EI331 Signals and Systems Lecture 9 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 9 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University March 26, 2019

Contents 1. Eigenvalues and Eigenfunctions 2. CT Fourier Series 3. Properties of CT Fourier Series 1/31

Recap R n DT signals C Z CT signals { δ k : k ∈ Z } { δ τ : τ ∈ R } 1 basis { e 1 , . . . , e n } � � � n general x x = x = x [ k ] δ k x = x ( τ ) δ τ d τ x k e k R k = 1 k ∈ Z Image under linear transformation basis { f 1 , . . . , f n } { h k : k ∈ Z } { h τ : τ ∈ R } � � � n general x Ax = T ( x ) = x [ k ] h k T ( x ) = x ( τ ) h τ d τ x k f k R k = 1 k ∈ Z LTI – T ( x ) = x ∗ h T ( x ) = x ∗ h 1 Not standard usage of term “basis”; understood in terms of integral representation on next line. 2/31

Eigenvalues and Eigenvectors In linear algebra, v ∈ R n is eigenvector of matrix A with associated eigenvalue λ , if Av = λ v Suppose A has n distinct eigenvalues λ 1 , . . . , λ n associated to eigenvectors v 1 , . . . , v n , respectively. • v 1 , . . . , v n form a basis of R n , i.e. ∀ x ∈ R n has following representation � n x = a k v k k = 1 • image under A � n � n Ax = a k Av k = a k λ k v k k = 1 k = 1 3/31

Change of Basis standard basis eigenbasis of A basis vectors { e 1 , . . . , e n } { v 1 , . . . , v n } � n � n decomposition x = x = x k e k a k v k k = 1 k = 1 � n � n image under A Ax = Ax = a k λ k v k x k f k k = 1 k = 1 Expansions of x and Ax • in standard basis: same coefficients { x k } , different basis vectors { e k } �→ { f k } • in eigenbasis: same basis vectors { v k } , different coefficients { a k } �→ { λ k a k } 4/31

Change of Basis standard basis eigenbasis basis vectors { δ k : k ∈ Z } ? � DT decomposition x = x [ k ] δ k k ∈ Z � image under T T ( x ) = x [ k ] h k k ∈ Z “standard basis” eigenbasis basis vectors { δ τ : τ ∈ R } ? � CT x = x ( τ ) δ τ d τ decomposition R � T ( x ) = x ( τ ) h τ d τ image under T R 5/31

Eigenvalues and Eigenfunctions of LTI Systems CT Exponential e st � � h ( τ ) e s ( t − τ ) d τ = e st h ( τ ) e − s τ d τ, T ( e st ) = R R e st is eigenfunction of T with associated eigenvalue � h ( τ ) e − s τ d τ H ( s ) = (system function) R DT Exponential z n � h [ k ] z n − k d τ = z n � T ( z n ) = h [ k ] z − k , k ∈ Z k ∈ Z z n is eigenfunction of T with associated eigenvalue � h [ k ] z − k H ( z ) = (system function) k ∈ Z 6/31

Response to Linear Combination of Eigenfunctions �� � � a k e s k t a k H ( s k ) e s k t CT = T k k �� � � a k z n a k H ( z k ) z n = DT T k k k k Input vs. output • linear combinations of same exponentials • different coefficients { a k } �→ { H ( s k ) a k } / { H ( z k ) a k } Questions • Which functions are linear combinations of exponentials? • How to find coefficients a k ? Fourier analysis focuses on e j ω t for CT and e j ω n for DT 7/31

Contents 1. Eigenvalues and Eigenfunctions 2. CT Fourier Series 3. Properties of CT Fourier Series 8/31

