0195136047.pdf

756 BASIC CONTROL SYSTEMS

+

++

+

DC

supply Load Vs

VG 3 , VG 4

−Vs

Vs

Q 1

VG 1 VG 3

Q 4

Q 2

Q 3

G

G

t

vo

(a) (b)

0

−

−

−−

2

TT

T

1

t

2

T

VG 1 , VG 2

T

1

0

0 t

2

T

Figure 16.1.9Single-phase dc–ac converter (inverter).(a)Circuit diagram.(b)Voltage waveforms.

EXAMPLE 16.1.1

Consider a diode circuit with anRLCload, as shown in Figure E16.1.1, and analyze it fori(t)

when the switchSis closed att=0. Treat the diode as ideal, so that the reverse recovery time

and the forward voltage drop are negligible. Allow for general initial conditions att=0 to have

nonzero current and a capacitor voltagevC=V 0.

Vs

Vo

vC

S

t = 0 D

C

L

R

+ i

−

+

−

Figure E16.1.1Diode circuit withRLCload.

Solution

The KVL equation for the load current is given by

L

di

dt

+Ri+

1

C

∫

idt+vC(att= 0 )=VS

Differentiating and then dividing both sides byL, we obtain

d^2 i

dt^2

+

R

L

di

dt

+

i

LC

= 0

Note that the capacitor will be charged to the source voltageVSunder steady-state conditions,

when the current will be zero. While the forced component of the current is zero in the solution