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Unformatted text preview: EXAMPLE 16.1 Find the Fourier series :or the periodic voltage shown in Fig. 16.5. SOLUTION When using Eqs. (16.3)—(l6.5) to ﬁnd av, ak, and bk, we may choose
the value of to. For the periodic voltage of Fig. 16.5, the best choice
for to IS zero. Any other choice makes the required integrations more Figure 16.5 The periodic voltage for Example 16.1. cumbersome. The expression for 110) between 0 and T is v0) = r. Vm The equation for an is T
_ .1. V \T
1 T V l A re (A a 2 n
m
au= —— tdt=—Vm. /}y”€féL I
TNT) 2  M...—
7// M5 3 av “ T This is clearly the average value of the waveform in Fig. 16.5. 1 ,
The equation for thMSef'dﬁ is ['k) L” , , ( ’ X) “.1: 2 T V
ak=——f ‘(l)tcoskw0tdt
T O T 2Vm 1 t .
= —— COS + ‘66 Sln T O l
= 2% [—(cos 277k — 1)] = 0 for all k. T2 mg 3/,
bk.
The equation for Wis
2 T V
bk=? 0 (Tﬂ)tsinkwotdt
T
2Vm 1 s' k t t c k t
=—— —— in a) —— 0s w
T2 kzwg 0 kwo 0
0
2Vm T
= 7 O———coserk
T~ kwo
—Vm r.
/ we)" st t 5 A “ka
The Fourier series forut is J ’ ‘1  v , .
() K'! _'L EﬂsanvLD V Vm °° 1
110‘) = 3r: — 71 ; ; sin moot
Vm Vm . Vm . Vm A
=—————sma)0t———Sin2th———Sln3wot—~
2 7t 27: 3.7 The effect of each type of symmetry on the Fourier coefﬁcients is
discussed in the following sections. EvenFunction Symmetry ..———' A function is deﬁned as even if
 ’
fCt) = f(—t). (16.13) Functions that satisfy Eq. (16.13) are said to be even because polyno
mial functions with only even exponents possess this characteristic.
For even periodic functions, the equations for the Fourier coefﬁcients reduce to 3
1
l (16.14)
‘ @ T/2
ak = / f(t) cos kwot dt, (16.15)
T 0
J (16.16) bk ; 0, for all k. Note that all the b coefﬁcients are zero if the periodic function is even.
Figure 16.6 illustrates an even periodic function. The derivations of Eqs. (l6.l4)—(16.16) follow directly from Eqs. (16.3)—(16.5). In each
derivation, w select to = ——T 2 nd then break the interval of inte gration into the range from ——T / 2 to 0 and 0 to T / 2, or 1 T/Z
up? / ftt)dt —T/2 1 T/Z T 0 f(t)dt. (16.17) 0
Figure 16.5 An even peritzudic function, f(t) = ___ l f t d;
f(r) T —T/2 f() + 1 Now we change the variable of integration in the ﬁrst integral on
the right—hand side of Eq. (16.17). Speciﬁcally, we let 1? = —x and 1 t . " ’ ‘ note that f(t) = f(—x) = f(x) because the function is even. We also
\ ' " observe thatx = T/2 when t = ——T/2 and dt = —dx. Then T/2 0 0
1;: T ] f(r)dz= f(x)(—dx)=/ f(x)dx, (16.18)
( —T/2 r/2 0 from 0 to T/2; therefore Eq. (16.17) is the same as Eq. (16.14). The ' which shows that the integration from —T / 2 to O is identical to that
\\ derivation of Eq. (16.15) proceeds along similar lines. Here, 2 O
ak = — / f(t) cos kwot dt
T 4/2 2 T/2
+ ‘T‘ f f0) COS kwot dt, (16.19)
0 (( but ' 0 0
i/ﬂjm cos kwot dt =/ f(x) cos(—kwox)(—dx)
. —T/2 T/ J / 7/2 = 1‘ f(x) cos kwox dx. (16.20) As before, the integration from —T / 2 to 0 is identical to that from 0
to T/2. Co:;1binii':__;Eq. (16.20) with Eq. (16.19) yields Eq. (16.15).
Aﬂthe b coefﬁcientsgrgazgro when f (t) is an even eriodic func
ti_o_n, because the integration fromd—T/ 2 to 0 is the exact negative of
the integration from O to T /2; that is. a} o o
/ f(t) sin kwot dt = f(x) sin(—kwox)(—dx)
_T/2 T/2 g = f(x) sin kwox dx. (16.21) When we use Eqs. (16.14) and (16.15) to ﬁnd the Fourier coefﬁcients,
the interval of integration must be between 0 and T / 2. OddFunction Symmetry A function is deﬁned as odd if f0) = —f(t) Functions that satisfy Eq. (16.22) are said to be odd because polyno
mial functions with only odd exponents have this characteristic. The
expressions for the Fourier coefﬁcients are (16.22) au 2 0;
ak = 0, fOr all 4o {/2
T 0 (16.24) bk = f(t) Sin kwot dt. (16.25) Note that all the a coefﬁcients are zero if the eriodic function is odd.
Figure 16.7 shows an odd periodic function. We use the same process to derive Eqs. (16.23)—(16.25) that we used to derive Eqa‘. (l6.l4)—(16.l6). We leave the derivations to you J _ in Problem 16.4. The evenness, or oddness, of a periodic function can be destroyed
by shifting the? function along the time axis. In other words, the judi
cious choice of where t = 0 may give a periodic function even or odd symmetry, For example, the triangular function shown in Fig. 16.8(a) — f {—r). fir): Figure 16.7 An odd periodic function f(t) = . _/
\‘v’ ,',[ \v"
(a)
f(t)
x” \
I; \ _ _f_\. _.. K. _+
\V’: —A[— \V/ \
lb) (a) Figure 16.8 How the choice of where t = O can
make a periodic function even, odd, or neither.
