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Home , Periodic function

11/16/99 (T.F. Weiss)

Lecture #18: Continuous time periodic signals Outline:

• of periodic functions Motivation: • Examples of Fourier series — periodic impulse train

• Representation of continuous time, periodic signals in the • Fourier transforms of periodic functions — relation to domain Fourier series

• Periodic signals occur frequently — motion of planets and • Conclusions their satellites, of oscillators, electric power dis- tribution, beating of the heart, vibration of vocal chords, etc.

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Fourier series coefficients Fourier series of a periodic The coefficients of the Fourier series can be found as follows.   Periodic time function T/2 T/2 ∞ 1 −j2πnt/T 1  j2πkt/T  −j2πnt/T x(t) is a periodic time function with period T . x(t)e dt = X[k]e e dt, T −T/2 T −T/2 −∞ k= ∞ 1 T/2 − 1 T/2 − x(t) x(t)e j2πnt/T dt = X[k] ej2π(k n)t/T dt . − − T T/2 k=−∞ T T/2 The integral can be evaluated as follows. 1 T/2 − 1ifk = n, 0 T t ej2π(k n)t/T dt = T −T/2 0ifk = n. The set of exponential time functions are said to be an orthonor- Such a periodic function can be expanded in an infinite series of mal basis. exponential time functions called the Fourier series, ∞ j2πnt/T The coefficients are x(t)= X[n]e . n=−∞ 1 T/2 − X[n]= x(t)e j2πnt/T dt. T −T/2

3 4 Examples of Fourier series of periodic time functions Definition of line spectra, Periodic impulse train The fundamental frequency fo =1/T . The Fourier series co- efficients plotted as a function of n or nfo is called a Fourier The periodic impulse train is an important periodic time function spectrum. and we derive its Fourier series coefficients. s(t) X[n] ∞ 1 1 1 1 1 1 1 s(t)= δ(t − nT ). n=−∞ 0 T t 0 1234 n ⇐ # ⇐ The Fourier series coefficients are found as follows 0 fo 2fo 3fo 4fo f Frequency 1 T/2 − S[n]= s(t)e j2πnt/T dt,

DC T −T/2 ∞ 1 T/2 − = δ(t − nT )e j2πnt/T dt, − T T/2 n=−∞ T/2 Fundamental 3rd harmonic 4th harmonic 1 − 1 2nd harmonic = δ(t)e j2πnt/T dt = . T −T/2 T

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Periodic impulse train, cont’d The Fourier series coefficients are 1 S[n]= . T Periodic impulse train, cont’d The time function and spectrum are shown below. The Fourier series of the periodic impulse train is ∞ ∞ − 1 j2πnt/T s(t) S[n] s(t)= δ(t nT )= e . n=−∞ T n=−∞ 1/T 1 1 1 1 1 1 1 It is not obvious that the two expressions are equal. To investi- gate this, we define the partial sum of the Fourier series, s (t), 0 T t 0123n N N 1 j2πnt/T sN (t)= e , To summarize, the periodic impulse train can be represented by T n=−N its Fourier series, and investigate its behavior as N →∞. ∞ ∞ 1 s(t)= δ(t − nT )= ej2πnt/T . n=−∞ T n=−∞

7 8 Periodic impulse train, cont’d Periodic impulse train, cont’d The partial sum of the Fourier series is 1 sin(2N +1)πt/T sN (t) s (t)= . N N N 1 j2πnt/T 1 j2πt/T n 4 N =2 T sin πt/T sN (t)= e = e . T =1 T n=−N T n=−N 2 Note that this function is pe- riodic with period T , and We can use the summation formula for a finite −2 −112 t (Lecture 10) to sum this series, −2 2N +1 sN (nT )= . 10 −N N+1 N =5 T j2πt/T − j2πt/T T =1 1 e e 5 The first zero of sN (t)isat sN (t)= , T 1 − ej2πt/T T 1 ej(2N+1)πt/T − e−j(2N+1)πt/T t = . s (t)= , 2N +1 N jπt/T − −jπt/T T e e 20 N =10 Thus, as N →∞, each lobe 1 sin(2N +1)πt/T T =1 gets larger and narrower. To sN (t)= . 10 T sin πt/T determine if each lobe acts as an impulse, we need to find its area. 9 10

