Document Sample

Learning Objectives C H A P T E R 21 ➣ Harmonic Analysis FOURIER ➣ Periodic Functions ➣ Trigonometric Fourier Series SERIES ➣ Alternate Forms of Trigonometric Fourier Series ➣ Certain Useful Integral Calculus Theorems ➣ Evalulation of Fourier Constants ➣ Different Types of Functional Symmetries ➣ Line or Frequency Spectrum ➣ Procedure for Finding the Fourier Series of a Given Function ➣ Wave Analyzer ➣ Spectrum Analyzer ➣ Fourier Analyzer ➣ Harmonic Synthesis With the help of Fourier Theorem, it is possible to determine the magnitude, order and phase of the several hormonics present in a complex periodic wave 780 Electrical Technology 21.1. Harmonic Analysis By harmonic analysis is meant the process of determining the magnitude, order and phase of the several harmonics present in a complex periodic wave. For carrying out this analysis, the following methods are available which are all based on Fourier theorem: (i) Analytical Method–the standard Fourier Analysis (ii) Graphical Method– (a) by Superposition Method (Wedgemore’ Method) (b) Twenty four Ordinate Method (iii) Electronic Method–by using a special instrument called ‘harmonic analyser’ We will consider the first and third methods only. 21.2. Periodic Functions A function f (t) is said to be periodic if f (t + T ) = f (t ) for all values of t where T is some positive number. This T is the interval between two successive repetitions and is called the period of f (t). A sine wave having a period of T = 2π / ω is a common example of periodic function. 21.3. Trigonometric Fourier Series Suppose that a given function f (t) satisfies the following conditions (known as Dirichlet conditions): 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 Fourier theorem, this function f (t) may be represented in the trigonometric form by the infinite series. f (t ) = a0 + a1 cos ω 0 t + a2 cos 2ω 0 t + a3 cos 3ω 0 t +...+ an cos nω 0 t + b1 sin ω 0 t + b2 sin 2ω 0 t + b3 sin 3ω 0 t +...+bn sin nω 0 t ∞ = a0 + ∑ (a n =1 n cos nω 0 t + bn sin nω 0 t ) ... (i) Putting ω 0 t = θ , we can write the above equation as under f (θ) = a0 + a1 cos θ + a2 cos 2θ + a3 cos 3θ+...+an cos nθ + b1 sin θ + b2 sin 2θ + b3 sin 3θ+...+bn sin nθ ∞ = a0 + ∑ (a n=1 n cos nθ + bn sin nθ) ... (ii) Since ω 0 = 2π / T , Eq. (i) above can be written as ∞ ∑Ga F 2πn 2πn IJ f (t ) = a0 + H n=1 n cos T t + bn sin T t K ... (iii) Fourier Series 781 where ω 0 is the fundamental angular frequency, T is the period and a0, an and bn are constants which depend on n and f (t). The process of determining the values of the constants a0, an and bn is called Fourier Analysis. Also, 0 2 /T 2 f0 where f0 is the fundamental frequency. It is seen from the above Fourier Series that the periodic function consists of sinusoidal components of frequency 0, ω 0 , 2ω 0 ....n ω 0 . This representation of the function f(t) is in the frequency domain. The first component a0 with zero frequency is called the dc component. The sine and cosine terms represent the harmonics. The number n represents the order of the harmonics. When n = 1, the component (a1 cos ω 0 t + b1 sin ω 0 t ) is called the first harmonic or the fundamental component of the waveform. When n = 2, the component (a2 cos 2ω 0 t + b2 sin 2ω 0 t ) is called the second harmonic of the waveform. The nth harmonic of the waveform is represented by (an cos nω 0 t + bn sin nω 0 t ) . It has a frequency of nω 0 i.e. n times the frequency of the fundamental component. 21.4. Alternate Forms of Trigonometric Fourier Series Eq. (i) given above can be written as f (t ) = a0 + (a1 cos ω 0 t + b1 sin ω 0 t ) + (a2 cos 2ω 0 t + b2 sin 2ω 0 t )+...+ (an cos nω 0 t + bn sin nω 0 t ) Let, an cos nω 0 t + bn sin nω 0 t = A n cos (nω 0 t − φ n ) = A n cos nω 0 t cos φ n + A n sin nω 0 t sin φ n ∴ an = An cos φ n and bn = An sin φ n ∴ An an 2 bn 2 and n tan 1 bn / an Similarly, let (an cos nω 0 t + bn sin nω 0 t = An sin (nω 0 t + Ψn )) = An sin nω 0 t cos Ψn + An cos nω 0 t sin φ n Fig. 21.1 As seen from Fig. 21.1, bn = An cos ψ n and an = An sin ψ n An = an + bn and ψ = tan −1 an / bn 2 2 ∴ The two angles φ n and ψ n are complementary angles. Hence, the Fourier series given in Art. 21.2 may be put in the following two alternate forms ∞ f (t ) = A0 + ∑A n=1 n cos (nω 0 t − φ n ) ∞ or f (t ) = A0 + ∑ A sin (nω t + ψ n=1 n 0 n) 782 Electrical Technology 21.5. Certain Useful Integral Calculus Theorems The Fourier coefficients or constants a0 , a1 , a2 ... an and b1 , b2 ..... bn can be evaluated by integration process for which purpose the following theorems will be used. (i) z0 2π sin nθdθ = 1 b 2π 1 cos nθ 0 = (1 − 1) = 0 n (ii) z 0 2π 1 1 sin nθdθ = − | cos nθ|2 π = − (0 − 0) = 0 n 0 n 2 1 2 1 1 (iii) sin 2 n d (1 cos 2n )d | 2 sin 2n |0 0 2 0 2 2n (iv) z 0 2π cos 2 nθdθ = 1 2 z 0 2π (cos 2nθ + 1)dθ = 1 1 2 2n sin 2nθ + θ = π 0 2π 2 1 2 (v) sin m cos n d {sin (m n) sin (m n) }d 0 2 0 2π 1 1 1 = − cos(m + n)θ − cos(m − n)θ = 0 2 m+n m−n 0 (vi) z 0 2π cos mθ cos nθdθ = 1 2 z 2π 0 {cos(m + n) θ + cos(m − n)θ}dθ 2π 1 1 1 = sin(m + n) θ + sin (m − n)θ = 0... for n ≠ m 2 m+n m−n 0 2 1 2 (vii) sin m sin n d {cos( m n) cos(m n) } d 0 2 0 2π 1 1 1 = sin(m − n) θ − sin(m + n)θ = 0... for n ≠ m 2 m−n m+n 0 where m and n are any positive integers. 21.6. Evaluation of Fourier Constants Let us now evaluate the constants a0, an and bn by using the above integral calculus theorems (i) Value of a0 For this purpose we will integrate both sides of the series given below over one period i.e. for θ = 0 to θ = 2π. f (θ) = a0 + a1 cos θ + a2 cos 2θ +...+ an cos nθ + b1 sin θ + b2 sin 2θ +...+bn sin nθ ∴ z0 2π f (θ)dθ = z0 2π a0 dθ + a1 z 2π 0 cos θdθ + a2 z 0 2π cos 2θdθ +...+ an z 2π 0 cos nθdθ + b1 z0 2π sin θdθ + b2 z 0 2π sin 2θdθ +...+bn z 0 2π sin nθdθ Fourier Series 783 = a0 | θ|0π +0 + 0 +...0 + 0 + 0 +....+0 = 2πa0 2 ∴ a0 = 1 2π z 0 2π f (θ)dθ or = 1 2π z −π π f (θ)dθ = mean value of f (θ) between the limits 0 to 2π i.e. over one cycle or period. 