CT Fourier Series Recall CT signal is periodic with period T if x ( t ) = x ( t + T ) , ∀ t ∈ R x = τ T x or • fundamental period T : smallest positive period • fundamental frequency ω 0 = 2 π T sin( ω 0 t ) , cos( ω 0 t ) , e j ω 0 t periodic with fundamental frequency ω 0 Fourier series represent periodic signals in terms of harmonically related sinusoids or complex exponentials � ∞ x ( t ) = a 0 + [ a k cos( k ω 0 t ) + b k sin( k ω 0 t )] k = 1 � ∞ c k e jk ω 0 t x ( t ) = k = −∞ 9/31

Joseph Fourier 1807, M´ emoire sur la propagation de la chaleur dans les corps solides ( Memoir on the propagation of heat in solid bodies ) 1822, Th´ eorie analytique de la chaleur ( The Analytical Theory of Heat ) Jean-Baptiste Joseph Fourier (from Wikipedia) ( https://gallica.bnf.fr ) 10/31

Harmonics ω ω 0 2 ω 0 3 ω 0 4 ω 0 5 ω 0 6 ω 0 0 | k | : harmonic # 0 3 5 6 1 2 4 D f s t f fi s u h o e f i C u t x n i c r h t d d r h o t a h n h h h a m d a h a r e r a m h r m n m r a o m t o r o a n m n o n l i c i / n i o c c i fi n c r i s c t h a r m o n i c 11/31

Harmonics ω ω 0 2 ω 0 3 ω 0 4 ω 0 5 ω 0 6 ω 0 0 t T = 2 π ω 0 t T = 2 π ω 0 12/31

Orthonormality of Harmonics Recall for k � = 0 , � b e jk ω 0 t dt = e jk ω 0 b − e jk ω 0 a jk ω 0 a Since e jk ω 0 t has period T , over any period � � t 0 + T 1 e jk ω 0 t dt = 1 e jk ω 0 t dt = δ [ k ] T T T t 0 Define inner product between two signals with period T by � � f , g � = 1 f ( t ) g ( t ) dt T T { e jk ω 0 t : k ∈ Z } is orthonormal system of functions � � e jk ω 0 t , e jm ω 0 t � = 1 e j ( k − m ) ω 0 t dt = δ km = δ [ k − m ] T T 13/31

Orthogonality of Harmonics For sines and cosines, � sin( k ω 0 t ) , sin( m ω 0 t ) � = 1 2 δ [ k − m ] − 1 2 δ [ k + m ] � cos( k ω 0 t ) , cos( m ω 0 t ) � = 1 2 δ [ k − m ] + 1 2 δ [ k + m ] � sin( k ω 0 t ) , cos( m ω 0 t ) � = 0 Proof. Use Euler’s formula and orthonormality of { e jk ω 0 t : k ∈ Z } 14/31

Orthogonality of Harmonics sin( ω 0 t ) t O T = 2 π sin( 2 ω 0 t ) ω 0 sin( ω 0 t ) t O T = 2 π cos( ω 0 t ) ω 0 15/31

Fourier Coefficients Suppose x has period T and Fourier series representation � ∞ ω 0 = 2 π c k e jk ω 0 t , x ( t ) = T k = −∞ Find Fourier coefficients using orthonormality of { e jk ω 0 t } � ∞ � x , e jm ω 0 t � = � c k e jk ω 0 t , e jm ω 0 t � k = −∞ � ∞ c k � e jk ω 0 t , e jm ω 0 t � = k = −∞ � ∞ = c k δ [ m − k ] = c m k = −∞ 16/31

Complex Fourier Series Synthesis equation � � ∞ ∞ x [ k ] e jk ω 0 t = x [ k ] e jk 2 π T t x ( t ) = ˆ ˆ k = −∞ k = −∞ N -th partial sum � N x [ k ] e jk ω 0 t S N ( x )( t ) = ˆ k = − N Analysis equation � � x [ k ] = � x , e jk ω 0 t � = 1 x ( t ) e − jk ω 0 t dt = 1 x ( t ) e − jk 2 π T t dt ˆ T T T T 17/31