(a) A periodic triangular wave that is neither
even nor odd. (b) The triangular wave of (a)
made even by shifting the function along the t
axis. (cT'fhe triangular wave of (a) made odd by
shifting the function along the z axis. ’— is neither even nor odd. However, we can make the function even, as
shown in Fig. 16.8(b), or odd, as shown in Fig. 16.8(c). HalfWave Symmetry A periodic function possesses halfwave symmetry if it satisﬁes the
constraint ftr) = —f(r — T/2). (16.25) gasm has li_r1:'—=.'~ ..‘. I: M mine— :r. . .r :a idﬁillli.'_1i to
_——I——’l'——_——'—‘——I__._—‘_
.n..1l lungnun Fur f.‘2.lllll‘i;’. ’:1.' I'..ll'.'L";l\')I]$ shtgxsn nil:31. in 7 and 10.6 have haltwave symmetry, whereas those H1 Figs. io.5 and
16.6 do not. Pint3 lli.:t 21.;Jllaaix'c ‘j. minen".
:0. ' If a periodic function has halfwave symmetry, both ak and bk
are zero for even values of k. Moreover, av also is zero because the
average value of a function with half—wave symmetry is zero. The expressions for the Fourier coefﬁcients are 5.4 (in: :i (unt'litin ;'Ir' where. av = 0, (16.27)
ak = O, for k even, (16.28)
772 _'__
ak = T f(t) cos kwot dt, for k odd, (16.29)
0
bk = O, for k even, (16.30)
4 7/2 X‘—
bk = ? f(z) sin kwot dt, for k odd. (16.31)
0 We derive Eqs. (16.27)—(l6.31) by starting with Eqs. (16.3)—(16.5)
and choosing the interval of integration as —T/2 to T /2. We then
divide this range into the intervals —T/2 to O and O to T/2. For
example, the derivation for ak is 2 F0+T
ak=——/ COS/(mot dl‘
T to T/Z
/ f(t) cos kwot dt 2
T T/2
2 0
= — / f(t) cos kwot dt
T 4/2 T ,"2 2
+ — f(t) cos kwot dt. 16.32
T O ( ) Now we change a variable in the ﬁrst integral on the righthand side
of Eq. (16.32). Speciﬁcally, we let tEx—Tﬂ.
Then x=T/2, whent=0,
x=0, whent=—T/2,
dz? =dx. We rewrite the ﬁrst integral as o
/ f(t) cos kwot dt —T/2 T/2 = f(x — T/2) cos kw0(x — T/Z) dx. (16.33)
0 Note that
cos kwo(x — T/Z) = cos(kw0x — kn) = cos k7: cos kwox and that, by hypothesis, f(x — T/Z) = *f(x)
Therefore Eq. (16.33) becomes 0
/ f0) cos kwot dt —T/2 T/2 . .
=] [—f(x)] cos kn cos kwox dx. (16.34)
0 Incorporating Eq. (16.34) into Eq. (16.32) gives 2 T/2 ak = —f(1 — cos kn) 0 f(t) cos kwot dt. (16.35) But cos k7: is 1 when k is even and —1 when k is odd. Therefore
Eq. (16.35) generates Eqs. (16.28) and (16.29). We leave it to you to verify that this same process can be used to
derive Eqs. (1630) and (16.31) (sec Problem 16.5). We summarize our observations by noting that the Fourier series ) fmiﬁﬁmWﬂlb
representation of a penOdlC funCtlon With half«wave symmetry has A
zero average, or dc, value and contains only odd harmonics. <4 t' Cecivw ‘y‘y’ “Va 54 Ck fungigym lﬁcrm ("E . I I I I
I r I Brﬂneces‘gauly' p ,
(Sjl/“wAe/h7l /§ KM; I  ‘ _ 1
’6‘ «av cm L « a? QWCGl wt\’)lt J
v x V V /
WV) The, 3 fvixmLuM m 7332, an? if” W" Cb Vt?“ i” Lt} flu":
QuarterWave Symmetry 5y Mum (1'7. arterwave symmetry describes a periodic function that
 = :':l m u Ina"m wmrnetg about the mid—
:u The function illustrated . .g ‘ ._ _ J 1.2;. about the midpoint ofthe
positive and negative half—cycles. The function in Fig. 16.9(b) does
not have quarterwave symmetry, although it does have halfwave
symmetry. t
A periodic function that lg; quarterwave symmetry can always be ,4 )4 Ag;
( m__a_d_e either even or odd by the propercwe point where t = 0. “ a b ﬂ L, (
For example, the function shown in Fig. 16.9(a) is odd and can be V made even by shifting the function T /4 units either right or left along
the 1 axis. However, the function in Fig. 16.9(b) can never be made
fit) either even or odd. To take advantage of quarterwave symmetry in
the calculation of the Fourier coefﬁcients, you must choose the point D
it here .' : !.I tn make the function either even or odd. 3! '?':° T:Ui!...!1<.‘i' a ma e even,
#
’— 1 The erm qu
1... .!. 1‘. '\ :_;Ii‘ . h. i n th ." _'.' :‘ltih. 1.1!. . \‘;J".\ \"I'I‘. T T/Z (a) en T/2 :I. : '1."_ because of halfwave symmetry, . «4 .x,;_ = I21 1' )r k even, because of halfwave symmetry,
.‘5 = g / fa) cos kwoz dt, for k odd, i’ . . (bl Figure 16.3 (a) A function that has quarterwave
symmetry. (b) A function that does not have
quarterwave symmetry. 91; z 41‘. fur all k‘, because the function is even. (16.36) Equations (16.36) result from the function’s quarterwave symmetry
in addition to its being even. Recall that quarterwave symmetry is
superimposed on half—wave symmetry, so we can eliminate av and ak
for k even. Comparing the expression for ak, k odd, in Eqs. (16.36)
with Eq. (16.29) shows that combining quarterwave symmetry with
evenness allows the shortening of the range of integration from 0 to T/2 to 0 to T/4. We leave the derivation of Eqs. (16.36) to you in
Problem 16.6. I the quarter wave symme ric :.:n.=.I=.:. a. _: ..: .. . av = 0, because the function 1 hid.
ak = 0, for all k, because the i'n:».:::..\n w (bk : O, for k even, because (if T/4
bk = — f f (t) sin kwot dt, for k odd.
