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Unformatted text preview: CHAPTER 16  FOURIER SERIES W0 List of topics for this chapter :
Trigonometric Fourier Series
Symmetry Considerations
Circuit Applications
Average Power and RMS Values
Exponential Fourier Series
Fourier Analysis with PSpice W6 TRIGONOMETRIC F OURIER SERIES Problem 16.1 [16.5] A voltage source has a periodic waveform deﬁned over its period
as V(t) = t(27r — t) V for 0 < t < 2n . Find the Fourier series for this voltage. 0<t<21r
w0=2n/T=1 1 1 7: 1 t3 f 2 H 2——J
a0 TE (t)dt 2nf(2nt t)dt 2n rm 3 a“ 21: 3 “ 3
21tt [(Znt — t2 ) cos(nt) dt = cos(nt) + — sin(nt)] 3" nn n
1 an = n n3 [2nt cos(nt) — 2 sin(nt) + nth Sin(nt)] V(t) = 21tt— t2,
T=21t 27:
0 3.
an—T 21:
0 1 y  4
n n3 [4m cos(21tn)] = r17 2(1 1)
a =_’ " —
n n2 bn = 52: {[(Znt — t2)sin(nt) d = % I(2nt — t2)sin(nt) dt 2nl bu = —n—n—2[sin(nt)—nt005(nt)] 7mg 7‘—
0 [2m sin(nt) + 2 cos(nt) — nzt2 cos(nt)] 3" 297 Hence, 27:2 °° 4
ft =——— —cos t
() 3 H; (n) Problem 16.2 Evaluate each of the following functions and determine if it is periodic. If it
is periodic, ﬁnd its period. (a) f(t) = cos(1tt/2) + sin(1rt) + «[3— cos(27tt) (b) y(t) = sin(\/—3_ nt)+cos(7tt)
(c) g(t) = 4 + sin(mt) (d) h(t) = 25in(5t)cos(3t) (e) z(t) = e‘tsin(1tt) (a) This is a periodic function with a period of 4 seconds. (b) This is a nonperiodic function since the ﬁrst term has an irrational
multiplier of at while the second has a rational multiplier. (c) The integral of this function goes to inﬁnity because of the dc function.
Thus this is a nonperiodic function._ (d) This is a periodic function with a period of 7: seconds.
(c) This is a nonperiodic function since it continuously changes as t goes to
inﬁnity. oma‘ SYMMETRY CONSIDERATIONS Problem 16.3 Determine the type of function represented by the signal in Figure 16.1.
Also, determine the Fourier series expansion. f(t) 0 1
V V 40* V V Figure 16.1 This is an odd function since f(t) = —f(~t). Therefore, a0 = 0 = an. 298 2 T .
:fj; f (t)sm(n0)ot)dt , where T = 1 sec and 030 = 2n rad/sec. 0"
:1
H For 0<t<1, f(t) = 20t—10 ll Solving for bn % £(20t — 10) sin(2n7rt)dt = 2[ £20t sin(2n7tt)dt — £10 sin(2nnt)dt:l l 1
= 2 [ 220 2 sin(2n1rt) —£cos(2nnt)) —[:—1—gcos(2nnt)]
4n 11: 2m: 0 2m: 0
20 20 — 10 — 20
= 2 0—0 ~—— l—O —— 1—1 =——————
[4n2112 ( ) 2n7t( ) 2n7t( ): 117:
—20 °° 1 ,
Therefore, f(t) = — —s1n(2n7tt)
7r “:1 11
Problem 16.4 [16.15] Calculate the Fourier coefﬁcients for the function in Figure 16.2.
f(t)
4
—5—4—3—2—1 0 1 2 3 4 5 t Figure 16.2 This is an even function, therefore bn = 0. In addition, T = 4 and 030 = (0/ 2 . 2 2
a0 =¥ /2f(t) dt=Z £4tdt=t2ig =1 4 /2 4
an = :f If f (t) cos(0)0nt) dt = Z x 4t cos(n1tt/ 2) dt a) 4 2t . 1
n — 4ln2 n2 cos(n7tt/ 2) + rm sm(n1tt/2)] 0 299 an = 2162 [cos(n7t/2)—1]+——8——sin(n7r/2)
11 1t nn Wm. CIRCUIT APPLICATIONS Problem 16.5
following values during that period, Vs(t) = 10 volts 0 < t < 1: msec = 0 nmsec<t<27r msec
100 k!)
