The exponential Fourier series for a periodic signal was developed in
Example 6.3 A half-rectified sine wave. Passing a sine wave of angular frequency N through a half-wave rectifier produces the signal shown in Figure 6.10
Fourier Series
Fourier Series: Half-wave Rectifier. ( ). (. ) π π π ω π ω π ω π ω ω π ω π ω π ω π. E. E t. E tdt. E dttu a. = -. -. = -. = = = ∫. ∫-. 0cos cos. 2 cos. 2 sin.
CHAPTER 4 FOURIER SERIES AND INTEGRALS
Example 1 Find the Fourier sine coefficients bk of the square wave SW(x). multiple of cosx is closest to f = cos3 x? 7. Sketch the 2π-periodic half wave ...
Essential Mathematical Methods for Physicists - Weber and Arfken.1.1
This is our Fourier exponential series [Eq. (14.32)]. Separating real and This is the output of a simple half-wave rectifier. It is also an ap ...
Table 15
Half wave rectified sine wave: 0. 2. T π ω After some algebra the delay can be represented as a phase shift in the Fourier series of the voltage waveform.
Fourier_Series_continuous_time_periodic_signal_and_ Fourier
c = Fourier coefficients of exponential form of Fourier series. When a waveform has half wave symmetry the Fourier series will consist of odd harmonic terms.
Unit 4 (Fourier Series & PDE with Constant Coefficient)
04-May-2020 n=1 n-1. So Half range sine series of f(x) on (0
Untitled
26-Apr-2007 Continuous-Time Complex-Exponential Fourier Series Complex exponential Fourier ... half-wave rectifier. Page 4. (b) Suppose the input sinusoid x ...
CHAPTER 3. SPECTRUM REPRESENTATION 58 - 3-4 Fourier Series
08-Sept-2012 ... exponential signal—the integral of a complex exponential ... EXERCISE 3.15: Find the Fourier Series coefficients of the half-wave rectified sine ...
Chapter 16 The Fourier Series
It is also useful to know the values of the cosine sine
Module 6 Introduction to Fourier series Objective:To understand
exponential fourier series. Problem 1:Find the Fourier series expansion of the half wave rectified sine wave shown in fig below. Solution :.
CHAPTER 3. SPECTRUM REPRESENTATION 58 - 3-4 Fourier Series
8 sept. 2012 half-wave rectified sine. Exploit complex exponential simplifications such as ej 2 k D 1 ej D 1
The exponential Fourier series for a periodic signal was developed in
Spectrum of a half-rectified sine wave. envelope of the amplitude lines - the dashed curve in the figure. Features to be noted here are: the uniform line
Fourier Series
Fourier Series: Half-wave Rectifier. • Ex. A sinusoidal voltage Esin?t is passed through a half-wave rectifier that clips the negative portion of the wave.
Table 15
Table 15.4-1 The Fourier Series of Selected Waveforms. Function. Trigonometric Fourier Series. Square wave: 0 Half wave rectified sine wave: 0.
Fourier Series and Fourier Transform
Fourier series is used to get frequency spectrum of a time-domain signal of the complex exponential Fourier series for a half wave rectified sine wave.
Half-Wave Rectifiers
wave rectifier circuit will enable the student to advance to the analysis of The Fourier series for the half-wave rectified sine wave for the voltage.
Fourier series & transform Representation of Continuous Time
Obtain the relation between trigonometric and exponential Fourier series The Fourier series expansion of half wave symmetry signal contains odd ...
Lecture 4&5 MATLAB applications In Signal Processing
Using Fourier series expansion a square wave Exponential Fourier Series. The coefficient c ... For the full-wave rectifier waveform shown in Figure
CHAPTER 3. SPECTRUM REPRESENTATION58
3-4 Fourier Series
The examples in Sec. 3-3 show that we can synthesizeperiodicwaveforms by using a sum ofharmonically relatedsinusoids. Now, we want to describe a general theory that shows howany periodic signal can besynthesized with a sum of harmonically related sinusoids,although the sum may need an infinite number of
terms. This is the mathematical theory ofFourier serieswhich uses the following representation: x.t/D1X kD1a kej.2=T0/kt(3.19)whereT0is the fundamental period of the periodic signalx.t/. Thekthcomplex exponential in (3.19) has a
frequency equal tofkDk=T0Hz, so all the frequencies are integer multiples of the fundamental frequency
f0D1=T0Hz.7There are two aspects of the Fourier theory: analysis and synthesis. Starting fromx.t/and calculatingfakg
is calledFourier analysis. The reverse process of starting fromfakgand generatingx.t/is calledFourier
synthesis. In this section, we will concentrate on analysis. The formula in (3.19) is the general synthesis formula. When the complex amplitudes areconjugate- symmetric,i.e.,akDa k, the synthesis formula becomes a sum of sinusoids of the form x.t/DA0C1X kD1A kcos..2=T0/ktCk/(3.20) whereA0Da0, and the amplitude and phase of thekthterm come from the polar form,akD12Akejk. In
other words, the conditionakDa kis sufficient for the synthesized waveform to be arealfunction of time. By appropriate choice of the complex amplitudesakin (3.19), we can represent a number of interestingperiodic waveforms, such as square waves, triangle waves, rectified sinusoids, and so on. The fact that a
discontinuous square wave can be represented with an infinite number of sinusoids was one of the amazing
claims in Fourier"s famous thesis of 1807. It took many years before mathematicians were able to develop a
rigorous convergence proof to support Fourier"s claim.3-4.1 Fourier Series: Analysis
How do we derive the coefficients for the harmonic sum in (3.19), i.e., how do we go fromx.t/toak? The
answer is that we use theFourier series integralto perform Fourier analysis. The complex amplitudes for any
periodic signal can be calculated with the Fourier integral a kD1T 0 T0Z 0 x.t/e j.2=T0/ktdt(3.21) whereT0is the fundamental period ofx.t/. A special case of (3.21) is thekD0case for the DC component a0which is obtained by
a 0D1T 0T 0Z 0 x.t/dt(3.22)7There are three ways to refer to the fundamental frequency: radian frequency!0in rad/sec, cyclic frequencyf0in Hz, or with
the periodT0in sec. Each one has its merits in certain situations. The relationship among these is!0D2f0D2=T0.
