Ultimo 23 supch16.pdf
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CHAPTER 16 - FOURIER SERIES 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 TRIGONOMETRIC FOURIER SERIES Problem 16.1 [16.5] A voltage source has a periodic waveform defined over its period as V ) t 2 ( t ) t ( v − π = for π < < 2 t 0 . Find the Fourier series for this voltage. 2 t t 2 ) t ( v − π = , π < < 2 t 0 π = 2 T 1 T 2 0 = π = ω π π − π π = − π π = = ∫ ∫ 2 0 3 2 2 0 2 T 0 0 3 t t 2 1 dt ) t t 2 ( 2 1 dt ) t ( f T 1 a 3 2 3 2 1 2 4 a 2 3 0 π = − π π = π π + π π = − π = ∫ 2 0 2 T 0 2 n ) nt sin( n t 2 ) nt cos( n 2 1 dt ) nt cos( ) t t 2 ( T 2 a π + − π = 2 0 2 2 3 n ] ) nt sin( t n ) nt sin( 2 ) nt cos( nt 2 [ n 1 - a 2 3 2 n n 4 - )] n 2 cos( n 4 [ n 1 ) 1 1 ( n 2 a = π π π − − = ∫ ∫ − π = − = dt ) nt sin( ) t nt 2 ( 1 dt ) nt sin( ) t nt 2 ( T 2 b 2 T 0 2 n π π − + π − − π = 2 0 2 2 3 0 2 n )] nt cos( t n ) nt cos( 2 ) nt sin( nt 2 [ n 1 )] nt cos( nt ) nt [sin( n 1 n 2 b 0 n 4 n 4 - b n = π + π =
Transcript of Ultimo 23 supch16.pdf
-
CHAPTER 16 - FOURIER SERIES
List of topics for this chapter :Trigonometric Fourier SeriesSymmetry ConsiderationsCircuit ApplicationsAverage Power and RMS ValuesExponential Fourier SeriesFourier Analysis with PSpice
TRIGONOMETRIC FOURIER SERIES
Problem 16.1 [16.5] A voltage source has a periodic waveform defined over its periodas V)t2(t)t(v = for
-
Hence,
=)t(f ====
1n2
2
)ntcos(n
43
2
Problem 16.2 Evaluate each of the following functions and determine if it is periodic. If itis periodic, find its period.
(a) f(t) = cos(t/2) + sin(t) + 3 cos(2t)(b) y(t) = sin( 3 t) + cos(t)(c) g(t) = 4 + sin(t)(d) h(t) = 2sin(5t)cos(3t)(e) z(t) = etsin(t)
(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 t 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 seconds.(e) This is a nonperiodic function since it continuously changes as t goes to
infinity.
SYMMETRY CONSIDERATIONS
Problem 16.3 Determine the type of function represented by the signal in Figure 16.1.Also, determine the Fourier series expansion.
Figure 16.1
This is an odd function since f(t) = f(t). Therefore, ao = 0 = an.
f(t)
t2101210
10
-
bn = T0 o dt)tnsin()t(fT2 , where T = 1 sec and o = 2 rad/sec.
For 0 < t < 1, f(t) = 20t 10
Solving for bn =
= 10 1010 dt)tn2sin(10dt)tn2sin(t202dt)tn2sin()10t20(12
=
1
0
1
022 )tn2cos(n2
10)tn2cos(n2
t20)tn2sin(n4202
=
=
n
20)11(n210)01(
n220)00(
n4202 22
Therefore, f(t) = ====
1n)tn2sin(
n
120
Problem 16.4 [16.15] Calculate the Fourier coefficients for the function in Figure 16.1.
Figure 16.1
This is an even function, therefore 0bn = . In addition, 4T = and 20 = .
==== 102102T00 tdtt442dt)t(fT2a 1 == 10
2T
0 0ndt)2tncos(t4
44
dt)ntcos()t(fT4
a
1022n )2tnsin(n
t2)2tncos(n
44a
+
=
t54321012345
4f(t)
-
=na )2nsin(n
8]1)2n[cos(n
1622
++++
CIRCUIT APPLICATIONS
Problem 16.5 Figure 16.1 and vs(t) is periodic with a period equal to 2 msec and has thefollowing values during that period,
Vs(t) = 10 volts 0 < t < msec = 0 msec < t < 2 msec
Figure 16.1
In addition, L = 1 H and C = 1 F. Determine the value of vo(t).