1 2 Also, a0 [net area]0 2 If we take the periodic function as f(t) and integrate over period T (which corresponds to 2π ), we get a0 = 1 T z0 T f (t )dt = 1 T z T /2 −T / 2 f (t )dt = 1 T z t1 +T t1 f (t )dt where t1 can have any value. (ii) Value of an For finding the value of an, multiply both sides of the Fourier Series by cos nθ and integrate between the limits 0 to 2 ∴ z0 2π f (θ) cos nθdθ = a0 z0 2π cos nθdθ + a1 z 0 2π cos θ cos nθd θ + a2 z0 2π cos 2θ cos nθdθ +an z 2π 0 cos 2 nθdθ + b1 z0 2π cos θ cos nθdθ + b2 z0 2π sin 2θ cos nθdθ +...+ bn z0 2π sin nθ cos nθdθ = 0 + 0 + 0 +...+ an z 0 2π cos 2 nθdθ + 0 + 0+...+0 = an z0 2π cos 2 nθdθ = πan ∴ an = 1 π z 0 2π f (θ) cos nθdθ = 2 × 1 2π z 2π 0 f (θ) cos nθ dθ = 2 × average value of f (θ) cos nθ over one cycle of the fundamental. 1 1 Also, an f ( ) cos n d 2 f ( )cos n d 2 If we take periodic function as f (t ), then different expressions for an are as under. an = 2 T z0 T f (t ) cos 2πn T t dt = 2 T z T /2 −T / 2 f (t ) cos 2πn T t dt Giving different numerical values to n, we get a1 = 2 × average of f (θ) cos θ over one cycle ....n = 1 a2 = 2 × average value of f (θ) cos 2θ over one cycle etc. .... n = 2 (iii) Value of bn For finding its value, multiply both sides of the Fourier Series of Eq. (i) by sin nθ and integrate between limits θ = 0 to θ = 2π. 784 Electrical Technology ∴ z0 2π f (θ) sin nθdθ = a0 z0 2π sin nθdθ + a1 z0 2π cos θ sin nθd θ + a2 z 2π 0 cos 2θ sin nθdθ +...+an z0 2π cos nθ sin nθdθ + b1 z 2π 0 sin θ sin nθdθ + b2 z0 2π sin 2θ sin nθdθ +....+ bn z0 2π sin 2 nθdθ = 0 + 0 + 0 +...+0 + 0 +...+ bn z0 2π sin 2 nθdθ = bn z 2π 0 sin 2 nθ dθ = bn π ∴ z0 2π f (θ)sin θdθ = bn × π ∴ bn = 1 π z 2π 0 f (θ) sin nθdθ = 2 × 1 2π z 0 2π f (θ) sin nθdθ = 2 × average value of f (θ) sin nθ over one cycle of the fundamental. ∴ b1 = 2 × average value of f (θ) sin θ over one cycle ... n = 1 b2 = 2 × average value of f (θ) sin 2θ over one cycle ... n = 2 Also, bn = 2 t z T 0 f (t )sin 2πn T t dt + 2 T z T /2 T /2 f (t )sin 2πn T t dt 2 T = T 0 z f (t )sin nω 0 t dt = 2 T /2 T T /2 f (t )sin nω 0 tdt z Hence, for Fourier analysis of a periodic function, the following procedure should be adopted: (i ) Find the term a0 by integrating both sides of the equation representing the periodic function between limits 0 to 2π or 0 to T or – T/2 to T/2 or t1 to (t1 + T). ∴ a0 = 1 2π z 2π 0 f (θ)dθ = 1 T z 0 T f (t )dt = 1 T z T /2 −T / 2 f (t )dt = 1 T z t1 +T t1 f (t )dt = average value of the function over one cycle. (ii) Find the value of an by multiplying both sides of the expression for Fourier series by cos 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). ∴ an = 1 π z 2π 0 f (θ) cos nθdθ = 2 × 1 2π z 0 2π f (θ) cos nθdθ Since π = T/2, we have an = 2 T z T 0 f (t ) cos 2πn T t dt = 2 T z T /2 −T / 2 f (t ) cos 2πn T t dt = 2 T z t1 t1 +T f (t ) cos 2πn T t dt = 2 T z T 0 f (t ) cos nω 0 tdt = 2 T z T /2 −T / 2 f (t ) cos nω 0 tdt = 2 T zt1 t1 +T f (t ) cos nω 0 tdt = 2 × average value of f (θ) cos nθ over one cycle of the fundamental. Fourier Series 785 Values 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 and integrating it between the limits 0 to 2π or 0 to T or – T/2 to T/2 or t1 to (t1 + T). ∴ bn = 1 π z0 2π f (θ) sin nθ = 2 × 1 2π z 0 2π f (θ) sin nθdθ = 2 T z 0 T f (t ) sin 2πn T t dt = 2 T z T /2 −T / 2 f (t )sin 2πn T t dt = 2 T z t1 t1 +T f (t )sin 2πn t t dt = 2 T z 0 T f (t )sin n ω 0 tdt = 2 T z T /2 −T / 2 f (t )sin nω 0 t dt = 2 T z t1 t1 + T f (t )sin nω 0 t dt 2πn = 2 × average value of f (θ) sin nθ of f (t) sin t or f (t ) sin nω 0 t over one cycle of T the 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 Symmetries A non-sinusoidal wave can have the following types of symmetry: 1. Even Symmetry The 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. along the + X-axis and –X-axis), we find the function to have the same value. For example in Fig. 21.2 (a), points A and B are equidistant from point O. Here the two function values are equal and 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 Electrical Technology cos 3t, sin2 5t (2 + t2 + t4) and a constant A because the replacement of t by (– t) does not change the value of any of these functions. For example, cos ωt = cos (−ωt ) . This type of symmetry can be easily recognised graphically because mirror symmetry exists about the vertical or f (t) axis. The function shown in Fig. 21.2 has even symmetry because when folded along vertical axis, the portions of the graph of the function for positive and negative time fit 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 has no sine terms. In general, b1, b2, b3 ... bn = 0. The Fourier series of an even function contains only a constant term and cosine terms i.e. ∞ ∞ 2πn f (t ) = a0 + ∑ n=1 an cos nω 0 t = a0 + ∑a n=1 n cos T t The value of an may be found by integrating over any half-period. ∴ 2 π an = π 0 2. Odd Symmetry z f (θ) cos nθdθ = 4 T z 0 T /2 f (t ) cos nωt dt A 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 find the function to be the same except for a reversal in sign. For example, in Fig. 21.3 the two points A and B are equidistant from point O. The two function values at A and B are equal in magnitude but opposite in sign. In other words, if we replace t by (– t), we obtained the negative of the given function. The X-axis divides an odd function into two halves with equal areas above and below the X-axis. Hence, a0 = 0. Examples of odd functions are: t, sin t, t cos 50 t (t + t 3 + t 5 ) and t (1 + t 2 ) A sine function is an odd function because sin (− ωt ) = − sin ωt . Fig. 21.3 An odd function has symmetry about the origin rather than about the f(t) axis which was the case for an even function. The effect of odd symmetry on a Fourier series is that it contains no constant term or consine term. It means that a0 = 0 and an = 0 i.e. a1, a2, a3....an = 0. The Fourier series expansion contains only sine terms. ∞ ∴ f (t ) = ∑ b sin nω t n =1 n 0 The value of bn may be found by integrating over any half-period. Fourier Series 787 ∴ bn = 2 π π 0 zf (θ) sin nθdθ = 4 T /2 T 0 z f (t ) sin nωt dt 3. Half-wave Symmetry or Mirror-Symmetry or Rotational Symmetry A function f (t) is said to possess half-wave symmetry if f (t) = − f (t ± T / 2) or − f (t) = f (t ± T / 2) . It means that the function remains the same if it is shifted to the left or right by half a period and then flipped over (i.e. multiplied by – 1) in respect to the t-axis or horizontal axis. It is called mirror symmetry because