Periodic impulse train, cont’d of a periodic function Fourier transform of a periodic impulse train We have two expressions for a periodic impulse train, The area of each period of sN (t) is simply ∞ ∞ 1 j2πnt/T T/2 T/2 N s(t)= δ(t − nT )= e . 1 j2πnt/T T Area = sN (t) dt = e dt n=−∞ n=−∞ −T/2 −T/2 T n=−N The Fourier transform of each expression is N T/2 1 j2πnt/T ∞ ∞ = e dt − 1 n − S(f)= e j2πnTf = δ f − . n=−N T T/2 n=−∞ T n=−∞ T The integral is zero except when n = 0 when it equals T so that Therefore, the Fourier transform of a periodic impulse train in Area = 1. Thus, each lobe of sN (t) becomes: tall, height is time is a periodic impulse train in frequency. sN (nT )=(2N +1)/T ; narrow, width is 2T/(2N + 1); and its area is 1. Thus, the partial sum approaches an infinite impulse s(t) S(f) train of unit area, 1 1 1 1 1 1 1 ⇐⇒F ∞ ∞ 1/T 1 s(t)= δ(t − nT )= ej2πnt/T . 0 T t 1 T 0 f n=−∞ n=−∞ T

11 12 Relation of Fourier series spectrum to Fourier transform of Fourier transform of an arbitrary periodic function a periodic impulse train Representation of a periodic function An arbitrary periodic function can be generated by convolving a pulse, xT (t), that represents one period of the periodic function S s(t) S[n] (f) with a periodic impulse train, s(t)= n δ(t − nT ), 1/T ∞ ∞ 1 1 1 1 1 1 1 1/T x(t)=xT (t) ∗ s(t)=xT (t) ∗ δ(t − nT )= xT (t − nT ). 0 T t 0123n 1 n=−∞ n=−∞ 0 f T xT (t) s(t) x(t)

Therefore, the Fourier transform of the periodic impulse train has ∗ 11111= an impulse at the frequency of each Fourier series component 00t − T 0 T t − T T t and the area of the impulse equals the Fourier series coefficient.

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Fourier transform of a periodic function The Fourier transform of the periodic function is F x(t)=xT (t) ∗ s(t) ⇐⇒ X (f)=XT (f) ×S(f).   ∞ ∞ n 1 n XT n X (f)=X (f) × δ f − =  T  δ f − . T Fourier transform of a periodic function, cont’d T n=−∞ T n=−∞ T T An important conclusion is that the Fourier transform of a pe- riodic function consists of impulses in frequency at multiples xT (t) s(t) x(t) of the fundamental frequency. Thus, periodic continuous time ∗ 1 1111= functions can be represented by a countably infinite number of 00t − T 0 T t − T T t complex exponentials. XT (f) S(f) X (f) 1 × = T −1 1 ff0 f T T

15 16 Fourier series coefficients The Fourier transform of the periodic function is Fourier series of a square — generation of the square   ∞ n wave XT n X (f)=  T  δ f − . We will find the Fourier series of a square wave by finding the T T n=−∞ Fourier transform of one period. Recall that T/2 −j2πft xT (t) x(t) XT (f)= xT (t)e dt −T/2 11 Therefore, 0 −T T − T 0 T t n 0 t X T/2 4 4 T T 1 −j2πnt/T = xT (t)e dt = X[n]. T T −T/2 Therefore, for an arbitrary periodic continuous time function, the x(t)=x (t) ∗ s(t) Fourier transform consists of impulses (located at the harmonic T ) whose areas are the Fourier series coefficients.