+
Vs“) a E C Vout(t)
Figure 16.3 In addition, L = 1 H and C = 1 uF. Determine the value of vo(t). Figure 16.3 and vs(t) is periodic with a period equal to 21: msec and has the The ﬁrst step is to ﬁnd the Fourier series for vs(t). an = 0 since this is an odd function. f(t) = a0 + an sin(n0)Ot)
n=l
T = 27mm“3 and (00 = 1000. 1 10“3 1:10'3 1 .
a0 = 2n10_3 [f 10dt+ fwd Odt]= 2n10_3(10t—0)' 1
1t10"3 2
27t10‘3 ll bn f f(t)sin(1000nt)dt = 300 7:10“3
0 = 5 volts [ f")— 105in(1000t)dt + 0] 1:10~3 ~ ~~—:31—310 00300000
7110 x10 11 0 _10
1th ll (cos(n7r) — 1) 22 for n = odd Thus, I bn = rm
0 for n = even °° 20 + ~———sin(l 000(1 + 2k)t) volts
k=1 (1 + 2k)n Therefore, Vs(t) = [5 Now let us look at the ﬁrst three terms. Clearly, for the dc term, Vo = 0 since the inductor looks like a short for dc. Forall the other
values of n, 30
v0 = co =1000n, for n = odd
105 + L/C 1(03L —l/(coC))
j((x)L —1/(0)C))
20L 20x106
= nth : n1:
j105(0)L —1/(0)C)) + L/C j105(1000n —1000/n) +1000
(1) i
For n = 1, a) = 1000. Therefore, V0 = 20/1t.
For n = 3, co = 3000. Therefore,
.the value olelC = —L/C— — lg)— : —j0.375§2 j(wL ~1/(03C)) — j2667 This value of impedance is so much smaller than the value of the resistor that we can neglect this
term and all of the others. Thus, vo(t) = Esin(1000t) volts
1: 301 Does this answer make any sense? If we look at this term and the values of L and C, we ﬁnd that
L and C are in parallel resonance when a) = 1000. Thus, this circuit is actually a ﬁlter that ﬁlters
out a single sine ane from the input signal. Problem 16.6 Refer to Figure 16.3. Change the value of L to (1/9) H. with everything else
remaining the same. Now solve for vo(t). Everything remains the same as Problem 16.5 up till
equation (a). The new value of L changes equation (a) as shown below. 20x106
. _ n 9
Thus, our new equation for V0 — 110 — — q + ——
9 n 9
For n = 1,
V _ 0.7074x106
° j105(111.11—1000) + 0.1111x106
6
E = j0.007958
— 3888.9x10
Clearly, this can be considered to be equal to zero.
For n = 3,
0.235 8x106 V = 2.122 volts For all other values of n, V() is essentially equal to zero. Therefore, vo(t) = 32.m'm(3000t) = 2.122sin(3000t) v 3n Problem 16.7 [16.25] If VS in the circuit of Figure 16.4 is the same as function f2 (t)
in Figure 16.5, determine the dc component and the ﬁrst three nonzero harmOnics of V0 (t) .