'\ O
A» can H N cram (Ude 7 ‘ Equations (16.37) are a direct consequence of quarterwave symmetry
Quﬁﬁ{ﬂt (AME [Mb TILYo and oddness. Again, quarterwave symmetry allows the shortening of the interval of integration from 0 to T/2 to 0 to T/4. We leave the
derivation of Eqs. (16.37) to you in Problem 16.7. Example 16.2 shows how to use symmetry to simplify the task of
ﬁnding the Fourier coefﬁcients. EXAMPLE 16.2 Find the Fourier series representation for the current waveform shown in Fig. 16.10. fit) I
S 0 L U T l 0 N .X __ '. f _‘_,___ \ _ _,r. ‘.. _
—— —T :\ f ’ 'I" ‘ ' \L E" ? '
We begin by looking for degrees of symmetry in the waveform. We r’ ﬁnd that the function is g‘édd and, in addition, h halfwave _a_n_d
quarterwave symmetry. Because the function is‘ all the a"'c'o
efﬁcients are zero; that is, av = O and ak = 0 for all k. Because the Figure 16.10 The periodic waveform for Exam
function has halfwave symmetry, bk = O for even values of k. Be— pie 16'2 cause the function ha; quarterwave symmetry, the expression for bk
for odd values of k is "— 8 T/4
bk = —/ i(t) sin kwot dl‘.
T 0
In the interval 0 5 t 5 T /4, the expression for i (t) is
. m
t = —t.
t( ) T
Thus W4 41 ~
bk = if —mt sin kwot dt
T 0 T . T/4
321m sm kwot ~ t cos kwot
_ T2 k2wg kwo 0
81 , k7: . '
= 7:21:12 3m 7 (k 18 odd). , The Fourier series representation of i(t) is 81m °° 1 . mt .
' =— ——s1n—smna)t
1m n2 2 n2 2 o
n=l,3,5.
81 1 l 1
m .  _ ‘ _ — . . . '
= —2— (Sm wot — 5 SID 3wot + 25 sm Swot 49 sm 76001 + > . 15.4 AN ALTERNATIVE TRIGONOMETRIC FORM
OF THE FOURIER SERIES In circuit applications of the Fourier series, we combine the cosine and sine terms in the series into a single term for convenience. Doing so allows the representation of each harmonic of v(t) or i(t) as a single phasor quantity. The cosine and sine terms may be merged in 00 either a cosine expression or a sine expression. Because we chose  v a at; 2: (an (OS Via/01' “1
the cosine format in the phasor method of analysis (see Chapter 9), 9 V '
we choose the cosine expression here for the alternative form of the
series. Thus we write the Fourier series in Eq. (16.2) as f(t) = a, + Z A, cos(nw0t — H), (16.38)
n=l where An and 6,, are deﬁned by the complex quantity (16.39) We derive 1H.)  '.'€ '1‘ .!':_i 5“, :.:>ilfg. !'J'I'.". phiu r tutmod to add
the cosine and sine terms in Eq. (16.2). We begin by expressing the
sine functions as cosine functions; that is, we rewrite Eq. (16.2) as 00
f(r) = a, + Z a, cos nwot + b, cos(nw0t — 90°). (16.40) n=1
Adding the terms under the summation sign by using phasors gives ﬂan cos moot} =anﬂ (16.41) and $91!)” cos(nwot — 900)} = b,[—90° = —jbn. {1642) 4‘» “5"0‘0‘5 4' 6” 5’“ “"56 Then ; I a;"":[ [L n (as "wat
W 4" 1 '5
*6»
fP{a,l cos nwot + b,l cosmth — 90°)} = an — an A” 4' 9’», 5,5 meat] remit—.61 . m , I
 v ' ,5
:AnL—ﬂ. (1543) Kg“), 9": 4‘" an an
Wh we inverse—transform Eq. (16.43), we get '
en '7 a, Cosnwl’ 1' 5n 5M "Wof'
an cos moot + b” cos(nw0t — 90°): CP"1{A,, [—Bn} : Aﬁicosgn as nwof
= An cos(na)ot — 9"). (16.44) + 9:“ 9” S‘ ” nw‘ij
Substituting Eq. (16.44) into Eq. (16.40) yields (16.38). Equa : An COS ( “moi _ 9" ‘)
tion (16.43) corresponds to Eq. (16.39). If the periodic function is w either even or odd, A” reduces to either an (even) or 19,, (odd), and 2 foe) : Q + 2 A“ £05 (n “gt_ 1%
9,, is either 0° (even) or 90° (odd). I I I V
The derivation of the alternative form of the Fourier series for a given periodic function is illustrated in Example 16.3. E X A M P L E 1 6 .3
a) Derive the expressions for ak and bk for the periodic function shown in Fig. 16.11. b) Write the ﬁrst four terms of the Fourier series representation of
v(t) using the format of Eq. (16.3%). SOLUTION a) The voltage v(t) is neither even nor odd, nor does it have half
wave symmetry. Therefore we use Eqs. (1614) and (16.5) to ﬁnd
ak and bk. Choosing to as zero, we obtain 2 r/4 T ak = — / Vm cos kwot dt + (0) cos kwot dt
T 0 T 4
/
_2Vm sinkwoflT/4_ Vm . kn
_ T kwo $0 —kﬂ $111— and 2 nT/4
Vm sin kwot dt U4
0 > 21',” I  cos kwot
kwo b) The average value of 110) is Vm(T/4) Vm
“"2 i T T' The values of ak — jb;c for k = 1, 2, and 3 are ‘V V 2v
A : ai“}b1=—m 'l=f m —45°,
1 7t 71 IT
.Vm Vm 0
A21 az—Jb2=0“1?7L—‘90’
—Vm Vm Jivm
, ._ =———— '—= —1350.
A3”? M 371' 371' 3:1 L— Thus the ﬁrst four terms in the Fourier series representation of v(t) are L’Vlﬁzcar V1,“th
V ‘ 2V ‘ V
v(t) = l +@ cos(a)0t — 45°) [email protected](2wot — 90°)
4 7: 71V
ﬁvm
03 t—135°+~~.
