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  • FOURIERSERIES21C H A P T E RLearning Objectives➣➣➣➣➣ Harmonic Analysis➣➣➣➣➣ Periodic Functions➣➣➣➣➣ Trigonometric FourierSeries➣➣➣➣➣ Alternate Forms ofTrigonometric FourierSeries➣➣➣➣➣ Certain Useful IntegralCalculus Theorems➣➣➣➣➣ Evalulation of FourierConstants➣➣➣➣➣ Different Types ofFunctional Symmetries➣➣➣➣➣ Line or FrequencySpectrum➣➣➣➣➣ Procedure for Finding theFourier Series of a GivenFunction➣➣➣➣➣ Wave Analyzer➣➣➣➣➣ Spectrum Analyzer➣➣➣➣➣ Fourier Analyzer➣➣➣➣➣ Harmonic SynthesisWith the help of Fourier Theorem, it ispossible to determine the magnitude, orderand phase of the several hormonicspresent in a complex periodic wave
  • 780 ElectricalTechnology21.1. Harmonic AnalysisBy harmonic analysis is meant the process of determining the magnitude, order and phase ofthe several harmonics present in a complex periodic wave.For carrying out this analysis, the following methods are available which are all based onFourier theorem:(i) Analytical Method–the standard Fourier Analysis(ii) Graphical Method– (a) by Superposition Method (Wedgemore’ Method) (b) Twenty fourOrdinate Method(iii) Electronic Method–by using a special instrument called ‘harmonic analyser’We will consider the first and third methods only.21.2. Periodic FunctionsA function f (t) is said to be periodic if f t T f t( ) ( )+ = for all values of t where T is somepositive number. This T is the interval between two successive repetitions and is called the periodof f (t). A sine wave having a period of T = 2π ω/ is a common example of periodic function.21.3. Trigonometric Fourier SeriesSuppose that a given function f (t) satisfies the following conditions (known as Dirichletconditions):1. f (t) is periodic having a period of T.2. f (t) is single-valued everywhere.3. In case it is discontinuous, f (t) has a finite number of discontinuities in any one period.4. f (t) has a finite number of maxima and minima in any one period.The function f(t) may represent either a voltage or current waveform. According to Fouriertheorem, this function f (t) may be represented in the trigonometric form by the infinite series.f t a a t a t a t a n tn( ) cos cos cos ... cos= + + + + +0 1 0 2 0 3 0 02 3ω ω ω ω+ b t b t b t b n tn1 0 2 0 3 0 02 3sin sin sin ... sinω ω ω ω+ + + += a a n t b n tn nn0 0 01+ +=∞∑( cos sin )ω ω... (i)Putting ω θ0t = , we can write the above equation as underf a a a a a n b b b b nn n( ) cos cos cos ... cos sin sin sin ... sinθ θ θ θ θ θ θ θ θ= + + + + + + + + + +0 1 2 3 1 2 32 3 2 3= a a n b nn nn01+ +=∞∑( cos sin )θ θ... (ii)Since ω π0 2= / T , Eq. (i) above can be written asf t a anTt bnTtn nn( ) cos sin= + +FHG IKJ=∞∑012 2π π... (iii)
  • Fourier Series 781where ω0 is the fundamental angular frequency, T is the period and a0, an and bn are constants whichdepend on n and f (t). The process of determining the values of the constants a0, an and bn is calledFourier Analysis. Also, 0 02 / 2T f where f0 is the fundamental frequency.It is seen from the above Fourier Series that the periodic function consists of sinusoidalcomponents of frequency 0, ω ω ω0 0 02, ....n . This representation of the function f(t) is in thefrequency domain. The first component a0 with zero frequency is called the dc component. Thesine and cosine terms represent the harmonics. The number n represents the order of theharmonics.When n = 1, the component ( cos sin )a t b t1 0 1 0ω ω+ is called the first harmonic or thefundamental component of the waveform.When n = 2, the component ( cos sin )a t b t2 0 2 02 2ω ω+ is called the second harmonic of thewaveform.The nth harmonic of the waveform is represented by ( cos sin )a n t b n tn nω ω0 0+ . It has afrequency of nω0 i.e. n times the frequency of the fundamental component.21.4. Alternate Forms of Trigonometric Fourier SeriesEq. (i) given above can be written asf t a a t b t a t b t a n t b n tn( ) ( cos sin ) ( cos sin ) ... ( cos sin )= + + + + + + +0 1 0 1 0 2 0 2 0 0 02 2ω ω ω ω ω ωnLet, a n t b n t n tn ncos sin cos( )ω ω ω φ0 0 0+ = −An n= A An n nn n t n tcos cos sin sinω φ ω φ0 0+∴ a A b An n n n n n= =cos sinφ φand∴ 2 2 1and tan /n n n n n nA a b b aSimilarly, let ( cos sin sin( ))a n t b n t A n tn n n nω ω ω0 0 0+ = + Ψ= A n t A n tn nsin cos cos sinω ω φ0 0Ψn n+As seen from Fig. 21.1, b A a An n n n n n= =cos sinψ ψand∴ A a b a bn n n n n= + = −2 2 1and ψ tan /The two angles φ ψn nand are complementary angles.Hence, the Fourier series given in Art. 21.2 may be put in the following two alternate formsf t A A n tn nn( ) cos( )= + −=∞∑0 01ω φor f t A A n tn nn( ) sin( )= + +=∞∑0 01ω ψFig. 21.1
  • 782 ElectricalTechnology21.5. Certain Useful Integral Calculus TheoremsThe Fourier coefficients or constants a a a a b b bn n0 1 2 1 2, , ... , .....and can be evaluated byintegration process for which purpose the following theorems will be used.(i) sin cos ( )n dbnnθ θ θπ π= = − =z 1 11 1 00202(ii) sin |cos | ( )n dnnnθ θ θππ= − = − − =z 1 10 0 00202(iii)2 22 200 01 1 1sin (1 cos2 ) | sin 2 |2 2 2n d n d nn(iv) cos (cos ) sin2020202122 112122n d n dnnθ θ θ θ θ θ πππ π= + = + =zz(v)2 20 01sin cos {sin( ) sin ( ) }2m n d m n m n d=121 1002−++ −−− =m nm nm nm ncos( ) cos( )θ θπ(vi) cos cos {cos( ) cos( ) }m n d m n m n dθ θ θ θ θ θππ= + + −zz 12 0202=121 1002m nm nm nm n n m++ +−− = ≠sin( ) sin( ) ...θ θπfor(vii)2 20 01sin sin {cos( ) cos( ) }2m n d m n m n d=121 1002m nm nm nm n n m−− −++ = ≠sin( ) sin( ) ...θ θπforwhere m and n are any positive integers.21.6. Evaluation of Fourier ConstantsLet us now evaluate the constants a0, an and bn by using the above integral calculus theorems(i) Value of a0For this purpose we will integrate both sides of the series given below over one period i.e.for θ θ π= =0 2to .f a a a a nn( ) cos cos ... cosθ θ θ θ= + + + +0 1 2 2 + b b b nn1 2 2sin sin ... sinθ θ θ+ + +∴02020 102202022π π π ππθ θ θ θ θ θ θ θ θz z z zz= + + + +f d a d a d a d a n dn( ) cos cos ... cos+ b d b d b n dn102202022π π πθ θ θ θ θ θz z z+ + +sin sin ... sin
  • Fourier Series 783= a a0 0200 0 0 0 0 0 2| | ... ....θ ππ+ + + + + + + =∴ a f d f d0021212= =−zzπθ θπθ θπππ( ) ( )or= mean value of f ( )θ between the limits 0 to 2π i.e. over one cycle or period.Also,20 01[netarea]2aIf we take the periodic function as f(t) and integrate over period T (which corresponds to 2π ),we get aTf t dtTf t dtTf t dtTTTtt T00 221 1 111= = =z z z−+( ) ( ) ( )//where t1 can have any value.(ii) Value of anFor finding the value of an, multiply both sides of the Fourier Series by cos nθ and integratebetween the limits 0 to 2∴ f n d a n d a n d( )cos cos cos cosθ θ θ θ θ θ θ θπ ππ020 10202z zz= + + a n d2022cos cosθ θ θπz+ +za n dn cos202θ θπb n d b n d1 202022cos cos sin cosθ θ θ θ θ θπ π+z z + + z... sin cosb n n dn θ θ θπ02= 0 0 0 0 0 0202202+ + + + + + + + = =z z... cos ... cosa n d a n d an n nθ θ θ θ ππ π∴ a f n d f n dn = = ×z z12120202πθ θ θπθ θ θπ π( )cos ( )cos= 2 × average value of f ( )θ cos nθ over one cycle of the fundamental.Also,1 1( )cos 2 ( )cos2na f n d f n dIf we take periodic function as f t( ), then different expressions for an are as under.aTf tnTt dtTf tnTt dtnTTT= =z z−2 2 2 20 22( )cos ( )cos//π πGiving different numerical values to n, we geta1 = 2 × average of f ( )θ cos θ over one cycle ....n = 1a2 = 2 × average value of f ( )θ cos 2θ over one cycle etc. .... n = 2(iii) Value of bnFor finding its value, multiply both sides of the Fourier Series of Eq. (i) by sin nθ andintegrate between limits θ θ π= =0 2to .
