^(tγ2). I solved the equation to find the values of the constants C1 and C2 which gave me the answer for the vC(t). The MATLAB program was used to calculate the value of vC(t) and it matched the answer from the equation. The MATLAB was then used to plot and shows the discharge curve of the voltage across the capacitor depending on time. The graph shows that there is a charge of 20V in the capacitor. When the switch is closed it moves from one part of the capacitor to the other making the voltage go down as the time extends. 5 | P a g e