23.6. The Complex Form. Introduction. Prerequisites. Learning Outcomes

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1 he Complex Form 3.6 Introduction In this Section we show how a Fourier series can be expressed more concisely if we introduce the complex number i where i =. By utilising the Euler relation: e iθ cos θ + i sin θ we can replace the trigonometric functions by complex exponential functions. By also combining the Fourier coefficients a n and b n into a complex coefficient c n through c n = (a n ib n we find that, for a given periodic signal, both sets of constants can be found in one operation. We also obtain Parseval s theorem which has important applications in electrical engineering. he complex formulation of a Fourier series is an important precursor of the Fourier transform which attempts to Fourier analyse non-periodic functions. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... know how to obtain a Fourier series be competent working with the complex numbers be familiar with the relation between the exponential function and the trigonometric functions express a periodic function in terms of its Fourier series in complex form understand Parseval s theorem HELM (8: Section 3.6: he Complex Form 53

2 . Complex exponential form of a Fourier series So far we have discussed the trigonometric form of a Fourier series i.e. we have represented functions of period in the terms of sinusoids, and possibly a constant term, using f(t = a + { ( ( } nπt nπt a n cos + b n sin. If we use the angular frequency ω = π we obtain the more concise form f(t = a + (a n cos nω t + b n sin nω t. We have seen that the Fourier coefficients are calculated using the following integrals: a n = f(t cos nω t dt n =,,,... ( b n = f(t sin nω t dt n =,,... ( An alternative, more concise form, of a Fourier series is available using complex quantities. his form is quite widely used by engineers, for example in Circuit heory and Control heory, and leads naturally into the Fourier ransform which is the subject of 4.. Revision of the exponential form of a complex number Recall that a complex number in Cartesian form which is written as z = a + ib, where a and b are real numbers and i =, can be written in polar form as z = r(cos θ + i sin θ where r = z = a + b and θ, the argument or phase of z, is such that a = r cos θ b = r sin θ. A more concise version of the polar form of z can be obtained by defining a complex exponential quantity e iθ by Euler s relation e iθ cos θ + i sin θ he polar angle θ is normally expressed in radians. Replacing i by i we obtain the alternative form e iθ cos θ i sin θ 54 HELM (8: Workbook 3: Fourier Series

3 ask Write down in cos θ±i sin θ form and also in Cartesian form (a e iπ/6 (b e iπ/6. Use Euler s relation: We have, by definition, ( π ( π (a e iπ/6 = cos + i sin = ( π ( π i (b e iπ/6 = cos i sin = i ask ( π Write down (a cos 6 ( π (b sin 6 in terms of e iπ/6 and e iπ/6. We have, adding the two results from the previous task ( π ( π e iπ/6 + e iπ/6 = cos or cos = ( e iπ/6 + e iπ/6 6 6 Similarly, subtracting the two results, ( π e iπ/6 e iπ/6 = i sin 6 (Don t forget the factor i in this latter case. or ( π sin = ( e iπ/6 e iπ/6 6 i Clearly, similar calculations could be carried out for any angle θ. he general results are summarised in the following Key Point. HELM (8: Section 3.6: he Complex Form 55

4 Key Point 8 Euler s Relations e iθ cos θ + i sin θ, e iθ cos θ i sin θ cos θ ( e iθ + e iθ sin θ ( e iθ e iθ i Using these results we can redraft an expression of the form a n cos nθ + b n sin nθ in terms of complex exponentials. (his expression, with θ = ω t, is of course the n th harmonic of a trigonometric Fourier series. ask Using the results from the Key Point 8 (with nθ instead of θ rewrite a n cos nθ + b n sin nθ in complex exponential form. First substitute for cos nθ and sin nθ with exponential expressions using Key Point 8: We have so a n cos nθ = a n a n cos nθ + b n sin nθ = a n ( e inθ + e inθ b n sin nθ = b n ( e inθ e inθ i ( e inθ + e inθ + b n ( e inθ e inθ i 56 HELM (8: Workbook 3: Fourier Series

5 Now collect the terms in e inθ and in e inθ and use the fact that i = i: We get ( a n + b n i or, since i = e inθ + ( a n b n e inθ i i i = i (a n ib n e inθ + (a n + ib n e inθ. Now write this expression in more concise form by defining c n = (a n ib n which has complex conjugate c n = (a n + ib n. Write the concise complex exponential expression for a n cos nθ + b n sin nθ: a n cos nθ + b n sin nθ = c n e inθ + c ne inθ Clearly, we can now rewrite the trigonometric Fourier series a + (a n cos nω t + b n sin nω t as a + ( cn e inωt + c ne inω t (3 A neater, and particularly concise, form of this expression can be obtained as follows: Firstly write a = c (which is consistent with the general definition of c n since b =. he second term in the summation c ne inωt = c e iωt + c e iωt +... can be written, if we define c n = c n = (a n + ib n, as c e iω t + c e iω t + c 3 e 3 iω t +... = Hence (3 can be written c + c n e inωt + n= n= c n e inω t c n e inω t or in the very concise form c n e inω t. HELM (8: Section 3.6: he Complex Form 57

