AC Waveforms
Sinusoidal Waveforms
Time Domain Descriptions
The basic unit of a sine wave is a cycle – the “distance” in time between any two repetitive
points on the waveform.
Period – T
The period T is the time required for 1 cycle – often expressed as msec or µsec.
Frequency – f
The frequency is a measure of the number of cycles that occur in 1 second. It was measured in
CPS (cycles/sec) but the modern unit is the Hertz (Hz) – often expressed as kHz or MHz.
f = 1/T
T = 1/f
Examples
What is the frequency of sine waves with the following periods T?
1. T = 1 msec
2. T = 1 µsec
3. T = 50 µsec
What is the period of sine waves with the following frequency f?
1. f = 25 kHz
2. f = 120 Hz
3. f = 200 kHz
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Amplitude Domain Descriptions
Commonly both voltage and current AC sinusoidal waveforms can be described using one or more of
three quantities.
Peak Value – VP
The peak value is the voltage from the 0 volt point to either the positive or negative maximums.
Peak-to-Peak Value – VP-P
The peak-to-peak value is the voltage from the positive to the negative maximum.
VP-P = 2 x VP
VP = VP-P/2
RMS (Root Mean Square) Value - VRMS
The RMS value of a sine wave is a value that indicates the power producing capability of the waveform.
For example a 100 V RMS sine wave can produce the same power in a load resistor as a 100 VD DC
voltage.
VRMS = VP/√2 = VP-P/2√2
VP = √2 x VRMS
VP-P = 2√2 x VRMS
√2 = 1.414
A voltmeter (DVM) set to AC volts measures an RMS Value
The factor √2 is derived by Integral Calculus.
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Instantaneous Value
There is actually a 4th way of describing AC sinusoidal waveforms but this is not used for their
measurement but as an algebraic description for theoretical derivations using mathematics. The
Instantaneous Value describes the value of the sine wave as a function of time and is thus always
changing.
v(t) = VMax sin(ωt)
Where v(t) is a function where v varies with t, VMAX is the peak value and ω is the frequency of the
waveform.
Average Value of a Sine Wave - VAVG
A voltmeter (DVM) set to DC volts measures Average Value. The average value of a +5 V DC voltage is
simply 5 V. For an AC sine wave the story is a bit different. Average value is actually a measure of the
total area under the curve for any AC waveform.
In the diagram shown, if one full cycle of the sine wave is considered, the area above the curve can be
considered positive and , the area below the curve can be considered negative. The total area consists
of these 2 equal, but opposite in sign, areas the sum of which is zero. Thus the average value of a sine
wave over one full cycle considered to be zero.
VAVG of a sine wave over 1 full cycle = 0
In higher level courses you will learn what happens to the average value if less than a full cycle is
considered.
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Circuit Laws for Sinusoidal Circuits
All circuit laws learned so far still apply to circuits with sinusoidal waveforms. This includes Ohm’s Law,
KVL, KCL, Voltage Divider etc. Careful attention must be paid to ensuring that units are consistent.
Examples
1. A resistor has 12 VP voltage across it and 100 mARMS flowing through it. What is the value of the
resistor?
R = V/I = 12 VP/ 100 mARMS = 8.49 VRMS/100 mARMS = 84.9 Ω
2. A 1 kΩ resistor is connected to a 12 VP-P voltage. What will a DVM set to AC volts measure?
A DVM set to AC volts measures 12 VP-P = 4.24 VRMS
3. A 1 kΩ resistor is connected to a 20 VP-P voltage. What are the peak and RMS values of the
current in the resistor?
I = V/R = 20 VP-P/1 kΩ = 20 mAP-P = 10 mAP = 7.07 mARMS
Power in Sinusoidal Circuits
The power dissipated in a resistor that is connected to a sinusoidal voltage source is
PD = VRMS x IRMS
Example
1. What power is dissipated in a 100 Ω resistor if V is 60 VP-P?
V = 60 VP-P = 21.2 VRMS
I = V/R = 21.2 VRMS/100 Ω = 212 mARMS
PD = 21.2 VRMS x 212 mARMS = 4.50 W
2. What DC voltage will produce the same power as in Question 1 in a 100 Ω resistor?
VDC = 21.2 V
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AC Waveforms with DC Levels or Offsets
More complex waveforms can be represented as a combination of an AC waveform plus a DC voltage
level. For example consider adding the DC level and sine wave signals shown.
+1 V DC
2 V P-P sine wave
2 V P-P sine wave
with a +1V DC Level or Offset
This waveform consists of an
AC voltage level plus a DC
voltage.
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Square Waves
A Square wave is generally seen as a rectangular waveform that varies with time.
Parameters
There are a number of parameters that can be specified and measured.
Frequency Domain Parameters
Period - T
Period T, as specified before, is the time for 1 cycle of the waveform.