Trigonometric Fourier Series For sines and cosines, � sin( k ω 0 t ) , sin( m ω 0 t ) � = 1 2 δ [ k − m ] − 1 2 δ [ k + m ] � cos( k ω 0 t ) , cos( m ω 0 t ) � = 1 2 δ [ k − m ] + 1 2 δ [ k + m ] � sin( k ω 0 t ) , cos( m ω 0 t ) � = 0 Synthesis equation � ∞ x ( t ) = a 0 + [ a k cos( k ω 0 t ) + b k sin( k ω 0 t )] k = 1 Analysis equation � � a k = 2 − δ [ k ] b k = 2 x ( t ) cos( k ω 0 t ) dt , x ( t ) sin( k ω 0 t ) dt T T T T 18/31

Equivalence of Two Forms Complex form � ∞ x [ k ] e jk ω 0 t x ( t ) = ˆ k = −∞ Trigonometric form � ∞ x ( t ) = a 0 + [ a k cos( k ω 0 t ) + b k sin( k ω 0 t )] k = 1 Conversion of coefficients (by Euler’s formula)     a 0 = ˆ x [ 0 ] x [ 0 ] = a 0 ˆ   x [ k ] = 1 a k = ˆ x [ k ] + ˆ x [ − k ] , k ≥ 1 ˆ 2 ( a k − jb k ) , k ≥ 1     x [ k ] = 1 b k = j (ˆ x [ k ] − ˆ x [ − k ]) , k ≥ 1 ˆ 2 ( a k + jb − k ) , k ≤ − 1 Negative frequencies in complex form introduced for mathematical convenience, no physical significance 19/31

Examples Example. x ( t ) = e j ω 0 t , ˆ x [ k ] = δ [ k − 1 ] 1 Spectrum ˆ x [ k ] − 3 − 2 − 1 0 1 2 3 k 2 e j ω 0 t + e − j φ Example. x ( t ) = cos( ω 0 t + φ ) = e j φ 2 e − j ω 0 t , x [ 1 ] = 1 x [ − 1 ] = 1 2 e j φ , 2 e − j φ , ˆ , ˆ x [ k ] = 0 , k � = ± 1 ˆ 1 1 2 2 Magnitude spectrum | ˆ x [ k ] | − 3 − 2 − 1 0 1 2 3 k φ − 1 Phase spectrum arg ˆ x [ k ] − 3 − 2 0 1 2 3 k − φ 20/31

Example: Triangle Wave In one period, x ( t ) = 1 − 2 | t | | t | ≤ T T , 2 x ( t ) t − T T 2 2 Fourier coefficients � � � T / 2 1 − 2 | t | x [ 0 ] = 1 dt = 1 k = 0 , ˆ T T 2 − T / 2 � � � T / 2 1 − 2 | t | x [ k ] = 1 2 e − jk ω 0 t dt = k � = 0 , k odd , ˆ π 2 k 2 T T − T / 2 k � = 0 , k even , x [ k ] = 0 ˆ 21/31

Example: Triangle Wave  1  2 , k = 0  ˆ x [ k ] = 2 π 2 k 2 , k � = 0 odd   0 , k � = 0 even Spectrum for fixed T , frequency spacing ∆ ω = 2 π T ω k = k 2 π 1 T 2 − 6 − 5 − 4 − 3 − 2 − 1 0 1 2 3 4 5 6 k 22/31

Example: Triangle Wave � 0 x [ k ] e jk ω 0 t S 0 ( x )( t ) = ˆ k = − 0 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 23/31

Example: Triangle Wave � 1 x [ k ] e jk ω 0 t S 1 ( x )( t ) = ˆ k = − 1 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 23/31

Example: Triangle Wave � 3 x [ k ] e jk ω 0 t S 3 ( x )( t ) = ˆ k = − 3 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 23/31

Example: Triangle Wave � 5 x [ k ] e jk ω 0 t S 5 ( x )( t ) = ˆ k = − 5 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 23/31

Example: Triangle Wave � 7 x [ k ] e jk ω 0 t S 7 ( x )( t ) = ˆ k = − 7 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 23/31