+ ———737rﬂ c s( we ) \I 511x“ 5T/4 3T/2 7T/4 2T 0 TM 772 3T/4 T Figure 16.11 The periodic function for Exam
ple 16.3. 15.5 AN APPLICATION Now we illustrate how to use a Fourier series re Lesentation of a
pgﬁgﬂigﬁmmmifu. action to ﬁnd the steadvstate response of a
WHe RC circuit shown in Fig. 16.12(a) will provide
our example. The circuit is energized with the periodic square—wave
voltage shown in Fig. 16. 12(b). The voltage across the capacitor is {he
dearEd response, or output, signal. The ﬁrst step in ﬁnding the steady—state response is to represent the
periodic excitation source with its Fourier series. After noting that the
source has o_dd, halfwave, and .; thil'T'JZ " ' ‘
the Fourier coefﬁcients reduce I_. .....
values:
9/ \ L‘. 531)". [ilci 8 T/4
bk 2 —— / Vm sin kwot dt T 0
4V = —’1 (k is odd). (16.45)
Ifk Then the Fourier series representation of 118 is
4V 1
m ~ sin moot. (16.46) (b) Figure 16.12 An RC circuit excited by a periodic
voltage. (a) The RC series circuit. (b) The square wave voltage. C V ~.——(/
. . . . (/gK) V ~ _——S——’ C Writing the series in expanded form, we have 17“) _; [ S J ' g {u .CLg ﬂ E
4V 4Vm . g t/ .( 2.1)
=—msinwt+——sm3wt ' g J
vg 7r 0 37: 0 ,5 a (d w) / —&—'gzc
4V 4V . 0 H'J
+ ——’“ sin 5th + J sin 76001? + . . . (16.47)
571 77r The voltage source expressed by Eq. (16.47) is the eguivalent of
inﬁnitel many seriesconnected sinusoidal smitten. each source hav
ing its own amplitude and frequency. ":a_. ; ’ "23117: i=1 :' suth
source to the output voltage, we use tl';.  For any one of the sinusoidal sources, the phasordomain expres
sion for the output voltage is ‘s...i'. Vs = ——.—. (16.48)
ll— JwRC V0 All the voltage sources are expressed as $12? functions, so we interpret
a phasor in terms of the sine instead of the cosine. In other words,
when we go from the phasor domain back to the time domain, we
simply write the timedomain expressions as sin(a)t + 9) instead of cos(a)r + 9). a The phasor output voltage owing to the fundamental frequency of
the smusordal source 15 (4Vm/7flﬂ
V = ————. .
"1 1+jcuoRC (1649)
Writing V01 in polar form gives
(7:;
(4V ) —ﬂ1
V01: _.___’"_L;, (15.50)
7r,/1+ ng2C2 *
where
ﬁ1=tan_1w0RC. (15.51) From Eq. (16.50), the timedomain expression for the fundamental
frequency component of 1),, is 4 Vm —————— sin(wot — ,31).
7TV/1 + cugRZC2 We derive the third—harmonic component of the output voltage in a
Similar manner. The thirdharmonic phasor voltage is (Wm/mg):
" 1+ j3woRC 4Vm
‘ ﬂ—ﬂ, 3m/l + 9w5R2C2 v01 = (16.52) V03 (16.53) where
53 =tan_13woRC. (15.54) . The timedomain expression for the thirdharmonic output voltage is 4Vm v03 = —————
3n,/1+ 9w5R2C2 Hence the expression for the kth—harmonic component of the output sin(3w0t —— 133). (16.55) voltage is 2
4Vm , _
vok = ————~——— sin(ka)ot — M) (k 13 odd), (15.56)
kin/1+ kzngzC2
where
at =tan‘1kw0RC (k is odd). (15.57) We now write down the Fourier series representation of the output
voltage: — ﬂ 4Vm I _ n
W = Z 7f n: y ny/l + (nwoRC)‘ The derivation of Eq. (16.58) was not difﬁcult. But, although we
have an analytic expression for the steadystate output, what 1100)
looks like is not immediately apparent from Eq. (16.58). As we men—
tioned earlier, this shortcoming is a problem with the Fourier series
approach. Equation (16.58) is not useless, however, because it gives
some feel for the steadystate waveform of 1200), if we focus on the
frequency response of the circuit. For example, if C is large, 1 /anC
is small for the higher order harmonics. Thus the capacitor short cir
“19 tilzitr':‘::.:1:_:::i.:. ' :Faa‘ :reforrn, and the
" ..._.;i:_11I“.';:. compared to
u'ics. Equation (lode, rcllects this condition oo
13 _:"‘_1.r“:n_‘r;'..: _: .':‘‘ .,\ .5... . '.
1; uidt, luf 144.0» 00 4Vm i .
m — Sin 1 — 90°
v0 eroRC "22:5,". n2 (MOO )
—4v °° 1
w m Z —2 cos moot. (16.59)
IrwoRC “:1” n Equation (1659) shows that the amplitude of the harmonic in the out " ‘ ' 5 " rupared with 1/ n for the input harmon
ica ff I'.‘ :5 ~  ..::;_ ‘. '. 1511;. 1:
then to a ﬁrst approximation I) a) (15 60)
% CO I . 00) 7rco0RC S 0 ’ and Fourier analvsis tells us that the S uarewave ' . e ormed iant. Now let’s see what happens as C —> O. The circuit shows that 110
and pg are the same when C = 0, because the capacitive branch looks
like an open circuit at all frequencies. Equation (16.58) predicts the
same result because, as C —> 0, 00
4V 1 ,
120:?“ E Esmnwot. (16.61)
n=l,3,5,... But Eq. (16.61) is identical to Eq. (16.46), and therefore v0 ——> vg as
C —> O. Thus Eq. (16.58) has proven useful because it enabled us to predic:
‘ that the output will be a highly distorted replica of the input waveform if C is large, and a reasonable replica if C is small. In Chapter 13, we
looked at the dgggion between the input and output in terms of how
much memory the system weighting functi_on_had. In the freguency . ‘—_—‘._—l—"'——__‘ . —" .
domain, we look at the distortion between the steadystate input anc
output in terms of '1 ' altered as they are 2 7.51.3 '1I‘_"E‘."1 . I ‘~ 19.! phase of the harmonics are
c :ircuit. When the network signiﬁcantly alters iiic' grim: relationships among the
harmonics at the output relative to that at the input, the output is 2': ,\
' : " inf)“: Thu: :'r:xl._l::rit:'I domain, Wt:
" ‘WJ P .t__._.l_ _ ac dtstwiwvt / For Lire circa: here, arripiimde
amplitudes of the input harmonics _... ..t. . . plitudes of the output harmonics decrease as ' aecause the
reas the am 1 . 1
n x/l + (nwoRC)2' This circuit also exhibits phase distortion because the phase angle 0:?
