EI331 Signals and Systems Lecture 18 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 18 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University April 25, 2019

Contents 1. DT Fourier Transform 2. Properties of DT Fourier Transform 1/33

DT Fourier Transform Let x be aperiodic DT signal with supp x ⊂ [ N 1 , N 2 ] x [ n ] n 0 Define periodic extension with period N > N 2 − N 1 + 1 , ∞ � x N [ n ] = x [ n + kN ] k = −∞ x N [ n ] n N 0 x is “periodic” with infinite period, i.e. x [ n ] = lim N →∞ x N [ n ] 2/33

DT Fourier Transform Expand x N into DT Fourier series x N [ n ] n 0 N N 2 N k � = 1 N k = 1 x N [ k ] = � x N , e j 2 π � x [ n ] e − j 2 π N X ( e jk ω 0 ) ˆ N n = N 1 where N 2 ∞ ω 0 = 2 π x [ n ] e − j ω n = � � X ( e j ω ) = x [ n ] e − j ω n , N n = N 1 n = −∞ X ( e j ω ) N ˆ x N [ k ] k ω 0 3/33

DT Fourier Transform As N increases, discrete frequencies sampled more densely x N [ n ] n 0 N N 2 N k � = 1 N k = 1 x N [ k ] = � x N , e j 2 π x [ n ] e − j 2 π � N X ( e jk ω 0 ) ˆ N n = N 1 where N 2 ∞ ω 0 = 2 π x [ n ] e − j ω n = � � X ( e j ω ) = x [ n ] e − j ω n , N n = N 1 n = −∞ X ( e j ω ) N ˆ x N [ k ] k ω 0 4/33

DT Fourier Transform x [ k ] approaches envelope X ( e j ω ) As N → ∞ , N ˆ x N [ n ] n N 0 N 2 N k � = 1 N k = 1 x N [ k ] = � x N , e j 2 π x [ n ] e − j 2 π � N X ( e jk ω 0 ) ˆ N n = N 1 where ∞ N 2 ω 0 = 2 π x [ n ] e − j ω n = X ( e j ω ) = � � x [ n ] e − j ω n , N n = N 1 n = −∞ X ( e j ω ) N ˆ x N [ k ] k ω 0 5/33

DT Fourier Transform Synthesis equation for DT Fourier series, N − 1 x [ k ] e jk ω 0 n = 1 � � X ( e jk ω 0 ) e jk ω 0 n x N [ n ] = ˆ N k ∈ [ N ] k = 0 Let ω k = k ω 0 , ∆ ω = ω k − ω k − 1 = ω 0 , N − 1 x N [ n ] = 1 � X ( e j ω k ) e j ω k n ∆ ω 2 π k = 0 As N → ∞ , ∆ ω → 0 , above Riemann sum becomes integral, � 2 π � x [ n ] = 1 X ( e j ω ) e j ω n d ω = 1 X ( e j ω ) e j ω n d ω 2 π 2 π 0 2 π Can integrate over any period of length 2 π 6/33

DT Fourier Transform Pair DT Fourier transform (analysis equation) ∞ � X ( e j ω ) = F ( x )( e j ω ) = x [ n ] e − j ω n n = −∞ • X ( e j ω ) called spectrum of x [ n ] , periodic with period 2 π DT Inverse Fourier transform (synthesis equation) � x [ n ] = F − 1 ( X )[ n ] = 1 X ( e j ω ) e j ω n d ω 2 π 2 π • Complex exponential at frequency ω has weight X ( e j ω ) d ω 2 π • Integrate over single period , i.e. only use distinct e j ω n 7/33

Duality between DTFT and CTFS DTFT pair CTFS pair • discrete time • continuous time • continuous frequency • discrete frequency analysis equation analysis equation ∞ � x [ k ] = 1 x ( t ) e − jk 2 π T t dt � ˆ X ( e j ω ) = x [ n ] e − j ω n T T n = −∞ synthesis equation synthesis equation ∞ � x [ n ] = 1 X ( e j ω ) e j ω n d ω � x [ k ] e jk 2 π T t x ( t ) = ˆ 2 π 2 π k = −∞ CTFS continuous variable − − − → doubly infinite sequences ← − − − periodic functions DTFT 8/33