The first step is to find the Fourier series for vs(t). an = 0 since this is an odd function.
f(t) = ao + =
1n
on )tnsin(b
T = 2x103 and o = 1000.
ao = ( ) 3333
1003
102
10
10
030t10
1021dt0dt10
1021
=
+
= 5 voltsbn =
+
=
310
03
T
030dt)t1000sin(10
101dt)nt1000sin()t(f
1022
100 k
+
vs(t) CL+
vout(t)
-
=
310
033 )t1000cos(10n10x10
1
= ( )1)ncos(n
10
Thus, bn = evennfor0
oddnforn
20
=
=
Therefore, vs(t) = volts)t)k21(1000sin()k21(205
1k
+
++
=
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. For all the othervalues of n,
vo =
+
))C/(1L(j
C/L
))C/(1L(jCL10
n
20
5, = 1000n, for n = odd
=
1000)n/1000n1000(10jn
10x20
C/L))C/(1L(10jCnL20
5
6
5 +
=
+
(1)
For n = 1, = 1000. Therefore, Vo = 20/.
For n = 3, = 3000. Therefore,
the value of L||C = ==
375.0j2667j1000
))C/(1L(jC/L
This value of impedance is so much smaller than the value of the resistor that we can neglect thisterm and all of the others. Thus,
vo(t) = volts)t1000sin(20
-
Does this answer make any sense? If we look at this term and the values of L and C, we find thatL and C are in parallel resonance when = 1000. Thus, this circuit is actually a filter that filtersout a single sine wave from the input signal.
Problem 16.6 Refer to Figure 16.1. Change the value of L to (1/9) H. with everything elseremaining the same. Now solve for vo(t). Everything remains the same as Problem 16.5 up tillequation (a). The new value of L changes equation (a) as shown below.
Thus, our new equation for Vo =
910
n
10009
n100010j9n
10x20
65
6
+
For n = 1,
007958.0j10x9.888j
10x7074.010x1111.0)100011.111(10j
10x7074.0V
5
6
65
6
o
=
+=
Clearly, this can be considered to be equal to zero.
For n = 3,
volts122.210x11111.0)3.3333.333(10j
10x2358.0V 656
o =
+=
For all other values of n, Vo is essentially equal to zero. Therefore,
vo(t) = V)t3000sin(122.2)t3000sin(320
====
Problem 16.7 [16.25] If sv in the circuit of Figure 16.1 is the same as function )t(f 2in Figure 16.2, determine the dc component and the first three nonzero harmonics of )t(vo .
Figure 16.1
+
vo
1 H
1
1
+
vs 1 F
-
Figure 16.2
The signal is even, hence, 0bn = . In addition, 3T = , 320 = .
1)t(vs = for all 1t0
-
Thus, so v3n2j1ZZ
v++
=
Simplifying, we get
s22o v)18n4(jn129j-
v+
=
For the dc case, 0n = and V43vs = and V832vv so == .
We can now solve for )t(vo
=)t(vo V3tn2
cosA83
1nnn
++++++++ ====
where =nA 22222 ]6)3n4[(n16)3n2sin()n6(++++
and =n
n23
3ntan90 1-
where we can further simplify nA to =nA81n4n
)sin944 ++++
32n(
AVERAGE POWER AND RMS VALUES
Problem 16.8 Given the signal shown in Figure 16.6, determine the exact value of the rms
t54321012
10 volts
v(t)
10
Figure 16.1
-
value of this wave shape. Using the Fourier series of the wave shape, calculate the estimated rmsvalue using all the terms up to and including n = 5.
We can use the definition of Vrms to calculate the rms value of the wave shape.
Vrms = T0 2 dt)t(vT1 where T = 2 sec.[ ]2
1
1
0
2
1
21
022
02 t100t100
21dt)10(dt)10(
21dt)t(v
21
+=
+=
= 0.5[100 0 + 200 100] = 100
Thus, Vrms = 100 = 10 volts.
We now proceed to the Fourier series. Please note, this is just the Fourier series of a standardsquare wave.
v(t) = =
1k
)tnsin(n
140, n = 2k 1.
For this problem, we want all the terms through and including n = 5 (k = 3).
For a Fourier series, we can solve for the rms value using,
Frms = ++1
2n
2n
2o )ba(2
1a
Thus, Vrms
++
=
+
+
251
911
2140
540
34040
21 222
= (40/)(0.7587) = 9.66 volts.
Although this answer is only within 5%, it is still significant enough for some cases. The reasonthat this is not closer to the actual value of 10 volts is that the coefficients for the Fourier series ofa square wave do not decrease in value as fast as they do for other signals.
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Problem 16.9 Given the triangular voltage wave shape shown in Figure 16.7, determine theexact value of the rms voltage. Then, calculate the approximate value of the rms value using theFourier terms up to and including n = 5.
First we will calculate the exact value using,
Vrms = 20 2 dt)t(v21 , 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.
2/1
0
32/1
022
02
3t800dtt400
24dt)t(v
21
== = (800/3)[(1/8) 0] = 100/3
Therefore, Vrms = 10/ 3 = 5.774 volts.
Now we can solve the Fourier series. The student can verify that the Fourier series for this waveshape is given by,
v(t) = =
1k
22 )tnsin(n180
, where n = 2k 1.