c J. H. McClellan, R. W. Schafer, & M. A. Yoder DRAFT, for ECE-2026 Fall-2012, September 8, 2012CHAPTER 3. SPECTRUM REPRESENTATION59
A common interpretation of (3.22) is thata0is simply the average value of the signal over one period.
The Fourier integral (3.21) is convenient if we have a formula that definesx.t/over one period. Twoexamples will be presented later to illustrate this point. On the other hand, ifx.t/is known only as a recording,
then numerical methods such as those discussed in Chapters 66 and??will be needed.3-4.2 Fourier Series Derivation
In this section, we present a derivation of the Fourier series integral formula (3.21). The derivation relies on a
simple property of the complex exponential signal-the integral of a complex exponential over any number of
complete periods is zero. In equation form, T 0Z 0 e j.2=T0/ktdtD0(3.23) whereT0is a period of the complex exponential whose frequency is!kD.2=T0/k, andkis a nonzero integer. Here is the integration: T 0Z 0 e j.2=T0/ktdtDej.2=T0/ktj.2=T 0/k T 0 0 D ej.2=T0/kT01j.2=T 0/kD0The numerator is zero becauseej2kD1for any integerk(positive or negative). Equation (3.23) can also be
justified if we use Euler"s formula to separate the integral into its real and imaginary parts and then integrate
cosine and sine separately-each one overkcomplete periods: T 0Z 0 e j.2=T0/ktdtDT 0Z 0 cos..2=T0/kt/dt CjT 0Z 0 sin..2=T0/kt/dtD0Cj0A key ingredient in the infinite series representation (3.19) is the form of the complex exponentials, which
all must repeat with the same period as the period of the signalx.t/, which isT0. If we definevk.t/to be the
complex exponential of frequency!kD.2=T0/k, then v k.t/Dej.2=T0/kt(3.24)Even though the minimum duration period ofvk.t/might be smaller thanT0, the following shows thatvk.t/
still repeats with a period ofT0: v k.tCT0/Dej.2=T0/k.tCT0/Dej.2=T0/ktej.2=T0/kT0
Dej.2=T0/ktej2k
Dej.2=T0/ktDvk.t/
c J. H. McClellan, R. W. Schafer, & M. A. Yoder DRAFT, for ECE-2026 Fall-2012, September 8, 2012CHAPTER 3. SPECTRUM REPRESENTATION60
where again we have usedej2kD1for any integerk(positive or negative).Next we can generalize the zero-integral property (3.23) of the complex exponential to involve two signals:
8Othogonality Property
T 0Z 0 v k.t/v `.t/dtD( T0ifkD`(3.25)
where the * superscript inv `.t/denotes the complex conjugate. Proof:Proving the orthogonality property is straightforward. We begin with T 0Z 0 v k.t/v `.t/dtDT 0Z 0 e j.2=T0/ktej.2=T0/`tdt D T 0Z 0 e j.2=T0/.k`/tdtThere are two cases to consider for the last integral: whenkD`the exponent becomes zero, so the integral is
T 0Z 0 e j.2=T0/.k`/tdtDT 0Z 0 e j0tdt D T 0Z 01dtDT0
T 0Z 0 e j.2=T0/.k`/tdtDT 0Z 0 e j.2=T0/mtdtD0 tion. If we assume that (3.19) is valid, x.t/D1X kD1a kej.2=T0/kt8The integral in (3.25) is called the "inner product" betweenvk.t/andv`.t/, sometimes denoted ashvk.t/;v`.t/i.
c J. H. McClellan, R. W. Schafer, & M. A. Yoder DRAFT, for ECE-2026 Fall-2012, September 8, 2012CHAPTER 3. SPECTRUM REPRESENTATION61
then we can multiply both sides by the complex exponentialv `.t/and integrate over the periodT0. T 0Z 0 x.t/ e j.2=T0/`tdtD T 0Z 0 1X kD1a kej.2=T0/kt! e j.2=T0/`tdt D 1X kD1a k0 @T 0Z 0 e j.2=T0/.k`/tdt1 A nonzero only whenkD`Da`T0(3.26)Notice that we are able to isolate one of the complex amplitudes.a`/in the final step by applying the orthogo-
nality property (3.25).The crucial step in (3.26) occurs when the order of the infinite summation and the integration is swapped.
