PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Fourier and Laplace 421

Then from Table 4.2

vt()
 etsin( t ut) ()

120

7

⋅⋅ 7

and therefore,

it

vt

R

etut

it Cdv t

dt

d

dt

e

R t

C

t

()

()

sin( ) ( )

() ()

6

7

7

1

40

120

7





⋅

⋅ ssin( ) ( )

() cos( ) ( )

7

3

7

77 7

3

7

tut

itC etttt te e



















sin  777 7

3

3

6

7

cos()sin() ( )

() () ()

()

tt

it it it

it e

LRC

L t













−

ut

 sin( 7 ) ( ) 7cos(7) sin(7) ( )

3

7

tute−t ttut

MATLAB Solution

% Script file: RLC _ parallel

t =-2:0.1:5;

t0 =0;

u = stepfun(t,t0);

vt =120/sqrt(7).*exp(-t).*sin(sqrt(7).*t).*u ;

figure(1)

subplot(2,2,1);

plot(t,vt);title(‘ v(t) vs.t (analytical)’);

ylabel(‘Amplitude (volts)’)

ir = 6/sqrt(7)*e x p(-t).*sin(sqrt(7).*t).*u;

subplot(2,2,2);

plot(t,ir);title(‘ ir(t) vs.t (analytical)’);

ylabel(‘ Amplitude (amps)’)

ic=3/sqrt(7).*exp(-t).*(sqrt(7).*cos(sqrt(7).*t)- sin(sqrt(7).*t)).*u;

subplot(2,2,3);

plot(t,ic);title(‘ ic(t) vs.t (analytical)’);

ylabel(‘Amplitude (amps)’);xlabel(‘time(sec)’);

subplot(2,2,4);

il = 3*ones(1,71)-ir-ic ;

plot(t,il);title(‘il(t) vs.t (analytical)’);

ylabel(‘Amplitude (amps)’) ;xlabel(‘time (sec)’);

figure(2)

subplot(2,2,1)

num = [0 0 120];

den = [1 2 8];

vti = impulse(num,den,t);vti=[zeros(1,20) vti’];

plot(t,vti);title(‘ v(t) vs. t (sym. Laplace)’);