High vs. Low Frequencies for DT Signals High frequencies around ( 2 k + 1 ) π , low frequencies around 2 k π x x x n n n φ 0 φ 1 φ 2 N [ n ] = cos( 0 · n ) = 1 N [ n ] = cos( π n / 4 ) N [ n ] = cos( π n / 2 ) x x x n n n φ 3 φ 4 φ 5 N [ n ] = cos( 3 π n / 4 ) N [ n ] = cos( π n ) N [ n ] = cos( 5 π n / 4 ) x x x n n n φ 6 φ 7 φ 8 N [ n ] = cos( 3 π n / 2 ) N [ n ] = cos( 7 π n / 4 ) N [ n ] = cos( 2 π n ) = 1 9/33

High vs. Low Frequencies for DT Signals Discrete frequencies of N -periodic signals • evenly spaced points on unit circle • low frequencies close to 1 ; high frequencies close to − 1 Im Im e j 1 2 π e j 2 2 π 5 8 e j 3 2 π e j 1 2 π 8 8 e j 2 2 π 5 e j 0 2 π e j 0 2 π 8 5 high high low low Re Re e j 4 2 π 8 e j 3 2 π 5 e j 5 2 π e j 7 2 π 8 8 e j 6 2 π e j 4 2 π 8 5 N = 8 N = 5 10/33

High vs. Low Frequencies for DT Signals x 1 [ n ] X 1 ( e j ω ) low ω n π − π − 4 π − 2 π 0 0 2 π 4 π x 2 [ n ] X 2 ( e j ω ) high ω n π − π − 4 π − 2 π 0 0 2 π 4 π 11/33

Example: One-sided Decaying Exponential 1 F x [ n ] = a n u [ n ] → X ( e j ω ) = ← − − 1 − ae − j ω , | a | < 1 a sin ω 1 | X ( e j ω ) | = arg X ( e j ω ) = − arctan √ 1 − 2 a cos ω + a 2 , 1 − a cos ω X 1 ( e j ω ) X 2 ( e j ω ) a = 0 . 5 a = − 0 . 5 1 1 1 − a 1 + a ω ω π π − π 1 − π 1 − 2 π 0 2 π − 2 π 0 2 π 1 + a 1 − a X 1 ( e j ω ) X 2 ( e j ω ) √ a arctan | a | √ arctan 1 − a 2 1 − a 2 ω ω π π − π − π − 2 π − 2 π 0 2 π 0 2 π √ a | a | − arctan √ − arctan 1 − a 2 1 − a 2 12/33

Example: Two-sided Decaying Exponential 1 − a 2 F x [ n ] = a | n | → X ( e j ω ) = ← − − 1 − 2 a cos ω + a 2 , | a | < 1 x 1 [ n ] x 2 [ n ] a = 0 . 5 a = − 0 . 5 n n X 1 ( e j ω ) X 2 ( e j ω ) 1 − a 1 + a 1 − a 1 + a 1 + a 1 − a 1 − a 1 + a ω ω π π − π − π − 2 π 2 π − 2 π 2 π 0 0 13/33

Example: Rectangular Pulse → X ( e j ω ) = sin( 2 N 1 + 1 ω ) F 2 x [ n ] = u [ n + N 1 ] − u [ n − N 1 − 1 ] ← − − sin ω 2 x [ n ] n − N 1 0 N 1 X ( e j ω ) ω π − π − 3 π − 2 π 0 2 π 3 π Dirichlet kernel (periodic) is DT counterpart of sinc (aperiodic) F → X ( e j ω ) = 1 NB. N 1 = 0 = ⇒ x [ n ] = δ [ n ] ← − − 14/33

Convergence of DTFT Let N � X N ( e j ω ) = x [ n ] e − j ω n n = − N Theorem. If x ∈ ℓ 1 , then X N converges to X uniformly, i.e. N →∞ � X − X N � ∞ = 0 lim Consequently, Fourier transform X is uniformly continuous. Theorem. If x ∈ ℓ 2 , then X N converges to X in L 2 , i.e. N →∞ � X − X N � 2 = 0 lim • CTFS: L 2 ( T ) → ℓ 2 is isomorphism between Hilbert spaces • DTFT: ℓ 2 → L 2 ( 2 π ) is also isomorphism 15/33

Example: Ideal Lowpass Filter ∞ x [ n ] = sin( Wn ) F → X ( e j ω ) = � ← − − [ u ( ω + W + 2 k π ) − u ( ω − W + 2 k π )] π n k = −∞ x [ n ] n 0 X ( e j ω ) 0 < W < π 1 ω π − π − 2 π − W 0 W 2 π NB. x / ∈ ℓ 1 and X discontinuous 16/33

