PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Fourier and Laplace 359

The process is illustrated as follows:

V

s

RI s LsI s LI

Is R Ls V

s

LI






() ()

() [ ]





0

0

Is

V

sR Ls

LI

RLs

Is

V

Lss RL

I

()

()

()











11

1

0

0



























 

11

0

sRL

Is

A

s

B

sRL

I

sRL























()

Solving for the constants A and B, and substituting in I(s), the following expression

is obtained:

Is

VR

s

VR

sRL

I

sRL

()
















0

Finally, transforming I(s) from the frequency domain back to the time domain, the

following solution for i(t) is obtained:

it I s

V

R

()

£^1 [()] ( 1 eutIeutRt L) () 0 Rt L()

R.4.111 As mentioned earlier, one of the most powerful applications of the LT is the solu-

tion of integrodifferential equations illustrated by the transient analysis of RC and

RL circuits of R.4.109 and R.4.110. These concepts are extended to include loop or

node equations.

The steps involved are summarized as follows:

a. Write the integrodifferential set of equations (loop or node equations) for a

given circuit.

b. Transform the integrodifferential equations of part a using Laplace into an alge-

braic set of equations, in which the initial conditions are automatically inserted.

c. The algebraic set of equations of part b are then solved for either the currents

{I(s)} (loop equations) or voltages {V(s)} (node equations).

d. Finally, the time solution is obtained by taking the ILT of the expressions obt-

ained in part c.

R.4.112 The example shown in the circuit diagram of Figure 4.18, is solved for the current

i(t), is used to illustrate the steps followed in the solution of an integrodifferential

system, in which each step is labeled according to R.4.111.

ANALYTICAL Solution

Step a

vt()
v tLR() v t()fort^0 (KVL)