• Lecture 8: RMS
ECEN 1400 Introduction to Analog and Digital Electronics
Lecture 8
Root mean square
•
•
•
•
Concept
RMS of a sinusoidal signal
Examples
Application to rectified signals
Robert R. McLeod, University of Colorado
http://www.sfu.ca/sonic-studio/handbook/Root_Mean_Square.html
91
• Lecture 8: RMS
ECEN 1400 Introduction to Analog and Digital Electronics
A useful number for describing
time-varying waveforms
Although peak or peak-to-peak amplitude might seem
like the most obvious descriptions of a time-varying
waveform, there are two reasons to want something
different.
1.  What we often care most about is the average power
we can extract, not instantaneous voltage or current.
2.  We would like a way to compare waveforms of
different shape on this basis (average power).
Consider a square
wave voltage with a
25% duty cycle (on for
only 25% of the
period). What average
power would it deliver
to a resistor?
Paverage =
T
4
(V
2
peak
)
@ D
V V
1.0
0.8
0.6
0.4
0.2
R + 34T (0 R )
T
0.01
0.02
0.03
0.04
0.05
@ D
t s
2
1 V peak (V peak 2 )
=
=
4 R
R
2
So we observe that this AC voltage would deliver the
same power as a DC voltage of Vpeak/2. This is the RMS
voltage description of this waveform.
Robert R. McLeod, University of Colorado
92
• Lecture 8: RMS
ECEN 1400 Introduction to Analog and Digital Electronics
Root mean square voltage of a
sinusoidal function
Q: What DC voltage would deliver the same average
power to a load as a sinusoidal AC voltage Vp sin(ωt)?
AC
DC
V (t ) = V p sin (2π t T )
V = VEquiv
P(t ) = V 2 (t ) R
2
PAverage = VEquiv
R
T
PAverage
1
= ∫ P(t )dt
T0
T
Setting them equal
[
]
VP2
2
= VEquiv
R
2R
1
2
= ∫ V p sin (2π t T ) R dt
T0
1 VP2 T
=
T R 2
VP2
=
2R
VEquiv
VP2 VP
=
=
2
2
VRMS
V (t )
VP
=
2
For a
sine
wave!
V 2 (t )
1.0
1.0
0.8
0.5
t
0.5
1.0
1.5
! 0.5
! 1.0
Robert R. McLeod, University of Colorado
VP2
2
0.6
2.0
0.4
0.2
0.5
1.0
1.5
2.0
t
93
• Lecture 8: RMS
ECEN 1400 Introduction to Analog and Digital Electronics
In general, then…
RMS stands for “root mean square”
because to calculate it, you take the
square Root of the Mean of the Square of
a signal. Why?
2
VRMS
Mean[V 2 (t )]
P
=
=
∴ V
= Mean[V 2 (t )]
Average
R
RMS
R
This works the same way for current
2
PAverage = I RMS
R = Mean[I 2 (t )]R ∴ I RMS = Mean[I 2 (t )]
So you can use RMS quantities as if they
were DC:
V
=I R
RMS
RMS
2
2
Paverage = VRMS I RMS = VRMS
R = I RMS
R
Neat! This is why wall voltage of 120 V
is actually 120 V RMS = 169.7 V peak
Robert R. McLeod, University of Colorado
94
• Lecture 8: RMS
ECEN 1400 Introduction to Analog and Digital Electronics
RMS voltage example
VP = 169.7 V
The multimeter reads RMS voltage of 120 V.
The oscilloscope reads VP = 169.7 V = 120 2
The average power delivered to the load is
PAverage
(
2
VRMS
VP2 VRMS 2
=
=
=
R
2R
2R
Robert R. McLeod, University of Colorado
)
2
1202
=
= 14.4 [W ]
1K
95
• Lecture 8: RMS
ECEN 1400 Introduction to Analog and Digital Electronics
RMS applied to rectifiers
Half-wave
VDC
T 2
T
VDC
1
1
1 T V
= ∫ V (t )dt = ∫ VP sin (2π t T )dt = VP = P
T0
T 0
T π π
T
VRMS
1
1
2
(
)
=
V
t
dt
=
T ∫0
T
T 2
2
(
)
[
]
V
sin
2
π
t
T
dt =
∫ P
0
1 2 T VP
VP =
T
4
2
Full-wave
VDC
T
VDC
T
1
1
1 2T 2
= ∫ V (t )dt = ∫ VP sin (2π t T ) dt = VP
= V
T0
T0
T
π π P
T
VRMS
T
1
1
1 2 T VP
2
2
(
)
(
)
[
]
=
V
t
dt
=
V
sin
2
π
t
T
dt
=
VP =
P
T ∫0
T ∫0
T
2
2
VDC of full-wave rectifier is twice that of half-wave rectifier
VRMS of full-wave rectified output is same as input, ignoring
diode forward bias drop.
Robert R. McLeod, University of Colorado
http://en.wikipedia.org/wiki/Rectifier
96
• Lecture 8: RMS
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 8.1
Q: Why is the RMS voltage of a rectified
waveform the same as the RMS voltage of
the original, bipolar waveform (assuming
the same peak voltage in each case)?
A: The instantaneous power delivered by the
bipolar waveform does not depend on the
sign of the voltage, so the same power is
delivered in the positive cycle as the negative
cycle. Changing the sign of the voltage does
not change the power, so the RMS voltages
are also the same.
Robert R. McLeod, University of Colorado
97
• Lecture 8: RMS
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 8.2
Q: Line voltage of 120 V RMS is
connected directly to a 5 Kohm resistor.
The average power dissipated and peak
current are?
A: Average P = (Vrms)2/R=2.88 W
Irms = Vrms/R = 24 mA
Ipeak = Sqrt(2) Irms = 33.9 mA
Robert R. McLeod, University of Colorado
98
• Lecture 8: RMS
ECEN 1400 Introduction to Analog and Digital Electronics
D2L 8.3
V
0.5
0.4
0.3
0.2
0.1
0.2
0.4
0.6
0.8
1.0

t tT
T
Q: The plot shows one period of a triangle wave with a peak
voltage of 0.5 V. What is the RMS voltage of this
waveform?
A: Since the waveform is symmetric, the
average of the first half of the period is the
same as the second half, which simplifies the
calculation. The equation for this time is
V=t, so we need the average of the function
t^2 over the period t=0 to .5. The integral of
t^2 is t^3/3, so the average = (.5^3/3-0)/.5 =
1/12. Finally, taking the square root, we get
1/(2 sqrt(3)) or about 0.3 V
Robert R. McLeod, University of Colorado
99