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# fourier_series - Fourier Series Do not distribute these...

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Unformatted text preview: Fourier Series Do not distribute these notes 147 Prof. R.T. M’Closkey, UCLA Periodic functions and Fourier Series Periodic signals, that is signals that satisfy u(t + T ) = u(t), (60) for all t and for some ﬁxed T > 0, may be expressed as an inﬁnite sum of sinusoids under some mild conditions: 1. u must be square integrable and Riemann integrable over one period, 2. u must have only a ﬁnite number of discontinuities in one period. The smallest T for which (60) is true is called the period of the signal. The Fourier series of the periodic signal, denoted uF S is 2 a0 + uF S (t) = T T ∞ (ak cos(kω0 t) + bk sin(kω0 t)) , (61) k =1 where ω0 = 2π/T is the fundamental frequency and the Fourier series coefﬁcients are given by, a0 := u(t)dt T ak := u(t) cos(kω0 t)dt, k = 1, 2, 3, . . . u(t) sin(kω0 t)dt, k = 1, 2, 3, . . . . T bk := T The reason we distinguish between the periodic signal u and its Fourier series uF S will be discussed later. The notation T denotes that the integral may be taken over any “T " seconds of the periodic waveform. For example, τ +T u(t) cos(kω0 t)dt, ak := k = 1, 2, 3, . . . τ Do not distribute these notes 148 Prof. R.T. M’Closkey, UCLA for any τ ∈ R. From the Fourier series expression it is seen that only discrete frequencies are required in the representation of u: {0, ω0 , 2ω0 , 3ω0 , . . . }. Furthermore, the sinusoids in the series satisfy the following: cos(kω0 t) cos(lω0 t)dt = T sin(kω0 t) sin(lω0 t)dt = T 0 k=l T 2 k=l 0 k=l , k=l T 2 , cos(kω0 t) sin(lω0 t)dt = 0 for all k, l T The Fourier series may be converted into a more compact form by introducing complex coefﬁcients. First, in (61) set 1 jkω0 t e + e−jkω0 t 2 1 jkω0 t e − e−jkω0 t , sin(kω0 t) = j2 cos(kω0 t) = so that a0 2 uF S (t) = + 2 T = = Deﬁne the new coefﬁcient a0 2 + 2 T a0 1 + 2 T ∞ (ak cos(kω0 t) + bk sin(kω0 t)) k =1 ∞ k =1 ∞ k =1 ak 1 jkω0 t 1 jkω0 t e + e−jkω0 t + bk e − e−jkω0 t 2 j2 , (62) (ak − jbk )ejkω0 t + (ak + jbk )e−jkω0 t . ck = ak − jbk = T u(t) cos(kω0 t)dt − j u(t) sin(kω0 t)dt T u(t)e−jkω0 t dt, = T Do not distribute these notes 149 Prof. R.T. M’Closkey, UCLA where k = 0, 1, 2, . . . . Suppose, however, we substitute “−k " into the expression for ck for any k = 0, 1, 2, . . . , c−k = u(t)ejkω0 t dt T = u(t) cos(kω0 t)dt + j T u(t) sin(kω0 t)dt T = ak + jbk = c∗ . k Thus, we may continue with (62), 1 c0 uF S (t) = + T T 1 = T ∞ ∞ k =1 ck ejkω0 t + c−k e−jkω0 t (63) ck ejkω0 t , k =−∞ where u(t)e−jkω0 t dt, ck = T k = . . . , −2, 1, 0, 1, 2, . . . . This is a concise expression for the Fourier series of a periodic signal u. Note that in practice we only need compute ck for k = 0, 1, 2, . . . because c−k = c∗ for any k . k Example. Consider the periodic signal shown below (red trace). The fundamental period is T = 1 second, so the fundamental frequency is ω0 = 2π rad/s. Thus, the frequencies present in the Fourier series are 1 HZ, 2 Hz, 3 Hz, etc. Also shown in this ﬁgure is the value c0 /T = 0.75 (the blue trace). Note that the simplest approximation of u is u(t) ≈ c0 , T ∀ t. Better approximations include more terms in the series. Do not distribute these notes 150 Prof. R.T. M’Closkey, UCLA 1.5 1.5 k=0 k = −1, 0, 1 periodic function 1 1 0.5 0.5 0 0 truncated Fourier series 0.5 2 1.5 1.5 1 0.5 0 0.5 seconds 1 1.5 0.5 2 2 1.5 k = −5, . . . , 0, . . . , 5 1 0.5 0 0.5 seconds 1 1.5 2 1 1.5 2 k = −40, . . . , 0, . . . , 40 0.5 0 1 1 0.5 1.5 0 0.5 2 1.5 1 Do not distribute these notes 