homework_4_sol

# homework_4_sol - EE40 P4.3 Homework#4 Solution The voltage...

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EE40 Homework #4 Solution P4.3 The voltage at t=+infinity is Vs=100V. The time constant of the circuit is 1 R Cm S τ = = So the general expression for the voltage across the capacitor would be ( ) 100 (100 ) , 0 t init Vt V e t =− where V init is the voltage across the capacitor at t=0. We can plug in its value in three cases, where Vinit=0V, Vinit=50V, Vinit=-50V, respectively. The similarities are: time constant is similar; the final value is similar. Below is the plot for Vinit=0V Below is the three plots together 1

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0 1 2 3 4 5 6 -50 0 50 100 voltage / Volts time / ms P4.16 P4.45 2
Note: For those interested students, Let’s carefully examine the following differential equation 1 () c c dV t CV t dt t + = Assume Vc(t)=f1(t) is one solution to the above differential equation, where f1(t) is 3

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some function of t. Also, let’s assume Vc(t)=f2(t) is one solution to the following differential equation, which is called the “homogeneous equation” 1 () () 0 c c dV t CV t dt + = The question is: Is Vc(t) = f1(t) + C 2 *f2(t) also a solution to the first differential equation (where C 2 is a constant)? To answer this question, let us plug Vc(t) into the left-hand-side (LHS) of the equation 12 2 11 1 2 1 2 2 2 1 2 2 ( () ) ( ) [ ( )] [ ( )] [( ) ] [ ( ) ] 0 c c dV t d f t C f t t C f t C f dt dt df t d C f t Cf t C C f t dt dt df t df t t C t dt dt tC t 2 2 t + += ++ =+ + + + + =+ ×= We found LHS = t, which means Vc(t) = f1(t) + C 2 f2(t) is also a solution to the first differential equation. Up to now, the attentive reader should generalize a general approach to solve the first differential equation: step1, obtain a solution f1(t) to the first differential equation above;
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homework_4_sol - EE40 P4.3 Homework#4 Solution The voltage...

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