
Let v(t) V
m
sin
ω
t V. A new voltage waveform is generated from this waveform by a
process called half-wave rectification, which is represented mathematically as
Find the rms value and cycle average value of v
r
(t).
SOLUTION
The area under squared v
r
(t) over one T will be half the area under squared v(t) since
one half-cycle is missing in v
r
(t). Therefore, rms value of v
r
(t) will be times the rms
value of v(t). Therefore, rms value of v
r
(t) 0.5 V
m
V.
Half-cycle average of v(t) is 2 V
m
/
π
. Therefore, half-cycle area
V
m
T/
π
. This area becomes the full cycle area in v
r
(t) since second half cycle is zero-
valued. Therefore, cycle average of