the negative portion of the wave is the mirror image of the positive portion 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 invert the negative half-cycle, we get the dashed ABC half-cycle which is exact duplicate of the positive half-cycle. Same is the case with the waveforms of Fig. 21.4 (b) and Fig. 21.4 (c). In case Fig. 21.4 of 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 is seen that the waveform of Fig. 21.4 (d) does not possess half-wave symmetry. It is so because when its negative half-cycle is inverted and shifted by half a period to the left it does not superimpose the positive half-cycle. It may be noted that half-wave symmetry may be present in a waveform which also shows either odd symmetry or even symmetry: For example, the square waveform shown in Fig. 21.4 (a) possesses even symmetry whereas the triangular waveform of Fig. 24.4 (b) has odd symmetry. All cosine and sine waves possess half- wave symmetry because 2πFG T IJ 2π FG IJ 2π 2π T FG2π IJ FG 2π IJ cos T H t± 2 K = cos T H K t ± π = − cos T t; sin T t± 2 = sin HT K H t ± π = − sin T t K It is worth noting that the Fourier series of any function which possesses half-wave symmetry has zero average value and contains only odd harmonics and is given by 788 Electrical Technology ∞ F 2πn 2πn I ∞ ∞ ∑G a cos T t + b sin T tJ = ∑(a cos nω t + b sin nω t) = ∑(a cos nθ + b sin nθ) f (t ) = H n=1 n K n n=1 n 0 n 0 n=1 n n odd odd odd where, an = 0 4 T z 2πn T T /2 dt = 2 π f (θ) cos nθdθ f (t ) cos π 0 z ... n odd bn = 4 T /2 T 0 4. Quarter-wave Symmetry z f (t )sin 2 πn T dt = 2 π π 0 f (θ)sin nθdθ z ... n odd An 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 / 2) and f (− t ) = − f (t ) Even quarter-wave f (t ) = − f (t + T / 2) and f (t ) = f (−t ) Fig. 21.5 Since each quarter cycle is the same in a way having quarter-wave symmetry, it is sufficient to 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, the Fourier series will contain only odd sine terms. ∞ 2πnt ∞ 1 2π ∴ f (t ) = ∑ bn sin or f (θ) = ∑ bn sin nθ, where bn = ∫ f (θ)sin nθd θ n =1 T n =1 π 0 odd odd It may be noted that in the case of odd quarter-wave symmetry, the integration may be carried over a quarter cycle. ∴ an = 4 π z0 π/2 f (θ) cos nθdθ ... n odd = z 8 T /4 T 0 f (t )sin nωtdt ... n odd (ii) If f(t) or f (θ) is even and, additionally, has quarter-wave symmetry, then a0 is 0 and bn is 0. Hence, the Fourier series will contain only odd cosine terms. Fourier Series 789 ∴ f (θ) = ∑ ∞ n =1 odd an cos nθdθ = ∞ ∑a n =1 odd n cos nω 0 tdt; where an = 1 π z 0 2π f (θ) cos nθdθ In this case an may be found by integrating over any quarter period. an = z 4 π/2 π 0 f (θ) sin nθdθ ... n odd = z 8 T /4 T 0 f (t ) cos nωt dt ... n odd 21.8. Line or Frequency Spectrum A plot which shows the amplitude of each frequency component in a complex waveform is called the line spectrum or frequency spectrum (Fig. 21.6). The amplitude of each frequency component is indicated by the length of the vertical line located at the corresponding frequency. Since the spectrum represents frequencies of the harmonics as discrete lines of appropriate height, it is also called a discrete spectrum. The lines decrease rapidly for waves having convergent series. Waves with discontinuities such as the sawtooth and square waves have spectra whose amplitudes decrease slowly because their series have strong high harmonics. On the other hand, the line spectra of waveforms without discontinuities and with a smooth appearance have lines Fig. 21.6 which decrease in height very rapidly. The harmonic content and the line spectrum of a wave represent the basic nature of that wave and never change irrespective of the method of analysis. Shifting the zero axis changes the symmetry of a given wave and gives its trigonometric series a completely different appearance but the same harmonics always appear in the series and their amplitude remains constant. Fig. 21.7 Fig. 21.7 shows a smooth wave alongwith its line spectrum. Since there are only sine terms in its trigonometric series (apart from a0 = π ), the harmonic amplitudes are given by bn. 790 Electrical Technology 21.9. Procedure for Finding the Fourier Series of a Given Function It is advisable to follow the following steps: 1. Step No. 1 If the function is defined by a set of equations, sketch it approximately and examine for symmetry. 2. Step No. 2 Whatever be the period of the given function, take it as 2π (Ex. 20.6) and find the Fourier series in the form f (θ) = a0 + a1 cos θ + a2 cos 2θ +...+ an cos nθ + b1 sin θ + b2 sin 2θ +...+bn sin nθ 3. Step No. 3 The value of the constant a0 can be found in most cases by inspection. Otherwise it can be found as under: a0 = 4. Step No. 4 1 2π z0 2π f (θ)dθ = 1 2π z −π π f (θ)dθ If there is no symmetry, then a0 is found as above whereas the other two fourier constants can be found by the relation. an = 1 π z 0 2π f (θ) cos nθdθ = 1 π z π −π f (θ) cos nθdθ bn = 5. Step No. 5 z 0 2π f (θ)sin nθdθ = 1 π z π −π f (θ)sin nθdθ If the function has even symmetry i.e. f (θ) = f (−θ) , then bn = 0 so that the Fourier series will have no sine terms. The series would be given by f (θ) = a0 + ∑a ∞ n =1 n cos nθdθ where an = 1 π z0 2π f (θ) cos nθ = 2 π z π 0 f (θ) cos nθdθ 6. Step No. 6 If the given function has odd symmetry i.e. f (−θ) = − f (θ) then a0 = 0 and an = 0. Hence, there would be no cosine terms in the Fourier series which accordingly would be given by z z ∞ ∑ b sin ω t ; where b 1 2π 2 π f (θ) = n 0 = f (θ) sin nθdθ = f (θ) sin nθdθ n n =1 π 0 π 0 7. Step No. 7 If the function possesses half-wave symmetry i.e. f (θ) = − f (θ ± π) or f (t ) = − f (t ± T / 2) , then a0 is 0 and the Fourier series contains only odd harmonics. The Fourier series is given by ∞ f (θ) = ∑ a (cos nθ + b sin nθ) n =1 n n odd Fourier Series 791 where an = 8. Step No. 8 1 π z 0 2π f (θ) cos nθdθ... n odd, bn = 2 π z 0 π f (θ) sin nθdθ ... n odd If the function has even quarter-wave symmetry then a0 = 0 and bn = 0. It means the Fourier series will contain no sine terms but only odd cosine terms. It would be given by 1 2 2 4 /2 f( ) an cos n ;where an f ( )cos n d f ( )cos n d f ( )cos n d 0 0 0 n 1 odd ... n odd 9. Step No. 9 If the function has odd quarter-wave symmetry, then a0 = 0 and an = 0. The Fourier series will contain only odd sine terms (but no cosine terms). 