4. chap4 bekg1123_sesi_1415_s3 (1)

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Transcript of 4. chap4 bekg1123_sesi_1415_s3 (1)

BEKG 1123PRINCIPLES OF ELECTRIC &

ELECTRONICS

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CHAPTER 4:

AC Sources

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Learning Outcome• In this chapter, we will cover on:

4.1 AC generator4.2 Signal generator 4.3 Waveform type4.4 AC characteristic

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• AC is an electrical current whose magnitude and direction vary sinusoidally with time.

• Such as current reverses at regular time intervals and has alternately positive and negative values.

INTRODUCTION

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• The circuits analysis is considering the time-varying voltage source or current source.

• Circuits driven by sinusoidal current or voltage sources are called ac circuits.

• A sinusoid can be express in either sine or cosine form.

INTRODUCTION..cont

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Difference between DC and AC

DC AC

V5V1kHz

R

I

V/I

t

V5V

IR

V/I

t

What is AC Generator?• Is a device that converts the mechanical

energy into electrical energy.

• Generators, dynamos and alternators are machines that convert mechanical power into electrical power. Basically it has a coil rotating in a magnetic field that will produce the electrical energy/power.

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GENERATING AC VOLTAGE

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• Sinusoidal voltages are produced by ac generators and electronic oscillators.

• One way to generate an ac voltage is to rotate a coil of wire at constant angular velocity in a uniform magnetic field.

• The magnitude of the resulting voltage is proportional to the rate at which flux lines are cut.(Faraday's Law)

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Generating AC Voltage…cont

BEKP 2323 10

Generating AC Voltage..cont

Function Generator

DENE1113 11

Function generators

Function selection

Frequency

Output level (amplitude)

DC offsetCMOS output

Range

Adjust

Duty cycle

Typical controls:

Outputs

Readout

Sine Sq ua re Tria ng le

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AC Characteristics• AC signals are generated by:

– AC generator– Electronic function generator

• Types of AC waveforms:– Sine wave– Square wave– Triangle wave– Saw-tooth wave

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Characteristics of sine wave• The sinusoidal waveform (sine wave) is the fundamental of

AC current and AC voltage waveform.

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Characteristics of sine wave• Sine waves are characterized by the amplitude and period

of waveform.• Amplitude:

– Is the maximum value of voltage or current.• Period:

– Is time interval for 1 complete cycle.

0 V

1 0 V

-1 0 V

1 5 V

-1 5 V

-2 0 V

t ( s )0 2 5 3 7 .5 5 0 .0

2 0 V

The amplitude (A) of this sine wave is

20 V

The period is 50.0 s

A

T

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Characteristics of sine wave• The period of a sine wave can be measured between any two

corresponding points on the waveform.

T T T T

T TA

By contrast, the amplitude of a sine wave is only measured from the center to the maximum point.

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Characteristics of sine wave• Frequency:

– Frequency ( f ) is the number of cycles that a sine wave completes in one second.

– Frequency is measured in hertz (Hz).• Example:

3.0 HzIf 3 cycles of a wave occur in one second, the frequency is

1.0 s

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Characteristics of sine wave• Relationship between period and frequency

– The period and frequency are reciprocals of each other.

– Frequency = 1/ time for 1 complete cycle

• Examples:– If the period is 50 s, what is the frequency?– If frequency is 60Hz, what is the period?

andf

T 1

Tf 1

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Waveform Terms & Definitions

Definitions:

Period – the time taken to complete a cycle, T (s)

Peak value – the maximum instantaneous value measured from its zero value, Vp @ Vm (V)

Peak-to-peak value – the maximum variation between the maximum positive instantaneous value and the maximum negative value, Vp-p (V)

Instantaneous voltage / current - has a value that corresponds to a specific time t. Every waveform has an infinity number of instantaneous values. Such a waveform is described as the parameter as a function of time. In the case of a voltage it will be written as v(t).

Cycle – the portion of a waveform contained in one period of time. For a sine wave, it is the complete event starting with a rise from zero energy to a maximum amplitude, its return to zero, the rise to a maximum in the opposite direction, and then its return to zero.