0.5 0 0.5 seconds 1 1.5 0.5 2 2 151 1.5 1 0.5 0 0.5 seconds Prof. R.T. M’Closkey, UCLA This example illustrates an interesting fact: u(t) = uF S (t). This may seem contradictory, however, u(t) = uF S (t) for any t where u is continuous. If u is not continuous at say ˜ t, then ˜ uF S (t) = 1 lim u(t) + lim u(t) . ˜ ˜ 2 t↑t t↓t For the current example, uF S (0) = 0.5 although limt↑0 u(t) = 0 and limt↓0 u(t) = 1. For a linear system driven by periodic inputs, however, the system makes no distinction between u and uF S ! Do not distribute these notes 152 Prof. R.T. M’Closkey, UCLA Spectrum of a periodic signal Let u be a periodic signal of period T and suppose uF S is its Fourier series 1 uF S (t) = T ∞ ck ejkω0 t , u(t)e−jkω0 t dt. ck = T k =−∞ It’s useful to deﬁne the “envelope" function u(t)e−jωt dt, c(ω ) = (64) T where ω is considered to be a continuous variable in contrast to the discrete frequencies present in the Fourier series, i.e. kω0 , k = 0, ±1, ±2, . . . . Note that ck = c(kω0 ). The spectrum of a periodic signal is just the set of Fourier series coefﬁcients with their corresponding frequencies. In fact, the most useful representation of the spectrum is just the plots of |ck | versus kω0 and ∠ck versus kω0 . This is analogous to the magnitude and phase plots associated with the frequency response of a linear system. The spectrum of a periodic signal, however, is discrete. In other words, it is deﬁned only at distinct frequencies that are determined by the period T of the signal. Do not distribute these notes 153 Prof. R.T. M’Closkey, UCLA Example. Consider the periodic signal shown on the next slide and given by 1 u(t) = t − 0.2, 2 t ∈ [0, 2], (65) for one period. The period is T = 2 so ω0 = π . The envelope function can be computed by applying teαt dt = − to 2 c(ω ) = 0 1 = 2 1 αt 1 αt e + te , α2 α 1 t − 0.2 e−jωt dt 2 1 −j 2 ω 2 e − 1 + j e−j 2ω ω2 ω 0.2 −j 2ω −j e −1 , ω (66) (ω = 0) The Fourier series coefﬁcients are obtained by “sampling" the envelope function, ck = c(kω0 ) 1 2 1 = e−j 2kπ − 1 + j e−j 2kπ 2 k2π2 kπ 1 =j , k=0 kπ c0 = 1, (computed separately) −j 0.2 −j 2kπ e −1 kπ (67) The amplitude and phase spectrum plots are shown below along with the envelope function. Do not distribute these notes 154 Prof. R.T. M’Closkey, UCLA 1 10 amplitude spectrum envelope function magnitude amplitude spectrum units 0 volts sec signal units 10 periodic signal Fourier series coefﬁcients 10 1 0.8 0.6 10 volts 1 2 5 4 3 2 1 0.4 0 150 0.2 1 2 3 4 5 1 2 3 4 5 200 100 0.4 3 2 1 0 sec 1 2 3 phase (degrees) 0.2 0 Hz phase spectrum 50 Fourier series coefﬁcients 0 50 100 150 200 5 Do not distribute these notes 155 4 3 2 1 0 Hz Prof. R.T. M’Closkey, UCLA Notion of Energy for Periodic Signals Consider a periodic signal u with period T seconds. The energy in one period, denoted u u E = T E, is deﬁned to be |u(t)|2 dt. Suppose the units of u are “volts". The energy unit is then V 2 · sec. If the unit of u is “m" (meters) then the energy unit is m2 · sec. The “s" unit is equivalent to 1/(1/sec) = 1/Hz, thus, for these two examples the energy units can also be equivalently expressed as V 2 /Hz and m2 /Hz. The energy in one period can also be expressed in terms of the Fourier series coefﬁcients of u: u E |u(t)|2 dt = T (u(t))∗ u(t)dt (since u is real) = T (uF S (t))∗ uF S (t)dt (can show u = T = T = = = 1 T2 1 T 1 T ∞ ∞ ck e jkω0 t ∗ k =−∞ 1 T ∞ E = uF S E) ck ejkω0 t dt k =−∞ T c∗ ck k k =−∞ ∞ c∗ ck k k =−∞ ∞ k =−∞ ω0 |ck |2 , 2π f0 where f0 is the fundamental frequency in Hz. Do not distribute these notes 156 Prof. R.T. M’Closkey, UCLA Its useful to determine the units associated with