1 2 2 4 /2 ∴ f( ) bn sin n ;where bn 0 f ( )sin n d 0 f ( )sin n d 0 f ( )sin n d n 1 odd ... n odd 10. Step No. 10 Having found the coefficients, the Fourier series as given in step No. 2 can be written down. 11. Step No. 11 The different harmonic amplitudes can be found by combining similar sine and cosine terms i.e. An = an + bn 2 2 where An is the amplitude of the nth harmonic. Table No. 21.1 Wave form Appearance Equation A. Sine wave f (t) = A = A sin ω t FG 1 + 1 sin ωt − 2 cos 2ωtIJ B. Half-wave rectified sine wave f (t ) = A Hπ 2 3π K 2 2 − cos 4ωt − cos 6 ωt 15π 35π 2 − cos 8ωt..... 63π 2A 2 C. Full-wave rectified f (t ) = (1 − cos 2ωt sine wave π 3 2 2 − cos 4ωt − cos 6ωt.....) 15 35 792 Electrical Technology 44 1 D. Rectangular or f (t ) = (sin ωt + sin 3ωt .....) π 3 square wave 1 1 + sin 5ωt + sin 7ωt +.... ) 5 7 4A 1 f (t ) = (cos ωt − cos 3ωt + π 3 1 cos 5ωt........) 5 A 2A 1 E. Rectangular or square f (t ) = + (sin ωt + sin 3ωt + 2 π 3 wave pulse 1 1 sin 5ωt + sin 7ωt +....) 5 7 8A 1 1 F. Triangular wave f (t ) = (sin ωt − sin 3ωt + π2 9 25 1 sin 5ωt − sin 7ωt +....) 49 8A 1 f (t ) = (cos ωt + cos 3ωt + π2 9 1 1 cos 5ωt + cos 7ωt +.....) 25 49 A 4A 1 G. Triangular pulse f (t ) = + (sin ωt − sin 3ωt 2 π2 9 1 1 + sin 5ωt − sin 7ωt +.....) 25 49 2A 1 H. Sawtooth wave f (t ) = (sin olegat − sin 2ωt + π 2 1 1 sin 3ωt − sin 4ωt +....) 3 4 A π FG 1 1 I. Sawtooth pulse f (t ) = π 2 H − sin ω 0 t − sin 2ωt − sin 3ωt 2 3 1 IJ − sin 4ωt 4 K Fourier Series 793 FG sin ωt − 1 sin 5ωt 3 3A π H J. Trapezoidal wave f (t ) = 2 25 + sin 7ωt......J 1 I 49 K 21.10. Wave Analyzer A wave analyzer is an instrument designed to measure the individual amplitude of each harmonic Fig. 21.8 component in a complex waveform. It is the simplest form of analysis in the frequency domain and can be performed with a set of tuned filters and a voltmeter. That is why such analyzers are also called frequency-selective voltmeters. Since such analyzers sample the frequency spectrum successively, i.e. one after the other, they are called non-real-time analyzers. The block diagram of a simple wave analyzer is shown in Fig. 21.8. It consists of a tunable fundamental frequency selector that detects the fundamental frequency f1 which is the lowest frequency contained in the input waveform. Once tuned to this fundamental frequency, a selective harmonic filter enables switching to multiples of f1. After amplification, the output is fed to an a.c. voltmeter, a recorder and a Wave analyzer frequency counter. The voltmeter reads the r.m.s amplitude of the harmonic wave, the recorder traces its waveform and the frequency counter gives its frequency. The line spectrum of the harmonic component can be plotted from the above data. For higher frequencies (MHz) heterodyne wave analyzers are generally used. Here, the input complex wave signal is heterodyned to a higher intermediate frequency (IF) by an internal local oscillator. The output of the IF amplifier is rectified and is applied to a dc voltmeter called heterodyned tuned voltmeter. 794 Electrical Technology The block diagram of a wave analyzer using the heterodyning principle is shown in Fig. 21.9. Fig. 21.9 The signal from the internal, variable-frequency oscillator heterodynes with the input signal in a mixer to produce output signal having frequencies equal to the sum and difference of the oscillator frequency f0 and the input frequency f1. Generally, the bandpass filter is tuned to the ’sum frequency’ 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 peak amplitudes of the fundamental component and other harmonic components can be calculated. 21.11. Spectrum Analyzer It is a real-time instrument i.e. it simultaneously displays on a CRT, the harmonic peak Fig. 21.10 values versus frequency of all wave components in the frequency range of the analyzer. The block diagram of such an analyzer is shown in Fig. 21.10. As seen, the spectrum analyzer uses a CRT in combination with a narrow-band superheterodyne receiver. The receiver is tuned by varying the frequency of the voltage-tuned variable-frequency oscillator which also controls the sawtooth generator that sweeps the horizontal time base of the Spectrum analyzer CRT deflection plates. As the oscillator is Fourier Series 795 swept through its frequency band by the sawtooth generator, the resultant signal mixes and beats with the input signal to produce an intermediate frequency (IF) signal in the mixer. The mixer output occurs only when there is a corresponding harmonic component in the input signal which matches with the sawtooth generator signal. The signals from the IF amplifier are detected and further amplified before applying them to the vertical deflection plates of the CRT. The resultant output displayed on the CRT represents the line spectrum of the input complex or nonsinusoidal waveform. 21.12. Fourier Analyzer A Fourier analyzer uses digital signal processing technique and provides information regarding the contents of a complex wave which goes beyond the capabilities of a spectrum analyzer. These analyzers are based on the calculation of the discrete Fourier transform using an algorithm called the fast Fourier transformer. This algorithm calculates the amplitude and phase of each harmonic component from a set of time domain samples of the input complex wave signal. Fig. 21.11 A basic block diagram of a Fourier analyzer is shown in Fig. 21.11. The complex wave signal applied to the instrument is first filtered to remove out-of-band frequency components. Next, the signal is applied to an analog/digital (A/D) convertor which samples and digitizes it at regular time intervals until a full set of samples (called a time record) has been collected. The microprocessor then performs a series of computations on the time data to obtain the frequency-domain results i.e. amplitude versus frequency relationships. These results which are stored in memory can be displayed on a CRT or recorded permanently with a recorder or plotter. Fourier Analyzer Since Fourier analyzers are digital instruments, they can be easily interfaced with a computer or other digital systems. Moreover, as compared to spectrum analyzers, they provide a higher degree of accuracy, stability and repeatability. 