Overview of CT Fourier Series Topics Orthogonality of CT complex - PowerPoint PPT Presentation

Overview of CT Fourier Series Topics • Orthogonality of CT complex sinusoidal harmonics • CT Fourier Series as a Design Task • Picking the frequencies • Picking the range • Finding the coefficients • Example J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 1

Motivation h ( t ) x ( t ) y ( t ) x ( t ) y ( t ) H ( jω ) e jωt → H ( jωt )e jωt X [ k ]e jω k t → � � X [ k ] H ( jω k )e jω k t k k � ∞ 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 2

CT Periodic Signals Design Task x ( t ) 1 t -2 -1 0 1 2 � 2( t + kT 0 ) kT 0 < t ≤ kT 0 + 0 . 5 x ( t ) = 1 kT 0 + 0 . 5 < t ≤ kT 0 + 1 • Suppose we have a CT periodic signal x ( t ) with fundamental period T • 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

DT Periodic Signals Design Task � X [ k ] e jω k t x ( t ) = ˆ k • The ˆ symbol indicates that the sum is an approximation (estimate) of x ( t ) • Enables us to calculate the system output easily • Must pick – The frequencies ω k – The range of the sum � k – The coefficients X [ k ] J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 4

Design Task: Picking the Frequencies � X [ k ] e jω k t x ( t ) = ˆ k • We know x ( t ) is periodic with some fundamental period T • If ˆ x ( t ) is to approximate x ( t ) accurately, it should also repeat every T seconds • In order for ˆ x ( t ) to be periodic with period T , every complex 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

Design Task: Picking the Range � X [ k ] e jkωt x ( t ) = ˆ k • Unlike DT complex sinusoids, e jkωt � = e jℓωt unless k = ℓ • Thus the range of the sum must be infinite to include all possible frequencies • This is different than the DT case J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 6

Design Task: Picking the Coefficients ∞ MSE = 1 � x ( t ) | 2 d t � X [ k ] e jkωt x ( t ) = ˆ | x ( t ) − ˆ T T k = −∞ • We would like to pick the coefficients X [ k ] so that ˆ x ( t ) is as close to x ( t ) as possible • But what is close? • One measure of the difference between two signals is the mean squared error ( MSE ) • 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

Orthogonality Two periodic signals x 1 ( t ) and x 2 ( t ) with the same period T are orthogonal if and only if � x 1 ( t ) x ∗ 2 ( t ) d t = 0 T � where T denotes an integral over any contiguous interval of duration T , � t 0 + T � x ( t ) d t = x ( t ) d t for any t 0 T t 0 J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 8

Orthogonality: Complex Sinusoids Consider two harmonic complex sinusoids x 1 ( t ) = e jk 1 ωt x 2 ( t ) = e jk 2 ωt Are they orthogonal? � � e jk 1 ωt e − jk 2 ωt d t x 1 ( t ) x ∗ 2 ( t ) d t = 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

Importance of Orthogonality Suppose that we know a signal is composed of a linear combination of harmonic complex sinusoids with fundamental period T ∞ � X [ k ] e jkωt x ( t ) = k = −∞ How do we solve for the coefficients X [ k ] for all k ? J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 10

Workspace J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 11

Design Task: Coefficient Optimization ∞ MSE = 1 � x ( t ) | 2 d t � X [ k ] e jkωt x ( t ) = ˆ | 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 12

CTFS Observations ∞ X [ k ] = 1 � x ( t )e − jkωt d t � X [ k ] e jkωt x ( t ) = T T k = −∞ • The first equation is called the synthesis equation • The second equation is called the analysis equation • The coefficients X [ k ] are called the spectral coefficients or Fourier series coefficients of x [ n ] • We denote the relationship of x ( T ) and X [ k ] by FS x ( t ) ⇐ ⇒ X [ k ] • 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

Discontinuities ∞ X [ k ] = 1 � x ( t )e − jkωt d t � X [ k ]e jkωt x ( t ) = ˆ T T k = −∞ • Just because MSE = 0 does not imply x ( t ) = ˆ x ( t ) • It does imply any differences occur only at a finite number of discrete (zero duration) points in time • In general, if t 0 is a point of discontinuity, then x ( t 0 ) = 1 ˆ 2 lim △→ 0 [ x ( t 0 + △ ) + x ( t 0 − △ )] • At all other points the signals are equal J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 14

Convergence ∞ X [ k ] = 1 � x ( t )e − jkωt d t � X [ k ]e jkωt x ( t ) = ˆ T T k = −∞ • Since the CTFS includes an infinite series, we must consider under what conditions it converges • An infinite sum is said to converge so long as it is bounded – Not infinite K X [ k ]e jkωt < ∞ � −∞ < lim K →∞ k = − K • Why didn’t we do this in the DT case? • Why don’t we have to consider convergence of the analysis equation? J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 15

Convergence Continued A sufficient condition for convergence (not proven) is � | x ( t ) | 2 d t < ∞ T • In other words, the signal has – Finite power – Finite energy over a single period • This is true of all signals you could generate in the lab • If x ( t ) is a continuous signal, then it is safe to assume that x ( t ) = x ( t ) ˆ • This is a stronger statement then merely stating that the CTFS converges • All of the periodic signals generated by a function generator have an equivalent FS representation J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 16

Dirichlet Conditions for Convergence The Fourier series representation of a periodic signal x ( t ) converges if all of the following conditions are met. � 1. T | x ( t ) | d t < ∞ 2. Finite number of discontinuities in a period T 3. Finite number of distinct maxima and minima in T 4. x ( t ) is single valued These are sufficient, but not necessary, conditions. J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 17

Example 4: CT Fourier Series Coefficients x ( t ) 1 t -2 -1 0 1 2 Find the Fourier series coefficients for the signal shown above. Plot k = − N X [ k ]e jkωt of the Fourier series for N = 1, x ( t ) ≈ � N partial sums ˆ 2, 5, 10, 50 & 100. t e at d t = e at d t = 1 a e at + C . 1 � a 2 e at ( at − 1) + C and � Hints: J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 18

Example 4: Workspace J. McNames Portland State University ECE 223 CT Fourier Series Ver. 1.07 19

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 20

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

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

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