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Fourier series of a square wave — Fourier transform of square wave Fourier series of a square wave — Fourier transform of one To obtain the Fourier transform of the square wave, we take the period of the square wave Fourier transform of one period of the square wave and multiply

F it by the Fourier transform of the periodic impulse train. x (t) ⇐⇒ X (f) ∞ T T T sin(πfT/2) 1 n X (f)=XT (f) ×S(f)= × δ f − , 2 πfT/2 T n=−∞ T T sin(πfT/2) ∞ X (f)= 1 sin(nπ/2) n T = δ f − . xT (t) 2 πfT/2 n=−∞ 2 nπ/2 T 1 T/2

XT (f) X −T T t (f) −2 2 4 4 f T T T/2 −9 −7 −5 −3 −1 1 3 5 7 9 0 f T T T T T T T T T T

19 20 Two-minute miniquiz solution

Problem 18-1 — Fourier series of the square wave Two-minute miniquiz problem

a) Since the Fourier transform of the square wave is Problem 18-1 — Fourier series of the square wave ∞ 1 sin(nπ/2) n X (f)= δ f − , n=−∞ 2 nπ/2 T the Fourier series coefficients are a) Determine the Fourier series coefficients of the square wave. 1 sin(nπ/2) X[n]= 2 nπ/2 b) From the Fourier series coefficients determine the average value of the square wave. x(t) 1 1 T/2 b) X[0] = x(t) dt =1/2. 1/2 T −T/2 − T 0 T t

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Fourier series of a square wave — synthesis of a square Fourier series of a square wave — synthesis of a square wave, cont’d wave To investigate the synthesis of the square wave, we consider the We can synthesize the square wave by adding complex exponen- partial sum of the Fourier series, tials weighted by their Fourier series coefficients, N ∞ ∞ 1 sin(nπ/2) j2πnt/T 1 sin(nπ/2) j2πnt/T x (t)= + cos(2πnt/T ). x(t)= X[n]e = e . N 2 nπ/2 n=−∞ n=−∞ 2 nπ/2 n=1 Note that the coefficients are even functions of n. Hence, we N=0 N=1 N=3 can rewrite the series as follows N=5 ∞ 1 1 sin(nπ/2) − x(t)= + ej2πnt/T + e j2πnt/T , 2 2 nπ/2 n=1 ∞ 1 sin(nπ/2) N=7 N=9 N=11 N=13 x(t)= + cos(2πnt/T ). 1 2 n=1 nπ/2 0.5 Note that all the even harmonics of x(t) are zero except for the term for n =0. −1 −0.5 0.5 1 t/T

21 22 Fourier series of a square wave — synthesis of a square wave, cont’d Demo illustrating the Gibbs’ phenomenon. Fourier series of a square wave — Gibbs’ phenomenon As harmonics are added to synthesize the square wave, the partial sum of the Fourier series converges to the square wave every- where except near the discontinuity where the partial sum takes on the value of 1/2. There are on either side of the discontinuity whose maximum over and undershoot approach 9% of the discontinuity independent of N. We can interpret the Gibbs’ phenomenon by examining the partial sum of the Fourier series as a filtering problem.

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Gibbs’ phenomenon interpretation as an ideal LPF of a Gibbs’ phenomenon — step response of an ideal LPF square wave The Gibbs’ phenomenon can be investigated further by examin- Truncation of the Fourier series of a square wave can be in- ing the step response of an ideal lowpass. terpreted in the as passing the square wave through an ideal lowpass filter that truncates the spectrum. That H(f) can be interpreted in the time domain as the of the x(t) xw(t) H(f) square wave with a sinc function. The response of the ideal low- − pass filter to the discontinuity of the square wave gives rise to W W f x(t) h(t) x (t) the oscillations. w ∗ = X (f) H(f) XN (f) × = t tt ff f Hence, x(t) h(t) x (t) N t sin(2πWt) ∗ = xw(t)=u(t)∗h(t)= h(τ) dτ where h(t)=2W . −∞ 2πWt tt t