This is a delicate manipulation that depends on convergence properties of the infinite series expansion. It
was also a topic of research that occupied mathematicians for a good part of the early 19th century. For our
purposes, if we assume thatx.t/is a smooth function and has only a finite number of discontinuities within
one period, then the swap is permissible.The final analysis formula is obtained by dividing both sides of (3.26) byT0and writinga`on one side of
the equation. Since`could be any index, we can replace`withkto obtainFourier Analysis Equation a kD1T 0T 0Z 0 x.t/e j.2=T0/ktdt(3.27) where!0D2=T0D2f0is the fundamental frequency of the periodic signalx.t/. This analysis formulagoes hand in hand with the synthesis formula for periodic signals, which isFourier Synthesis Equation
x.t/D1X kD1a kej.2=T0/kt(3.28)3-5 Spectrum of the Fourier Series
When we discussed the spectrum in Section 3-1, we described a graphical procedure for drawing the spectrum
whenx.t/is composed of a sum of complex exponentials. By virtue of the synthesis formula (3.28), the Fourier
series coefficientsakare, in fact, the complex amplitudes that define the spectrum ofx.t/. In order to illustrate
this general connection between the Fourier series and the spectrum, we use the "sine-cubed" signal. First, we
derive theakcoefficients forx.t/Dsin3.3t/, and then we sketch its spectrum. c J. H. McClellan, R. W. Schafer, & M. A. Yoder DRAFT, for ECE-2026 Fall-2012, September 8, 2012 CHAPTER 3. SPECTRUM REPRESENTATION62CDROMExample 3-7: Fourier Series withoutIntegration
Determine the Fourier series coefficients of the signal: x.t/Dsin3.3t/Solution:There are two ways to get theakcoefficients: plugx.t/into the Fourier integral (3.27), or use
the inverse Euler formula to expandx.t/into a sum of complex exponentials. It is far easier to use the latter
approach. Using the inverse Euler formula for sin./, we make the following expansion of the sine-cubed
function: x.t/D ej3tej3t2j 3 D 18j ej9t3ej6tej3tC3ej3tej6tej9t D j8 ej9tC3j8 ej3tC3j8 ej3tCj8 ej9t(3.29) Weseethat(3.29)isthesumoffourtermswithfrequencies:!D 3and!D 9rad/s. Sincegcd.3;9/D3, the fundamental frequency is!0D3rad/sec. The Fourier series coefficients are indexed in terms of the
fundamental frequency, so a kD8:0forkD0
j38 forkD 10forkD 2
j18 forkD 30forkD 4;5;6;:::(3.30)
This example shows that it is not always necessary to evaluate an integral to obtain thefakgcoefficients.Now we can draw the spectrum (Fig. 3-13) because we know that we have four nonzeroakcomponents
located at the four frequencies:!D f9;3; 3; 9grad/sec. We prefer to plot the spectrum versus frequency in hertz in this case, so the spectrum lines9are atfD 4:5;1:5; 1:5;and 4.5 Hz. The second
harmonic is missing and the third harmonic is at 4.5 Hz.CDROMEXERCISE 3.5:Use the Fourier integral to determine all the Fourier series coefficients of the "sine-
cubed" signal. In other words, evaluate the integral a kD1T 0T 0Z 0 sin3.3t/ej.2=T0/ktdt
for allk.Hints: Find the period first, so that the integration interval is known. In addition, you might find it easier
to convert the sin3./function to exponential form (via the inverse Euler formula for sin./) before doing the
Fourier integral on each of four different terms. If you then invoke the orthogonality property on each integral,
you should get exactly the same answer as (3.30).9Repeating this footnote from Section 3.1 here for convenience:Spectra of signals comprised of individual sinusoids are often
called "line spectra". The term "line" seems appropriate for us here because we plot the components as vertical lines positioned at the
individual frequencies. However, the term originated in physics where lines are observed in emission or absorption spectra formed with
optical prisms (which serve as optical spectrum analyzers). These lines correspond to energy being emitted or absorbed at wavelengths
that are characteristic of atoms or ions. c J. H. McClellan, R. W. Schafer, & M. A. Yoder DRAFT, for ECE-2026 Fall-2012, September 8, 2012 CHAPTER 3. SPECTRUM REPRESENTATION634.5 1.5 1.5 4.50 fquotesdbs_dbs5.pdfusesText_9[PDF] exponential function to log calculator
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