1 Q 1 H
2 mo
+
vs C 1 F 1 Q V0
Figure 16.4 302 f2“) —2—1012345 t Figure 16.5 The signal is even, hence, bn = 0. In addition, T = 3 , (00 = 27t/ 3 . Vs(t)=1 fora110<t<1
Vs(t)=2 foralll<t<1.5 a0 =%[£ldt+ f5 2dtl=g . an = :1 {cos(2n1tt/3) dt + f5 2cos(2n7tt/3) dt] 3
4T 3 . 1 6 . 151 _2 .
an  3 L 21m Sln(2n7rt/3)’0 + 2m sm(2n1rt/3)ll J— rm s1n(2nn/3) 4 2 °° 1 ,
vS (t) = g ~ ; l~I;s1n(2n1r/3) cos(2n1tt/3) Now consider this circuit, 1 Q j2n7t/3 L, Z_[;j_3 1‘3
e _ 2m: 1—j3/2n7t _2n1t——j3 303 Z Thus: Vo Vs Simplifying, we get
_ V° :12n7t+j(4n27r2—18)Vs Forthe dc case, n = 0 and Vs =3/4V and V0 = VS/2=3/8V. We can now solve for v0 (t) Vo (t) = [3+ 2A“ cos[21;nt + @n)] V
n=l (6/n 7t) sin(2mt/ 3) where An = —————————————
‘/16n2 n2 + [(4n2 n2/3). 6]2 and (9“ = 90° —— tan“ — 3 2m: where we can further simplify An to An M
n7t‘l4n4 7:4 + 81 W0 AVERAGE POWER AND RMS VALUES Problem 16.8 Given the signal shown in Figure 16.6, determine the exact value of the rms V(t)
10 volts —2—10123:45 t
_10I_I_II__II_ Figure 16.6 304 value of this wave shape. Using the Fourier series of the wave shape, calculate the estimated rms
. value using all the terms up to and including 11 = 5. We can use the deﬁnition of VIms to calculate the rrns value of the wave shape. ’1
V,ms = —ffV2(t)dt where T = 2sec. é. fvz(t)dt=%[1:(10)2dt+ f(—10)zdt]=%[100t; +100tIf] = 0.5[100—0+200— 100] = 100 Thus, V,ms = 100 = 10 volts. We now proceed to the Fourier series. Please note, this is just the Fourier series of a standard
square wave. v(t) = ~49 lsin(n1tt), n = 2k— 1.
TC k=1n . For this problem, we want all the terms through and including 11 = 5 (k = 3). For a Fourier series, we can solve for the rms value using, Fm = /a3 +33% +bg)
1 Th v ~ errata “5’ “"5’211'37: 51: 1:2 925 = (40/1t)(0.7587) = 9.66 volts. Although this answer is only Within 5%, it is still signiﬁcant enough for some cases. The reason
that this is not closer to the actual value of 10 volts is that the coefﬁcients for the Fourier series of a square wave do not decrease in value as fast as they do for other signals. 305 Problem 16.9 Given the triangular voltage wave shape shown in Figure 16.7, determine the .
exact value of the rms voltage. Then, calculate the approximate value of the rms value using the
Fourier terms up to and including n = 5. t
10 volts V( ) —2 —1 o 1 2 t
/ v 4 10 v
Figure 16.6 First we will calculate the exact value using, 1 .
,igfvzamt, where v(t) = 20t for 0<t<1/2. Note that due to symmetry, we only need to use the range, 0 < t < 1/2. Vrms 1] 1/2 0 O = (800/3)[(1/8)—0] = 100/3 800t3 1 2 _4 /2 2 _
5fv(t)dt—§£ 400tdt— Therefore, V,ms = 10/ x/g = 5.774 volts. Now we can solve the Fourier series. The student can verify that the Fourier series for this wave
shape is given by, 80 °° 1 .
v(t) = —2 —2s1n(n1rt), where n = 2k— 1.
7T k=1 11 Through n = 5 we get, v(t) 5 QKSinwt) + lsin(37:t) + —1— sin(57tt) volts.