21.13. Harmonic Synthesis It is the process of building up the shape of a complex waveform by adding the instantaneous values of the fundamental and harmonics. It is a graphical procedure based on the knowledge of the different components of a complex waveform. 796 Electrical Technology Example 21.1. A complex voltage waveform contains a fundamental voltage of r.m.s. value 220 V and frequency 50 Hz alongwith a 20% third harmonic which has a phase angle lagging by 3π / 4 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 angular velocity is ω = 2π × 50 = 100π rad/s. Hence, the fundamental voltage is represented by 310 sin 100π t. The amplitude of the third harmonic = 20% of 310 = 62V. Its frequency is 3 × 50 = 150 Hz. Hence, its angular frequency is = 2π × 150 = 300 π rad/s. Accordingly, the third harmonic voltage can be represented by the equation 62 sin (300 πt − 3π / 4) . The equation of the complex voltage is given by υ = 310 sin 100 πt + 62 sin(300 πt − 3π / 4) . Fig 21.12 In Fig. 21.12 are shown one cycle of the fundamental and three cycles of the third harmonic component initially lagging by 3 π / 4 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 the positive half-cycle. This is a feature of waveforms possessing half-wave symmetry which contains the fundamental and odd harmonics. Example 21.2. For the nonsinusoidal wave shown in Fig. 21.13, determine (a) Fourier coefficients a0, a3 and b4 and (b) frequency of the fourth harmonic if the wave has a period of 0.02 second. Solution. The function f (θ) has a constant value from θ = 0 to θ = 4 π / 3 radian and 0 value from θ = 4 π / 3 radian to θ = 2π radian. 1 ( net area per cycle) 2π = 1 6 × 4 π = 4 FG IJ (a) a0 = 2π 0 2π 3 H K a3 = 1 π z 0 2π f (θ) cos 3θdθ = 1 π z 0 4π/3 6 cos 3θdθ Fig. 21.13 Fourier Series 797 4π/3 6 sin 3θ 2 = = (sin 4 π) = 0 π 3 0 π b4 = 1 π z0 2π f (θ)sin 4θdθ = 1 π z0 4π/3 6 sin 4θdθ 6 − cos 4θ 4π /3 FG 3 cos 16 π −3 IJ 9 = π 4 0 =− 2π 3 H − cos 0 = 2π (−0.5 − 1) = K 4π (b) Frequency of the fourth harmonic = 4 f0 = 4 / T = 4 / 0.02 = 200 Hz. Example 21.3. Find the Fourier series of the “half sinusoidal” voltage waveform which represents 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 and ω 0 = 2 f 0 = 10π rad/s. Moreover, the function has even symmetry. Hence, the Fourier series will 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 = – 0.5 to t = 0.15 second in order to get fewer equations and hence fewer integrals. The function can be written as f (t ) = Vm cos 10πt −0.05 < t < 0.05 =0 0.05 < t < 0.15 a0 = 1 T z 0.15 −0.05 f (t )dt = 1 0.2 LM N z 0.05 −0.05 Vm cos 10 πtdt + z 0.15 0.05 0. dt OP Q = Vm sin 10 πt 0.2 10π 0.05 −0 .05 = Vm π 2V an = m 0.2 z 0.05 −0.05 cos 10 πt.cos 10 πnt dt The expression we obtain after integration cannot be solved when n = 1 although it can be solved when n is other than unity. For n = 1, we have a1 = 10Vm z 0.05 −0.05 cos 2 10 πt dt = Vm 2 When n ≠ 1 , then an = 10Vm z 0 .05 −0.05 cos 10 πt.cos 10 π ntdt = 10Vm 2 z 0 .05 −0.05 [cos 10 π(1 + n)t cos 10 π(1 − n)t ]dt = 2Vm cos(πn / 2) π . (1 − n2 ) ... n ≠ 1 2Vm cos π 2Vm −1 2Vm 2V cos 3π / 2 2V cos 2π 2V a2 = . = . = ; a3 = m . = 0; a4 = m . =− m π 1− 4 π −3 2π π 1− 3 2 π 1− 4 2 15π 2Vm cos 3π 2Vm a5 = 0; a6 = . = and so on π 1 − 62 35π Substituting the values of a0, a1, a2, a4 etc. in the standard Fourier series expression given in Art. 20.3. we have f (t ) a0 a1 cos 2 0t a2 cos 2 0t a4 cos 4 0t a6 cos 6 0t .... 798 Electrical Technology Fig. 21.14 Vm Vm 2V 2V 2V = + cos 10πt + m cos 20 πt − m cos 40πt + m cos 60 πt.... π 2 3π 15π 35 FG 1 + 1 cos ω t + 2 cos 2ω tπ − 2 cos 4ω t + 2 cos 6ω t...IJ = Vm Hπ 2 0 3π 15π 0 35π K 0 0 The line spectrum which is a plot of the harmonic amplitudes versus frequency is given in Fig. 21.14(b). Example 21.4. Determine the Fourier series for the square voltage pulse shown in Fig. 21.15 (a) and plot its line spectrum. (Network Theory, Nagpur Univ. 1992) FG 2πt IJ Solution. The wave represents a periodic function of θ or ωt or HTK having a period extending over 2π radians or T seconds. The general expression for this wave can be written as f (θ) = a0 + a1 cos θ + a2 cos 2θ + a3 cos 3θ +... ...+ b1 sin θ + b2 sin 2θ + b3 sin 3θ +... Fig. 21.15 (i) Now, f (θ) = V; θ = 0 to θ = π; f (θ) = 0, from θ = π to θ = 2π ∴ a0 = 1 2π z 0 2π (θ)d θ = 1 2π R S T z 0 π f (θ)dθ + z π 2π f (θ)dθ U V W Fourier Series 799 or a0 = 1 2π R S T z z π 0 Vdθ + 2π π 0dθ = U V W V π 2π | θ|0 +0 = V2 π × π = V 2 (ii) an = 1 π z 2π 0 f (θ) cos nθdθ = 1 π R S T z0 π V cos nθdθ + z 2π π 0 × cos nθdθ U V W = V π z π 0 cos nθdθ + 0 = V nπ π |sin nθ|0 = 0 ... whether n is odd or even (iii) bn = 1 π z 0 2π f (θ) sin nθdθ = 1 π R S T z π 0 V sin nθdθ + z π 2π 0 × sin nθdθ U V W = V π z π 0 sin nθdθ + 0 = V − cos nθ π n π 0 = V nπ (− cos nπ + 1) Now, when n is odd, (1 − cos nπ ) = 2 but when n is even, (1 − cos nπ ) = 0. 2V V V 2V ∴ b1 = ... n = 1; b2 = × 0 = 0 ... n = 2; b3 = ×2 = ... n = 3 and so on. π 2π 3π 3π Hence, substituting the values of a0 , a1 , a2 etc. and b1 , b2 etc. in the above given Fourier series, we get V 2V 2V 2V E 2V 1 1 FG IJ f (θ) = + 2 π sin θ + 3π sin 3θ + 5π sin 5θ +... = + 2 π sin ω 0 t + sin 3ω 0 t + sin 5ω 0 t +... 