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h ( t ) x ( t ) y ( t ) x ( t ) y ( t ) H ( j ) CT Fourier Series as - PowerPoint PPT Presentation

Oct 21, 2022 •160 likes •339 views

Overview of CT Fourier Series Topics Motivation Orthogonality of CT complex sinusoidal harmonics h ( t ) x ( t ) y ( t ) x ( t ) y ( t ) H ( j ) CT Fourier Series as a Design Task Picking the frequencies e jt H ( jt )e jt

  1. Overview of CT Fourier Series Topics Motivation • Orthogonality of CT complex sinusoidal harmonics h ( t ) x ( t ) y ( t ) x ( t ) y ( t ) H ( jω ) • CT Fourier Series as a Design Task • Picking the frequencies e jωt → H ( jωt )e jωt • Picking the range X [ k ]e jω k t → � � X [ k ] H ( jω k )e jω k t • Finding the coefficients k k � ∞ • Example h ( t )e − jωt d t H ( jω ) = F { h ( t ) } = −∞ • For now, we restrict out attention to CT periodic signals: x ( t + T ) = x ( t ) , T > 0 • Would like to represent x ( t ) as a sum of complex sinusoids • Why? Gives us insight and simplifies computation J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 1 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 2 CT Periodic Signals Design Task DT Periodic Signals Design Task � X [ k ] e jω k t x ( t ) = ˆ x ( t ) 1 k • The ˆ symbol indicates that the sum is an approximation t -2 -1 0 1 2 (estimate) of x ( t ) � • Enables us to calculate the system output easily 2( t + kT 0 ) kT 0 < t ≤ kT 0 + 0 . 5 x ( t ) = • Must pick 1 kT 0 + 0 . 5 < t ≤ kT 0 + 1 – The frequencies ω k • Suppose we have a CT periodic signal x ( t ) with fundamental – The range of the sum � period T k – The coefficients X [ k ] • The signal is applied at the input of an LTI system • We would like to estimate the signal as a sum of complex sinusoids � X [ k ] e jω k t x ( t ) = ˆ k J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 3 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 4

  2. Design Task: Picking the Frequencies Design Task: Picking the Range � � X [ k ] e jω k t X [ k ] e jkωt x ( t ) = ˆ x ( t ) = ˆ k k • We know x ( t ) is periodic with some fundamental period T • Unlike DT complex sinusoids, e jkωt � = e jℓωt unless k = ℓ • If ˆ x ( t ) is to approximate x ( t ) accurately, it should also repeat • Thus the range of the sum must be infinite to include all possible every T seconds frequencies • In order for ˆ x ( t ) to be periodic with period T , every complex • This is different than the DT case sinusoid must also be periodic • Only a harmonic set of complex sinusoids have this property • Thus ω k = kω where ω = 2 π T � X [ k ] e jkωt x ( t ) = ˆ k J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 5 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 6 Design Task: Picking the Coefficients Orthogonality ∞ Two periodic signals x 1 ( t ) and x 2 ( t ) with the same period T are MSE = 1 � x ( t ) | 2 d t � X [ k ] e jkωt x ( t ) = ˆ | x ( t ) − ˆ orthogonal if and only if T T k = −∞ � x 1 ( t ) x ∗ 2 ( t ) d t = 0 • We would like to pick the coefficients X [ k ] so that ˆ x ( t ) is as close T to x ( t ) as possible � where T denotes an integral over any contiguous interval of duration T , • But what is close? � t 0 + T � • One measure of the difference between two signals is the mean x ( t ) d t = x ( t ) d t for any t 0 squared error ( MSE ) T t 0 • There are other measures, but this is a convenient one because we can differentiate it • If MSE = 0 , does this imply x ( t ) = ˆ x ( t ) ? • Since the signal is periodic, the MSE is calculated over a single fundamental period of T • How do we pick the coefficients X [ k ] to minimize the MSE ? J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 7 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 8

  3. Orthogonality: Complex Sinusoids Importance of Orthogonality Consider two harmonic complex sinusoids Suppose that we know a signal is composed of a linear combination of harmonic complex sinusoids with fundamental period T x 1 ( t ) = e jk 1 ωt x 2 ( t ) = e jk 2 ωt ∞ � X [ k ] e jkωt Are they orthogonal? x ( t ) = k = −∞ � � e jk 1 ωt e − jk 2 ωt d t x 1 ( t ) x ∗ 2 ( t ) d t = How do we solve for the coefficients X [ k ] for all k ? T T � e j ( k 1 − k 2 ) ωt d t = T ? = 0 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 9 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 10 Workspace Design Task: Coefficient Optimization ∞ MSE = 1 � x ( t ) | 2 d t � X [ k ] e jkωt ˆ x ( t ) = | x ( t ) − ˆ T T k = −∞ • We’ve already solved for the coefficients using orthogonality • It turns out (see advanced texts or classes) that this solution results in MSE = 0 • But unlike the DT case, this does NOT imply ˆ x ( t ) = x ( t ) necessarily • But the squared error has zero area, so any difference is probably negligible J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 11 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 12