each input harmonic is zero, whereas that of the nth harmonic in the
output signal is —— tan“1 nwoRC. M An Application of the Direct approach to the
SteadyState Response f“ M For the simple RC circuit shown in Fig. 16.l2(a). we can derive
the expression for the steadystate response without resorting to the
Fourier series representation of the excitation function. Doing this
extra analysis here adds to our understanding of the Fourier series
approach. To ﬁnd the steady—state expression for no by straightforward circuit
analysis, we reason as follows. The squarewave excitation function
alternates between charging =he LT:3;‘.:ictmr +Vm and —Vm. Af
ter the circuit reaches arr: 'i‘.'<I:1re ranInuit]. Cl'H'i alternate chargiri—g
Worries periodic. We know from the analysrs of the single time
constant RC circuit (Chapter 7) that the response to abrupt changes in
the driving voltage is exponential. Thus the steadystate waveform of
the voltage across the capacitor in the circuit shown in Fig. l6.12(a)
is as shown in Fig. 16.13. The analytic expressions for 1100‘) in the time intervals 0 5 t g T / 2
and T/2 51‘s T are Toward — V," Toward — Vm
v0 = Vm + (V: ~ nae—“RC, 0 :2 s T/2, (18.62)
Figure 16.13 The steadystate waveform of 12,, for ' _[1_(T mn/RC
the circui in Fig. 16.12(a). v0 = —m + (V2 + lee /‘ , T/2 S t s T. (15.83) t ~(t~'{)/rzc
~zvmlp e i We derive Eqs. (16.62) and (16.63) by using the methods of Chapter 7,
as summarized by Eq. (7.60). We obtain the values of V1 and V2 by
noting from Eq. (16.62) that V2 = Vm + (V1 — tamWm,  (16.64)
and from Eq. (16.63) that
V1: Vm + (V; + nae“MC. (16.65)
Solving Eqs. (16.64) and (16.65) for V1 and V2 yields
Vm(1_ e—T/ZRC)
V = —V = ——————— 16.£‘6
Substituting Eq. (16.66) into Eqs. (16.62) and (16.63) gives
2V", _ .
U0=Vm— W8 t/RC, f
and
v — —V + ——2Y’”———e‘[’—(T/2)1/RC r/2 < t < T (16 468)
0" m 1+e—T/2RC ’ — — ' ' Equations (16.67) and (16.68) indicate that'vo(t) has halfwave sym
metry and that therefore the average value of no is .2339. This result
agrees with the Fourier series solution for the steadystate response—
namely, that because the excitation function has no zero frequency
cmt, the response can have no such component. Equatian
(16.67) and (16.68) also show the effect of changing the size of the
capacitor. If C is small, the exponential functions quickly vam'sh,
v0 = Vm between 0 and T / 2, and 110 = —Vm between T /2 and T. In
other words, no ——> pg as C —> 0. If C is large, the output waveformbe
comes triangular in shape, as Fig. 16.14 shows. Note tha@$
we may approximate the exponential terms e"/RC and e‘["(T/2)l/RC
by the linear terms 1 — (t/RC) and 1 — [   ' "‘ tively. Equation (16.59) gives the waveform. M Figure 16.14 summarizes the results. The dashed line in Fig. 16.14
is the input voltage, the solid colored line depicts the output voltage
when C is small, and the solid black line depicts the output voltage
when C is large. Finally, we verify that the steady—state response of Eqs. (16.67)
and (16.68) is equivalent to the Fourier series solution in Eq. (16.58).
To do so we simply derive the Fourier series representation of the (x. Figure 16.14 The effect of capacitor size on the
steadystate response. periodic function described by Eqs. (16.67) and (16.68). We have
already noted that the periodic voltage response has halfwave sym
metry. Therefore the Fourier series contains only odd harmonics. For k odd, 4 T/2 2Vm8—t/RC
Gk = fI/O (Vm — COS [C6001 dt
_ —8RCV,,,
‘ T[1+(kwoRC)2] 4 772 2Vme"/RC ..
bk 2 ]—‘V/Ov (Vm "‘ Sin kaI (it 4v,,, _ Skwova3C2 _ krr T[1+(kaRC)2] (k is odd), (16.69) (k is odd). (16.70) To show that the results obtained from Eqs. (16.69) and (16.70) are
consistent with Eq. (16.58), we must prove that . 4v 1
ag+b2—— —’"— (16.71) 'k’ kn \/1+(kw0RC)2’ and that a_k = —ka)oRC. (16.72) bk We leave you to verify Eqs. (16.69)—(16.72) in Problems 16.19 and
16.20. Equations (16.71) and (16.72) are used with Eqs. (16.38) and
(16.39) to derive the Fourier series expression in Eq. (16.58); we leave
the details to you in Problem 16.21. With this illustrative circuit, we showed how to use the Fourier se
ries in conjunction with the p’ri‘ngiple of superposition to obtain the
steadystate response to _a periodic driving function. Again, the prin
cipal shortcoming of the Fourier series approach is the difﬁculty of
ascertaining the waveform of the response. However, by thinking in
terms of a Circuit’s frequency response, we can deduce a reasonable
approximation of the steadystate response by using a ﬁnite number
of appropriate terms in the Fourier series representation. (See Prob—
lems 16.25 and 16.27.) 3 "5 11‘7me E—FK'E‘GU‘ETW‘rv—TmT — —— — PERIODIC FUNCTIONS If we have the Fourier series representation of the voltage and current
at a pair of terminals in a linear 1urnped~parameter circuit, we can
easily express the average power at the terminals as a function of the
harmonic voltages and currents. Using the trigonometric form of the
Fourier series expressed in Eq. (1638), we write the periodic voltage
and current at the terminals of a network as 00
v = VdC + Z V” COS(nw0t — elm) n=l (16.73) 00
i = 1dc + Z 1,, cos(nw0t — 9,"). (16.74) n=1 The notation used in Eqs. (16.73) and (16.74) is deﬁned as follows: Vdc = the amplitude of the _dc voltage component,
V" = the amplitude of the nthharmonic voltage,
9U” = the phase angle of the nth~harmonic voltage.