  • 784 ElectricalTechnology∴ f n d a n d a n d( )sin sin cos sinθ θ θ θ θ θ θ θπ ππ= +z zz00210202+ a n d a n n dn202022cos sin ... cos sinθ θ θ θ θ θππ+ + zz+ b n d b n d b n dn1 202022022sin sin sin sin .... sinθ θ θ θ θ θ θ θπ π π+ + +z z z= 0 0 0 0 0 202202+ + + + + + + = =z z... ... sin sinb n d b n d bn n nθ θ θ θ ππ π∴ f d bn( )sinθ θ θ ππ02z = ×∴ b f n d f n dn = = ×z z12120202πθ θ θπθ θ θπ π( )sin ( )sin= 2 × average value of f ( )θ sin nθ over one cycle of the fundamental.∴ b1 = 2 × average value of f ( )sinθ θ over one cycle ... n = 1b2 = 2 × average value of f ( )sinθ θ2 over one cycle ... n = 2Also, btf tnTt dtTf tnTt dtnTTT= +z z2 2 2 20 22( )sin ( )sin//π π=2 200220Tf t n t dtTf t n tdtTTTz z=( )sin ( )sin//ω ωHence, for Fourier analysis of a periodic function, the following procedure should beadopted:(i) Find the term a0 by integrating both sides of the equation representing the periodicfunction between limits 0 to 2π or 0 to T or – T/2 to T/2 or t1 to (t1 + T).∴ a f dTf t dtTf t dtTf t dttt TTTT002220121 1 111= = = =z zzz +−πθ θπ( ) ( ) ( ) ( )//= average value of the function over one cycle.(ii) Find the value of an by multiplying both sides of the expression for Fourier series bycos nθ and then integrating it between limits 0 to 2π or 0 to T or – T/2 to T/2 or t1 to (t1 + T).∴ a f n d f n dn = = ×z z12120202πθ θ θπθ θ θπ π( )cos ( )cosSince π = T/2, we haveaTf tnTtdtTf tnTt dtTf tnTtdtnTTTtt T= = =z z z−+2 2 2 2 2 20 2211( )cos ( )cos ( )cos//π π π=2 2 20 0 0220 11Tf t n tdtTf t n tdtTf t n tdttt TTTT( )cos ( )cos ( )cos//ω ω ω= =+− zzz= 2 × average value of f n( )cosθ θ over one cycle of the fundamental.
  • Fourier Series 785Values of a1, a2, a3 etc. can be found from above by putting n = 1, 2, 3 etc.(iii) Similarly, find the value of bn by multiplying both sides of Fourier series by sin nθ andintegrating it between the limits 0 to 2π or 0 to T or – T/2 to T/2 or t1 to (t1 + T).∴ b f n f n dn = = ×z z12120202πθ θπθ θ θπ π( )sin ( )sin=2 2 2 2 2 20 2211Tf tnTtdtTf tnTt dtTf tntt dtTTTtt T( )sin ( )sin ( )sin//π π πz z z= =−+=2 2 200022011Tf t n tdtTf t n t dtTf t n t dtTTTtt T( )sin ( )sin ( )sin//ω ω ω= =z z z−+= 2 × average value of f ( )θ sin nθ of f (t) sin2πnTt f tor ( ) sin n tω0 over one cycle ofthe fundamental.Values of b1, b2, b3 etc. can be found from above by putting n = 1, 2, 3 etc.21.7. Different Types of Functional SymmetriesA non-sinusoidal wave can have the following types of symmetry:1. Even SymmetryThe function f (t) is said to possess even symmetry if f (t) = f (– t).It means that as we travel equal amounts in time to the left and right of the origin (i.e. alongthe + X-axis and –X-axis), we find the function to have the same value. For example inFig. 21.2 (a), points A and B are equidistant from point O. Here the two function values are equaland positive. At points C and D, the two values of the function are again equal, though negative.Such a function is symmetric with respect to the vertical axis. Examples of even function are: t2,Fig. 21.2
  • 786 ElectricalTechnologycos 3t, sin25t (2 + t2+ t4) and a constant A because the replacement of t by (– t) does not changethe value of any of these functions. For example, cos cos( )ω ωt t= − .This type of symmetry can be easily recognised graphically because mirror symmetry existsabout the vertical or f (t) axis. The function shown in Fig. 21.2 has even symmetry because whenfolded along vertical axis, the portions of the graph of the function for positive and negative timefit exactly, one on top of the other.The effect of the even symmetry on Fourier series is that the constant bn = 0 i.e. the wave hasno sine terms. In general, b1, b2, b3 ... bn = 0. The Fourier series of an even function contains onlya constant term and cosine terms i.e.f t a a n t a anTtn nnn( ) cos cos= + = +=∞=∞∑∑0 0 0112ωπThe value of an may be found by integrating over any half-period.∴ a f n dTf t n t dtnT= = zz2 4020πθ θ θ ωπ( )cos ( )cos/2. Odd SymmetryA function f (t) is said to possess odd symmetry if f t f t( ) ( )− = − .It means that as we travel an equal amount in time to the left or right from the origin, we findthe function to be the same except for a reversal in sign. For example, in Fig. 21.3 the two pointsA and B are equidistant from point O. The two function values at A and B are equal in magnitudebut opposite in sign. In other words, if we replace t by (– t), we obtained the negative of the givenfunction. The X-axis divides an odd function into two halves with equal areas above and below theX-axis. Hence, a0 = 0.Examples of odd functions are: t, sin t, t cos 50 t t t t t t( ) ( )+ + +3 5 21andA sine function is an odd function because sin ( ) sin− = −ω ωt t .Fig. 21.3An odd function has symmetry about the origin rather than about the f(t) axis which was thecase for an even function. The effect of odd symmetry on a Fourier series is that it contains noconstant term or consine term. It means that a0 = 0 and an = 0 i.e. a1, a2, = 0. The Fourierseries expansion contains only sine terms.∴ f t b n tnn( ) sin==∞∑10ωThe value of bn may be found by integrating over any half-period.