7: 9 25
Therefore, 1+ —— + —] = 5.771 volts. Clearly, this compares very favorably to the exact value of 5.774. The reason for this is because
the Fourier series for a triangular wave shape converges very quickly. . 306 . Problem 16.10 [16.31] The voltage across the terminals of a circuit is
v(t) = 30 + 20 cos(1201tt + 45°) + IOCos(1207tt — 45°) V The current entering the terminal at higher potential is i(t) = 6 + 4cos(1201tt + 10°) — 2003(I’120nt — 60°) A Find:
(a) the rms value of the voltage,
(b) the rms value of the current,
(0) the average value of the power absorbed by the circuit. (a) V,ms = 1F13, +%Z(a§ +133) = 1[(30)2 43902 +102) = 33.91 V
n=l ——
(b) 1ms =1/62 +G)(42 + 22) = 6.782 A 1
(c) P = V... I... +~2—ZV. I. cos<®.. —<I>..) . P = (30)(60) + (0.5) [(20)(4) cos(45° —10°)— (10)(2)cos(45° + 60°)]
P = 180 + 32.76 — 9.659 = 203.1 w Problem 16.11 Determine the rms value of a triangular wave shape with a peaktopeak
value of 40 volts. If this wave shape is placed across a 10ohm resistor, determine the average power dissipated by that resistor. As we saw in problem 16.9, the rms value of a triangular wave shape is given by, vrms = vim/‘5 = zen/3 = 11.547 volts. Average power = Vmsz/R = (11.547)2/10 = 13.333 watts. Wan—mo EXPONENTIAL FOURIER SERIES Problem 16.12 Given the sawtooth voltage wave shape shown in Figure 16.8, ﬁnd its
. exponential (complex) Fourier series. 307 t
10 V V0
..—2 —1 0 1 2 t
v v _r_10 v v
Figure 16.7
c1] = % fv(t)e”j“‘°°'dt , where T = 1 and V(t) = (20t — 10) for 0 < t < 1. SinceT = 1,0)0 = 27:. Therefore, 0n = % £(20t  10) e"j2"“‘dt = 20 Ite'jz’m‘dt 10 J: e'J'Z’m‘dt ~j27mt 1 te—j21mt e e
—10 ~j27mt
= 20 ~ "*7‘”?
_]27tn (—127tn) 0 —j21m j21m
20[ e + e 1 — j27rn O — 10 [6pm _ 21m 27m
_10[_~1___J_] ' 10
21m 21m H —j2nn 47:2n2 — 41r2n2 : J— 27m ll j 1 1
20 ~——+ —
[27m 41t2n2 47th2 In addition, 00 = 0. = Z _._l_1_0_ej2n7tt Thus,
n=~oo m"
n¢0
Problem 16.13 [16.3 7] Determine the exponential Fourier series for f (t) = t2 , 1t<t<1r,with f(t+27tn)=f(t). (no =27t/T=1 308 1 7r . c = — t2 e'J‘“ dt 0 n 27¢
Integrating by parts twice gives, 0n = 2cos(n7r/n2) = (2)(1“/n2) , n at 0 For n=0, 1 1: 2
00:57; mtz dt=zt3— Hence,
l “2 'n
f(t) __+n=E— __._2_el t n¢0 W FOURIER ANALYSIS WITH PSPICE Problem 16.14 [16.51] Calculate the Fourier coefﬁcients of the signal in Figure 16.8
using PSpice. f(t)
4 —4 ——2 0 2 4 6 8t Figure 16.8 The Schematic is shown below. In the Transient dialog box, we type “Print step = 0.01s, Final
time = 36s, Center frequency = 0.1667, Output vars = v(1),” and click Enable Fourier. 309 Prob. 16.51, After simulation, the output ﬁle includes the following Fourier components, F OURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 2.000396E+00 _ HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
(HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) .
n
u 310 ...
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This note was uploaded on 09/06/2010 for the course EE 50145 taught by Professor Chi during the Spring '10 term at UCSF.
 Spring '10
 Chi

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