[dc = the amplitude of the dc current component,
1,, z: the amplitude of the nthharmonic current,
em = the phase angle of the nthharmonic current. We assume that the current reference is in the direction of the
reference voltage drop across the terminals (using the passive sign
convention), so that the instantaneous power at the terminals is vz. The average ower is 1 t0+T 1 t0+T
=?/ pdt=—T—/ uidt.
to to To ﬁnd the expression for the average power, we substitute Eqs. (16.73) and (16.74) into Eq. (16.75) and integrate. At ﬁrst glance,
tit}; appear: to be a formidable task. because the product vi requires (16.75) uI_'.'.' E“ u  _.I_;:" t tan: tr..'*.:'.:=_ "l‘.!'.'i_‘ . (Fir1;" 1:1. Uhnn _1J'r." 2.5:; CI'i'1_)'::L'i' ‘\!i._::.' .mt': _'. =_. =. '' uen A revrew or Eqs. (lam—(10.10) should Nuance ,cu oi the
validity of this observation. Therefore Eq. (16.75) reduces to [0+T
/ VnIn cos(nwor — 9m) P = ——Vdcldct[ +2: ? r
0 n=1 0
X cos(nwot — 6)") dt. (16.76)
Now, using the trigonometric identity
1 1
cos oz cos )3 = E cos(a — 6) + E cos(oz + )3),
we simplify Eq. (16.76) to (GMT/m1
1 00 ann [gelT M
P = VacIda + T 2 [costevn — an)
+ cos(2nw0t — 6W, — 610] dt. (16.77) 5":1223 tr. ZCT’T. so ii‘.Ll=.‘f i:'._' The second ‘_ (16.78) 4, FZVL Equation (16.78) is particularly important because it states that in
the case of an interaction between a periodic voltage and the corre—
sponding periodic current, the total average power is the sum of the ‘ average powers obtained from the interactholtages
oWncy Currents and voltages of different frequencies
do not interact to produce average power. Therefore, in average—power
calculations involving periodic functions, the total average power is
the superposition of the average powers associated with each har—
monic voltage and current. Example 16.4 illustrates the computation ~ of average power involving a periodic voltage. \(nIn_ \Q.I__r\:ﬂ/n 2, r 01 v2 WL‘RN :Ph 00
P: P¢¢+éfpn (’65 ( 87% “(96/13 Vdcldc: rPdo ) EXAMPLE16.4 ,2“ L {[H‘
 / \ \
Assume that the periodic squarewave voltage in Example 16.3 is A Jr 1 t .
applied across the terminals of a 15 S2 resistor. The value of Vm is A 3/ 5'32 3 I R L 60 v, and that O@ a) Write the first ﬁ ve nonzero terms of the Fourier series representa— ]
. tion of 00‘). Use the trigonometric form given in Eq. (16.38). b) Calculate the average power associated with each term in (a).
c) Calculate the total average power delivered to the 15 Q resistor. d) What percentage of the total power is delivered by the ﬁrst ﬁve
terms of the Fourier series? SOLUTION a) The dc component of v(t) is = <60><T/4>r av = 15 V.
T :i m 29? .'
From Example i' 6.3 we have
V ’ — _ — I  —_
A1=~/§60/7r=27.01V, 4, 3 1 1r J7r 71' L.
61:45°,
A2 = 60/7: = 19.10 v,
62 = 90°, '
A3 = 20ﬁ/n = 9.00 v,
63 =1350,
A4 =0,
94 = 0°,
A5 = 5.40 V,
95 =45°,
2 27f 1000
we = 37.1 = —(5——) = 400:: rad/s. Thus, using the ﬁrst ﬁve nonzero terms of the Fourier series, 0(1) = 15 + 27.01 cos(400m — 45°)
+ 19.10 cos(800nt — 90°)
+ 9.00 cos(12007rt — 135°)
+ 5.40 005(2000m — 45°) + . . . V, b) '1 he voltage is applied to the terrriimle of a resistor, so we can ﬁnd the power associated with each term as follows: ,7
1 '2
Pdc = 1—35 = 15 w,
< a l’\
k L 2 f > 1 1 27.01 4' ‘ P1 — — == 24.32 W, n i V (1/) . p 2 15
’ K ‘ 1 19.102 ‘
P2= _ ==12.16W [73 V313"
2 13 r" 7’
I 92 . L
_ __ = '7 P3—215 2..0W, pm : Vn
1 5.42
*' P5 = —— = 0.97 W. 41K
2 l
C) To obtain the total average power delivered to the 15 8'2 resistor,
we ﬁrst calculate the rms value of v(t): 8‘41 fox Ulve fnm)
60 2 T 4
V“: I guizrmm
+ The total average power delivered to the 15 Q resistor is
V (M  '302
A ' — (l )9: T=——=:60W.
15
V 1 " d) The total power delivered by the ﬁrst ﬁve nonzero terms is
{L MS 3 
’ T =Pdc+P1+P2+I3+P5=53.15W. This is (55.15/60)(100), or 91.92% of the total. 5"
J) )9” {7110+ Zion (09(Qvn “(57514) 5)“, /P Vex/way” {Zests’t’l‘b’g N 0:721) ow: 16.7 THE RMS VALUE OF A PERIODIC FUNCTION The rms value of a periodic function can be expressed in terms of the
Fourier coefﬁcients; by deﬁnition, 1 10—11"
Fm: \/— I my d:. (16.79)
T .0 Representing f (t) by its Fourier series yields 2 1 t0+T 0°
Frms = ? / av + E An cos(nwot — 9,.) dl‘. (16.80)
to n=1
The integral of the squared time function simpliﬁ 95 because the only
terms to survive integration over a period are the product of the dc
term and the harmonic products of the same frequency. All other products integrate to zero. Therefore Eq. (16.80) reduces to 1 2 I 00 T 2
Fm: ? avT—tZEAn
\ n=l
.00 ’
An
=\a%+Z—2
n=1 1‘ z 7. I 6?” 4:5,]
.00 A 2 A
: ag+z<7£> (16.81)
\ n=1 2 Equation (16.81) states that the rms value of a periodic function is
the square root of the sum obtained by adding the square of the rrns
value of each harmonic to the square of the dc value. For example,
let’s assume that a periodic voltage is represented by the ﬁnite series 1) = 10 + 30 cos(a)0t — 91) + 20 c030.th — 92)
+ 5 CUJ(3U)0t — 93) + 2 COS(5a)0t —~ 95). The rms value of this voltage is v = V/102 + <30N§>2 + (20%)2 + (s/Jiﬂ + <2N§>2
= M = 27.65 v. Usually, inﬁnitely many terms are required to represent a periodic .