  • Fourier Series 787∴ b f n dTf t n t dtnT= =z z2 40 02πθ θ θ ωπ( )sin ( )sin/3. Half-wave Symmetry or Mirror-Symmetry or Rotational SymmetryA function f (t) is said to possess half-wave symmetry if f t f t T f t f t T( ) ( / ) ( ) ( / )= − ± − = ±2 2or .It means that the function remains the same if it is shifted to the left or right by half a periodand then flipped over (i.e. multiplied by – 1) in respect to the t-axis or horizontal axis. It is calledmirror symmetry because the negative portion of the wave is the mirror image of the positiveportion of the wave displaced horizontally a distance T/2.In other words, a waveform possesses half symmetry only when we invert its negative half-cycle and get an exact duplicate of its positive half-cycle. For example, in Fig. 21.4 (a) if we invertthe negative half-cycle, we get the dashed ABC half-cycle which is exact duplicate of the positivehalf-cycle. Same is the case with the waveforms of Fig. 21.4 (b) and Fig. 21.4 (c). In caseFig. 21.4of doubt, it is helpful to shift the inverted half-cycle by a half-period to the left and see if it super-imposes the positive half-cycle. If it does so, there exists half-wave symmetry otherwise not. It isseen that the waveform of Fig. 21.4 (d) does not possess half-wave symmetry. It is so becausewhen its negative half-cycle is inverted and shifted by half a period to the left it does notsuperimpose the positive half-cycle.It may be noted that half-wave symmetry may be present in a waveform which also showseither odd symmetry or even symmetry:For example, the square waveform shown in Fig. 21.4 (a) possesses even symmetry whereasthe triangular waveform of Fig. 24.4 (b) has odd symmetry. All cosine and sine waves possess half-wave symmetry becausecos cos cos ;sin sin sin222 2 222 2π πππ π πππTtTTtTtTtTTtTt±FHG IKJ = ±FHG IKJ = − ±FHG IKJ = ±FHG IKJ = −It is worth noting that the Fourier series of any function which possesses half-wave symmetryhas zero average value and contains only odd harmonics and is given by
  • 788 ElectricalTechnologyf t anTt bnTt a n t b n t a n b nn nnoddn nnoddn nnodd( ) cos sin ( cos sin ) ( cos sin )= +FHG IKJ = + = +=∞=∞=∞∑ ∑ ∑2 210 01 1π πω ω θ θwhere, aTf tnTdt f n dnT= =z z4 2 2020( )cos ( )cos/ ππθ θ θπ... n oddbTf tnTdt f n dnT= =z z4 2 2020( )sin ( )sin/ ππθ θ θπ... n odd4. Quarter-wave SymmetryAn odd or even function with rotational symmetry is said to possess quarter-wave symmetry.Fig. 21.5 (a) possesses half-wave symmetry as well as odd symmetry. The wave shown in Fig.21.5 (b) has both half-wave symmetry and even symmetry.The mathematical test for quarter-wave symmetry is as under:Odd quarter-wave f t f t T f t f t( ) ( / ) ( ) ( )= − + − = −2 andEven quarter-wave f t f t T f t f t( ) ( / ) ( ) ( )= − + = −2 andFig. 21.5Since each quarter cycle is the same in a way having quarter-wave symmetry, it is sufficientto integrate over one quarter period i.e. from 0 to T/4 and then multiply the result by 4.(i) If f(t) or f ( )θ is odd and has quarter-wave symmetry, then a0 is 0 and an is 0. Hence, theFourier series will contain only odd sine terms.∴201 12 1( ) sin or ( ) sin , where ( )sinn n nn nodd oddntf t b f b n b f n dT∞ ∞ π= =π= θ = θ = θ θ θπ∑ ∑ ∫It may be noted that in the case of odd quarter-wave symmetry, the integration may be carriedover a quarter cycle.∴ a f n dn = z402πθ θ θπ( )cos/... n odd=804Tf t n tdtT( )sin/ωz ... n odd(ii) If f(t) or f ( )θ is even and, additionally, has quarter-wave symmetry, then a0 is 0 and bnis 0. Hence, the Fourier series will contain only odd cosine terms.
  • Fourier Series 789∴ f a n d a n tdt a f n dn nnoddnnodd( ) cos cos ; ( )cosθ θ θ ωπθ θ θπ= = ==∞=∞∑∑ z101021whereIn this case an may be found by integrating over any quarter period.a f n dn = z402πθ θ θπ( )sin/... n odd=804Tf t n t dtT( )cos/ωz ... n odd21.8. Line or Frequency SpectrumA plot which shows the amplitude of each frequency component in a complex waveform iscalled the line spectrum or frequency spectrum (Fig. 21.6). The amplitude of each frequencycomponent is indicated by the length ofthe vertical line located at thecorresponding frequency. Since thespectrum represents frequencies of theharmonics as discrete lines of appropriateheight, it is also called a discretespectrum. The lines decrease rapidly forwaves having convergent series. Waveswith discontinuities such as the sawtoothand square waves have spectra whoseamplitudes decrease slowly because theirseries have strong high harmonics. Onthe other hand, the line spectra ofwaveforms without discontinuities andwith a smooth appearance have lineswhich decrease in height very rapidly.The harmonic content and the line spectrum of a wave represent the basic nature of that waveand never change irrespective of the method of analysis. Shifting the zero axis changes thesymmetry of a given wave and gives its trigonometric series a completely different appearance butthe same harmonics always appear in the series and their amplitude remains constant.Fig. 21.7Fig. 21.7 shows a smooth wave alongwith its line spectrum. Since there are only sine termsin its trigonometric series (apart from a0 = π ), the harmonic amplitudes are given by bn.Fig. 21.6
  • 790 ElectricalTechnology21.9. Procedure for Finding the Fourier Series of a Given FunctionIt is advisable to follow the following steps:1. Step No. 1If the function is defined by a set of equations, sketch it approximately and examine forsymmetry.2. Step No. 2Whatever be the period of the given function, take it as 2π (Ex. 20.6) and find the Fourierseries in the formf a a a a n b b b nn n( ) cos cos ... cos sin sin ... sinθ θ θ θ θ θ θ= + + + + + + + +0 1 2 1 22 23. Step No. 3The value of the constant a0 can be found in most cases by inspection. Otherwise it can befound as under:a f d f d0021212= =z z−πθ θπθ θπππ( ) ( )4. Step No. 4If there is no symmetry, then a0 is found as above whereas the other two fourier constants canbe found by the relation.a f n d f n dn = =z z−1 102πθ θ θπθ θ θπππ( )cos ( )cosb f n d f n dn = =z z−( )sin ( )sinθ θ θπθ θ θπππ02 15. Step No. 5If the function has even symmetry i.e. f f( ) ( )θ θ= − , then bn = 0 so that the Fourier serieswill have no sine terms. The series would be given byf a a n d a f n f n dn nn( ) cos ( )cos ( )cosθ θ θπθ θπθ θ θπ π= + = ==∞∑ z z010201 2where6. Step No. 6If the given function has odd symmetry i.e. f f( ) ( )− = −θ θ then a0 = 0 and an = 0. Hence,there would be no cosine terms in the Fourier series which accordingly would be given byf b tnn( ) sinθ ω==∞∑ 01; where b f n d f n dn = =z z1 2020πθ θ θπθ θ θπ π( )sin ( )sin7. Step No. 7If the function possesses half-wave symmetry i.e. ( ) ( ) or ( ) ( / 2)f f f t f t Tθ = − θ ± π = − ± ,then a0 is 0 and the Fourier series contains only odd harmonics. The Fourier series is given byf a n b nn nnodd( ) (cos sin )θ θ θ= +=∞∑1
