AVERAGE VALUES (1)
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Average value of waveform is the average of all its
values over a period of time
Computing the average over time involves adding all
the values that occur in a specific time interval and
dividing that sum by time
This is done by computing the area of waveform over a
period of time
Area above time axis is a positive area
Area below time axis is a negative area
Algebraic signs must be taken into account when
computing total (net) area
Time interval is the period T
Find the average value of the following AC waveform
v(t)
A1
+15V
A2
0.2
0.6
0.8
0
1.2
t(s)
-3V
T
7. Measuring AC
Waveforms
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AVERAGE VALUES (2)
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Sine waves are symmetrical about time axis, thus
average is zero: positive area cancels negative area
Must use calculus to compute average, and we take
into consideration the area of a half cycle of a
sinusoidal waveform, called a pulse
Using calculus it can be shown that:
Area = 2Vp/ω V-s or 2Ip/ω A-s
What is the average of the following waveform
i(t)
0.4A
0
0.1
0.2
0.3
7. Measuring AC
Waveforms
0.4
0.5
t (s)
2
AVERAGE VALUES (3)
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Average value of waveform also called dc value
Knowledge of average value important in the design of
high power devices such as power supplies and power
amplifiers
Waveforms in circuits regarded as sine waves with dc
offset
DC level (offset) is simply added to ac waveform
Equation for ac voltage with dc component Vdc is
v(t) = Vdc + Vp sin(ωt + ϕ)
Vdc can either be positive or negative
Minimum v(t) = Vdc – Vp volts
Maximum v(t) = Vdc + Vp volts
Find the average value of v(t), and derive mathematical
expression as a function of time
+12V
Vdc
0
2
10
t (µs)
-2V
7. Measuring AC
Waveforms
3
EFFECTIVE (RMS) VALUES (1)
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Average values not useful
for comparing sinusoidal ac
waveforms because
average value is zero
Use measure called
effective or root mean
square (rms) value
Measure shows how
effective the waveform
produces heat in a
resistance
RMS measure eliminates
consideration of waveform
polarity
The DC voltage that
causes the same heating in
resistance R as an ac
voltage is the effective
value (rms value) of the ac
voltage (likewise for
current)
2
Vp ≈ 0.707Vp
2
2
I eff =
I p ≈ 0.707I p
2
Vp = 2Veff ≈ 1.414Vp
Veff =
I p = 2I eff ≈ 1.414I p
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Find the effective value of
the ac current i(t) =
4.2sin(5000t + 45°)A
The ac voltage supplied at
a terminal has a frequency
of 60Hz and effective value
of 120V rms. Derive the
mathematical expression
for the voltage at the
terminal, assuming zero
phase angle
7. Measuring AC
Waveforms
4
EFFECTIVE (RMS) VALUES (2)
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Relationship between peak and rms values given in
previous equations valid only for sinusoids
It is possible to compute effective value of some
nonsinusoidal waveforms using root mean square
method
Square the waveform, find its average, then square root
the answer
√[average(waveform)2]
Find the effective value of the voltage of this waveform
v(t)
+20V
2
0
1
4
7
5
9
6
12
10 11
t (s)
-12V
7. Measuring AC
Waveforms
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AC VOLTAGE AND CURRENT IN
RESISTANCE
• Ohm’s Law can be applied to an ac circuit
containing a resistance to determine ac current in
a resistance when ac voltage is applied across it
• At every instant in time, the current in the resistor
is the voltage at that instant divided by the
resistance, i.e. instantaneous current is
instantaneous voltage divided by the resistance
v(t )
R
i(t ) =
• For a sinusoidal voltage
i(t ) =
Vp
R
sin(ωt + φ )
• Ip = Vp/R
• The voltage across and the current through the
resistor have the same phase angle, they are in
phase
7. Measuring AC
Waveforms
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AC VOLTAGE AND CURRENT IN
RESISTANCE: EXAMPLES
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The current in a 2.2 kΩ resistor is:
i(t) = 5sin(2π × 100t + 45°) mA
Write the mathematical expression for the voltage
across the capacitor
What is the effective value of the resistor voltage?
What is the instantaneous value of the resistor at t =
0.4ms?
The ac voltage across a 150Ω resistor is:
39sin(2π × 103t) V. At what value of t does the current
through the resistor equal -0.26A?
7. Measuring AC
Waveforms
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CAPACITORS AND AC (1)
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AC current through a capacitor depends not only on the
voltage across it, but also on the frequency of that
voltage
The property of a capacitor that causes it to resist the
flow of ac current through it is called capacitive
reactance, denoted by XC. Its units are also Ohms
XC = 1/ωC = 1/2πfC Ohms
XC is inversely proportional to frequency
The greater the frequency, the smaller the reactance
and thus the greater the current through the capacitor
The lower the frequency, the greater the reactance
Graph of reactance XC v frequency f
XC
XC = 1/2πfC
Capacitor is an open
circuit when dc voltage
is connected across it
i.e. f = 0
Ip =
0
f
Vp volts
= amps
X C ohms
I p = VpωC = Vp 2πfC
7. Measuring AC
Waveforms
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CAPACITORS AND AC (2)
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Example: The ac voltage across a 0.5µF capacitor is:
v(t) = 16sin(2 × 103t) V
What is the capacitive reactance of the capacitor?
What is the peak value of the current through it?
Example: The ac current through a 20µF capacitor is
i(t) = 3sin(800t) A. What is the peak voltage across the
capacitor?
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The peak current through a capacitor does not occur at
the same instant of time that the voltage across it
reaches its peak value
The current through a capacitor leads the voltage
across it by 90°
vC(t) = Vpsin(ωt + ϕ)V
iC(t) = Ip sin(ωt + ϕ + 90°)A
Example: The voltage across a 0.01µF capacitor is
vC(t) = 240sin(1.25×104t -30°)V. Write the mathematical
expression for the current through it.
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7. Measuring AC
Waveforms
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INDUCTORS AND AC
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Inductance resists or impedes the flow of ac current
Property called inductive reactance XL where:
XL = ωL = 2πfL ohms; waveform:
XL (Ω)
XL
f
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XL directly proportional to frequency
XL decreases with frequency, approaching 0Ω as
frequency approaches zero (dc)
Inductor is a short circuit when a dc voltage is
connected across it
Ohm’s Law: Ip = Vp/XL amperes
Voltage across inductor leads current through it by 90°
iL(t) = Ipsin(ωt + ϕ); vL(t) = Vpsin(ωt + ϕ + 90°)
Example: The current through an 80mH is 0.1sin(400t25°)A. Write the mathematical expression for the
voltage
7. Measuring AC
Waveforms
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AVERAGE POWER
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The power at any instant in
time, instantaneous
power:
p(t) = v(t)i(t) = [i(t)]2R =
[v(t)]2/R
More useful measure is
average power, which is
the average of the
instantaneous power over
a period of time
Example: Find the
average power dissipated
through a 50Ω with a
voltage of 12sin(377t)V
across it. Use all the
equations on the left.
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Pavg = I p2 R watts
2
2
Pavg = I eff
R watts
Pavg =
Vp2
watts
2R
Veff2
Pavg =
watts
R
Vp I p
Pavg =
watts
2
Pavg = Veff I eff watts
7. Measuring AC
Waveforms
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