the square of magnitude of the coefﬁcients, i.e. determine the units of |ck |2 . For example, suppose the unit associated with u is “volts," then the energy in one period has unit V 2 · sec and since the 1/T factor in front of the summation has unit 1/sec, the units of |ck |2 must be V 2 · sec2 or, equivalently, V 2 /Hz2 . There is a useful graphical representation of u E that shows how the energy in one period is distributed over frequency. This is explored in the next examples. Example. Graphing |ck |2 over an interval of f0 Hz which is centered at kf0 Hz yields a graph with whose area is equal to u E . Considering the previous example we have the following result: Square of magnitude of Fourier series coefficients Square of signal 0.6 square of FS coefﬁcient magnitude square of envelope function magnitude 0.4 0.3 0.2 0.5 volts2 |ck|2 (V2/Hz2) 0.5 0.6 ∞ ￿ area = k=−∞ f0 |ck |2 area = ￿ T 0.3 |u(t)|2 dt areas0.2 under red graphs are equal! 0.1 0 3 0.4 0.1 2 1 0 Hz 1 2 3 0 3 2 1 0 sec 1 2 3 The equality between the areas is called Parseval’s Theorem. Do not distribute these notes 157 Prof. R.T. M’Closkey, UCLA Example. Consider the following four signals with period T = 2 seconds: 1 u1 (t) = √ 2 u2 (t) = cos(πt) u3 (t) = 1 √ 2 1 − √2 u4 (t) = 2 t ∈ [0, 0.25] 0 t ∈ [0.25, 2] t ∈ [−0.5, 0.5] t ∈ [0.5, 1.5] Note that u1 is periodic with any period, however, we will analyze it from the perspective of having period T = 2 seconds since that matches the periods of the other signals. See the ﬁgures on the following page. The energy in one period for all of these signals is 1. This is calculated by squaring the signals and measuring the area in one period (see the second set of ﬁgures). Finally, the energy spectral density for each signal is just a plot of the square of the absolute value of the Fourier series coefﬁcients versus frequency as shown in the third slide from this one. The area under the energy spectral density plots is also 1 but note that the signals distribute the energy differently: the constant signal concentrates all of the energy at a single frequency (0 Hz), whereas the periodic pulse spreads the energy around at many frequencies. Do not distribute these notes 158 Prof. R.T. M’Closkey, UCLA Constant signal 3 2.5 Square wave 3 u1 2.5 1.5 1 1 volts 2 1.5 volts 2 u3 0.5 0.5 0 0 0.5 0.5 1 1 1.5 1.5 2 3 2 1 0 seconds 1 2 2 3 3 2 1 Sinusoidal signal 3 2.5 1 2 3 1 2 3 Periodic pulse 3 u2 2.5 1.5 1.5 1 u4 2 1 volts 2 volts 0 seconds 0.5 0.5 0 0 0.5 0.5 1 1 1.5 1.5 2 3 2 Do not distribute these notes 1 0 seconds 1 2 2 3 3 159 2 1 0 seconds Prof. R.T. M’Closkey, UCLA Constant signal 5 4.5 4 Square wave 5 u2 1 4.5 4 3 volts 3.5 3 volts 3.5 u2 3 2.5 2 2.5 2 1.5 1.5 area = 1 1 area = 1 1 0.5 0.5 0 3 2 1 0 seconds 1 2 0 3 3 2 1 Sinusoidal signal 5 4.5 4 1 2 3 1 2 3 Periodic pulse 5 u2 2 4.5 4 3 u2 4 area = 1 3.5 3 volts 3.5 volts 0 seconds 2.5 2 2.5 2 area = 1 1.5 1.5 1 1 0.5 0.5 0 3 2 Do not distribute these notes 1 0 seconds 1 2 0 3 3 160 2 1 0 seconds Prof. R.T. M’Closkey, UCLA Spectral density of constant signal Spectral density of square wave signal 4 3.5 3.5 3 area = 1 2.5 2 1.5 All energy concentrated at 0 Hz 1 Energy Spectral Density (volt2/Hz2) Energy Spectral Density (volt2/Hz2) 4 0.5 0 3 3 Energy is spread at different frequencies but still concentrated at -0.5 and 0.5 Hz 2.5 2 1.5 1 area = 1 0.5 2 1 0 Hz 1 2 0 3 3 2 Spectral density of sinusoidal signal 1 2 3 4 All energy concentrated at -0.5 and 0.5 Hz 3.5 Energy Spectral Density (volt2/Hz2) Energy Spectral Density (volt2/Hz2) 0 Hz Spectral density of periodic pulse signal 4 3 2.5 2 area = 1 1.5 1 0.5 0 3 1 3.5 Energy