function by a Fourier series, and therefore Eq. (16.81) yields an esti mate of the true rms value. We illustrate this result in Example 16.5. EXAMPLE153 Use Eq. (16.81) to estimate the rms value of the voltage in Exam
ple 16.4. SOLUTlON From Example 16.4, VdC = 15 V, V1 = 27.01 /«/2 V, the rms value of the fundamental, V2 = 19.10/«/2 the rms value of the second harmonic,
V3 = 9.00M? v, the rms value of the third harmonic, V5 = 5.40 / J2 V, the _rms value of the ﬁfth harmonic. Therefore, 1s2+<a5>2+<e>2+<a>2+<129>2 = 28.76 V. From Example 164, ﬁe ‘trg rms value is 30 V. We approach this
value by including more and more harmonics in Eq. (16.81). For
example, if we include the harmonics through k = 9, the equation
yields a value of 29.32 V. 15.3 TH E! EXPO MERIT] FORM OF TH E FOU REﬁ which completes the derivation of Eq. (1683). To complete the deri
. vation of Eq. (16.82), we ﬁrst observe from Eq. (16.88) that S ERIES
1 1 foiT
The exponential form of the Fourier series is of interest because it ‘3 C0 = — f f (t) dt 2 av. (16.89)
allows us to express the series concise1y. The exponential form of the T t0
genes 15 1 Next we note that
00 1
= w jnmol < 1 _ t0+T , 1
f“) 2 We ’ “632) C) = — / f(r)e1”“’0‘ dt = c; = —(a,. + jbn). (16.90)
n=—oo _ T )0 2
where S . . . .
ubstituting Eqs. (16.87), (16.89), and (16.90) into Eq. (16.86) yields
1 10+T . t 00
C = — 1‘: ‘1’“) dt. 16.83 » . .
" T f.) f )8 ( ) ftr) = C0 + Z (CneJWO’ + Cze‘m‘)
n21
To derive Eqs. (16.82) and. (16.83), we return to Eq. (16.2) and replace Do 00
the cosine and sine functions with their exponential equivalents: : C” 61'”th + Z czejnwot. (16.91)
ejnwcit + ejnwot :0 ":1
003 "(00’ = (1584) Note that the second summation on the right—hand side of Eq. (16.91)
ejwot # [ﬂaw is equivalent to summing Cnejna’o‘ from —1 to —00; that is,
sin moot = (16.85) 00 ‘ _OO 1
 Z cge‘lmo‘ = Z CneJWO‘. (16.92)
Substituting Eqs. (16.84) and (1685) into Eq. (16.2) gives 71:1 n=—l
00 a I I b . V Because the summation from —1 to —00 is the same as the summation
f(t) = av + Z él(elm’o' + e’lnw‘”) + 2—;(ejnw0t  eﬂwor) from —00 to —l, we use Eq. (16.92) to rewrite Eq. (16.91):
:1
noo . . 0° . '1 .
= av + 2CD; _ Jig) ejnwot + e—jnwoz (16.86) = Z Cnelnwot + z Cnejnwot
n=1 2 2 ":0 —Oo
00 1
Now we deﬁne C" as = Z Cnejnmot, (16.93)
—OO
1  A" e 16 87
Cu = Etan — an) : ‘2' ‘ m ” =1’2v3W ' ‘ ~ ( ' ) which completes the derivation of Eq. (16.82).
_——’—~ We may also express the rrns value of a periodic function in terms
From the deﬁnition of C”, of the complex Fourier coefﬁcients. From Eqs. (16.81), (16.87), and
(16.89),
1 2 t0+T 2 to+T
Cn = — —/ f(t) cos moot dt — j—/ f(t) sin moot dt
2 T ’0 T 2‘0 2 a2 + by:
Frms = a); + Z 2 , 1 to” . . =1
= . ./ f(t)(cos moot — J sm moot) dz N n
T to [12 + b2
(Cm: ———” ", . (16.95) 1 [0” 1' 0’d (1688)
_’ — I 8— mo ti '
T [.0 ﬂ ) C2=a2 (16.96) Substituting Eqs. (16.95) and (16.96) into Eq. (16.94) yields the de sired expression:
N
Fm; = C5 + 2 2 10,12. (16.97) \J . n=l Example 16.6 illustrates the process of ﬁnding the exponential
Fourier series representation of a periodic function.
E X A M P L E 1 6 .6
Find the exponential Fourier series for the periodic voltage shown in
Fig. 16.15 V“)
Vm
S 0 L U T I 0 N
— I _ . —1/2 0 7/2 7—7/2 T T+1l2 t
Using ——r/2 as the starting pomt for the integration, we have, from
Eq. (16.83), Figure 16.15 The periodic voltage for Exam
ple 16.6. {/2 r
C” = 4:] Vme—jnwot df
1 —r/2
Vm (e‘jWO’) “2
T ——]na)0 _r/2
_j Vm (ejnwor/Z _ ejnwor/Z) ‘ ‘
non *5 [VM T: VMT/ Sm “Wot/2
2V , .__. E ‘ ' r r : ————— E
=—”—L—s1nnwor/2.  .C 5’“ ’V‘u’o /Z __‘ [AU/Z} .z/L
moor _ mm
2 /
H t! m ' 5 ’ “ 1
Here, because v(t) has even symmetry, [9,, = 0 for an n, and hence a 5 Ha . V
we expect C" to be real. Moreover, the amplitude of Cn follows a ‘ X (sin x) /x distribution, as indicated when we rewrite C _ er sin(nwor/2)
"— T moor/2 We say more about this subject in Section 16.9. The exponential
series representation of u(t) is _d
n:=—oo : er i sin(nwot/2) ejnwoz
T moor/2 7L=——OO (>0 .
