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4.3 Fourier Series

Denition 4.41.

Exp onentialF ourierseries : Let the (real or complex) signalr(t) be aperiodicsignal with periodT0. Suppose the followingDirichletconditions are satised: (a)r(t) is absolutely integrable over its period; i.e.,RT0

0jr(t)jdt <1.

(b) The n umberof maxima and minima of r(t) in each period is nite. (c) The n umberof discon tinuitiesof r(t) in each period is nite. Thenr(t) can be \expanded" into a linear combination of the complex exponential signalsej2(kf0)t1 k=1as ~r(t) =1X k=1c kej2(kf0)t=c0+1X k=1 c kej2(kf0)t+ckej2(kf0)t (37) where f 0=1T 0and c k=1T 0+T0Z r(t)ej2(kf0)tdt;(38) for somearbitrary.

We give some remarks here.

~r(t) =r(t);ifr(t) is continuous att r(t+)+r(t)2 ;ifr(t) is not continuous att Although ~r(t) may not be exactly the same asr(t), for the purpose of our class, it is sucient to simply treat them as being the same (to avoid having two dierent notations). Of course, we need to keep in mind that unexpected results may arise at the discontinuity points. The parameterin the limits of the integration (38) is arbitrary. It can be chosen to simplify computation of the integral. Some references simply writeck=1T 0R T

0r(t)ejk!0tdtto emphasize that we only need

to integrate over one period of the signal; the starting point is not important. 54
The coecientsckare called the (kth)Fourier (series) coecients of (the signal)r(t). These are, in general, complex numbers. c0=1T 0R T

0r(t)dt= average or DC value ofr(t)

The quantityf0=1T

0is called thefundamental frequencyof the

signalr(t). Thekth multiple of the fundamental frequency (for positive k's) is called thekthharmonic. ckej2(kf0)t+ckej2(kf0)t= thekthharmonic componentofr(t). k= 1)fundamental componentofr(t).

4.42.Being able to writer(t) =P1

k=1cnej2(kf0)tmeans we can easily nd the Fourier transform of any periodic function: r(t) =1X k=1c kej2(kf0)tF*)F1R(f) = The Fourier transform of any periodic function is simply a bunch of weighted delta functions occuring at multiples of the fundamental frequency f 0.

4.43.Formula (38) for nding the Fourier (series) coecients

c k=1T 0+T0Z r(t)ej2(kf0)tdt(39) is strikingly similar to formula (5) for nding the Fourier transform:

R(f) =1

Z 1 r(t)ej2ftdt:(40)

There are three main dierences.

We have spent quite some eort learning about the Fourier transform of a signal and its properties. It would be nice to have a way to reuse those concepts with Fourier series. Identifying the three dierences above lets us see their connection. 55

4.44.Getting the F ourierco ecientsf romthe F ouriertransform :

Step 1

Consider a restricted v ersionrT0(t) ofr(t) where we only considerr(t) for one period.1 6 4 6 4 D

PStep 2Find the F ouriertransfor mRT0(f) ofrT0(t)

Step 3The Fourier coecients are simply scaled samples of the

Fourier transform:

c k=1T

0RT0(kf0):

Example 4.45.

T rainof Impulses : Find the Fourier series expansion for the train of impulses (T0)(t) =1X n=1(tnT0) drawn in Figure 21. This innite train of equally-spaced -functions is usually denoted by the Cyrillic letter (shah). 1 P

1 1 1 1 1 1 1Figure 21: Train of impulses

56

4.46.The Fourier series derived in Example 4.44 gives an interesting

Fourier transform pair:

1 X n=1(tnT0) =1X k=11T

0ej2(kf0)tF*)F1(41)1

P

1 1 1 1 1 1 1

B B 4 B 4 B 4 B 4 B 4 B 4 B

4A special case whenT0= 1 is quite easy to remember:

1 X n=1(tn)F*)F11 X k=1(fk) (42) 1 P

1 1 1 1 1 1 1

B

1 1 1 1 1 1 1Once we remember (42), we can easily use the scaling properties of the

Fourier transform (21) and the delta function (18) to generalize the special case (42) back to (41): 1 X n=1(atn) =x(at)F*)F11jajXfa =1jaj1 X k=1fa k 1jaj1 X n=1 tna

F*)F11jajjaj1X

k=1(fka) 1 X n=1 tna

F*)F1jaj1X

k=1(fka)

At the end, we plug-ina=f0= 1=T0.