is spread at different frequencies that are multiples of 0.5 Hz 3 2.5 2 1.5 1 area = 1 0.5 2 Do not distribute these notes 1 0 Hz 1 2 0 3 3 161 2 1 0 Hz 1 2 3 Prof. R.T. M’Closkey, UCLA Linear systems and periodic signals The periodic portion of a linear system’s response to a periodic input u can be calculated using Fourier series. Suppose an input to a linear system, denoted u, has period T and Fourier series representation, 1 uF S (t) = T ∞ ck ejkω0 t k =−∞ Remember that we refer to the set of Fourier series coefﬁcients, along with the frequency associated with the coefﬁcient, as the spectrum of the signal. Furthermore, recall that the sinusoidal response of a linear system with frequency response function H (jω ) to a given sinusoidal input, say ck ejkω0 t , is simply, y (t) = H (jkω0 ) ck ejkω0 t . u(t) Invoking superposition, the system’s response to the sum of sinusoids representing the Fourier series of the input is sum of the system’s response to each sinusoidal component in uF S . In other words, 1 yF S (t) = T ∞ H (jkω0 )ck ejkω0 t , k =−∞ (68) ck ˜ is the Fourier series representation of the periodic portion of the output with coefﬁcients denoted by ck . If it is ˜ necessary to solve an initial value problem then homogeneous terms with parameters can be added to (68). Note that the spectrum of the output is the set {ck }. In other words, the input spectrum is multiplied by the system’s ˜ frequency response function to produce the output spectrum. In fact, this idea can be extended: if any two of the following quantities are given, then the third can be found. In other words, {ck }, {ck }, ˜ H, 1. If the periodic input and frequency response function are given, then the periodic portion of the output can be determined from ck = H (jkω0 )ck , ˜ Do not distribute these notes k = . . . , −2, −1, 0, 1, 2, . . . 162 Prof. R.T. M’Closkey, UCLA where {ck } is the spectrum of the periodic input signal and H is the frequency response function. 2. If the periodic input and periodic output are given (say from a test of an asymptotically stable system), then the frequency response function can be estimated at those frequencies where the input Fourier series coefﬁcients are non-zero, H (jkω0 ) = ck ˜ , ck k = . . . , −2, −1, 0, 1, 2, . . . This process is called system identiﬁcation. 3. If the periodic output and frequency response function are given, then the periodic input can be computed ck = ck ˜ , H (jkω0 ) k = . . . , −2, −1, 0, 1, 2, . . . assuming H is not zero at the Fourier series frequencies. This process is called deconvolution. If any two of these three quantities are known, then the third can be found Periodic input spectrum {ck } Do not distribute these notes Linear System with Freq. Resp. Function H 163 ˜ Periodic output spectrum {ck } Prof. R.T. M’Closkey, UCLA Example of determining the periodic output given a periodic input and frequency response function. Consider the linear system described by the second order ODE (damped spring-mass system), 2 2 x + 2ξωn x + ωn x = ωn u. ¨ ˙ This ODE is subjected to the input (65). The frequency response function is 2 ωn H (jω ) = 2 2 s + 2ξωn s + ωn = s=jω 2 ωn 2 −ω 2 + j 2ξωn ω + ωn . The Fourier series of the periodic portion of the output is given by 1 xF S (t) = T = 1 2 ∞ k =−∞ ∞ k =−∞ H (jkω0 t)ck ejkω0 t 2 ωn · ck ejπkt , 2 π 2 + j 2ξω kπ + ω 2 −k n n where the ck are given by (67). Thus, the Fourier series coefﬁcients of the periodic portion of the response due to the periodic input u are: 2 ωn 1 ck = ˜ ·j , 2 −k 2 π 2 + j 2ξωn kπ + ωn kω0 c0 = 1 · c0 = 1 ˜ k=0 Note that the envelope function can also be multiplied by