This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: v 1 (0) = 12 = K 1 + K 2 (0) = K 1 = 12 (9) We also know that v 2 (0) = 0 and v 2 is given by Eq. 6. Solving this gives K 2 = 48. Hence v 1 = 12 e 2 t + 48 te 2 t (10) Also, from the figure, i = 2 v 1 4 , hence i = 6 e 2 t + 24 te 2 t A . (11) (3) Problem 7.24. Solution: R1 3 R2 3 L1 1H C1 1F + V2 3V + V1 I1 2A 2t+1V i i2 i3 i Figure 1: Problem 7.24. We are supposed to find only the forcing solution i f ( t ) given the input as shown in Figure 1. Differential equation of the circuit We will use mesh analysis with the meshes as showin in Figure 1. We know that the meshes with currents i and i 2 share a current source hence we have a supermesh. The constraint equation for this supermesh is i 2 1 = 2 , or i 2 = 2 + i (12) Applying KVL around the supermesh and using Eq. 12, we obtain 1 di dt + 3 i 2 + integraldisplay t ( i 2 i 3 ) d = 2 t + 1 (13) d 2 i dt 2 + 3 di 2 dt + i 2 i 3 = 2 (by differentiating once w.r.t. t ) (14) Page 3 Applying KVL around mesh 3, we obtain 3 i 3 + integraldisplay t ( i 3 i 2 ) d = 3 3 i 3 integraldisplay t ( i 2 i 3 ) d = 3 Substituting the expression for integraltext t ( i 2 i 3 ) from Eq. 13 into the above equation yields 3 i 3 + di dt + 3 i 2 2 t 1 = 3 (15) and by using the expression for i 3 from Eq. 14 into the above equation and simplifying, we obtain 3 d 2 i dt 2 + 10 di 2 dt + 6 i 2 = 2 t + 4 , or d 2 i dt 2 + 10 3 di 2 dt + 2 i 2 = 2 3 t + 4 3 and using Eq. 12, the above equation becomes d 2 i dt 2 + 10 3 di dt + 2 i = 2 3 t 8 3 (16) Forced Response From the above equation the forcing function is f ( t ) = 2 3 t 8 3 , and we know that the forced solution, also called the particular solution, is of the form (see [1], Table 7.2, p. 287) i f = K 1 + tK 2 (17) If this is a solution to Eq. 16 then it must satisfy that differential equation. Substituting this guess solution into the differential equation, we obtain (0) + 10 3 ( K 2 ) + (2 K 1 + 2 tK 2 ) = 2 3 t 8 3 , (18) and simplifying this (by equating the coefficients of t on both sides of the equation and by equating the remaining terms), we obtain the values of the constants K 2 = 1 3 and k 2 = 17 9 . Hence the forced solution is i f = 1 3 t 17 9 . (19) (4) Problem 7.25. We are to find the voltage of the top node, which is essential the voltage across the capacitor C 1 , for all time, given the current source to be i = e 2 t cos3 t A. Solution: Page 4 C1 0.25 L1 0.5 R1 1 2 1 I? Figure 2: Problem 7.25. The problem here is at t < 0, the circuit is an RC circuit and at t 0 it becomes an RLC circuit. In the former case, it is a first order system and in the latter case, it is a second order system. We will solve for the forced response for these two cases separately....
View
Full
Document
This note was uploaded on 05/06/2010 for the course ENG ECSE210 taught by Professor Levine during the Fall '09 term at McGill.
 Fall '09
 Levine

Click to edit the document details