Generation of Sinusoidal Voltage Waveform (AC) and Some Fundamental Concepts

 

In this lesson, firstly, how a sinusoidal waveform (ac) is generated, is described, and then the terms, such as average and effective (rms) values, related to periodic voltage or current waveforms, are explained. Lastly, some examples to find average and root mean square (rms) values of some periodic waveforms are presented.

Keywords: Sinusoidal waveforms, Generation, Average and RMS values of Waveforms.

After going through this lesson, the students will be able to answer the following questions:

 

1. What is an ac voltage waveform?

 

2. How a sinusoidal voltage waveform is generated, with some detail?

 

3. For periodic voltage or current waveforms, to compute or obtain the average and rms values, and also the time period.

 

4. To compare the different periodic waveforms, using above values.

 

Generation of Sinusoidal (AC) Voltage Waveform

A multi-turn coil is placed inside a magnet with an air gap as shown in Fig. 12.1. The flux lines are from North Pole to South Pole. The coil is rotated at an angular speed, nπω2=(rad/s).

πω2=n= speed of the coil (rev/sec, or rps)

= speed of the coil (rev/min, or rpm) nN⋅=60

l = length of the coil (m)

b = width (diameter) of the coil (m)

T = No. of turns in the coil RN BωASFig. 12.1 Schematic diagram for single phase ac generation

B = flux density in the air gap () 2/mWb

nbvπ= = tangential velocity of the coil (m/sec)

B

At a certain instant t, the coil is an angle (rad), tωθ= with the horizontal (Fig. 12.2). The emf (V) induced on one side of the coil (conductor) isθsinvlB,

θ can also be termed as angular displacement.

The emf induced in the coil (single turn) is θπθsin2sin2nblBvlB=

The total emf induced or generated in the multi-turn coil is

θθπθπθsinsin2sin2)(mETnblBnblBTe===

This emf as a function of time, can be expressed as, tEtemωsin)(=. The graph of or )(te)(θe, which is a sinusoidal waveform, is shown in Fig. 12.4a

Area of the coil blam==)(2

Flux cut by the coil (Wb) = BblBa==φ

Flux linkage (Wb) = blBTT==φψ

It may be noted these values of flux φ and flux linkage ψ, are maximum, with the coil being at horizontal position, 0=θ. These values change, as the coil moves from the horizontal position (Fig. 12.2). So, also is the value of induced emf as stated earlier.

The maximum value of the induced emf is,

dtdnTnTblBnEmθψψωψπφππ=====222

Determination of frequency (f) in the ac generator

In the above case, the frequency (Hz) of the emf generated is Magnetic Field b A oθL

nf==)2/(πω, no. of poles being 2, i.e. having only one pole pair.

In the ac generator, no. of poles = p, and the speed (rps) = n, then the frequency in Hz or cycles/sec, is

f = no. of cycles/sec = no. of cycles per rev × no. of rev per sec = no. of pairs of poles × no. of rev per sec = np)2/(

or, πω22120⋅==pNpf

Example

For a 4-pole ac generator to obtain a voltage having a frequency of 50 Hz,

the speed is, 2545022=×==pfn rps = 500,16025=⋅ rpm

For a 2-pole (p = 2) machine, the speed should be 3,000 rpm.

Similarly, the speed of the machine having different no. of poles, required to generate a frequency of 50 Hz can be computed.

Sinusoidal voltage waveform having frequency, with time period (sec), ffT1=

Periodic Voltage or Current Waveform

Average value

The current waveform shown in Fig. 12.3a, is periodic in nature, with time period, T. It is positive for first half cycle, while it is negative for second half cycle.

The average value of the waveform, is defined as )(ti

∫∫===2020)(2)(21TTavdttiTdttiTcyclehalfofperiodTimecyclehalfoverAreaI

Please note that, in this case, only half cycle, or half of the time period, is to be used for computing the average value, as the average value of the waveform over full cycle is zero (0).

If the half time period (T/2) is divided into 6 equal time intervals (TΔ),

cyclehalfofperiodTimecyclehalfoverAreaiiiiTTiiiiIav=+++=Δ⋅Δ+++=6)(6)(63216321􀀢􀀢

Please note that no. of time intervals is n = 6.

Root Mean Square (RMS) value

For this current in half time period subdivided into 6 time intervals as given above, in the resistance R, the average value of energy dissipated is given by

Riiii⎥⎦⎤⎢⎣⎡+++∝6)(26232221􀀢

The graph of the square of the current waveform, is shown in Fig. 12.3b. Let I be the value of the direct current that produces the same energy dissipated in the resistance R, as produced by the periodic waveform with half time period subdivided into time intervals, )(2tin

RTnTiiiiRIn⎥⎦⎤⎢⎣⎡Δ⋅Δ+++=)(22322212􀀢

cyclehalfofperiodTimecyclehalfovercurveiofAreaTnTiiiiIn22232221)(=Δ⋅Δ+++=􀀢

∫∫==202202221TTdtiTdtiT

This value is termed as Root Mean Square (RMS) or effective one. Also to be noted that the same rms value of the current is obtained using the full cycle, or the time period.