Through n = 5 we get,
v(t)
++
)t5sin(
251)t3sin(
91)tsin(802 volts.
Therefore,
Vrms
++
6251
8111
2180
2 = 5.771 volts.
Clearly, this compares very favorably to the exact value of 5.774. The reason for this is becausethe Fourier series for a triangular wave shape converges very quickly.
10
10 voltsv(t)
t21012
Figure 16.0
-
Problem 16.10 [16.31] The voltage across the terminals of a circuit is
V)45t120cos(10)45t120cos(2030)t(v +++=
The current entering the terminal at higher potential is
A)60t120cos(2)10t120cos(46)t(i ++=
Find:(a) the rms value of the voltage,(b) the rms value of the current,(c) the average value of the power absorbed by the circuit.
(a) =+
+=++=
=
)1020(21)30()ba(
21
aV 2221n
2n
2n
20rms V91.33
(b) =+
+= )24(
21
6I 222rms A782.6
(c) += )cos(IV21
IVP nnnndcdc)]6045-cos()2)(10()1045cos()4)(20[()5.0()60)(30(P ++=
=+= 659.976.32180P W1.203
Problem 16.11 Determine the rms value of a triangular wave shape with a peak-to-peakvalue of 40 volts. If this wave shape is placed across a 10-ohm resistor, determine the averagepower dissipated by that resistor.
As we saw in problem 16.9, the rms value of a triangular wave shape is given by,
Vrms = Vpeak/ 3 = 20/ 3 = 11.547 volts.
Average power = Vrms2/R = (11.547)2/10 = 13.333 watts.
EXPONENTIAL FOURIER SERIES
Problem 16.12 Given the sawtooth voltage wave shape shown in Figure 16.8, find itsexponential (complex) Fourier series.
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cn = T0 tjn dte)t(vT1 o , where T = 1 and v(t) = (20t 10) for 0 < t < 1.Since T = 1, o = 2.
Therefore, cn = = 10 nt2j10 nt2j10 nt2j dte10dtte20dte)10t20(11
=
1
0
nt2j1
02
nt2jnt2j
n2je10)n2j(
e
n2jte20
=
+
n2j
n2j
e10n4
1n4
e
n2je20 n2j2222
n2jn2j
=
+
n2j
n2j10
n41
n41
n2j20 2222 = n2
10j
In addition, co = 0.
Thus, v(t) = ====
0n
n
tn2jen
10j
Problem 16.13 [16.37] Determine the exponential Fourier series for 2t)t(f = ,
-
=-
jnt-2n dtet2
1c
Integrating by parts twice gives,
)n1-)(2()nncos(2c 2n2n == , 0n
For 0n = ,
3dtt21
c2
-
20
=
=
Hence,
=)t(f ====
++++
0n-
n-1)
n
jnt2
2
en
)(2(3
FOURIER ANALYSIS WITH PSPICE
Problem 16.14 [16.51] Calculate the Fourier coefficients of the signal in Figure 16.1using PSpice.
Figure 16.1
The Schematic is shown below. In the Transient dialog box, we type Print step = 0.01s, Finaltime = 36s, Center frequency = 0.1667, Output vars = v(1), and click Enable Fourier.
2
f(t)
4 0 2 4 6 t8
4
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After simulation, the output file includes the following Fourier components,
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 2.000396E+00
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZEDNO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)1 1.667E01 2.432E+00 1.000E+00 8.996E+01 0.000E+002 3.334E01 6.576E04 2.705E04 8.932E+01 6.467E013 5.001E01 5.403E01 2.222E01 9.011E+01 1.801E+024 6.668E01 3.343E04 1.375E04 9.134E+01 1.813E+025 8.335E01 9.716E02 3.996E02 8.982E+01 1.433E016 1.000E+00 7.481E06 3.076E06 9.000E+01 3.581E027 1.167E+00 4.968E02 2.043E02 8.975E+01 2.173E018 1.334E+00 1.613E04 6.634E05 8.722E+01 2.748E+009 1.500E+00 6.002E02 2.468E02 9.032E+01 1.803E+02
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PrefaceChapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of AnalysisChapter 4 Circuit TheoremsChapter 5 Operational AmplifierChapter 6 Capacitors and InductorsChapter 7 First-Order CircuitsChapter 8 Second-Order CircuitsChapter 9 Sinusoids and PhasorsChapter 10 Sinusoidal Steady-State AnalysisChapter 11 AC Power AnalysisChapter 12 Three-Phase CircuitsChapter 13 Magnetically Coupled CircuitsChapter 14 Frequency ResponseChapter 15 Laplace TransformChapter 16 Fourier Series Trigonometric Fourier SeriesSymmetry ConsiderationsCircuit Applications Average Power and RMS ValuesExponential Fourier Series Fourier Analysis with PSpice
Chapter 17 Fourier TransformChapter 18 Two-Port Networks
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