25 26 Gibbs’ phenomenon — step response of an ideal LPF, Gibbs’ phenomenon — step response of an ideal LPF, cont’d cont’d We evaluate the integral The integral function, t sin y t sin(2πWτ) Si(t)= dy, xw(t)= 2W dτ, 0 y −∞ 2πWτ is plotted below. by changing variables y =2πWt which yields 2πWt sin y Si(t) xw(t)= dy. −∞ y 1.5 π/2 1 sin t This is closely related to a tabulated function called the sine 0.5 t integral function Si(t) defined as t −20 −10 10 20 -0.5 t sin y Si(t)= dy. -1 0 y −π/2 -1.5

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Conclusions Gibbs’ phenomenon — step response of an ideal LPF, Time Function Fourier Transform Fourier Series X (f) cont’d x(t) We express the step with the truncated spectrum in terms of the sine integral function as t X f 1 1 x(t) (f) X[n] xw(t)= + Si(2πWt), 2 π which is shown plotted below for several values of W . t f n

1 1 F 0.5 W=1 0.5 W=2 • Aperiodic, continuous functions of time ⇐⇒ aperiodic, con- tinuous functions of frequency. 1 1 W=3 W=4 F 0.5 0.5 • Periodic, continuous functions of time ⇐⇒ impulses whose −3−2−1 1 2 3 −3−2−1 1 2 3 areas are the Fourier series coefficients and located at dis- crete frequencies.

29 30 Joseph Fourier, continued Historical perspective Jean Baptiste Joseph Fourier (1768-1830) • Born of humble origins — his father was the town tailor, he was orphaned at age 9, and raised by a neighbor.

• Went to military school and discovered mathematics at age 13. Joseph Fourier was born in Aux- erre, France on March 21, 1768 • and died in Paris on May 4, Taught mathematics, rhetoric, philosophy, and history in a Benedictine school in his home town. 1830. This is one of two por- traits that has survived. • Became active in local politics and social causes. His political honesty got him in hot water — he was arrested first by the Robespierre regime and later by the enemy post-Robespierre regime. He was just barely saved from the guillotine twice.

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Joseph Fourier, continued

Joseph Fourier, continued

• Taught at the Ecole Polytechnique where he developed an excellent reputation as a lecturer and began publishing math- ematical research. He taught with Lagrange and Monge — • In 1802, Napoleon appointed Fourier as the prefect of Is`ere, a two eminent mathematicians. French department whose center was Grenoble. This was an important administrative/political position somewhat akin to the governor of a US state. Fourier always hoped to complete • When Napoleon Bonaparte was put in charge of a French ex- his tasks and return to a scholarly life, but he continued to pedition to Egypt, Fourier was chosen as a scientific member. take on important administrative posts throughout his life. He held political and administrative positions in Egypt and His mathematical and scientific studies were largely part-time proved to be an able administrator. When the French occu- efforts. pation of Egypt ended, Fourier returned to teaching mathe- matics in France.

33 34 Joseph Fourier, continued Joseph Fourier, continued Fourier’s work and that of those that followed him have had a profound effect on science and mathematics. Fourier transforms • In 1807, Fourier submitted a paper on the use of trigonomet- are common tools in many different fields of science. ric series to solve problems in heat conduction to the Institute 12 of France. It was reviewed by four famous French mathe- 50 10 maticians — Lagrange, Laplace, Lacroix, and Monge. Three 100 150 8

200 of the four referees voted to accept the paper, Lagrange was 6 250

opposed. Lagrange had been on one side of a raging con- 300 4

350 troversy on the representation of functions by trigonomet- 2 400

ric series. Lagrange did not believe that arbitrary functions 450 0 500 -2 could be expanded in trigonometric series as Fourier’s paper 550 claimed. Fourier’s original paper was never published, but in 100 200 300 400 500 1822 he published his work in a book The Analytical Theory Fourier Fourier transformed of Heat in 1822. Although he did not originate trigonometric series, nor did he determine the precise conditions for their The image of Fourier (left) is 582 by 582 pixels. The logarithm validity, he did use them to solve problems in heat conduction of the magnitude of the Fourier transform of this image is plotted thus illustrating their utility. on the right. 35 36