PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

358 Practical MATLAB® Applications for Engineers

Then, applying Kirchhoff’s voltage law to the loop of the circuit diagram, shown in

Figure 4.15, and performing some algebraic manipulation the current I(s) can be solved,

and an expression for i(t) (ILT) can be obtained, as indicated in the following:

V

s

RI s

Is

sC

V

s

IsR

Is

sC

V

s

V

s

Is R

sC

VV







()

()

()

()

()

0

0

(^10)

 


 ss
Is
VV
RsRC








⋅





()
()


0 1
1
and the current i(t) is therefore given by
it I s

VV

R

()

£() euttRC()

1  0

[]









R.4.110 Now consider the simple RL circuit shown in Figure 4.16, and let us solve for i(t),
assuming that the initial inductor’s current is IL(0) = I 0 = 10 (amp), in a counter-
clockwise direction.

ANALYTICAL Solution

The time domain circuit diagram of Figure 4.16 is transformed using the LT into the

frequency domain circuit diagram shown in Figure 4.17.

Then the analytical solution leading to i(t) is obtained by the following steps:

a. Write the resulting equation by applying Kirchhoff’s voltage law around the

loop of the circuit shown in Figure 4.17

b. Solve for I(s) (using partial fractions expansion if necessary)

c. Obtain i(t) by taking the ILT of I(s)

V

R

sw closes at t = 0

i(t)

L

FIGURE 4.16

Network of R.4.110.

V/s

Vr = I(s)R

sw closes at t = 0

I(s)

Vl(s) = Ls I(s) − LI 0

I 0

FIGURE 4.17

Frequency domain representation of the

network of Figure 4.16.