V—\ V r sm(nwor/2) n
v<r>= .> w w: AL’IPLlTUDE AND PHASE SPECTRA
A periodic time function is deﬁned by its Fourier coefﬁcients and its
period. In other words, when we know av, an, bn, and T, we can con ..
struct f (t), at least theoretically. When we know an and b", We also ’
know the amplitude (An) and phase angle (—6”) of each harmonic.
Again, we cannot, in general, visualize what the periodic function
looks like in the time domain from a description of the coefﬁcients and phase angles; nevertheless, we recognize that these quantities
characterize the periodic function completely. Thus, with sufﬁcient
computing time, we can synthesize the timedomain waveform from
the amplitude and phase angle data. Also, when a periodic driving
function is exciting a circuit that is highly frequency selective, the
Fourier series of the steady—state response is dominated by just a few
terms. Thus the description of the response in terms of amplitude and _, g ‘C r /
phase may provide an understanding of the output waveform. m " ‘7': ’ 5/ l
in terms of the am litude and hasc an 1e of each term in the Fourier  % J”
. P P g :)C_551hD—e5(/)
h We can present graphically the description of a periodic function
series of f (t). The plot of the amplitude of each term versus the . ~ ‘5’— T
frequency is called the amplitude spectrum of f (t), and the plot of the phase angle versus the frequency is called the phase spectrum of f (of) f (t). Because the amplitude and phase angle data occur at discrete
values of the frequency (that is, at we, 2w0, 3600, . . .), these plots also
are referred to as line spectra. An Illustration of Amplitude and Phase Spectra . nth. (‘ﬂzo mace ?,ir§i2/T,r3r5 Amplitude and phase spectra plots are based on either Eq. (16.38) (An
and —9,,) or Eq. (16.82) (Cn). We focus on Eq. (16.82) and leave the plots based on Eq. (16.38) to Problem 16.45. To illustrate the W
amplitude and phase spectra, which are based on the “Wm to,
of the Fourier series, we use the periodic voltage of xample 16.6. ’ L To aid the discussion, we assume that Vm = 5V and t : T/S. From
Example 16.6, V 1: sin(na)0t 2 _ _ _ _ a _ 4 _ _ 
anm /)’ 109876543276121234567891:
T moor/2
0.4 which for the assumed values of Vm and r reduces to Figure 1615 The plot of _Cssn_ versus H . < /5) . when r = T/5, for the periodic voltage for Sin rm 1 16.6. C,, =1————. (16.99) Exam" 8
INT/5 Figure 16.16 illustrates the plot of the magnitude of C, from Eq.
(16.99) for values of n ranging from —10 to +10. The ﬁgure clearly
shows that the amplitude spectrum is bounded by the envelope of
the 1(sin x) /xl function. We used the order of the harmonic as the
frequency scale because the numerical value of T is not speciﬁed.
When we know T, we also know too and the frequency corresponding to each harmonic.
Figure 16.17 provides the plot of l(sin x)/x[ versus x, Where x is ‘2” “1'5” '7 *05” 0 1 05" " 1°" 27' in radians. It shows that the function goes through zero whenever x is
an integral multiple of mFrom Eq. (16.98), Figure 16.17 The plot of (sin x)/x versus x. 1 111 :1 H11 _ the. I..." UﬂL.a._Y_+.+ _+_+_+_§,1_.J_LL1_+_+_+_+_1_L n —15 —13 —11 —9 —7 —5 3 —T 1 2 3 4 5 6 7 8 910111213141516
14 12 10 8 —6 —4 2 (16.100) From Eq. (16.100), we deduce that the amplitude spectrum goes
through zero whenever nr/ T is an integer. For example, in the
plot, t/ T is 1/5, and therefore the envelope goes through zero at
n = 5, 10, 15, and so on. In other words, the ﬁfth, tenth, ﬁfteenth, . . .
harmonics are all zero. As the reciprocal of r/T becomes an in
creasingly larger integer, the number of harmonics between every 71'
radians increases. If mr/ T is not an integer, the amplitude spectrum
still follows the [(sin x‘)/ x] envelope. However, the envelope is not
zero at an integral multiple of too. Because C" is real for all n, the phase angle associated with C n is
either zero or 180°, depending on the algebraic sign of (sin nrt/S)/
(mt/5). For example, the phase angle is zero for n = 0, i1, i2, 21:3,
and i4. It is not deﬁned at n = :1:5, because Cis is zero. The phase
angle is 180° at n 2 11:6, i7, i8, and i9, and it is not deﬁned at
21:10. This pattern repeats itself as n takes on larger integer values.
Figure 16.18 shows the phase angle of Cn given by Eq. (16.98). Now, what happens to the amplitude and phase spectra if f (t) is
shifted along the time axis? To ﬁnd out, we shift the periodic voltage
in Example 16.6 to units to the right. By hypothesis, 0C)
, .
v(t) = :4 Cnel’m’o’. n=—oo (16.101) Therefore 00 00
tie—:0): Z Cnej"w°(t_t°):= Z cneJ’wotoeiwot, (16.102) n=OO Viz—00 which indicates that shifting the origin has no effect on the amplitude
spectrum, because 1Cn1=lCne""“’0‘°l. (16.103) However, reference to Eq. (1687) reveals that the phase spectrum has
changed to —(9n + ntwoto) rads. For example, let’s shift the periodic voltage in Example 1.6.6 r/ 2 units to the right. As before, we assume
that r = T / 5; then the new phase angle 6;, is 9;, = ~(6n—1—mt/5). (16.104) We have plotted Eq. (16.104) in Fig. 16.19 for n ranging from —8
to +8. Note that no phase angle is associated with a zero amplitude
coefﬁcient. You may wonder Why we have devoted so much attention to the
amplitude spectrum of the periodic pulse in Example 16.6. The reason
is that this particular periodic waveform provides an excellent way to
illustrate the transition from the Fourier series representation of a peri
odic function to the Fourier transform representation of a nonperiodic
function. We discuss the Fourier transform in Chapter 17. —216' m Figure 16.19 The plot of egsn versus n for
Eq. (16.104). ...
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