  • Fourier Series 791where a f n d nn = z102πθ θ θπ( )cos ... odd, b f n dn = z20πθ θ θπ( )sin ... n odd8. Step No. 8If the function has even quarter-wave symmetry then a0 = 0 and bn = 0. It means the Fourierseries will contain no sine terms but only odd cosine terms. It would be given by2 / 20 0 011 2 4( ) cos ;where ( )cos ( )cos ( )cosn nnoddf a n a f n d f n d f n d... n odd9. Step No. 9If the function has odd quarter-wave symmetry, then a0 = 0 and an = 0. The Fourier series willcontain only odd sine terms (but no cosine terms).∴2 /20 0 011 2 4( ) sin ;where ( )sin ( )sin ( )sinn nnoddf b n b f n d f n d f n d... n odd10. Step No. 10Having found the coefficients, the Fourier series as given in step No. 2 can be written down.11. Step No. 11The different harmonic amplitudes can be found by combining similar sine and cosine termsi.e. A a bn n n= +2 2where An is the amplitude of the nth harmonic.Table No. 21.1Wave form Appearance EquationA. Sine wave f (t) = A = A sin ω tB. Half-wave rectified f t A t t( ) sin cos= + −FHG IKJ1 12232πωπω− −21542356πωπωcos cost t−2638πωcos .....tC. Full-wave rectified f tAt( ) ( cos= −21232πω− −21542356cos cos .....)ω ωt tsine wavesine wave
  • 792 ElectricalTechnologyD. Rectangular or f t t t( ) (sin sin= +44 133πω ω .....)+ + +155177sin sin ....ω ωt t )f tAt t( ) (cos cos= − +4 133πω ω155cos ........)ωtE. Rectangular or square f tA At t( ) (sin sin= + + +22 133πω ω155177sin sin ....)ω ωt t+ +F. Triangular wave f tAt t( ) (sin sin= − +8 1931252πω ωsin sin ....)51497ω ωt t− +f tAt t( ) (cos cos= + +8 1932πω ω12551497cos cos .....)ω ωt t+ +G. Triangular pulse f tA At t( ) (sin sin= + −24 1932πω ω+ − +12551497sin sin .....)ω ωt tH. Sawtooth wave f tAolegat t( ) (sin sin= − +2 122πω133144sin sin ....)ω ωt t− +I. Sawtooth pulse f tAt t t( ) sin sin sin= − − −FHGππω ω ω21221330−IKJ144sin ωtsquare wavewave pulse
  • Fourier Series 793J. Trapezoidal wave f tAt t( ) sin sin= −FHG3 3 12552πω ω+IKJ1497sin ......ωt21.10. Wave AnalyzerAwaveanalyzerisaninstrumentdesignedtomeasuretheindividualamplitudeofeach harmonicFig. 21.8component in a complex waveform. It is the simplest form of analysis in the frequency domain andcan be performed with a set of tuned filters and a voltmeter. That is why such analyzers are alsocalled frequency-selective voltmeters. Sincesuch analyzers sample the frequency spectrumsuccessively, i.e. one after the other, they arecalled non-real-time analyzers.The block diagram of a simple waveanalyzer is shown in Fig. 21.8. It consists of atunable fundamental frequency selector thatdetects the fundamental frequency f1 which isthe lowest frequency contained in the inputwaveform.Once tuned to this fundamental frequency,a selective harmonic filter enables switching tomultiples of f1. After amplification, the outputis fed to an a.c. voltmeter, a recorder and afrequency counter. The voltmeter reads ther.m.s amplitude of the harmonic wave, the recorder traces its waveform and the frequency countergives its frequency. The line spectrum of the harmonic component can be plotted from the abovedata.For higher frequencies (MHz) heterodyne wave analyzers are generally used. Here, the inputcomplex wave signal is heterodyned to a higher intermediate frequency (IF) by an internal localoscillator. The output of the IF amplifier is rectified and is applied to a dc voltmeter calledheterodyned tuned voltmeter.Wave analyzer
  • 794 ElectricalTechnologyThe block diagram of a wave analyzer using the heterodyning principle is shown in Fig. 21.9.Fig. 21.9The signal from the internal, variable-frequency oscillator heterodynes with the input signalin a mixer to produce output signal having frequencies equal to the sum and difference of theoscillator frequency f0 and the input frequency f1. Generally, the bandpass filter is tuned to the ’sumfrequency’ which is allowed to pass through. The signal coming out of the filter is amplified,rectified and then applied to a dc voltmeter having a decibel-calibrated scale. In this way, the peakamplitudes of the fundamental component and other harmonic components can be calculated.21.11. Spectrum AnalyzerIt is a real-time instrument i.e. it simultaneously displays on a CRT, the harmonic peakFig. 21.10values versus frequency of all wavecomponents in the frequency range of theanalyzer. The block diagram of such ananalyzer is shown in Fig. 21.10.As seen, the spectrum analyzer uses aCRT in combination with a narrow-bandsuperheterodyne receiver. The receiver istuned by varying the frequency of thevoltage-tuned variable-frequency oscillatorwhich also controls the sawtooth generatorthat sweeps the horizontal time base of theCRT deflection plates. As the oscillator isSpectrum analyzer
  • Fourier Series 795swept through its frequency band by the sawtooth generator, the resultant signal mixes and beatswith the input signal to produce an intermediate frequency (IF) signal in the mixer. The mixeroutput occurs only when there is a corresponding harmonic component in the input signal whichmatches with the sawtooth generator signal. The signals from the IF amplifier are detected andfurther amplified before applying them to the vertical deflection plates of the CRT. The resultantoutput displayed on the CRT represents the line spectrum of the input complex or nonsinusoidalwaveform.21.12. Fourier AnalyzerA Fourier analyzer uses digital signal processing technique and provides informationregarding the contents of a complex wave which goes beyond the capabilities of a spectrumanalyzer. These analyzers are based on the calculation of the discrete Fourier transform using analgorithm called the fast Fourier transformer. This algorithm calculates the amplitude and phase ofeach harmonic component from a set of time domain samples of the input complex wave signal.Fig. 21.11A basic block diagram of a Fourier analyzer isshown in Fig. 21.11. The complex wave signalapplied to the instrument is first filtered to