3 5 H K It is seen that the Fourier series contains a constant term V/2 and odd harmonic components whose amplitudes are as under: 2V Amplitude of fundamental or first harmonic = π 2V Amplitude of second harmonic = 3π 2V Amplitude of third harmonic = and so on. 5 The plot of harmonic amplitude versus the harmonic frequencies (called line spectrum) is shown in Fig. 21.15 (b). Example 21.5. Obtain the Fourier series for the square wave pulse train indicated in Fig. 21.16. (Network Theory and Design, AMIETE June, 1990) Solution. Here T = 2 second, ω 0 = 2π / T = π rad/s. The given function is defined by f (t ) = 1 0 < t < 1 = 0 and = 0; 1<t<2 a0 = 1 T z T 0 f (t )dt = 1 2 z 2 0 1. dt = 1 2 LM N z z 1 0 1. dt + 1 2 OP 0. dt = Q 1 2 Even otherwise by inspection a0 = (1 + 0)/2 = 1/2 an = 2 T z0 T f (t ) cos nω 0 tdt = z 2 2 2 0 1. cos nπtdt = LM N z 0 1 z 1.cos nπtdt + (0). cos nπtdt 1 2 OP Q 800 Electrical Technology 1 1 sin nπt = ∫0 cos nπtdt = nπ 0 =0 bn = 2 T z 0 T z f (t )sin nω 0 tdt = [ 1.sin nπt dt ] + 0 1 z 1 2 (0) sin nπtdt ] = z 0 1 sin nπt dt 1 − cos nπt 1 − cos nπ = = nπ 0 nπ ∴ bn = 2 / nπ ..... when n is odd; = 0 ... when n is even Fig. 21.16 2 1 1 2 1 1 ∴ f (t ) a0 sin 3n t sin t sin 3 t sin 5 t , etc. n 1 n 2 3 5 odd Example 21.6. Find the trigonometric Fourier series for the square voltage waveform shown in 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 also possesses 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 by f (θ) = odd ∞ ∑ b sin nθt n =1 n where bn = 1 π z0 2π f (θ) sin nθdθ = 1 π R S T z0 π V sin nθdθ + z π 2π − V sin nθdθ = U V W V πn π − cos nθ 0 + V πn 2π cos nθ π V = {(− cos nπ + cos 0) − cos πn} πn V 2V = {(1 − cos nπ ) + (1 − cos nπ)} = (1 − cos nπ) πn πn Now, 1 – cos nπ = 2 when n is odd Fourier Series 801 and =0 when n is even 2V 4V ∴ b1 = ×2= ... putting n = 1; b2 = 0 ... putting n = 2 π π 2V 4V b3 = ×2= ... putting n = 3; b4 = 0... putting n = 4 π3 3π 2V 4V b5 = ×2 = .. putting n = 5 and so on. π.5 5π Fig. 21.17 Hence, the Fourier series for the given waveform is 4V 4V 4V f( ) sin sin 3 sin 5 ... 3 5 4V FG 1 1 4V 2π 1 IJ6π 1FG 10 π IJ = π H sin ω 0 t + 3ω 0 t + sin 5ω 0 t +... = 3 5 π sin T t + sin 3 KT 5 H t + sin T t +... K The line spectrum of the function is shown in Fig. 21.17 (b). It would be seen that the harmonic amplitudes decrease as 1/n, that is, the third harmonic amplitude is 1/3 as large as the fundamental, 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 to illustrated a singhtly different technique. As seen from Fig. 21.17(a) T = 2π , hence, ω 0 = 2πf0 = 2π / T = 2π / π = 1. Over one period the function can be defined as f (t ) = V 0 < t < 5 = −V , π < t < 2π 1 2 1 2 1 1 bn f (t)sin n 0tdt f (t )sin ntdt f (t)sin ntdt f (t )sin ntdt 0 0 0 = 1 π z0 π V sin ntdt + 1 π z π 2π (−V ) sin ntdt = V − cos nt π n π 0 + V cos nt π n 2π π 802 Electrical Technology V V = − (cos nπ − cos 0) + (cos 2nπ − cos nπ) nπ nπ 2V Since cos 0 is 1 and cos 2nπ = 1 ∴ bn = (1 − cos nπ) nπ When n is even, cos nπ = 1 ∴ bn = 0 2V 4V When n is odd, cos nπ = – 1 ∴ bn (1 1) , n 1, 3, 5... n n Substituting the value of bn, the Fourier series become 4V F I ∞ ∞ ∑ ∑ n sin nt = π G sin t + 3 sin 3t + 5 sin 5t +...J 4V 4V 1 1 1 f (t ) = n =1 nπ sin nt = π n =1 H K odd odd Since ω 0 = 1 , the above expression in general terms becomes 4V FG 1 1 IJ f (t ) = π H sin ω 0 t + sin 3ω 0 t + sin 5ω 0 t +... 3 5 K The line spectrum is as shown in Fig. 21.17 (b). Example 21.8. Determine the Fourier series of the square voltage waveform shown in Fig. 21.18. Solution. As compared to Fig. 21.17 (a) given above, the vertical axis of figure has been shifted by π / 2 radians. Replacing t by (t + π / 2) in the above equation, the Fourier series of the waveform shown in Fig. 21.18 becomes Fig. 21.18 4V LM FG IJπ 1 π FG IJ 1 π FG IJ OP f (t ) = π H K sin t + N 2 + sin 3 t + 3 2 H K + sin 5 t + +... 5 2 H K Q 4V F = π GH cos t − 1 cos 3t + 1 cos 5t...IJK 3 5 Example 21.9. Determine the trigonometric Fourier series for the half-wave rectified sine wave form shown in Fig. 21.19 (a) and sketches line spectrum. Fig. 21.19 Fourier Series 803 Solution. The given waveform shows no symmetry, hence its series would contain both sine and cosine terms. Moreover, its average value is not obtainable by inpsection, hence it will have to be found by integration. Here, T 2 0 2 /T 1 . Hence, equation of the given waveform is V = Vm sin ωt = Vm sin t . The given waveform is defined by f (t ) = Vm sin t, 0 < t < π = 0, π < t < 2π a0 = 1 2π z 2π 0 f (t )dt = 1 2π LM N z 0 π Vm sin tdt + zπ 2π dt = 1 2π z 0 π Vm sin tdt OP Q Vm π V V = |− cos t|0 = − m (cos π − cos 0) = m 2π 2π π an = 1 π z 2π 0 f (t ) cos ntdt = 1 π z 0 π Vm sin t cos ntdt = Vm 2π z π 0 [sin(n + 1)t − sin(n − 1)t ]dt = Vm − cos(n + 1)t cos(n − 1)t 2π n+1 + n −1 π 0 Vm − LM cos(n + 1)π cos(n − 1)π + + 1 − 1 OP = 2π N n+1 n−1 n+1 n−1 Q when n is even, cos (n + 1)π = −1 and cos(n − 1)π = −1 an = Vm 1 − 1LM+ 1 − 1 =− 2Vm OP ∴ N 2π n + 1 n − 1 n + 1 n − 1 π(n 2 − 1) Q ... n = 2, 4, 6 etc. when n is odd and ≠ 1, cos (n + 1)π = 1 and cos (n − 1)π = 1 Vm 1 FG 1 1 1 IJ ∴ an = 2π − + H+ − n +1 n −1 n +1 n −1 =0 K ... n = 3, 5, 7 etc. 1 Vm Vm when n 1, a1 Vm sin t.cos tdt sin t cos tdt sin 2tdt 0 0 0 2 0 Hence, an = 0 ...n = 1, 3, 5... bn = 1 π z 0 2π f (t ) sin ntdt = 1 π LM N z 0 π Vm sin t sin ntdt + z π 2π (0) sin ntdt OP Q Vm π = π 0 z sin t sin ntdt = 0 for n = 2, 3, 4, 5 etc. However, the expression indeterminate for n = 1 so that b1 has to be evaluated separately. b1 = 1 π π 0 z V π V Vm sin t.sin tdt = m sin 2 tdt = m π 0 2 The required Fourier series for the half-wave rectified voltage waveform is z Vm Vm 2V ∑G n ∞ F cos nt IJ f (t ) = π + 2 sin t − m π H n =1 − 1K 2 even Vm π FG 2 2 2 IJ = π 2 H 1 + sin t − cos 2t − cos 4t − 3 15 35 cos 6t... K 804 Electrical Technology FG Vm π 2 2 2 IJ = Hπ 1 + sin ω 0 t − cos 2ω 0 t − cos ω 0 t − 2 3 15 35 cos 6ω 0 t −... K V F π = T H G1 + 2 sin θ − 2 cos 2θ − 15 cos 4θ − 35 cos 6θ−...IJK m 3 2 2 The line spectrum is shown in Fig. 21.19 (b) which has a strong fundamental term with rapidly decreasing amplitudes of the higher harmonics. Example 21.10. Find the trigonometrical Fourier series for the full wave rectified voltage sine wave shown in Fig. 21.20. Solution. Since the given function has even symmetry, bn = 0 i.e. it will contain no sine terms in its series. The equation of the sinusoidal sine wave given by V = Vm sin θ . In other words, f (θ) = Vm = sin θ . a0 = 1 2π 2π 0 z f (θ)dθ = 1 2π 2π 0 z 2V π Vm sin θdθ = m 2π 0 sin θdθ It is so because the two parts 0 − π and π − 2 are identical. z Vm π 2V ∴ a0 = |− cos θ|0 = m π π an = 1 π z0 2π f (θ) cos nθdθ = 2Vm π z 0 π sin θ cos nθdθ 1 Now, sin A cos B = [sin( A + B) + sin( A − B)] 2 ∴ an = 2Vm π z0 π [sin (1 + n)θ + sin (1 − n)θ]dθ Fig. 21.20 Vm cos(1 + n)θ cos(1 − n)θ = − + π (1 + n) (1 − n) LM −Vm cos(1 + n)π cos(1 − n)π + − 1 − 1 OP ... when n is odd =0= π 1+ n N 1− n 1+ n 1− n Q Fourier Series 805 However, when n is even, then Vm 1 1 1 1 2Vm 1 1 4Vm an 1 n 1 n 1 n 1 n 1 n 1 n ( n 2 1) ∞ cos 2θ ∑ (n 4Vm ∴ f (θ) = a0 − π n=2 2 − 1) even 2Vm 4Vm 1 1 FG 1 IJ ∴ f (θ) = π − π 3 15 