Transcript Intro to AC and Sinusoids

Lesson 14:
Introduction to AC and
Sinusoids
1
Learning Objectives
• Compare AC and DC voltage and current sources as defined
by voltage polarity, current direction and magnitude over time.
• Define the basic sinusoidal wave equations and waveforms,
and determine amplitude, peak to peak values, phase, period,
frequency, and angular velocity.
• Determine the instantaneous value of a sinusoidal waveform.
• Graph sinusoidal wave equations as a function of time and
angular velocity using degrees and radians.
• Define effective / root mean squared values.
• Define phase shift and determine phase differences between
same frequency waveforms.
2
Direct Current (DC)
REVIEW
• DC sources have fixed polarities and magnitudes.
• DC voltage and current sources are represented by
capital E and I.
3
Alternating Current (AC)
• A sinusoidal AC waveform starts at zero then:
−
−
−
−
−
Increases to a positive maximum…
Decreases to zero…
Changes polarity…
Increases to a negative maximum…
Returns to zero.
• Variation is called a cycle.
• AC sources have a sinusoidal
waveform.
• AC sources are represented by
lowercase e(t) or i(t).
4
Generating AC Voltage
• Rotating a coil in fixed magnetic field generates sinusoidal
voltage.
5
Sinusoidal AC Current
• AC current changes direction each cycle with the
source voltage.
6
Time Scales
• Horizontal scale can represent degrees or time.
7
Frequency
• Frequency (f) is the number of cycles per second
of a waveform.
• Unit of frequency is hertz (Hz).
• 1 Hz = 1 cycle per second.
8
Period
• Period of a waveform:
− Time it takes to complete one cycle.
• Time is measured in seconds.
• The period (T) is the reciprocal of frequency:
1

f
(s)
9
Amplitude and
Peak-to-Peak Value
• Amplitude of a sine wave is the distance from its
average to its peak.
• We use Em for amplitude
• Peak-to-peak voltage is measured between
minimum and maximum peaks
• We use Epp or Vpp
Amplitude
Peak-to-Peak
10
Example Problem 1
What is the waveform’s period, frequency, Vm and VPP?
Amplitude
Peak-to-Peak
T = 0.4s
Vm  8V

1
1
=> f =
f
T
V pp  8V  (8V )  16V
f=
1
 2.5Hz
0.4 s
11
The Basic Sine Wave Equation
• The equation for a sinusoidal source is given:
e  Em sin( ) V
where Em is peak coil voltage and  is the angular position.

12
Instantaneous Value
• The instantaneous value is the value of the voltage at a
particular instant in time.

• The instantaneous value of the waveform can be determined
by solving the equation for a specific value of .
• For example, if  =37⁰ and amplitude were 10V, then the
instantaneous value at that point would be:
e(37)  10sin(37) V = 6.01 V
13
Example Problem 2
A sine wave has a value of 50V at  = 150˚. What is the
value of Em (the amplitude)?

e  Em sin( ) V
e(150)  Em sin(150) =50V
50V
Em 
=100V
sin(150)
14
Radian Measure
• Conversion for radians to degrees:
2 radians = 360º
 Radians = 180º
/2 radians = 90º
1 radian = 57.296º
15
Angular Velocity
• The rate that the generator coil rotates is called its
angular velocity ().
• Angular position can be expressed in terms of angular
velocity and time.
 =  t (radians)
• Rewriting the sinusoidal equation:
e (t) = Em sin  t (V)
16
Relationship Between
, T and f (NOT WTF!!)
• Conversion from frequency (f) in Hz to angular
velocity () in radians per second
 = 2 f
(rad/s)
• In terms of the period (T)
2
  2 f 
T
(rad/s)
17
Sinusoids as Functions of Time
• Voltages can be expressed as a function of time in
terms of angular velocity ():
e  t   Em sin(t ) V 
• Or in terms of the frequency (f):
e  t   Em sin((2f )t ) V 
• Or in terms of Period (T):
2
e(t )  Em sin(( )t ) (V)
T
18
  2 f 
2
T
(rad/s)
Example Problem 3
A waveform has a frequency of 100 Hz, and has an instantaneous value of
100V at 1.25 msec.
Determine the sine wave equation. What is the voltage at 2.5 msec?
e  t   Em sin 2 ( ft ) V 
e 1.25ms   Em sin[2 (100 Hz )(1.25ms ) ]  100V
100V
Em 
 141V
sin[2 (100 Hz )(1.25ms ) ]
Now, calculate the voltage at 2.5 msec:
e  2.5ms   141V sin[2 (100 Hz )(2.5ms) ]  141V
19
Phase Shifts
• A phase shift occurs when e(t) does not pass through
zero at t = 0 sec.
• If e(t) is shifted left (leading), then:
e = Em sin ( t + )
• If e(t) is shifted right (lagging), then:
e = Em sin ( t - )
20
Phase shift
• The angle by which the wave LEADS or LAGS the
zero point can be calculated based upon the Δt.
 10  s 
 t 
    360  
 360  36
T 
 100  s 
• The phase angle is written in DEGREES.
21
Phase Relationships
i leads v by 80°
i leads v by 110°
V and i are in phase
22
Effective (RMS) Values
• Effective values tell us about a waveform’s ability to do work.
• An effective value is an equivalent DC value.
• It tells how many volts or amps of DC that an AC
waveform supplies in terms of its ability to produce the
same average power.
• They are “Root Mean Squared” (RMS) values:
• The terms RMS and effective are synonymous.
Vm
Vrms 
 0.707Vm
2
Im
I rms 
 0.707 I m
2
23
Example Problem 4
Tie it all together:
The 120VDC source shown delivers 3.6 W to the load. Determine the peak
values of the sinusoidal voltage and current (Em and IM) such that the AC
source delivers the same power to the load.
PDC  VDC * I DC
I DC
I m  I DC 2  I RMS 2  I eff 2
I m  30mA 2  42.43mA
Vrms
PDC 3.6W


 30mA
VDC 120V
Vm

 Vm  EDC 2
2
Vm  120V 2 =169.68V
24
QUESTIONS?
25