removeout-of-band frequency components. Next, the signalis applied to an analog/digital (A/D) convertorwhich samples and digitizes it at regular timeintervals until a full set of samples (called a timerecord) has been collected. The microprocessor thenperforms a series of computations on the time datato obtain the frequency-domain results i.e.amplitude versus frequency relationships. Theseresults which are stored in memory can bedisplayed on a CRT or recorded permanently with arecorder or plotter.Since Fourier analyzers are digitalinstruments, they can be easily interfaced with a computer or other digital systems. Moreover, ascompared to spectrum analyzers, they provide a higher degree of accuracy, stability and repeatability.21.13. Harmonic SynthesisIt is the process of building up the shape of a complex waveform by adding the instantaneousvalues of the fundamental and harmonics. It is a graphical procedure based on the knowledge of thedifferent components of a complex waveform.Fourier Analyzer
  • 796 ElectricalTechnologyFig. 21.13Example 21.1. A complex voltage waveform contains a fundamental voltage of r.m.s. value220 V and frequency 50 Hz alongwith a 20% third harmonic which has a phase angle lagging by3 4π / radian at t = 0. Find an expression representing the instantaneous complex voltage v.Using harmonic synthesis, also sketch the complex waveform over one cycle of the fundamental.Solution. The maximum value of the fundamental voltage is = 200 2× = 310 V. Its angularvelocity is ω π= ×2 50 = 100π rad/s. Hence, the fundamental voltage is represented by 310 sin100π t.The amplitude of the third harmonic = 20% of 310 = 62V. Its frequency is 3 × 50 = 150 Hz.Hence, its angular frequency is = 2 150 300π π× = rad/s. Accordingly, the third harmonic voltagecan be represented by the equation 62 sin ( / )300 3 4π πt − . The equation of the complex voltageis given by υ = 310 sin 100 62 300 3 4π π πt t+ −sin( / ) .Fig 21.12In Fig. 21.12 are shown one cycle of the fundamental and three cycles of the third harmoniccomponent initially lagging by 3 4π / radian or 135°. By adding ordinates at different intervals,the complex voltage waveform is built up as shown.Incidentally, it would be seen that if the negative half-cycle is reversed, it is identical to thepositive half-cycle. This is a feature of waveforms possessing half-wave symmetry which containsthe fundamental and odd harmonics.Example 21.2. For the nonsinusoidal wave shown in Fig. 21.13, determine (a) Fouriercoefficients a0, a3 and b4 and (b) frequency of the fourth harmonic if the wave has a period of 0.02second.Solution. The function f ( )θ has a constant value fromθ = 0 to θ = 4 3π / radian and 0 value from θ π= 4 3/radian to θ π= 2 radian.(a) a012=π( )net area per cycle 02 126434πππ= ×FHG IKJ =a f d d30204 31316 3= =z zπθ θ θπθ θπ π( )cos cos/
  • Fourier Series 797=6 3324 004 3πθπππsin(sin )/= =b f d d40204 31416 4= =z zπθ θ θπθ θπ π( )sin sin/=6 44321630320 5 19404 3πθπππ ππ−= − −FHG IKJ =−− − =cos coscos ( . )/(b) Frequency of the fourth harmonic = 4 4 4 0 020f T= =/ / . = 200 Hz.Example 21.3. Find the Fourier series of the “half sinusoidal” voltage waveform whichrepresents the output of a half-wave rectifier. Sketch its line spectrum.Solution. As seen from Fig. 21.14 (a), T = 0.2 second, f0 = 1/T = 1/0.2 = 5 Hz andω0 = 02 f = 10π rad/s. Moreover, the function has even symmetry. Hence, the Fourier serieswill contain no sine terms because bn = 0.The limits of integration would not be taken from t = 0 to t = 0.2 second, but from t = – 0.5to t = 0.15 second in order to get fewer equations and hence fewer integrals. The function can bewritten asf t V tm( ) cos= 10π − < <0 05 0 05. .t= 0 0 05 015. .< <taTf t dt V tdt dtm00 050 150 050 150 050 051 10 210 0= = +LNM OQP− −z zz( ).cos .......π=V t VaVt nt dtm mnm0 2101020 210 100 050 050 050 05.sin.cos .cos....ππ ππ π−−= = zThe expression we obtain after integration cannot be solved when n = 1 although it can besolved when n is other than unity. For n = 1, we havea V t dtVmm120 050 0510 102= =−z cos..πWhen n ≠ 1 , then a V t ntdtn m=−z10 10 100 050 05cos .cos..π π=10210 1 10 12 2110 050 052Vn t n t dtV nnnm m[cos ( ) cos ( ) ] .cos( / )( ).....π πππ+ − =−≠−zaV V VaVaV Vm m m m m m2 3 2 4 221 42 13222 3 21 302 21 4215=−=−−= =−= =−= −πππ π πππππ.cos. ; .cos /; .cosa aV Vm m5 6 202 31 6235= =−=; .cosπππand so onSubstituting the values of a0, a1, a2, a4 etc. in the standard Fourier series expression given inArt. 20.3. we have0 1 0 2 2 0 4 0 6 0( ) cos 2 cos cos 4 cos6 ....f t a a t a t a t a t
  • 798 ElectricalTechnology=V VtVtVtVtm m m m mππππππ π+ + − +21023202154023560cos cos cos cos ....= V t t t tm1 12232215423560 0 0 0πωπω ππωπω+ + − +FHG IKJcos cos cos cos ...The line spectrum which is a plot of the harmonic amplitudes versus frequency is given inFig. 21.14(b).Example 21.4. Determine the Fourier series for the square voltage pulse shown inFig. 21.15 (a) and plot its line spectrum. (Network Theory, Nagpur Univ. 1992)Solution. The wave represents a periodic function of θ or ωπttTor2FHG IKJ having a periodextending over 2π radians or T seconds. The general expression for this wave can be written asf a a a a( ) cos cos cos ...θ θ θ θ= + + + +0 1 2 32 3...+ b b b1 2 32 3sin sin sin ...θ θ θ+ + +Fig. 21.15(i) Now, f V f( ) ; ; ( ) ,θ θ θ π θ θ π θ π= = = = = =0 0 2to from to∴ a d f d f d002 201212= = +RSTUVWz zzπθ θπθ θ θ θππππ( ) ( ) ( )Fig. 21.14
  • Fourier Series 799or a Vd dVVV0020 2120202= +RSTUVW= + = × =z zπθ θπθ ππππππ| |(ii) a f n d V n d n dn = = + ×RSTUVWz zz1 1002 20πθ θ θπθ θ θ θππππ( )cos cos cos=Vn dVnnπθ θπθππcos |sin |+ = =z 0 000 ... whether n is odd or even(iii) b f n d V n d n dn = = + ×RSTUVWz zz1 1002 20πθ θ θπθ θ θ θππππ( )sin sin sin=Vn dV nnVnnπθ θπθππππsincos( cos )+ =−= − +z 0 100Now, when n is odd, ( cos )1− nπ = 2 but when n is even, ( cos )1− nπ = 0.∴ bVn bV1 22120 0= = = × =π π... ; ... n = 2; bV Vn33223= × =π π... = 3 and so on.Hence, substituting the values of a a a0 1 2, , etc. and b b1 2, etc. in the above given Fourierseries, we getfV V V V E Vt t t( ) sin sin sin ... sin sin sin ...θπθπθπθπω ω ω= + + + + = + + + +FHG IKJ22 23325522 1331550 0 0It is seen that the Fourier series contains a constant term V/2 and odd harmonic componentswhose amplitudes are as under:Amplitude of fundamental or first harmonic =2VπAmplitude of second harmonic =23VπAmplitude of third harmonic =25Vand so on.The plot of harmonic amplitude versus the harmonic frequencies (called line spectrum) isshown in Fig. 21.15 (b).Example 21.5. Obtain the Fourier series for the square wave pulse train indicated inFig. 21.16. (Network Theory and Design, AMIETE June, 1990)Solution. Here T = 2 second, ω π0 2= / T = π rad/s. The given function is defined byf t t( ) = < <1 0 1 = 0 and= 0; 1 < t < 2aTf t dt dt dt dtT00 1201021 121121 012= = = +LNM OQP=z zzz( ) . . .Even otherwise by inspection a0 = (1 + 0)/2 = 1/2aTf t n tdt n tdt n tdt n tdtnT= = = +LNM OQPzzzz2 221 1 001201020( )cos .cos .cos ( ).cosω π π π