6 he complex Fourier coefficients c n can be readily obtained as follows using ( and ( for a n, b n. Firstly c = a = f(t dt For n =,, 3,... we have c n = (a n ib n = Also for n =,, 3,... we have f(t(cos nω t i sin nω t dt i.e. c n = (4 f(te inω t dt (5 c n = c n = (a n + ib n = f(te inω t dt his last expression is equivalent to stating that for n =,, 3,... c n = f(te inω t dt he three equations (4, (5, (6 can thus all be contained in the one expression (6 c n = f(te inω t dt for n =, ±, ±, ±3,... he results of this discussion are summarised in the following Key Point. Key Point 9 Fourier Series in Complex Form A function f(t of period has a complex Fourier series f(t = c n e inω t where c n = f(te inωt dt For the special case = π, so that ω =, these formulae become particularly simple: f(t = c n e int c n = π f(te int dt. π π 58 HELM (8: Workbook 3: Fourier Series

7 3. Properties of the complex Fourier coefficients Using properties of the trigonometric Fourier coefficients a n, b n we can readily deduce the following results for the c n coefficients:. c = a is always real.. Suppose the periodic function f(t is even so that all b n are zero. hen, since in the complex form the b n arise as the imaginary part of c n, it follows that for f(t even the coefficients c n (n = ±, ±,... are wholly real. ask If f(t is odd, what can you deduce about the Fourier coefficients c n? Since, for an odd periodic function the Fourier coefficients a n (which constitute the real part of c n are zero, then in this case the complex coefficients c n are wholly imaginary. 3. Since c n = f(te inω t dt then if f(t is even, c n will be real, and we have two possible methods for evaluating c n : ( (a Evaluate the integral above as it stands i.e. over the full range,. Note carefully that the second term in the integrand is neither an even nor an odd function so the integrand itself is ( even function ( neither even nor odd function = neither even nor odd function. hus we cannot write c n = / (b Put e inω t = cos nω t i sin nω t so f(te inω t dt f(te inω t = f(t cos nω t if(t sin nω t = ( even( even i( even( odd Hence c n = f(t cos nω t dt = a n. = ( even i( odd. 4. If f(t + = f(t then of course only odd harmonic coefficients c n (n = ±, ±3, ±5,... will arise in the complex Fourier series just as with trigonometric series. HELM (8: Section 3.6: he Complex Form 59

8 Example 4 Find the complex Fourier series of the saw-tooth wave shown in Figure 4: f(t A t Figure 4 Solution We have f(t = At < t < f(t + = f(t he period is in this case so ω = π. Looking at the graph of f(t we can say immediately (a the Fourier series will contain a constant term c (b if we imagine shifting the horizontal axis up to A the signal can be written f(t = A + g(t, where g(t is an odd function with complex Fourier coefficients that are purely imaginary. Hence we expect the required complex Fourier series of f(t to contain a constant term A and complex exponential terms with purely imaginary coefficients. We have, from the general theory, and using < t < as the basic period for integrating, c n = At e inω t dt = A We can evaluate the integral using parts: te inω t dt = [ te inω t ( inω ] + inω = e inω ( inω ( inω te inω t dt [ e inω t e inω t dt ] 6 HELM (8: Workbook 3: Fourier Series

9 Solution (contd. But ω = π so e inω = e inπ = cos nπ i sin nπ Hence the integral becomes Hence = i = ( e inω inω ( inω c n = A ( Note that c n = Also c = inω = ia πn ia π( n = ia πn = c n At dt = A as expected. n = ±, ±,... as it must Hence the required complex Fourier series is f(t = A + ia e inω t π n n which could be written, showing only the constant and the first two harmonics, as f(t = A {... i e iω t ie iωt + π + ie iωt + i e } iω t π he corresponding trigonometric Fourier series for the function can be readily obtained from this complex series by combining the terms in ±n, n =,, 3,... For example this first harmonic is A { ie iω t + ie } iω t = A π π { i(cos ω t i sin ω t + i(cos ω t + i sin ω t} = A π ( sin ω t = A π sin ω t Performing similar calculations on the other harmonics we obtain the trigonometric form of the Fourier series f(t = A A sin nω t. π n HELM (8: Section 3.6: he Complex Form 6

