Electrical Circuits II (ECE233b) - Instructional Web...
Transcript of Electrical Circuits II (ECE233b) - Instructional Web...
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Electrical Circuits II (ECE233b)
The University of Western Ontario Faculty of Engineering Science
Anestis Dounavis
Fourier Analysis Techniques (Part 1)
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BackgroundWe have discussed the following solution methodologies
Phasor transforms
Laplace transforms
Steady state; Sinusoidal analysis
Transient and steady state; Any type of forcing function
The periodic signal is expresses as a summation of sinusoids with harmonically related frequencies.
Fourier transforms Steady state; non-sinusoidal but periodic signals.
Fourier analysis techniques
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Fourier SeriesA periodic function satisfies the following relationship
)nTf(tf(t) 0
where T0 is the period 0 is the fundamental frequency (rad/s) 0=2/T0
Examples of periodic signals
3,......2,1,n
T0 2T0 3T0 t
A
f(t)
T0 2T0 3T0 t
A
f(t)sawtooth waveform square waveform
Can be expressed as
1n
)tcos(nDaf(t) n0n0 (phasor notation)
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Fourier SeriesFourier series can be expressed in the following forms
Exponential Fourier Series
Trigonometric Fourier Series
n
tjnω
0nn
tjnω0
00 eeaf(t) nn cc
1n
t))sin(nbt)cos(n(aaf(t) 0n0n0
The two forms are similar since sincos jje
where n is an integer a0, an, bn and cn are the Fourier coefficients 20, 30, …. , n0 are the harmonic frequencies 0 is the fundamental frequency
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Fourier SeriesRelationship between exponential and trigonometric form
0nn
tjnω0
0eaf(t) nc
11 nn
tjnωtjnω0
00 eea nn cc
tjnω-tjnω0
00 eea *nn cc
1n
1n
)tjnω0
0eRe(2a nc
1n
)tjnωnn0
0)eRe((Da
1n
)tjnωnn0
0)ejb-Re((aa
1n
t))sin(nbt)cos(n(aa 0n0n0
* = complex conjugate
where
nnn jba nc2nD
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Fourier Series
tt 2*3sin312sin
t2*3sin31
t2sin
Approximating a square wave
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Fourier Series
tnnn
2*)12(sin)12(
18
1
Approximating a square wave
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Fourier SeriesDetermining the coefficients for exponential Fourier seriesAny physical realizable periodic signal may be represented over the interval t1<t<t1+T0 by the exponential Fourier series
n
tjnω0ef(t) nc
Multiplying both sides by e-jk 0t and integrating over the interval t1 to t1+T0 yields
0k
Tt
t
tjkω-
n
tjnωn
Tt
t
tjkω- Tcdteecdtf(t)e01
1
00
01
1
0
since
kn forTkn for0
dte0
Tt
t
tk)ω-j(n-01
1
0
The Fourier coefficients are defined as
01
1
0
Tt
t
tjnω-
0n dtf(t)e
T1c
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Example 1Determine the exponential Fourier series for the periodic voltage waveform shown
T/4
V
v(t)
T/2-T/2
-T/4
-V
Fourier SeriesStrategy to determine Fourier Coefficients
1. Identify T, 0 and defining equation of function2. Determine coefficient by appropriate integration
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Fourier SeriesDetermining the coefficients for trigonometric Fourier series
1n
t))sin(nbt)cos(n(aaf(t) 0n0n0
01
1
0
Tt
t
tjnω-
0n dtf(t)e
T1c
Using the result obtained for the exponential Fourier series
andnn jba nc2
)sin()cos( tjtt00
-jn nne 0
yields the following Fourier coefficients for a0, an and bn
01
1
Tt
t0
0n t)dtf(t)cos(nω
T2a
01
1
Tt
t0
0n t)dtf(t)sin(nω
T2b
01
1
Tt
t00 f(t)dt
T1a
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Fourier Series
If a signal exhibits certain symmetrical properties, we can take advantage of these properties to simplify the calculations of the Fourier coefficients
There are 3 types of symmetry: 1) Even-function symmetry2) Odd-function symmetry3) Half-wave symmetry
Trigonometric Fourier series
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Fourier Series Even-function symmetry: A function is said to be even if
f(-t)f(t) Example of even functions
Even functions can only be approximated with even basis functions eg) cos(not)
1) Therefore can only have “an” coefficients since bn=02) Can integrate over half cycle
2/0T
00
0n t)dtf(t)cos(nω
T4a
2/0T
000 f(t)dt
T2a
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Fourier SeriesOdd-function symmetry: A function is said to be odd if
f(-t)f(t) Example of odd functions
Odd functions can only be approximated with odd basis functions eg) sin(not)
1) Therefore can only have “bn” coefficients since an=02) Can integrate over half cycle
2/0T
00
0n t)dtf(t)sin(nω
T4b
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Fourier SeriesHalf-Wave symmetry: A function is said to posses half-wave symmetry if
2T-tff(t) 0
Example of Half-Wave symmetry
This equation states that each half-cycle is an inverted version of the adjacent half-cycle; that is, if the waveform from –T0/2 to 0 is inverted, it is identical to the waveform from 0 to T0/2.
