hw5_sol_sp08

# hw5_sol_sp08 - i ECE220 Homework Assignment#5 Solutions...

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Unformatted text preview: i ECE220 Homework Assignment #5 Solutions Problem 7.5 Consider the differential equation dv ( t ) dt + 2 . × 10 6 v ( t ) = 10 u ( t ) (0.1) with the initial condition v (0) = 0 . (a) Find v c ( t ) . (b) Find v p ( t ) . (c) Find the total solution v ( t ) . (d) Plot all three solutions v c ( t ) , v p ( t ) , v ( t ) and the driving force. When plotting the complementary solution, use the constant you derived in part (c). Solution (a) The complementary solution, v c ( t ) , is the solution to the equation dv c ( t ) dt + 2 . × 10 6 v c ( t ) = 0 . We can write dv c ( t ) v c ( t ) =- 2 . × 10 6 dt Integrating both sides, we get ln( v c ( t )) =- 2 . × 10 6 t + c that is, v c ( t ) = e- 2 . × 10 6 t + c = Ce- 2 . × 10 6 t Therefore, the complementary solution is v c ( t ) = Ce- 2 . × 10 6 t u ( t ) . Note that we may also derive this directly from equation (7.35), page 341. (b) Since the driving force is in the form of a unit step function, we may use the method in section 3.2.1, page 344 (or equation 7.47, Theorem 7.4, page 348). ii Here, we have v s ( t ) = 10 u ( t ) ; therefore, A = 10 and a = 2 . × 10 6 . We may guess v p ( t ) = B · u ( t ) , and we need to determine the value of the unknown constant B . From equation (7.39), we have v p ( t ) = 10 2 . × 10 6 = 1 2 . × 10 5 u ( t ) (c) The total solution is given by v ( t ) = v c ( t ) + v p ( t ) → = Ce- 2 . × 10 6 t u ( t ) + 1 2 . × 10 5 u ( t ) We can determine C from the boundary condition, v (0) = C + 1 2 . × 10 5 = 0 C =- 1 2 . × 10 5 Therefore, v ( t ) =- 1 2 . × 10 5 e- 2 . × 10 6 t u ( t ) + 1 2 . × 10 5 u ( t ) = 1 2 . × 10 5 (1- e- 2 . × 10 6 t ) u ( t ) (d) The following matlab script produces the graphs shown in figure 0.1. clear % Define the time interval tmin=-10^-6; tmax=5*10^-6; dt=6*10^-6/1000; t=tmin:dt:tmax; % Define the driving force, total, particular % and complementary solution signals vs=10.*(t>0); v=1/(2*10^5)*(1-exp(-2.0*10^6*t)).*(t>0); vp=1/(2*10^5).*(t>0); vc=1/(2*10^5)*(-exp(-2.0*10^6*t)).*(t>0); iii % Plot the signals subplot( 2 , 1 , 1 ) plot ( t , vs ) grid on xlabel(’t’) ylabel(’v_s(t)’) axis([-10^-6,5*10^-6,-1,11]) subplot( 2 , 1 , 2 ) plot( t , vc, t, vp, t, v ) legend(’v_c(t)’, ’v_p(t)’, ’v(t)’) grid on xlabel(’t’) title(’v_c(t), v_p(t), v(t)’) axis([-10^-6,5*10^-6,-6*10^-6,6*10^-6])-1 1 2 3 4 5 x 10-6 2 4 6 8 10 t v s (t)-1 1 2 3 4 5 x 10-6-6-4-2 2 4 6 x 10-6 t v c (t), v p (t), v(t) v c (t) v p (t) v(t) Figure 0.1. Problem 7.5, part d. iv Problem 7.7 Consider the differential equation dy ( t ) dt + 10 y ( t ) = 2cos(2 π 1700 t + π/ 3) u ( t ) (0.2) with the initial condition y (0) = 1 ....
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hw5_sol_sp08 - i ECE220 Homework Assignment#5 Solutions...

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