Average and RMS Values of Sinusoidal Voltage Waveform

As shown earlier, normally the voltage generated, which is also transmitted and then distributed to the consumer, is the sinusoidal waveform with a frequency of 50 Hz in

this country. The waveform of the voltage, and the square of waveform, , are shown in figures 12.4a and12.4b respectively. )(tv)(2tv

Time period, ωπ/)2(/1==fT; in angle (πω2=T)

Half time period, ωπ/)2/(12/==fT; in angle (πω=2/T)

0)(sin)(;0sin)(≤≤=≤≤=tfortVtvforVvmmωπωθπθθ

mmmmavVVVdVdvV637.02cossin1)(1000=====∫∫πθπθθπθθππππ

2102210222102)2cos1(21sin11⎥⎦⎤⎢⎣⎡−=⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡=∫∫∫πππθθπθθπθπdVdVdvVmm

mmmmVVVV707.022)2sin21(22122102==⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡−=ππθθππ

or, VVm2=

If time t, is used as a variable, instead of angleθ,

mmmmavVVtVdttVdttvV637.02)(cossin)(1000=====∫∫πωωπωωπωωππωπωπ

In the same way, the rms value, V can be determined.

If the average value of the above waveform is computed over total time period T, it comes out as zero, as the area of first (positive) half cycle is the same as that of second (negative) half cycle. However, the rms value remains same, if it is computed over total time period.

The different factors are defined as:

Form factor 11.1637.0707.0===mmVVvalueAveragevalueRMS

Peak factor 414.1707.0===mmVVvalueAveragevalueMaximum

Note: The rms value is always greater than the average value, except for a rectangular waveform, in which case the heating effect remains constant, so that the average and the rms values are same.

Example

The examples of the two waveforms given are periodic in nature.

1. Triangular current waveform (Fig. 12.5)

Time period = T

0)(≤≤=tTforTtItim

mmmTmTmTavIITTItTIdtTtITdttiTI5.02221)(12202200======∫∫

21221332210332210222210233311⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡⋅=⎥⎥⎦⎤⎢⎢⎣⎡⋅=⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡=∫∫mmTmTmTITTItTIdtTtITdtiTI

mmII57735.03==

Two factors of the waveform are:

Form factor 1547.15.057735.0===mmIIvalueAveragevalueRMS

Peak factor 0.25.0===mmIIvalueAveragevalueMaximum

To note that the form factor is slightly higher than that for the sinusoidal waveform, while the peak factor is much higher.

2. Trapezoidal voltage waveform (Fig. 12.6)

Time period (T) = 8 ms

Half time period (2T) ms428== 0 T 2T ti Im Fig. 12.5 Triangular current waveform v t (sec)Fig. 12.6 Trapezoidal voltage waveform 0 1 2 3 4 5 6 7 8 +5 -5

34)4(5)(;135)(;015)15()(≤≤−=≤≤=≤≤===tforttvtfortvtfortttmtv

Please note that time, t is in ms, and slope, m is in V/ms. Also to be noted that, as in the case of sinusoidal waveform, only half time period is taken here for the computation of the average and rms values.

VtttdttdtdttdttvTVTav75.341525)13(52541)4(2552541)4(55541)(213423110243311020==⎥⎦⎤⎢⎣⎡+−+=⎥⎦⎤⎢⎣⎡−++=⎥⎦⎤⎢⎣⎡−++==∫∫∫∫VtttdttdtdttdtvTVT0825.467.16350325)13(2532541)4(3252532541)4(5)5()5(41212121343311032143231210221202===⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛+−+=⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛−++=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−++=⎥⎥⎦⎤⎢⎢⎣⎡=∫∫∫∫

Two factors of the waveform are:

Form factor 0887.175.30825.4===valueAveragevalueRMS

Peak factor 3333.175.30.5===valueAveragevalueMaximum

To note that the both the above factors are slightly lower than those for the sinusoidal waveform.

Similarly, the average and rms or effective values of periodic voltage or current waveforms can be computed.

In this lesson, starting with the generation of single phase ac voltage, the terms, such as average and rms values, related to periodic voltage and current waveforms are explained with examples. In the next lesson, the background material required - the representation of sinusoidal voltage/current as phasors, the rectangular and polar forms of the phasors, as complex quantity, and the mathematical operations - addition/subtraction and multiplication/division, using phasors as complex quantity, are discussed in detail with numerical examples. In the following lessons, the study of circuits fed from single phase ac supply, is presented.

Problems

12.1 What is the speed in rpm of an ac generator with 4 poles, to produce a voltage with a frequency of 50 Hz

(a) 3000 (b) 1500 (c) 1000 (d) 750

12.2 Determine the No. of poles required in an ac generator running at 1,000 rpm, to produce a voltage with a frequency of 50 Hz.

(a) 2 (b) 4 (c) 6 (d) 8

12.3 Calculate the speed in rpm of an ac generator with 24 poles, to produce a voltage with a frequency of 50 Hz.

(a) 300 (b) 250 (c) 200 (d) 150

12.4 Determine the average and root mean square (rms) values of the following waveforms.

+V 0 T/2 T 3T/2 2T5T/2 3T (a)t v V -V v 0 2T/3 t T 5T/3 2T 8T/3(b)3T 11T/3 4T

List of Figures

Fig. 12.1 Schematic diagram for single phase ac generation

Fig. 12.2 (a) Coil position for Fig. 12.1, and (b) Details

Fig. 12.3 Periodic current (i) waveform

(a) Current (i), (b) Square of current () 2i

Fig. 12.4 Sinusoidal voltage waveform

(a) Voltage (v), (b) Square of voltage () 2v

Fig. 12.5 Triangular current waveform

Fig. 12.6 Trapezoidal voltage waveform