H cos 2 θ − cos 4θ + 35 cos 6θ +... K Example 21.11. Determine the Fourier series for the waveform shown in Fig. 21.21 (a) and sketch its line spectrum. Solution. Its is seen from Fig. 21.21 (a) that the waveform equation is f (θ) = (Vm / π)θ . The given function f (θ) is defined by Vm f( ) , 0 = 0, 2 Since the function possesses neither even nor odd symmetry, it will contain both sine and cosine terms. Average value of the wave over one cycle is Vm/4 or a0 = Vm/4. It is so because the average value over the first half cycle is Vm/2 and over the second half cycle is 0 hence, the average value (Vm / 2) + 0 Vm for full cycle is = = 2 4 an = 1 π z 0 2π f (θ)nθdθ = 1 π LM N z 0 π (Vm / π )θ cos nθdθ + zπ 2π (0) cos nθdθ OP Q = Vm π2 z 0 π θ cos nθdθ = Vm cos nθ π2 n2 + θ n π V sin nθ = 2m 2 (cos nπ − 1) 0 π n ∴ an = 0 when n is odd because cos nπ − 1 = 0 = −2Vm / π 2n2 when n is even 1 2π 1 π 2π bn = π ∫0 f (θ) sin θd θ = π ∫0 (Vm / π)θ sin nθd θ + ∫π (0) sin nθd θ = 1 π z π 0 FG V IJ θ sin nθdθ = V H πK m π m 2 sin nθ n 2 θ − cos nθ n π 0 = −Vm πn cos nπ ∴ bn = −Vm | πnwhenn is even bn = +Vm | π n when n is odd Substituting the values of various constants in the general expression for Fourier series, we get 806 Electrical Technology Vm 2Vm 2Vm 2Vm f (θ) = − 2 cos θ − cos 3θ − cos 5θ... 4 π (3π) 2 (5π) 2 Vm V V + sin θ − m sin 2θ + m sin 3θ.... π 2π 3π Fig. 21.21 The amplitudes of even harmonics are given directly by bn but amplitudes of odd harmonics are given by An = an + bn 2 2 (Art. 21.4) For example, A1 = (2Vm / π 2 ) 2 + (Vm / π) 2 = 0.377 Vm F 2V I + FG V IJ = 0.109 V 2 2 A3 = GH (3π) JK H 2π K m 2 m m F 2V I + FG V IJ = 0.064 V and so on. =G 2 2 H (5π) JK H 5π K m m A5 2 m 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 or f (t) = t. T = 2, ω 0 = 2π / t = 2π / 2 = π By inspection it is clear that a0 = 2/2 = 1, an = 2 T z 0 T z 2 f (t ) cos nω 0 t = t.cos nπt. dt 0 Since we have to find the integral of two functions, we use the technique of integration by parts i.e. z z z FGH z uvdx = u vdx − du dx vdx dx IJ K Fourier Series 807 ∴ an = t. z0 2 cos nπtdt − z FGH z 0 2 1. 2 0 cos nπtdt dt IJ K 2 2 t cos nπt 1 = sin nπt + = 0+ (cos 2nπ − cos 0) nπ 0 (nπ ) 2 0 (nπ ) 2 Since cos 2nπ = cos 0 for all values of n, hence an = 0 Fig. 21.22 bn = 2 T z 0 T f (t ) sin nω 0 tdt = z 0 T t sin nπtdt Employing integration by parts, we get, z 2 bn = t sin nπt dt − 0 z z 2 0 FG1. H 0 2 sin nπtdt dt IJ K z 2 2 2 − cos nπt 2 − cos nπt −t sin nπt sin 2nπ 2 = t. dt − = .cos nπt + = − cos 2nπ nπ 0 0 nπ nπ 0 (nπ ) 2 0 (nπ ) 2 nπ The sine term is 0 for all values of n because sign of any multiple of 2π is 0. Since value of cosine term is 1 for any multiple of 2π , we have, bn = −2 / nπ . ∞ ∞ ∑ b sin nω t = a ∑ 2 1 ∴ f (t ) = a0 + n 0 0 − sin nπt n=1 π n=1 n 2 FG 1 1 IJ = 1− π H sin πt + sin 2πt + sin 3πt +... 2 3 K The 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 waveform shown in Fig. 21.23. Solution. Since the waveform possesses odd symmetry, hence a0 = 0 and an = 0 i.e. there would 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. Since the waveform possesses quarter-wave symmetry, it is necessary to integrate over only one quarter period of finding the Fourier coefficients. 808 Electrical Technology Fig. 21.23 ∞ f (t ) = ∑ b sin nω t n =1 n 0 odd where bn = 8 T z t/4 0 f (t )sin nω 0 t The quarter-wave of the given waveform can be represented by equation of a straight line. Slope of the straight line is Vm /(T / 40) 4Vm / T . Hence, using Y = mx, we have f (t ) = FG 4V IJ t 0 < t < T / 4 ∴ b HTK m n = 8 T z T /4 0 FG 4V IJ t.sin nω t = 32V HTK m T 0 2 m z 0 T /4 t.sin nω 0 tdt Using the theorem of integration by parts, we have bn = 32Vm T 2 LMt N z 0 t/4 sin nω 0 t dt − z z T /4 0 FG1. H T /4 0 sin nω 0 t dt dt IJ OP K Q T /4 32Vm t cos n 0T /4 sin n 0t 32Vm T cos n 0T / 4 sin n 0T / 4 2 2 2 2 T 4 n 0 (n 0) T 4 n 0 (n 0) Now, ω 0 = 2π / T or ω 0T = 2π ∴ nω 0 T / 4 = nπ / 2 ∴ cos nω 0 T / 4 = cos nπ / 2 = 0 when n is odd 32Vm T 32V nπ 8Vm nπ ∴ bn = sin nω 0 = 2 m 2 sin = 2 2 sin n 2 ω 2T 2 0 4 n (2 π ) 2 n π 2 8Vm −8V ∴ bn = ....n = 1, 5,9,13 .... bn = 2 m .... n = 3, 7,11,15,... n 2 π2 n π2 Substituting this value of bn, the Fourier series for the given waveform becomes 8Vm 1 1 1 f (t ) 2 sin 0t 2 3 0t 2 sin 5 0t sin 7 0t ... 3 5 72 Example 21.14. Determine the Fourier series of the triangular waveform shown in Fig. 21.24. Fourier Series 809 Solution. Since the function has even symmetry, bn = 0. Moreover, it also has half- wave symmerty, hence, a0 = 0. The Fourier series ∞ can be written as f (t ) = ∑a n =1 n cos nω 0 t where 2 T an f (t ) cos n 0 t dt. T 0 Fig. 21.24 4Vm T The function is given by the relation f (t ) t T 4 It is so because for the interval 0 ≤ t ≤ T / 2, the slope of the line is 4 Vm / T . 2 T 4 T /4 ∴ an f (t )cos n 0t dt f (t )cos n 0t dt T 0 T 0 16Vm T /2 T 16Vm T /2 4Vm T /2 t cos n 0t dt t cos n 0t dt cos n 0t dt T2 0 4 T2 0 T 0 T /2 T /2 −16Vm 1 t 4V sin nω 0 t = ⋅ cos nω 0 t + sin nω 0 t + T2 4n ω 0 2 2 nω 0 T nω 0 0 0 Substituting ω = 2π / T , we get 16Vm T2 T2 2V an 2 2 2 (cos n 1) (sin n ) sin n T 4 n 4 n n Now, sin nπ = 0 for all values of n, cos nπ = 1 when n is even add – 1 when n is odd. 8Vm ∴ an = π 2 n2 ∞ ∞ cos nω 0 t ∑ ∑ 8Vm 8Vm ∴ f (t ) = cos nω 0 t = n= 0 π n 2 2 π 2 n =1 n2 odd odd 8Vm 1 1 1 = 2 cos 0t cos 3 0t .cos 5 0t cos 7 0t ... 9 25 49 Alternative Solution We can deduce the Fourier series from Fig. of Ex. 21.11 by shifting the vertical axis by π / 2 radians to the right. Replacing t by (t + π / 2) in the Fourier series of Ex. 21.11, we get 8Vm 1 1 1 n f(t) 2 sin 0 t 2 sin 3 0 t 2 sin 5 0 t 2 sin 7 0 t ... 2 3 2 5 2 7 2 8Vm 1 1 1 2 cos 0t cos 3 0t cos 5 0t cos 0t ... 9 25 49 810 Electrical Technology Example 21.15. Obtain the Fourier series representation of the sawtooth waveform shown in Fig. 21.25 (a) and plot its spectrum. Fig. 21.25 Solution. By inspection, we know that the average value of the wave is zero over a cycle because 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 contain only sine terms. 