Symmetric around horizontal axis Independent of where t=0
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Fourier SeriesExample of Half-Wave symmetry
1. The average of the function is zero, therefore a0=02. an=bn=0 if n is even and
2/0T
00
0n t)dtf(t)cos(nω
T4a
2/0T
00
0n t)dtf(t)sin(nω
T4b
if n is odd
Also if the function is odd then an=0and if the function is even then bn=0
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Example 2Determine the trigonometric Fourier series for the periodic voltage waveform shown and compare results with example 1.
T/4
V
v(t)
T/2-T/2
-T/4
-V
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Example 3Determine the trigonometric Fourier series for the periodic voltage waveform shown.
V
-T T 2T
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Example 4Determine the type of symmetry exhibited by the waveform.
2
1
-1
-2
-1
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Time-ShiftingTime shifting of a periodic waveform f(t) is defined as
n
tjnω0ef(t) nc
n
t-(tjnω0
00e)t-f(t )nc
time shifting f(t)
n
tjnω
n
tjnωtjnω0
0000 e)ee)t-f(t nn k(c
where 00tjnωenn ck
Therefore, time shift in the time domain corresponds to a phase shift in the frequency domain
To compute the phase shift in degrees
Phase shift (degrees) = 0td = (3600)td/T0
td=time delay
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Example 5
T/4
V
v(t)
T/2-T/2
-T/4
-V
If the wave form in example 1 is shifted by a quarter period as shown below, determine the Fourier coefficients
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Frequency SpectrumFrequency Spectrum: The frequency spectrum of f(t) expressed as a Fourier series consists of an amplitude spectrum and a phase spectrum
Amplitude Spectrum: Plot of the amplitude of the harmonics versus frequency.
Phase Spectrum: Plot of the phase of the harmonics versus frequency.
Since the frequency components are discrete, the spectra is called line spectra
The plots are based on the following equations
0nn
tjnω0
0eaf(t) nc
1n
)tjnωnn0
0)eRe((Da
where nnn jba nc2nD
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Example 6The Fourier series for the waveform shown is given by
10
T/2-T/2
-10
odd nn
v(t)1
0220 )cos(40)sin(20 tnn
tnn
Plot the first four terms of the amplitude and phase spectra for this signal
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Steady-State Network ResponseIf a periodic signal is applied to a network the steady state response of the circuit can be found as follows:
If the input forcing function for a network is a voltage, the input can be expressed as
1. Represent the periodic forcing function by a Fourier series
....(t)v(t)vv v(t) 210
+-
+-
+-
Network(t)v1
0v
(t)v2
2. Use phasor analysis in the frequency domain to determine the network response due to each source3. The network response due to each source is transformed to the time domain
4. Add time domain solutions due to each source (Principle of Superposition) to obtain Fourier series for the total steady- state network response
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Example 7Determine the steady-state voltage vo(t) in for the circuit shown if the input voltage v(t) is given by the expression
odd nn
v(t)1
0220 )cos(40)sin(20 tnn
tnn
+ - 11F
+vo(t)
-
2
v(t)
1
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Average PowerRecall that average power is defined as
T
0
v(t)i(t)dtT1P
If v(t) and i(t) are periodic functions, the signals can be expressed as:
1n
v0ndc )θtcos(nVVv(t)n
1n
i0ndc )θtcos(nIIi(t)n
Note to calculate the average power involves the product of two infinite series
However, only products at the same frequency survive integration over one period (due to orthogonal properties of basis functions)
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Average PowerNote:
T
000 nmT/2
nm0t)dtt)cos(mcos(n
Therefore we can show that the average power can be expressed as:
1n
ivnn
dcdc )θcos(θ2IVIVP
nn
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Example 8For the network shown, the input voltage is
+-v(t)
100F 20mH
16
i(t)
)00 20-12cos(754t)3016cos(377t42v(t)
Compute the current i(t) and determine the average power absorbed by the network