Unit 3                                                       

2270                              PRACTICE EXAM SOLUTION Prob 2                                     

 

2.      (45 points)

 

The current source is a dc current source. After being open for a long time, the switch is closed at t = 0.

a.   Write a numerical time-domain expression for v(t).

b.   From the Laplace transform of v(t), find the numerical values of v(t) for t = 0+ and t → ∞.

 

ans:  a)     v(t > 0) = 2.003 e-0.4kt - 0.003 e-249.6kt V

         b)     v(t = 0+) = 2V, v(t∞) = 0V

 

sol'n:  (a)   First we find initial conditions for L and C. (We need these for s-domain models of L and C.)

For t = 0-, L acts like short, C acts like open circuit.

 

When we close the switch, we short out the first R.

s-domain model:

Note:     A DC source corresponds to a step function even if there is no switch and the source has the same output for all time. Thus, we have Ig/s as the source in the s-domain. (Conceptually, we only need the current source for t > 0 because the initial conditions on L and C account for what the current source did for t < 0.)

Note:     We may choose either a series sL and V-source for L or a parallel sL and I‑source for L. Here, the parallel I-source model is more convenient. The same applies to the C.

Normally, we might use superposition at this point, turning on the I-sources one at a time and then summing currents or voltages to get a final answer.

Here, however, we have parallel I-sources that sum:

Combining the parallel impedances and using V = Iz, we have

To compute the parallel z value, we factor out 1/sC to remove fractions:

Now multiply through by sC:

Factor out the denominator again:

Multiply through by s2LC + 1:

Now divide top and bottom by RLC to make the coefficient of the highest power of s in the denominator equal to unity.

Check:  Using the numerator and the first term in the denominator, we have the following units analysis:

Thus, we have an impedance as we should have. The other terms in the denominator have the same units as s2 since the units of s are, ironically, 1/sec or 1/s.

Now we plug in numbers to compute V(s):

Find poles for quadratic term in preparation for partial fractions:

s1,2 = -125k ± 75k rad/s         (based on 52 − 42 = 32 pythagorean triple)

s1 = −50k, s2 = −200 rad/s

Now use partial fractions:

Use the standard inverse Laplace transform term:

This gives the final answer:

v(t > 0) = 2.003 e-0.4kt - 0.003 e-249.6kt V

 

sol'n:  (b)  Use the initial value theorem to find v(t=0+):

The largest power of s dominates in the numerator and in the denominator.

Note:     We expect v(0+) = 2V since this is the initial capacitor voltage.

Use the final value theorem to find v(t→∞):

Note:     We expect v(t) to decay, since L becomes a short.