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WAVEFORM TERMS & DEFINITIONS contd.

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Sinusoids• Consider the expression of a sinusoidal voltage

where

( ) sinmv t V t

= the amplitude of the sinusoid = the angular frequency in radians/s = the argument of the sinusoid

mV

t

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• The sinusoid repeats itself every T seconds, thus T is called the period of the sinusoid or the time taken to complete one cycle. (s)

SINUSOIDS contd.

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• The number of cycles per second is called frequency, f. (Hz)

• Angular frequency, ω. (rad/s)

• An important value of the sinusoidal function is its RMS (root-mean-square) value.

SINUSOIDS contd.

1fT

2 f

2m

RMS dcVV V

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• Note: Radian measure– ω is usually expressed in radian/s– 2 radians = 360– to convert from degrees to radians, multiply by /180.– to convert from radians to degrees, multiply by 180/.

• From the general expression of the sinusoidal voltage, we can find the value of voltage at any given instant of time.

SINUSOIDS contd.

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• If the waveform does not pass through zero at t=0, it has a phase shift.

• For a waveform shifted left,

• For waveform shifted right,

where

( ) sinmv t V t

= phase angle of the sinusoid function

SINUSOIDS contd.

( ) sinmv t V t

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SINUSOIDS contd.

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Example:1. Find the amplitude, phase, period and

frequency of the sinusoid

Solution:

Amplitude, Vm = 12VPhase, = 10˚Angular frequency, ω = 50rad/s

thus the period, T=

The frequency, f =

SINUSOIDS contd.

2 250

0.1257s

7.9581 zT

H

( ) 12sin 50 10v t t

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2. A sinusoidal voltage is given by the expression V = 300 cos (120t + 30).

a) What is the frequency in Hz?b) What is the period of the voltage in miliseconds?c) What is the magnitude of V at t = 2.778ms?d) What is the RMS value of V?

Solution:a) Given ω = 120 = 2f, thus f = 60Hzb) T = 1/f = 16.67msc) V = V = 300 cos (120x2.778m + 30)

= 300 cos (60 + 30) = 0Vd) Vrms = 300/√2 = 212.13V

SINUSOIDS contd.

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• Consider the following:SINUSOIDS contd.

1( ) sinmv t V t 2( ) sinmv t V t

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• The v2 is occurred first in time.

• Thus it can be said that v2 leads v1 by or v1 lags v2 by .

• If ≠ 0 we can say v1 and v2 are out of phase.

• If = 0 we can say v1 and v2 are in phase.

• v1 and v2 can be compared in this manner because they operate at the same frequency (do not need to have the same amplitude).

SINUSOIDS contd.

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• Transformation between cosine and sine form and Converting from negative to positive magnitude

SINUSOIDS contd.

)90cos(sin

)90sin(cos

)180cos(cos

)180sin(sin

o

o

o

o

tt

tt

tt

tt

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Example:1. For the following sinusoidal voltage, find the

value v at t = 0s and t = 0.5s.

v = 6 cos (100t + 60˚)

Solution:

Note: both ωt and must be in same unit before adding them up.

SINUSOIDS contd.

at t = 0.5sv = 6 cos (50 rad +60˚) = 4.26V

at t = 0s v = 6 cos (0+60˚) = 3V

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2. Calculate the phase angle between v1 = -10 cos (ωt + 50)v2 = 12 sin (ωt - 10)

State which sinusoid is leading.

Solution:

In order to compare v1 and v2, we must express them in the same form (either in cosine or sine function) with positive magnitude. Note: the value of must be between 0 to 180

v1 = -10 cos (ωt + 50) = 10 cos (ωt + 50 - 180) = 10 cos (ωt - 130)

SINUSOIDS contd.

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and v2 = 12 sin (ωt - 10) = 12 cos (ωt - 10 - 90)

= 12 cos (ωt - 100)

the equation v2 can be written in the following form

v2 = 12 cos (ωt - 130 + 30)

‘+30’ in the above expression means v2 leads v1 by 30

SINUSOIDS contd.