  • 800 ElectricalTechnology=1100sincos 0n tn tdtnππ = =π∫bTf t n tdt n t dtnT= = zz210010( )sin [ .sin ]ω π + ( )sin ] sin00112n tdt n t dtπ π= zz=−=−cos cosn tnnnππππ011∴ b nn = 2 / π ..... when n is odd; = 0... when n is evenFig. 21.16∴ 012 1 1 2 1 1( ) sin3 sin sin3 sin5 , .2 3 5noddf t a n t t t t etcnExample 21.6. Find the trigonometric Fourier series for the square voltage waveform shownin Fig. 21.17(a) and sketch the line spectrum.Solution. The function shown in Fig. 21.17 (a) is an odd function because at any time f(– t)= – f (t). Hence, its Fourier series will contain only sine terms i.e. an = 0. The function alsopossesses half-wave symmetry, hence, it will contain only odd harmonics.As seen from Art. 21.7 (2) the Fourier series for the above wave is given byf b n tnnodd( ) sinθ θ==∞∑1where b f n dn = z102πθ θ θπ( )sin=1 20 02πθ θ θ θπθπθπππ πππV n d V n dVnnVnnsin sin cos cos+ −RSTUVW= − +zz=Vnn nππ π{( cos cos ) cos }− + −0=Vnn nVnnππ πππ{( cos ) ( cos )} ( cos )1 121− + − = −Now, 1 – cosnπ = 2 when n is odd
  • Fourier Series 801and = 0 when n is even∴ bV V1224= × =π π... putting n = 1; b2 = 0 ... putting n = 2bV V323243= × =π π... putting n = 3; b4 = 0... putting n = 4bV V525245= × =π π... putting n = 5 and so on.Fig. 21.17Hence, the Fourier series for the given waveform is4 4 4( ) sin sin3 sin5 ...3 5V V Vf=4 1331550 0 0Vt t tπω ω ωsin sin ...+ + +FHG IKJ =4 2 136 1510VTtTtTtππ π πsin sin sin ...+ + +FHG IKJThe line spectrum of the function is shown in Fig. 21.17 (b). It would be seen that theharmonic amplitudes decrease as 1/n, that is, the third harmonic amplitude is 1/3 as large as thefundamental, the fifth harmonic is 1/5 as large and so on.Example 21.7. Determine the Fourier series for the square voltage waveform shown in Fig.21.17 (a). Plot its line spectrum.Solution. This is the same question as given in Ex. 21.6 but has been repeated toillustrated a singhtly different technique. As seen from Fig. 21.17(a) T = 2π, hence,ω π π π π0 02 2 2 1= = = =f T/ / . Over one period the function can be defined asf t V t( ) = < <0 5= − < <V t, π π22 200 0 01 1 1 1( )sin ( )sin ( )sin ( )sinnb f t n tdt f t ntdt f t ntdt f t ntdt=1 1 20 02π π π πππππππV ntdt V ntdtV ntnV ntnsin ( )sincos cos+ − =−+zz
  • 802 ElectricalTechnology= − − + −VnnVnn nππππ π(cos cos ) (cos cos )0 2Since cos 0 is 1 and cos 2 1nπ = ∴ bVnnn = −21ππ( cos )When n is even, cosnπ = 1 ∴ bn = 0When n is odd, cos nπ = – 1∴2 4(1 1) , 1,3,5...nV Vb nn nSubstituting the value of bn, the Fourier series becomef tVnntVnntVt t tnoddnodd( ) sin sin sin sin sin ...= = = + + +FHG IKJ=∞=∞∑ ∑4 4 1 4 1331551 1π π πSince ω0 1= , the above expression in general terms becomesf tVt t t( ) sin sin sin ...= + + +FHG IKJ4 1331550 0 0πω ω ωThe line spectrum is as shown in Fig.21.17 (b).Example 21.8. Determine theFourier series of the square voltagewaveform shown in Fig. 21.18.Solution. As compared to Fig. 21.17(a) given above, the vertical axis of figurehas been shifted by π / 2 radians.Replacing t by ( / )t + π 2 in the aboveequation, the Fourier series of thewaveform shown in Fig. 21.18 becomesf tVt t t( ) sin sin sin ...= +FHG IKJ+ +FHG IKJ+ +FHG IKJ+LNM OQP4213321552ππ π π=4 133155Vt t tπcos cos cos ...− +FHG IKJExample 21.9. Determine the trigonometric Fourier series for the half-wave rectified sinewave form shown in Fig. 21.19 (a) and sketches line spectrum.Fig. 21.18Fig. 21.19
  • Fourier Series 803Solution. The given waveform shows no symmetry, hence its series would contain both sineand cosine terms. Moreover, its average value is not obtainable by inpsection, hence it will haveto be found by integration.Here, 02 2 / 1T T . Hence, equation of the given waveform is V = Vm sinωt V tm= sin .The given waveform is defined by f t V t t tm( ) sin , ,= < < = < <0 0 2π π πa f t dt V tdt dt V tdtm m002020121212= = + =LNM OQPz zzzπ π ππ ππππ( ) sin sin=VtV Vm m m2 200π ππππ| cos | (cos cos )− = − − =a f t ntdt V t ntdtn m= =z z1 1020π ππ π( )cos sin cos=Vn t n t dtV n tnn tnm m21 12111100π ππ π[sin( ) sin( ) ]cos( ) cos( )+ − − =− +++−−z=V nnnn n nm211111111ππ π−+++−−++−−LNM OQPcos( ) cos( )when n is even, cos ( ) cos( )n n+ = − − = −1 1 1 1π πand∴ aVn n n nVnnm m=+−−++−−LNM OQP= −−211111111212π π( )... n = 2, 4, 6 etc.when n is odd and ≠ 1, cos ( )n +1 π = 1 and cos ( )n −1 π = 1∴ aVn n n nnm= −++−++−−FHG IKJ =2111111110π... n = 3, 5, 7 etc.when 10 0 011, sin .cos sin cos sin 2 02m mmV Vn a V t tdt t tdt tdtHence, an = 0 ...n = 1, 3, 5...b f t ntdt V t ntdt ntdtn m= = +LNM OQPz zz1 1002 20π πππππ( )sin sin sin ( )sin=Vt ntdtmππsin sin =z 00for n = 2, 3, 4, 5 etc.However, the expression indeterminate for n = 1 so that b1 has to be evaluated separately.b V t tdtVtdtVmm m102012= = =z zπ ππ πsin .sin sinThe required Fourier series for the half-wave rectified voltage waveform isf tV VtV ntnm m mneven( ) sincos= + −−FHG IKJ=∞∑π π22121=Vt t t tmππ1223221542356+ − − −FHG IKJsin cos cos cos ...
  • 804 ElectricalTechnology=Vt t t tmππω ω ω ω1223221523560 0 0 0+ − − − −FHG IKJsin cos cos cos ...=VTm1223221542356+ − − − −FHG IKJπθ θ θ θsin cos cos cos ...The line spectrum is shown in Fig. 21.19 (b) which has a strong fundamental term withrapidly decreasing amplitudes of the higher harmonics.Example 21.10. Find the trigonometrical Fourier series for the full wave rectified voltagesine wave shown in Fig. 21.20.Solution. Since the given function has even symmetry, bn = 0 i.e. it will contain no sine termsin its series.The equation of the sinusoidal sine wave given by V Vm= sinθ . In other words,f Vm( ) sinθ θ= = .a f d V dVdmm002020121222= = =zz zπθ θπθ θπθ θππ π( ) sin sinIt is so because the two parts 0 − π and π − 2 are identical.∴ aV Vm m0 02= − =πθππ| cos |a f n dVn dnm= =z z1 2020πθ θ θπθ θ θπ π( )cos sin cosNow, sin A cos [sin( ) sin( )]B A B A B= + + −12∴ aVn n dnm= + + −z21 10πθ θ θπ[sin( ) sin( ) ]Fig. 21.20= −+++−−V nnnnmπθ θcos( )( )cos( )( )1111= 0 =− +++−−−+−−LNM OQPV nnnn n nmππ πcos( ) cos( )11111111... when n is odd