Frequency
Frequency f represents the number of cycles that occur in time (usually 1 second). Frequency is related
to period f = 1/T.
Amplitude
The peak amplitude indicates the size of the waveform. In the diagram above
VP = +5 V.
The Peak-to Peak value is not used in the diagram above. There is a RMS for a square wave but it is not
commonly used.
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Duty Cycle
Square waves can also vary with respect to how much time that they are at the high level voltage during
a cycle.
Time High (THIGH) and Time Low (TLOW)
Describing Duty Cycle requires some new terminology for the waveform - Time High (THIGH) and Time
Low (TLOW).
The period of the waveform T is now
T = THIGH + TLOW
Frequency
Frequency f represents the number of cycles that occur in time (usually 1 second). Frequency is related
to period f = 1/T.
Duty Cycle
The Duty Cycle D.C. is described as
D.C. = THIGH / (THIGH + TLOW) x 100%
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Average Value
The average value of the waveform is controlled by the Duty Cycle.
The average value VAVG for a square wave where THIGH = TLOW is
VAVG = (+5 V + 0)/2 = +2.5 V
In general the average value VAVG for a square wave is
VAVG = Duty Cycle x High level voltage
Examples
Calculate the period, frequency, duty cycle and average value for the following 0 to +5 V square waves.
1. THIGH = 250 µsec TLOW = 250 µsec
Period T = THIGH + TLOW = 250 µsec + 250 µsec = 500 µsec
Frequency f = 1/T = 1/500 µsec = 2000 Hz
Duty Cycle = THIGH / (THIGH + TLOW) x 100% = 250 µsec/(250 µsec + 250 µsec) x 100 % = 50%
Average value VAVG = 0.50 x +5 V = 2.5 V
2. THIGH = 200 µsec TLOW = 50 µsec
Period T = THIGH + TLOW = 200 µsec + 50 µsec = 250 µsec
Frequency f = 1/T = 1/250 µsec = 4 kHz
Duty Cycle = THIGH / (THIGH + TLOW) x 100% = 200 µsec/(200 µsec + 50 µsec) x 100 % = 80%
Average value VAVG = 0.80 x +5 V = 4.0 V
3. THIGH = 2.5 msec TLOW = 7.5 msec
Period T = THIGH + TLOW = 2.5 msec + 7.5 msec = 10 msec
Frequency f = 1/T = 1/10 msec = 100 Hz
Duty Cycle = THIGH / (THIGH + TLOW) x 100% = 2.5 msec/(2.5 msec + 7.5 msec) x 100 % = 25%
Average value VAVG = 0.25 x +5 V = 1.25 V
The results are plotted
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Average Voltage (V)
4
3
2
1
20
40
60
80
Duty Cycle (%)
DC Motor Speed Control
The speed of a DC motor can be controlled by the DC voltage connected to it. For example you could
connect a DC motor to a DC power supply and vary the speed by adjusting the voltage of the power
supply. It is not practical to use a variable DC power supply to vary the speed of a DC motor connected
to a computer port.
Pulse Width Modulation
In Pulse Width Modulation the speed of a DC motor for example is controlled by the average DC voltage
of a variable duty cycle square wave produced by a computer at an output port – the higher the duty
cycle => the higher the average DC voltage => the faster the motor turns. The relationship has a linear
shape but can become non-linear at higher speeds.
Motor Speed (rpm)
25
50
75
Duty Cycle (%)
From the examples 1, 2 ,, and 3 above, which case will cause the motor to run the fastest and the
slowest?
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Square Wave - Most General Case
The square wave looked at earlier is a specific case – the most general case is shown.
THIGH
TLOW
+ V2
-V1
Period
The period of the waveform T is still
T = THIGH + TLOW
Frequency
Frequency f represents the number of cycles that occur in time (usually 1 second). Frequency is related
to period f = 1/T.
Peak Value – VP
There are two peak values for this waveform – positive peak and a negative peak and in general they are
not the same value.
Peak-to-Peak Value – VP-P
The peak-to-peak value is the voltage from the positive to the negative maximum.
Duty Cycle
The Duty Cycle (D.C.) is still described as
D.C. = THIGH / (THIGH + TLOW) x 100%
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Average Value
The average value of the waveform is controlled by the Duty Cycle.
The average value VAVG this generalized square wave is now modified
In general the average value VAVG for a square wave is
VAVG = (D. C. x High level voltage) + ((1 – D. C.) x Low Level Voltage)
Example
A square has a positive peak voltage V2 of +10 V and a negative peak voltage V1 of -5 V. The duty cycle is
60%. What is the average value?
VAVG = (.60 x 10 V) +((1 - .60) x -5 v) = 6 V + (.40 x – 5 V) = 6 V – 2 V = 4 V
Application Note
Written by David Lloyd
Computer Engineering Program
Humber College
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