  4. CTFS Observations Discontinuities ∞ ∞ X [ k ] = 1 � X [ k ] = 1 � x ( t )e − jkωt d t x ( t )e − jkωt d t � X [ k ] e jkωt � X [ k ]e jkωt x ( t ) = x ( t ) = ˆ T T T T k = −∞ k = −∞ • Just because MSE = 0 does not imply x ( t ) = ˆ x ( t ) • The first equation is called the synthesis equation • It does imply any differences occur only at a finite number of • The second equation is called the analysis equation discrete (zero duration) points in time • The coefficients X [ k ] are called the spectral coefficients or • In general, if t 0 is a point of discontinuity, then Fourier series coefficients of x [ n ] x ( t 0 ) = 1 • We denote the relationship of x ( T ) and X [ k ] by ˆ 2 lim △→ 0 [ x ( t 0 + △ ) + x ( t 0 − △ )] FS x ( t ) ⇐ ⇒ X [ k ] • At all other points the signals are equal • Both are complete representations of the signal: if we know one, we can compute the other • X [ k ] is a function of frequency ( kω ) J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 13 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 14 Convergence Convergence Continued A sufficient condition for convergence (not proven) is ∞ X [ k ] = 1 � x ( t )e − jkωt d t � X [ k ]e jkωt x ( t ) = ˆ � T | x ( t ) | 2 d t < ∞ T k = −∞ T • Since the CTFS includes an infinite series, we must consider under • In other words, the signal has what conditions it converges – Finite power • An infinite sum is said to converge so long as it is bounded – Finite energy over a single period – Not infinite • This is true of all signals you could generate in the lab K X [ k ]e jkωt < ∞ � • If x ( t ) is a continuous signal, then it is safe to assume that −∞ < lim K →∞ x ( t ) = x ( t ) ˆ k = − K • This is a stronger statement then merely stating that the CTFS • Why didn’t we do this in the DT case? converges • Why don’t we have to consider convergence of the analysis • All of the periodic signals generated by a function generator have equation? an equivalent FS representation J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 15 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 16

  5. Dirichlet Conditions for Convergence Example 4: CT Fourier Series Coefficients The Fourier series representation of a periodic signal x ( t ) converges if x ( t ) all of the following conditions are met. 1 � 1. T | x ( t ) | d t < ∞ t -2 -1 0 1 2 2. Finite number of discontinuities in a period T Find the Fourier series coefficients for the signal shown above. Plot 3. Finite number of distinct maxima and minima in T k = − N X [ k ]e jkωt of the Fourier series for N = 1, x ( t ) ≈ � N partial sums ˆ 4. x ( t ) is single valued 2, 5, 10, 50 & 100. t e at d t = e at d t = 1 a e at + C . � a 2 e at ( at − 1) + C and 1 � Hints: These are sufficient, but not necessary, conditions. J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 17 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 18 Example 4: Workspace Example 4: Fourier Series Coefficients Fourier Series Coefficients 0.8 0.6 |X[k]| 0.4 0.2 0 −25 −20 −15 −10 −5 0 5 10 15 20 25 4 2 ∠ X[k] 0 −2 −4 −25 −20 −15 −10 −5 0 5 10 15 20 25 kth (harmonic) J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 19 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 20

  6. Example 4: Partial Fourier Series N =1 Example 4: Partial Fourier Series N =2 Fourier Series Approximation (N=1) Fourier Series Approximation (N=2) 1.2 1.2 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 −0.2 −0.2 −0.4 −0.4 −1.5 −1 −0.5 0 0.5 1 1.5 −1.5 −1 −0.5 0 0.5 1 1.5 Time (sec) Time (sec) J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 21 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 22 Example 4: Partial Fourier Series N =5 Example 4: Partial Fourier Series N =10 Fourier Series Approximation (N=5) Fourier Series Approximation (N=10) 1.2 1.2 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 −0.2 −0.2 −0.4 −0.4 −1.5 −1 −0.5 0 0.5 1 1.5 −1.5 −1 −0.5 0 0.5 1 1.5 Time (sec) Time (sec) J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 23 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 24

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