  • Fourier Series 805However, when n is even, then22 41 1 1 1 1 11 1 1 1 1 1 ( 1)m m mnV V Van n n n n n n∴ f aVnmneven( )cos( )θπθ= −−=∞∑0 224 21∴ fV Vm m( ) cos cos cos ...θπ πθ θ θ= − − + +FHG IKJ2 4 13115413562Example 21.11. Determine the Fourier series for the waveform shown in Fig. 21.21 (a) andsketch its line spectrum.Solution. Its is seen from Fig. 21.21 (a) that the waveform equation is f Vm( ) ( / )θ π θ= . Thegiven function f ( )θ is defined by( ) , 0mVf= 0, 2Since the function possesses neither even nor odd symmetry, it will contain both sine andcosine terms.Average value of the wave over one cycle is Vm/4 or a0 = Vm/4. It is so because the averagevalue over the first half cycle is Vm/2 and over the second half cycle is 0 hence, the average valuefor full cycle is =( / )V Vm m2 02 4+=a f n d V n d n dn m= = +LNM OQPz zz1 1002 20πθ θ θππ θ θ θ θ θππππ( ) ( / ) cos ( )cos=Vn dV nn nnVnnm m mπθ θ θπθ θθπππ π2 0 2 202 21coscossin (cos )z = + = −∴ an = 0 when n is odd because cosnπ − =1 0= −2 2 2V nm / π when n is even2 20 01 1( )sin ( / ) sin (0)sinn mb f d V n d n dπ π ππ= θ θ θ = π θ θ θ + θ θπ π∫ ∫ ∫=10 2 20π πθ θ θπθ θθπππ πVn dV nn nnVnnm m mFHG IKJ = − =−z sinsincos cos∴ b V nwhenn b Vn m n m n= − = +| |π πiseven when n is oddSubstituting the values of various constants in the general expression for Fourier series, weget
  • 806 ElectricalTechnologyfV V V Vm m m m( ) cos( )cos( )cos ...θπθπθπθ= − − −42 2332552 2 2+V V Vm m mπθπθπθsin sin sin ....− +2233Fig. 21.21The amplitudes of even harmonics are given directly by bn but amplitudes of odd harmonicsare given by A a bn n n= +2 2 (Art. 21.4)For example,A V Vm m12 2 22= +( / ) ( / )π π = 0.377 VmAV Vm m3 22 223 2=FHGIKJ +FHG IKJ( )π π= 0.109 VmAV VVm mm5 22 225 50 064=FHGIKJ +FHG IKJ =( ).π πand so on.The line spectrum is as shown in Fig. 21.21 (b).Example 21.12. Find the Fourier series for the sawtooth waveform shown in Fig. 21.22 (a).Sketch its line spectrum.Solution. Using by the relation y = mx, the equation of the function becomes f (t) = 1, t orf (t) = t.T t= = = =2 2 2 20, / /ω π π πBy inspection it is clear that a0 = 2/2 = 1, aTf t n t t n t dtnT= =z z200 02( )cos .cos .ω πSince we have to find the integral of two functions, we use the technique of integration byparts i.e.uvdx u vdxdudxvdx dx= −FHG IKJzzzz
  • Fourier Series 807∴ a t n tdt n tdt dtn = −FHG IKJzzz. cos . cosπ π1020202=tnn tn tn nnππππ ππsincos( ) ( )(cos cos )022022012 0+ = + −Since cos cos2 0nπ = for all values of n, hence an = 0Fig. 21.22bTf t n tdt t n tdtnT T= =z z2000( )sin sinω πEmploying integration by parts, we get,b t n t dt n tdt dtn = −FHG IKJz zzsin . sinπ π0202021= tn tndtn tntnn tn tnnn nn.cos cos.cossin( )sin( )cos−−−=−+ = −zππππ ππππππ ππ02020220222 22The sine term is 0 for all values of n because sign of any multiple of 2π is 0. Since value ofcosine term is 1 for any multiple of 2π , we have, b nn = −2 / π .∴ f t a b n t ann tnn n( ) sin sin= + = −=∞=∞∑ ∑010 012 1ωππ= 12 122133− + + +FHG IKJππ π πsin sin sin ...t t tThe line spectrum showing the amplitudes of various harmonics is shown in Fig. 21.22 (b).Example 21.13. Determine the trigonometric Fourier series of the triangular waveformshown in Fig. 21.23.Solution. Since the waveform possesses odd symmetry, hence a0 = 0 and an = 0 i.e. therewould be no cosine terms in the series. Moreover, the waveform has half-wave symmetry. Hence,series will have only odd harmonics. In the present case, there would be only odd sine terms. Sincethe waveform possesses quarter-wave symmetry, it is necessary to integrate over only one quarterperiod of finding the Fourier coefficients.
  • 808 ElectricalTechnologyFig. 21.23f t b n tnnodd( ) sin==∞∑ ω01where bTf t n tnt= z8004( )sin/ωThe quarter-wave of the given waveform can be represented by equation of a straight line.Slope of the straight line is /( / 40) 4 /m mV T V T .Hence, using Y = mx, we havef tVTt t Tm( ) /=FHG IKJ < <40 4 ∴ bTVTt n tVTt n tdtnmTmT=FHG IKJ =z z8 4 32040 2 004/ /.sin .sinω ωUsing the theorem of integration by parts, we havebVTt n t dt n t dt dtnmt T T= −FHG IKJLNM OQPz z z3212 04004040/ / /sin . sinω ω/40 0 0 02 2 2 20 00 032 cos / 4 sin 32 cos / 4 sin / 44 4( ) ( )Tm mV n T n t V n T n Tt Tn nT n T nNow, ω π0 2= / T or ω π0 2T = ∴ n T nω π0 4 2/ /=∴ cos / cos /n T nω π0 4 2 0= = when n is odd∴ bVn TnT Vnn Vnnnm m m= = =324322 282202 2 0 2 2 2 2ωωππππsin( )sin sin∴ 2 2 2 28 8.... 1,5,9,13.... .... 3,7,11,15,...m mn nV Vb n b nn n−= = = =π πSubstituting this value of bn, the Fourier series for the given waveform becomes0 0 0 02 2 2 28 1 1 1( ) sin 3 sin5 sin7 ...3 5 7mVf t t t t tExample 21.14. Determine the Fourier series of the triangular waveform shown in Fig. 21.24.
  • Fourier Series 809Solution. Since the function has evensymmetry, bn = 0. Moreover, it also has half-wave symmerty, hence, a0 = 0. The Fourier seriescan be written as f t a n tnn( ) cos==∞∑10ω where002( )cos .Tna f t n t dtTThe function is given by the relation4( )4mV Tf t tTIt is so because for the interval 0 2≤ ≤t T / , the slope of the line is 4 / .mV T∴/40 00 02 4( )cos ( )cosT Tna f t n t dt f t n t dtT T/2 /2 /20 0 02 20 0 016 16 4cos cos cos4T T Tm m mV V VTt n t dt t n t dt n t dtTT T=−⋅ + +16 1442 202 0000200 02VT nn ttnn tVTn tnmT Tωωωωωωcos sinsin/ /Substituting ω π= 2 / ,T we get2 22 2 216 2(cos 1) (sin ) sin44mnV T T Va n n nn nT nNow, sinnπ = 0 for all values of n n, cos π = 1 when n is even add – 1 when n is odd.∴ aVnnm=82 2π∴ f tVnn tV n tnnoddm mnodd( ) coscos= ==∞=∞∑ ∑02 2 0 21028 8πωπω= 0 0 0 028 1 1 1cos cos3 .cos5 cos7 ...9 25 49mVt t t tAlternative SolutionWe can deduce the Fourier series from Fig. of Ex. 21.11 by shifting the vertical axis by π / 2radians to the right. Replacing t by ( / )t + π 2 in the Fourier series of Ex. 21.11, we getf(t) 0 0 0 02 2 2 28 1 1 1sin sin3 sin5 sin7 ...2 2 2 23 5 7mV nt t t t0 0 0 028 1 1 1cos cos3 cos5 cos ...9 25 49mVt t t tFig. 21.24