Convergence as Generalized Function Complex exponential x [ n ] = e j ω 0 n not in ℓ 1 or ℓ 2 . (Shifted) Dirichlet kernel ∞ X N ( e j ω n ) = sin( 2 N 1 + 1 ( ω − ω 0 )) � 2 → 2 π δ ( ω − ω 0 − 2 ℓπ ) sin ω − ω 0 2 ℓ = −∞ in the sense of distribution (generalized function), i.e. ∞ F � x [ n ] = e j ω 0 n → X ( e j ω ) = 2 π ← − − δ ( ω − ω 0 − 2 ℓπ ) ℓ = −∞ Formal verification using synthesis equation � π � π x [ n ] = 1 X ( e j ω ) e j ω n d ω = δ ( ω − ω 0 ) e j ω n d ω = e j ω 0 n 2 π − π − π NB. Can also obtain by sampling e j ω 0 t with T = 1 , ω s = 2 π NB. x [ n ] not necessarily periodic! DC corresponds to ω 0 = 0 . 17/33

Example: Sinusoids x [ n ] = cos( ω 0 n + φ ) = e j φ 2 e j ω 0 n + e − j φ 2 e − j ω 0 n ∞ � X ( e j ω ) = π e j φ δ ( ω − ω 0 − 2 ℓπ ) + π e − j φ δ ( ω + ω 0 − 2 ℓπ ) ℓ = −∞ | X ( e j ω ) | π π ω ω 0 π − π − ω 0 − 3 π − 2 π 0 2 π 3 π arg X ( e j ω ) φ − ω 0 ω ω 0 π − π − 3 π − 2 π 0 2 π 3 π − φ 18/33

DTFT for Periodic Signals x N [ n ] n N 0 x [ n ] n 0 Recall DTFS coefficient ˆ x N [ k ] is amplitude at frequency k ω 0 ω 0 = 2 π x N [ k ] = 1 N X ( e jk ω 0 ) , ˆ where X = F { x } , N X ( e j ω ) N ˆ x N [ k ] k ω 0 19/33

DTFT for Periodic Signals x N [ n ] n N 0 x [ n ] n 0 DTFT ( 2 π ) − 1 X ( e j ω ) is density at frequency ω ∞ ∞ � � X ( e j ω ) = 2 π X ( e jk ω 0 ) δ ( ω − k ω 0 ) ˆ x N [ k ] δ ( ω − k ω 0 ) = ω 0 k = −∞ k = −∞ ω 0 X ( e j ω ) X N ( j ω ) k ω 0 20/33

DTFT for Periodic Signals Recall DTFS of x N N − 1 x N [ k ] e j 2 k π � N n x N [ n ] = ˆ k = 0 Take DTFT of both sides, use linearity and DTFT of complex exponentials, N − 1 N − 1 ∞ 2 πδ ( ω − 2 k π � x N [ k ] F { e j 2 k π � � X N ( e j ω ) = N n } = ˆ x N [ k ] ˆ N − 2 ℓπ ) k = 0 k = 0 ℓ = −∞ ∞ N − 1 x N [ k ] δ ( ω − 2 ( k + ℓ N ) π � � = 2 π ˆ ) N ℓ = −∞ k = 0 ∞ x N [ k ] δ ( ω − 2 k π � = 2 π ˆ N ) (Recall ˆ x N [ k ] = ˆ x N [ k + ℓ N ] ) k = −∞ 21/33

DTFT for Periodic Signals Can also verify formally using synthesis equation � 2 π − π � 2 π − π ∞ x N [ k ] δ ( ω − 2 k π 1 N N � X ( e j ω ) e j ω n d ω = N ) e j ω n d ω ˆ 2 π − π − π k = −∞ N N Only terms with k = 0 , 1 , . . . , N − 1 lies in interval of integration � 2 π − π N − 1 N − 1 x N [ k ] δ ( ω − 2 k π N N n = x N [ n ] � � x N [ k ] e j 2 k π N ) e j ω n d ω = ˆ ˆ − π k = 0 k = 0 N where last equality is DTFS synthesis equation Thus verified � x N [ n ] = 1 X ( e j ω ) e j ω n d ω 2 π 2 π 22/33

Example: DT Periodic Impulse Train ∞ ∞ → X ( e j ω ) = 2 π δ ( ω − 2 π k F � � x [ n ] = δ [ n − kN ] ← − − N ) N k = −∞ k = −∞ x N [ n ] 1 n 0 N X ( e j ω ) 2 π N ω π − π − 3 π − 2 π 0 2 π 2 π 3 π N 23/33