Definition of the Complex Fourier Series 2.22 2000/07/24 2007/05/10 15:56:05.880 GMT-5 Don Johnson dhj@rice.edu Don Johnson dhj@rice.edu Shawn Stewart mrshawn@alumni.rice.edu Benjamin Fite bfite@rice.edu complex Fourier series conjugate symmetry Euler relations Fourier coefficients Fourier series orthogonality Parseval's theorem periodic signal power pulse spectrum square wave Using Euler's relations to define the complex Fourier series. We can greatly simplify the expressions for a signal's Fourier series by using Euler's relations. In this way, when we combine cosine and sine terms at the same frequency a k 2 k t T b k 2 k t T 1 2 a k 2 k t T b k 2 k t T 1 2 a k 2 k t T b k 2 k t T By defining k k 0 c k 1 2 a k b k and c 0 a 0 , we can rewrite the Fourier series of a square wave as s t k c k 2 k t T with k k 0 c - k c k 1 2 a k b k . In this way, we have the complex Fourier series. The positive index terms correspond to the first portion of the common-frequency term and the negative indexed terms to the second. Thus, the complex Fourier series and the sine-cosine series are identical, each representing a signal's spectrum in slightly different ways. Manipulations and calculations are more streamlined with complex Fourier series, and it is our representation of choice. To aid in finding Fourier coefficients, we note the orthogonality property t 0 T 2 k t T 2 l t T T k l 0 k l We can find a signal's complex Fourier spectrum with c k 1 T t 0 T s t 2 k t T The complex Fourier series for the square wave is sq t k k … -3 -1 1 3 … 2 k 2 k t T What is the complex Fourier series for a sinusoid? Because of Euler's relation, 2 f t 1 2 2 f t 1 2 2 f t Thus, c 1 1 2 , c − 1 1 2 , and the other coefficients are zero. A signal's Fourier series spectrum c k has interesting properties. If s t is real, c − k c k (real-valued periodic signals have conjugate-symmetric spectra). This result follows from the integral that calculates the c k from the signal. Furthermore, this result means that c k c − k : The real part of the Fourier coefficients for real-valued signals is even. Similarly, c k c − k : The imaginary parts of the Fourier coefficients have odd symmetry. Consequently, if you are given the Fourier coefficients for positive indices and zero and are told the signal is real-valued, you can find the negative-indexed coefficients, hence the entire spectrum. This kind of symmetry, c − k c k , is known as conjugate symmetry. If s t s t , which says the signal has even symmetry about the origin, c − k c k . Given the previous property for real-valued signals, the Fourier coefficients of even signals are real-valued. A real-valued Fourier expansion amounts to an expansion in terms of only cosines, which is the simplest example of an even signal. If s t s t , which says the signal has odd symmetry, c − k c k . Therefore, the Fourier coefficients are purely imaginary. The square wave is a great example of an odd-symmetric signal. The spectral coefficients for the periodic signal s t τ are c k 2 k τ T , where c k denotes the spectrum of s t . Thus, delaying a signal by τ seconds results in a spectrum having a linear phase shift of 2 k τ T in comparison to the spectrum of the undelayed signal. Note that the spectral magnitude is unaffected. Showing this property is easy. 1 T t 0 T s t τ 2 k t T 1 T t τ T τ s t 2 k t τ T 1 T 2 k t T t τ T τ s t 2 k t T At this point, the range of integration extends over a period of the integrand. Consequently, it should not matter how we integrate over a period, which means that t τ T τ · t 0 T · , and we have our result. The Fourier series obeys Parseval's Theorem, one of the most important results in signal analysis. This general mathematical result formalizes what we have already seen: you can calculate a signal's power in either the time domain or the frequency domain. Parseval's Theorem Power calculated in the time domain equals the power calculated in the frequency domain. 1 T t 0 T s t 2 k c k 2 This result is a (simpler) re-expression of how to calculate a signal's power than with the real-valued Fourier series expression for power. Let's calculate the spectrum of the periodic pulse signal shown here.

Periodic pulse signal.

The pulse width is Δ , the period T , and the amplitude A . The complex Fourier spectrum of this signal is given by c k A T t 0 Δ 2 k Δ T A 2 k 2 k Δ T 1 At this point, simplifying this expression requires knowing an interesting property. 1 θ θ 2 θ 2 θ 2 θ 2 2 θ 2 Armed with this result, we can simply express the Fourier series coefficients for our pulse sequence. c k A k Δ T k Δ T k Because this signal is real-valued, we find that the coefficients do indeed have conjugate symmetry: c k c − k . The periodic pulse signal has neither even nor odd symmetry; consequently, no additional symmetry exists in the spectrum. Because the spectrum is complex valued, to plot it we need to calculate its magnitude and phase. c k A k Δ T k c k k Δ T neg k Δ T k sign k The function neg · equals -1 if its argument is negative and zero otherwise. What is the value of c 0 ? Recalling that this spectral coefficient corresponds to the signal's average value, does your answer make sense? c 0 A Δ T . This quantity clearly corresponds to the periodic pulse signal's average value. The somewhat complicated expression for the phase results because the sine term can be negative; magnitudes must be positive, leaving the occasional negative values to be accounted for as a phase shift of .

Periodic Pulse Sequence The magnitude and phase of the periodic pulse sequence's spectrum is shown for positive-frequency indices. Here Δ T 0.2 and A 1 .

Also note the presence of a linear phase term (the first term in c k is proportional to frequency k T ). Comparing this term with that predicted from delaying a signal, a delay of Δ 2 is present in our signal. Advancing the signal by this amount centers the pulse about the origin, leaving an even signal, which in turn means that its spectrum is real-valued. Thus, our calculated spectrum is consistent with the properties of the Fourier spectrum. Investigate the half-wave rectified sine wave's spectrum for a linear phase term. If one is present, show how to delay or advance the signal to create an even or odd signal. If one is not present, convince yourself that no delay would yield a signal having even or odd symmetry. A half-wave rectified sine wave occurs when Δ T 2 . Thus, c k k 2 neg k 2 k , which corresponds to a linear function of k . The linear phase can be removed by delaying the signal by 1 4 of a period. The phase plot shown in requires some explanation as it does not seem to agree with what suggests. There, the phase has a linear component, with a jump of every time the sinusoidal term changes sign. We must realize that any integer multiple of 2 can be added to a phase at each frequency without affecting the value of the complex spectrum. We see that at frequency index 4 the phase is nearly . The phase at index 5 is undefined because the magnitude is zero in this example. At index 6, the formula suggests that the phase of the linear term should be less than (more negative) than . In addition, we expect a shift of in the phase between indices 4 and 6. Thus, the phase value predicted by the formula is a little less than 2 . Because we can add 2 without affecting the value of the spectrum at index 6, the result is a slightly negative number as shown. Thus, the formula and the plot do agree. In phase calculations like those made in MATLAB, values are usually confined to the range by adding some (possibly negative) multiple of 2 to each phase value.