  • 810 ElectricalTechnologyExample 21.15. Obtain the Fourier series representation of the sawtooth waveform shownin Fig. 21.25 (a) and plot its spectrum.Fig. 21.25Solution. By inspection, we know that the average value of the wave is zero over a cyclebecause the height of the curve below and above the X-axis is the same hence, a0 = 0. Moreover,it has odd symmetry so that an = 0 i.e. there would be no cosine terms. The series will containonly sine terms.∴2011( ) sin where ( )sinn nnf b n b f n dThe slope of curve is m V= / π∴ we get, f V( ) ( / ) .θ π θ=If we are the limit of integration form − +π πto then2 21 1 2sin sin cos cosnV V Vb n d n n nn nnThe above result has been obtained by making use of integration by parts as explained earlier.cosnπ is positive when n is even and is negative when n is odd and thus the signs of thecoefficients alternate. The required Fourier series is2 1 1 1( ) sin sin 2 sin3 sin 4 ...2 3 4Vfor 0 0 0 02 1 1 1( ) sin sin 2 sin3 sin 4 ...2 3 4Vf t t t t tAs seen, the coefficients decrease as 1/n so that the series converges slowly as shown by theline spectrum of Fig. 21.25 (b).The amplitudes of the fundamental of first harmonic, second harmonic, third harmonic andfourth harmonic are ( / ),( / ),( / )2 2 2 2 3π π πV V and ( / )2 4V π respectively.Tutorial Problem. 21.11. Determine the Fourier series for the triangular waveform shown in Fig. 21.26 (a)(Network Theory and Design, AMIETE June 1990)
  • Fourier Series 811Fig. 21.26⎡ ⎤− ω + ω + ω +⎢ ⎥π⎣ ⎦0 0 02 2 21 4 1 1cos cos3 cos5 ....2 3 5t t t2. Find the values of the Fourier coefficients a0, an and bn for the function given in fig. 21.26 (b).n⎡ π π ⎤⎛ ⎞= = = −⎜ ⎟⎢ ⎥π π⎝ ⎠⎣ ⎦0 n n2 7 2 n 7 2a ;a sin ;b 1 cos3 n 3 n 33. Determine the trigonometric series of the triangular waveform shown in Fig. 21.27. Sketch its linespectrum.Fig. 21.273m m0 0 02 2 2V 4V 1 1cos t cos t cos5 t ...2 3 54. Determine the Fourier series for the sawtooth waveform shown in Fig. 21.28.Fig. 21.28 Fig. 21.29
  • 812 ElectricalTechnology⎡ ⎤= + ω + ω + ω +⎢ ⎥π⎣ ⎦m m0 0 0V V 1 1f(t) (sin t sin2 t sin3 t ....2 2 35. Represent the full-wave rectified voltage sine waveform shown in Fig. 21.29 by a Fourier series.⎡ ⎤⎛ ⎞= + + ω − ω + ω⎜ ⎟⎢ ⎥π ⎝ ⎠⎣ ⎦m2V 2 2 2f(t) 1 cos2 t cos4 t cos6 t...3 15 356. Obtain trigonometric Fourier series for the wave form shown in Fig. 21.30.Fig. 21.30 Fig. 21.31⎡ ⎤⎛ ⎞= − ω + ω + ω + ω⎜ ⎟⎢ ⎥π ⎝ ⎠⎣ ⎦m0 0 0 02V 1 1 1f(t) sin t sin2 t sin3 t sin4 t...2 3 47. Find the Fourier series for the sawtooth waveform shown in Fig. 21.31.⎡ ⎤⎛ ⎞= − π + π − π +⎜ ⎟⎢ ⎥π⎝ ⎠⎣ ⎦2 1 1f(t) sin2 t sin4 t sin6 t ...2 38. For the waveform of Fig. 21.32, find the Fourierseries terms up to the 5th harmonic.(Network Theory Nagpur Univ. 1993)⎡ ⎤⎛ ⎞= + + +⎜ ⎟⎢ ⎥π ⎝ ⎠⎣ ⎦16 2 1V(t) sint sin3t sin5t ...3 59. Determine Fourier series of a repetitive triangular wave as shown in Fig. 21.33.Fig. 21.33(a) What is the magnitude of d.c. component?(b) What is the fundamental frequency?Fig. 21.32
  • Fourier Series 813(c) What is the magnitude of the fundamental?(d) Obtain its frequency spectrurm. (Network Theory Nagpur Univ.1993)[( ) ( ) ( ) / ]a b c5V 1 Hz 10 voltπ10. Determine the Fourier series of voltage responses obtained at the o/p of a half wave rectifiershown in Fig. 21.34. Plot the discrete spectrum of the waveform.(Elect. Network Analysis Nagpur Univ. 1993)Fig. 21.34⎡ ⎤= + ω + π − π + π⎢ ⎥π π π⎣ ⎦m m m m m0V V 2V 2V 2VV(t) cos t cos10 t cos20 t cos30 t...2 3 15 3511. Determine the Fourier coefficients and plot amplitude and phase spectral.(Network Analysis Nagpur Univ. 1993)⎡= = = =⎢ π⎣m3 n 1 24Va 0,b 0,a ,a 0− ⎤= = = ⎥π π ⎦m m3 4 54V 4Va ,a 0,a3Fig. 21.35OBJECTIVE TESTS – 211. A given function f (t) can be represented bya Fourier series if it(a) is periodic(b) is single valued(c) has a finite number of maxima andminima in any one period(d) all of the above.2. In a Fourier series expansion of a periodicfunction, the coefficient a0 represents its(a) net area per cycle(b) d.c value
  • 814 ElectricalTechnology(c) average value over half cycle(d) average a.c value per cycle3. If in the Fourier series of a periodic function,the coefficient a0 is zero, it means that thefunction has(a) odd symmetry(b) even quarter-wave symmetry(c) odd quarter-wave symmetry(d) any of the above.4. A periodic function f (t) is said to possessodd quarter-wave symmetry if(a) f (t) = f (–t)(b) f (–t) = –f (t)(c) f (t) = –f (t + T/2)(a) both (b) and (c).5. If the average value of a periodic functionover one period is zero and it consists ofonly odd harmonics then it must bepossessing _______ symmetry.(a) half-wave(b) even quarter-wave(c) odd quarter-wave(d) odd.6. If in the Fourier series of a periodic function,the coefficient a0 = 0 and an = 0, then it mustbe having _______ symmetry.(a) odd(b) odd quarter-wave(c) even(d) either (a) or (b).7. In the case of a periodic function possessinghalf-wave symmetry, which Fouriercoefficient is zero ?(a) an (b) bn(c) a0 (d) none of above.8. A periodic function has zero average valueover a cycle and its Fourier series consists ofonly odd cosine terms. What is the symmetrypossessed by this function.(a) even(b) odd(c) even quarter-wave(d) odd quarter-wave9. Which of the following periodic functionpossesses even symmetry ?(a) cos 3t (b) sin t(c) t . cos 50 t (d) (t + t2+ t5).10. If the Fourier coefficient bn of a periodicfunction is zero, then it must possess ______symmetry.(a) even(b) even quarter-wave(c) odd(d) either (a) and (b).11. A complex voltage waveform is given byV = 120 sin ω ω πt t+ + +36 3 2 12( / ) sin( ).5ω πt + It has a time period of T seconds.The percentage fifth harmonic contents in thewaveform is(a) 12 (b) 10(c) 36 (d) 512. In the waveform of Q. 11 above, the phasedisplacement of the third harmonic representsa time interval of ___ seconds.(a) T/12 (b) T/3(c) 3T (d) T/3613. When the negative half-cycle of a complexwaveform is reversed, it becomes identical toits positive half-cycle. This feature indicatesthat the complex waveform is composed of(a) fundamental(b) odd harmonics(c) even harmonics(d) both (a) and (b)(e) both (a) and (c)14. A periodic waveform possessing half-wavesymmetry has no(a) even harmonics(b) odd harmonics(c) sine terms(d) cosine terms15. The Fourier series of a wave form possessingeven quarter-wave symmetry has only(a) even harmonics(b) odd cosine terms(c) odd sine terms(d) both (b) and (c).16. The Fourier series of a waveform possessingodd quarter-wave symmetry contains only(a) even harmonics(b) odd cosine terms(c) odd cosine terms(d) none of aboveANSWERS1. (d) 2. (b) 3. (d) 4. (d) 5. (a) 6. (d) 7. (c) 8. (c) 9. (a) 10. (d)11. (b) 12. (a) 13. (d) 14. (a) 15. (b) 16. (c)