57
Example 4.47.Find the Fourier coecients of thesquare pulse periodic signal[6, p 57].1 F6 4 v 6 4 F6 4 PN P 5 4

B: the scaled

Fourier transformof

the restricted(one period) version of ݎ P. period 0 0c k=1T

0RT0(kf0) =1T

0 T 02 sinc 2T04 (f) f=kf0! 12 sinc k2 12 sink2 k 2 =sink2 k kk2sin k2 c k=sin(k2 )k 0 1 2 3 4 5 58

Remarks:

(a) Multiplication b ythis signal is equiv alentto a switc hing(ON-OFF) operation. (Same as periodically turning the switch on (letting another F6 4 v 6 4 PPI P I PF6

4(b)This signal can b eexpressed via a cosine function with the same p eriod:

r(t) = 1[cos(2f0t)0] =1;cos(2f0t)0;

0;otherwise.

1 F6 4 v 6 4 PN P F6

4(c)A duty cycleis the percentage of one period in which a signal is

\active". Here, duty cycle = proportion of the \ON" time = widthperiod

In this example, the duty cycle is

T0=2T

0= 50%. When the duty cycle is1n

thenth harmonic (cn) along with its nonzero multiples are suppressed. 59

4.48.Parseval's Identity:Pr=D

jr(t)j2E 1T 0R T

0jr(t)j2dt=1P

k=1jckj2. 4.49. F ourierseries expansion for real v aluedfunction : Suppose r(t) in the previous section is real-valued; that isr=r. Then, we have c k=ckand we provide here three alternative ways to represent the Fourier series expansion: ~r(t) =1X k=1c nej2kf0t=c0+1X k=1 ckej2kf0t+ckej2kf0t(43a) =c0+1X k=1(akcos(2kf0t)) +1X k=1(bksin(2kf0t)) (43b) =c0+ 21X k=1jckjcos(2kf0t+\ck) (43c) where the corresponding coecients are obtained from c k=1T 0+T0Z r(t)ej2kf0tdt=12 (akjbk) (44) a k= 2Refckg=2T 0Z T

0r(t)cos(2kf0t)dt(45)

b k=2Imfckg=2T 0Z T

0r(t)sin(2kf0t)dt(46)

2jckj=qa

2k+b2k(47)

\ck=arctanbka k (48) c 0=a02 (49)

The Parseval's identity can then be expressed as

P r=D jr(t)j2E =1T 0Z T

0jr(t)j2dt=1X

k=1jckj2=c20+ 21X k=1jckj2 60

4.50.To go from (43a) to (43b) and (43c), note that when we replaceck

byck, we have c kej2kf0t+ckej2kf0t=ckej2kf0t+ckej2kf0t =ckej2kf0t+ckej2kf0t = 2Re ckej2kf0t: Expression (43c) then follows directly from the phasor concept: Re ckej2kf0t=jckjcos(2kf0t+\ck):

To get (43b), substituteckby Refckg+jImfckg

andej2kf0tby cos(2kf0t) +jsin(2kf0t). Example 4.51.For the train of impulses in Example 4.44, 1 X n=1(tn) =1X k=11T

0ej2(kf0)t=1T

0+2T 01 X k=1cosk!0t(50) Example 4.52.For the rectangular pulse train in Example 4.46,1 F6 4 v 6 4 P

Fourier series expansion:

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