the frequency response function: 2 ωn 1 H (jω )c(ω ) = 2 −ω 2 + j 2ξπω + ωn 2 H (0)c(0) = 1 2 1 −jω e − 1 + j e−jω , ω2 ω (ω = 0) In order to graph these relationships let’s assume ξ = 0.2 and ωn = 2π rad/sec. The amplitude and phase plots of the output Fourier series are shown on the next slide. Note that the system attenuates the input sinusoidal components for |k | > 2, thus the periodic response is dominated by the terms corresponding to k = 0, ±1, ±2. Do not distribute these notes 164 Prof. R.T. M’Closkey, UCLA 1 periodic output + transient stable system periodic input 1.5 periodic input periodic output magnitude (various units) 10 amplitude spectrum input FS system freq resp output FS systemʼs freq resp 0 10 input spectrum 10 1 volts 1 10 0.5 2 200 0 output spectrum 5 0 Hz phase spectrum input FS system freq resp output FS 150 2 1 0 sec 1 2 100 3 phase (degrees) 0.5 3 5 50 0 50 100 150 200 5 Do not distribute these notes 165 0 Hz 5 Prof. R.T. M’Closkey, UCLA Example of computing a frequency response function from knowledge of the periodic input and periodic output This example discusses a useful method for estimating the frequency response function of a linear system under test. The idea is to change our perspective from one of computing the output spectrum given the input spectrum and frequency response function to computing the frequency response function from the spectra of the periodic input and periodic output. In other words, when testing an asymptotically system whose “dynamics" we wish to determine, we can apply a periodic input and then record the periodic response (once transients have decayed). The frequency response function of the system can be estimated at those frequencies represented in the Fourier series. Speciﬁcally, if u has the Fourier series representation, 1 uF S (t) = T ∞ ck ejkω0 t , k =−∞ where {ck } is the input spectrum, and y is measured after the transients have decayed so it is periodic with Fourier series, 1 yF S (t) = T ∞ ck ejkω0 t , ˜ k =−∞ where {ck } is the output spectrum, then the frequency response function is estimated to be ˜ H (jkω0 ) = ck ˜ , ck k = 0, ±1, ±2, . . . . It is assumed that ck = 0, for any given k . If this is not the case then H is simply not calculated at that frequency. Note that H is estimated at grid of frequencies and that the grid spacing is determined by ω0 , which is inversely proportional to T . Thus, if a grid of, say, at least 0.2 Hz is desired, then the periodic input must be chosen so that T ≥ 5 seconds. The following shows a pictorial representation of this process. Do not distribute these notes 166 Prof. R.T. M’Closkey, UCLA OBJECTIVE: estimate the frequency response of the “unknown” linear system known periodic input (applied by engineer) periodic input (T=0.5) 2 known periodic output (measured by engineer) periodic output 2 1.5 1.5 1 1 Asymptotically Stable Linear System -unknown ODE- 0.5 0 0.5 0.5 0 0.5 1 1 1.5 1.5 2 1 0.5 0 seconds 0.5 2 1 1 0.5 calculate Fourier Series 10 10 2 10 Frequency response estimate 2 magnitude 10 Frequency response estimated at integer multiples of the fundamental frequency 10 3 1 10 4 0 10 1 2 10 10 3 10 magnitude magnitude 10 1 spectrum of periodic output 0 10 ck 1 0.5 calculate Fourier Series spectrum of periodic input 0 10 0 seconds output is periodic after the transient due to initial conditions as decayed (this is why asymptotic stability is important) ck ˜ ck 1 Hz 10 10 10 10 0 10 10 10 2 ck ˜ 1 2 3 4 0 10 1 2 10 10 3 10 Hz 3 0 10 1 2 10 10 3 10 Hz Do not distribute these notes 167 Prof. R.T. M’Closkey, UCLA ...
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