1 2 ∴ f( ) bn sin n where bn f ( )sin n d 0 n 1 The slope of curve is m = V / π ∴ we get, f (θ) = (V / π) θ. If we are the limit of integration form − π to + π then 1 V V 1 2V bn sin n d 2 2 sin n cos n cos n n n n The above result has been obtained by making use of integration by parts as explained earlier. cos nπ is positive when n is even and is negative when n is odd and thus the signs of the coefficients alternate. The required Fourier series is 2V 1 1 1 f( ) sin sin 2 sin 3 sin 4 ... 2 3 4 2V 1 1 1 or f (t ) sin 0tsin 2 0t sin 3 0t sin 4 0t ... 2 3 4 As seen, the coefficients decrease as 1/n so that the series converges slowly as shown by the line spectrum of Fig. 21.25 (b). The amplitudes of the fundamental of first harmonic, second harmonic, third harmonic and fourth harmonic are (2 / π),(2V / 2π),(2V / 3π) and (2V / 4 π ) respectively. Tutorial Problem. 21.1 1. Determine the Fourier series for the triangular waveform shown in Fig. 21.26 (a) (Network Theory and Design, AMIETE June 1990) Fourier Series 811 Fig. 21.26 ⎡1 4 1 1 ⎤ ⎢ 2 − π2 cosω0 t + 32 cos3ω0 t + 52 cos5ω0 t + ....⎥ ⎣ ⎦ 2. Find the values of the Fourier coefficients a0, an and bn for the function given in fig. 21.26 (b). ⎡ 2 7 2πn 7 ⎛ 2πn ⎞ ⎤ ⎢a0 = 3 ;an = nπ sin 3 ;bn = nπ ⎜ 1 − cos 3 ⎟ ⎥ ⎣ ⎝ ⎠⎦ 3. Determine the trigonometric series of the triangular waveform shown in Fig. 21.27. Sketch its line spectrum. Fig. 21.27 Vm 4Vm 1 1 2 cos 0 t cos 3 0 t 2 cos5 0 t ... 2 32 5 4. Determine the Fourier series for the sawtooth waveform shown in Fig. 21.28. Fig. 21.28 Fig. 21.29 812 Electrical Technology ⎡ Vm Vm 1 1 ⎤ ⎢f(t) = 2 + π (sinω0 t + 2 sin2ω0t + 3 sin3ω0t + ....⎥ ⎣ ⎦ 5. Represent the full-wave rectified voltage sine waveform shown in Fig. 21.29 by a Fourier series. ⎡ 2Vm ⎛ 2 2 2 ⎞⎤ ⎢f(t) = π + ⎜ 1 + 3 cos2ωt − 15 cos4ωt + 35 cos6ωt ... ⎟⎥ ⎣ ⎝ ⎠⎦ 6. Obtain trigonometric Fourier series for the wave form shown in Fig. 21.30. Fig. 21.30 Fig. 21.31 ⎡ 2Vm ⎛ 1 1 1 ⎞⎤ ⎢f(t) = − π ⎜ sinω0 t + 2 sin2ω0 t + 3 sin3ω0 t + 4 sin4ω0 t ... ⎟⎥ ⎣ ⎝ ⎠⎦ 7. Find the Fourier series for the sawtooth waveform shown in Fig. 21.31. ⎡ 2⎛ 1 1 ⎞⎤ ⎢f(t) = π ⎜ −sin2πt + 2 sin4πt − 3 sin6πt + ... ⎟⎥ ⎣ ⎝ ⎠⎦ 8. For the waveform of Fig. 21.32, find the Fourier series terms up to the 5th harmonic. (Network Theory Nagpur Univ. 1993) ⎡ 16 ⎛ 2 ⎞ 1 ⎤ ⎢V(t) = π ⎜ sint + 3 sin3t ⎟ + 5 sin5t + ...⎥ ⎣ ⎝ ⎠ ⎦ Fig. 21.32 9. 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? Fourier Series 813 (c) What is the magnitude of the fundamental? (d) Obtain its frequency spectrurm. (Network Theory Nagpur Univ.1993) [(a) 5 V (b) 1 Hz (c) 10 / π volt ] 10. Determine the Fourier series of voltage responses obtained at the o/p of a half wave rectifier shown in Fig. 21.34. Plot the discrete spectrum of the waveform. (Elect. Network Analysis Nagpur Univ. 1993) Fig. 21.34 ⎡ Vm Vm 2Vm 2Vm 2Vm ⎤ ⎢ V(t) = π + 2 cosω0 t + 3π cos10πt − 15 cos20πt + 35π cos30πt ...⎥ ⎣ ⎦ 11. Determine the Fourier coefficients and plot amplitude and phase spectral. (Network Analysis Nagpur Univ. 1993) ⎡ 4Vm ⎢a3 = 0,bn = 0,a1 = π ,a2 = 0 ⎣ −4Vm 4V ⎤ a3 = ,a4 = 0,a5 = m ⎥ 3π π ⎦ Fig. 21.35 OBJECTIVE TESTS – 21 1. A given function f (t) can be represented by (d) all of the above. a Fourier series if it 2. In a Fourier series expansion of a periodic (a) is periodic function, the coefficient a0 represents its (b) is single valued (a) net area per cycle (c) has a finite number of maxima and (b) d.c value minima in any one period 814 Electrical Technology (c) average value over half cycle 10. If the Fourier coefficient bn of a periodic (d) average a.c value per cycle function is zero, then it must possess ______ 3. If in the Fourier series of a periodic function, symmetry. the coefficient a0 is zero, it means that the (a) even function has (b) even quarter-wave (a) odd symmetry (c) odd (b) even quarter-wave symmetry (d) either (a) and (b). (c) odd quarter-wave symmetry 11. A complex voltage waveform is given by (d) any of the above. V = 120 sin ωt + 36(3ωt + π / 2) + 12 sin 4. A periodic function f (t) is said to possess (5ωt + π ). It has a time period of T seconds. odd quarter-wave symmetry if The percentage fifth harmonic contents in the (a) f (t) = f (–t) waveform is (b) f (–t) = –f (t) (a) 12 (b) 10 (c) f (t) = –f (t + T/2) (c) 36 (d) 5 (a) both (b) and (c). 12. In the waveform of Q. 11 above, the phase 5. If the average value of a periodic function displacement of the third harmonic represents over one period is zero and it consists of a time interval of ___ seconds. only odd harmonics then it must be (a) T/12 (b) T/3 possessing _______ symmetry. (c) 3T (d) T/36 (a) half-wave 13. When the negative half-cycle of a complex (b) even quarter-wave waveform is reversed, it becomes identical to (c) odd quarter-wave its positive half-cycle. This feature indicates (d) odd. that the complex waveform is composed of 6. If in the Fourier series of a periodic function, (a) fundamental the coefficient a0 = 0 and an = 0, then it must (b) odd harmonics be having _______ symmetry. (c) even harmonics (a) odd (d) both (a) and (b) (b) odd quarter-wave (e) both (a) and (c) (c) even 14. A periodic waveform possessing half-wave (d) either (a) or (b). symmetry has no 7. In the case of a periodic function possessing (a) even harmonics half-wave symmetry, which Fourier (b) odd harmonics coefficient is zero ? (a) an (b) bn (c) sine terms (c) a0 (d) none of above. (d) cosine terms 8. A periodic function has zero average value 15. The Fourier series of a wave form possessing over a cycle and its Fourier series consists of even quarter-wave symmetry has only only odd cosine terms. What is the symmetry (a) even harmonics possessed by this function. (b) odd cosine terms (a) even (c) odd sine terms (b) odd (d) both (b) and (c). (c) even quarter-wave 16. The Fourier series of a waveform possessing (d) odd quarter-wave odd quarter-wave symmetry contains only 9. Which of the following periodic function (a) even harmonics possesses even symmetry ? (b) odd cosine terms (a) cos 3t (b) sin t (c) odd cosine terms (c) t . cos 50 t (d) (t + t2 + t5). (d) none of above ANSWERS 1. (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)

DOCUMENT INFO

Shared By:

Categories:

Tags:
basic, electrical, engineering

Stats:

views: | 47 |

posted: | 10/2/2012 |

language: | English |

pages: | 36 |

OTHER DOCS BY hamada1331

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.