10 ask Find the complex Fourier series of the periodic function: f(t = e t π < t < π f(t + π = f(t f(t π π 3π t Firstly write down an integral expression for the Fourier coefficients c n : We have, since = π, so ω = c n = π π π e t e int dt Now combine the real exponential and the complex exponential as one term and carry out the integration: We have c n = π π π e ( int dt = [ ] e ( int π = ( e ( inπ ( e inπ π ( in π π ( in 6 HELM (8: Workbook 3: Fourier Series

11 Now simplify this as far as possible and write out the Fourier series: e ( inπ = e π e inπ = e π (cos nπ i sin nπ = e π cos nπ e ( inπ = e π e inπ = e π cos nπ Hence c n = π ( in (eπ e π cos nπ = sinh π ( + in cos nπ π ( + n Note that the coefficients c n n = ±, ±,... have both real and imaginary parts in this case as the function being expanded is neither even nor odd. Also c n = sinh π ( in sinh π ( in cos( nπ = π ( + ( n π ( + n cos nπ = c n as required. his includes the constant term c = sinh π. Hence the required Fourier series is π f(t = sinh π n ( + in ( π ( + n e int since cos nπ = ( n. HELM (8: Section 3.6: he Complex Form 63

12 4. Parseval s theorem his is essentially a mathematical theorem but has, as we shall see, an important engineering interpretation particularly in electrical engineering. Parseval s theorem states that if f(t is a periodic function with period and if c n (n =, ±, ±,... denote the complex Fourier coefficients of f(t, then f (t dt = c n. In words the theorem states that the mean square value of the signal f(t over one period equals the sum of the squared magnitudes of all the complex Fourier coefficients. Proof of Parseval s theorem. Assume f(t has a complex Fourier series of the usual form: ( f(t = c n e inω t ω = π where hen Hence c n = f(te inω t dt f (t = f(tf(t = f(t c n e inω t = c n f(te inω t f (t dt = = cn f(te inω t dt cn f(te inω t dt = c n c n = c n which completes the proof. Parseval s theorem can also be written in terms of the Fourier coefficients a n, b n of the trigonometric Fourier series. Recall that so c = a c n = a n ib n n =,, 3,... c n = a n + ib n n =,, 3,... so c n = a n + b n 4 n = ±, ±, ±3, HELM (8: Workbook 3: Fourier Series

13 c n = a 4 + a n + b n 4 and hence Parseval s theorem becomes f (tdt = a 4 + (a n + b n (7 he engineering interpretation of this theorem is as follows. Suppose f(t denotes an electrical signal (current or voltage, then from elementary circuit theory f (t is the instantaneous power (in a ohm resistor so that f (t dt is the energy dissipated in the resistor during one period. Now a sinusoid wave of the form A cos ωt ( or A sin ωt has a mean square value A A so a purely sinusoidal signal would dissipate a power in a ohm resistor. Hence Parseval s theorem in the form (7 states that the average power dissipated over period equals the sum of the powers of the constant (or d.c. components and of all the sinusoidal (or alternating components. ask he triangular signal shown below has trigonometric Fourier series f(t = π 4 cos nt. π n ( odd n [his was deduced in the ask in Section 3.3, page 39.] π f(t π π t Use Parseval s theorem to show that (n odd n 4 = π4 96. HELM (8: Section 3.6: he Complex Form 65

14 First, identify a, a n and b n for this situation and write down the definition of f(t for this case: We have a = π 4 n =, 3, 5,... a n = n π n =, 4, 6,... Also b n = n =,, 3, 4,... f(t = t f(t + π = f(t π < t < π Now evaluate the integral on the left hand side of Parseval s theorem and hence complete the problem: 66 HELM (8: Workbook 3: Fourier Series

15 We have f (t = t so f (t dt = π π π t dt = π [ t 3 he right-hand side of Parseval s theorem is a 4 + a n = π n 4 π Hence π 3 = π π (n odd n 4 (n odd 3 ] π 8 π π = π 3 (n odd n 4 = π (n odd n 4 = π4 96. Exercises Obtain the complex Fourier series for each of the following functions of period π.. f(t = t π t π. f(t = t t π 3. f(t = e t π t π s. i ( n n eint (sum from to excluding n =.. π + i n eint (sum from to excluding n =. 3. sinh π π ( n ( + in ( + n eint (sum from to. HELM (8: Section 3.6: he Complex Form 67