EI331 Signals and Systems Lecture 14 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 14 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University April 11, 2019

Contents 1. CT Fourier Transform 2. Fourier Transform of L 1 Signals 3. Fourier Transform of More General Functions 4. Fourier Transform of Periodic Signals 1/30

CT Fourier Transform Fourier transform (analysis equation) � ∞ x ( t ) e − j ω t dt X ( j ω ) = F { x } ( j ω ) = −∞ X ( j ω ) called spectrum of x ( t ) Inverse Fourier transform (synthesis equation) � ∞ x ( t ) = F − 1 { X } ( t ) = 1 X ( j ω ) e j ω t d ω 2 π −∞ Superposition of complex exponentials at continuum of 1 frequencies; frequency ω has density 2 π X ( j ω ) 2/30

CT Fourier Transform Two equivalent representations of same signal F • time domain vs. frequency domain: x ( t ) ← − − → X ( j ω ) Also widely used elsewhere. In probability theory, • characteristic function of random variable X with density p ( x ) � ∞ ϕ X ( t ) = E [ e jtX ] = p ( x ) e jxt dx −∞ In quantum mechanics • position representation: wave function ψ ( x ) ◮ | ψ ( x ) | 2 probability density of finding particle at position x • momentum representation: Ψ( p ) � ∞ 1 ψ ( x ) e − jpx / � dx √ Ψ( p ) = 2 π � −∞ ◮ | Ψ( p ) | 2 probability density of finding particle with momentum p 3/30

Contents 1. CT Fourier Transform 2. Fourier Transform of L 1 Signals 3. Fourier Transform of More General Functions 4. Fourier Transform of Periodic Signals 4/30

Fourier Transform of L 1 Signals Recall from calculus, for real-valued x , improper integral � T 2 � x ( t ) dt � lim x ( t ) dt T 1 , T 2 →∞ R − T 1 � is well-defined if x ∈ L 1 ( R ) , i.e. � x � 1 = R | x ( t ) | dt < ∞ Same is true for complex-valued x = u + jv , since | u | , | v | ≤ | x | and � � � x ( t ) dt � u ( t ) dt + j v ( t ) dt R R R For signal x ∈ L 1 ( R ) • x ( t ) e − j ω t ∈ L 1 ( R ) , since | x ( t ) e − j ω t | = | x ( t ) | � ∞ • Fourier transform X ( j ω ) = −∞ x ( t ) e − j ω t dt is well-defined improper integral for each ω 5/30

Fourier Transform of L 1 Signals Theorem. If x ∈ L 1 , then X is bounded, in fact � X � ∞ ≤ � x � 1 Proof. Consequence of following lemma. Lemma. For complex-valued function f of real variable t , � � � � � � f ( t ) dt � ≤ | f ( t ) | dt � � � � � Proof. If | f ( t ) | dt = ∞ , trivial. Assume | f ( t ) | dt < ∞ . Let � � � f ( t ) dt | e j φ . Then phase of f ( t ) dt be φ , i.e. f ( t ) dt = | � � � � � � = e − j φ e − j φ f ( t ) dt � � f ( t ) dt f ( t ) dt = � � � Taking real part � � � � � � e − j φ f ( t ) dt = Re [ e − j φ f ( t )] dt ≤ | e − j φ f ( t ) | dt � � f ( t ) dt � = Re � � � 6/30

Fourier Transform of L 1 Signals Theorem. If x ∈ L 1 , then X is uniformly continuous Proof. For any T > 0 , � | x ( t ) e − j ω 1 t − x ( t ) e − j ω 2 t | dt | X ( j ω 1 ) − X ( j ω 2 ) | ≤ R � � � � 2 sin ∆ ω t � � = | x ( t ) | · (∆ ω = ω 1 − ω 2 ) � dt � � 2 R � � ≤ 2 | x ( t ) | dt + | ∆ ω | T | x ( t ) | dt | t | > T | t |≤ T � | x ( t ) | dt + | ∆ ω | T � x � 1 � I 1 + I 2 ≤ 2 | t | > T Given ǫ > 0 , since x ∈ L 1 ( R ) , exists T > 0 s.t. I 1 < ǫ/ 2 . Fix such T . For | ∆ ω | ≤ ǫ/ ( 2 T � x � 1 ) , I 2 < ǫ/ 2 . Thus | X ( j ω 1 ) − X ( j ω 2 ) | < ǫ . 7/30

Example: Right-sided Decaying Exponential x ( t ) | X ( j ω ) | x ( t ) = e − at u ( t ) , a > 0 1 1 / a √ 1 2 e 2 a � ∞ t 0 1 / a e − ( a + j ω ) t dt X ( j ω ) = 0 ∞ 1 1 a + j ω e − ( a + j ω ) t � = − 0 = � ω a − a 0 a + j ω � arg X ( j ω ) NB. Above formula for X also works for complex a with Re a > 0 . π 2 π 1 4 For a > 0 , √ a | X ( j ω ) | = a 2 + ω 2 , − a ω − π 4 arg X ( j ω ) = − arctan ω − π a 2 8/30

Example: Two-sided Decaying Exponential x ( t ) x ( t ) = e − a | t | , a > 0 1 1 / e � ∞ e − a | t | e − j ω t dt X ( j ω ) = −∞ � 0 � ∞ t − 1 / a 0 1 / a e ( a − j ω ) t dt + e − ( a + j ω ) t dt = −∞ 0 1 1 X ( j ω ) = a − j ω + a + j ω 2 a 2 / a = a 2 + ω 2 1 / a NB. Above formula for X also works for complex a with Re a > 0 . ω a − a 0 9/30

Example: Gaussian For a > 0 , � π ae − ω 2 F x ( t ) = e − at 2 ← − − → X ( j ω ) = 4 a x ( t ) X ( j ω ) √ a = 1 / 2 2 π a = 1 a = 2 1 ω t √ √ 2 ω 2 = F In particular, x ( t ) = e − 1 2 t 2 2 π e − 1 ← − − → X ( j ω ) = 2 π x ( ω ) , √ i.e. F { x } = 2 π x 10/30

Example: Gaussian For a > 0 , � π ae − ω 2 x ( t ) = e − at 2 F ← − − → X ( j ω ) = 4 a Proof. � d � ∞ � ∞ dte − at 2 � d e − at 2 ( − jt ) e − j ω t dt = j e − j ω t dt d ω X ( j ω ) = 2 a −∞ −∞ e − at 2 � d � ∞ � = − j dte − j ω t (integration by parts) dt 2 a −∞ � ∞ = − ω e − at 2 e − j ω t dt = − ω 2 aX ( j ω ) 2 a −∞ � π d ω 2 ⇒ X ( j ω ) = X ( j 0 ) e − ω 2 ae − ω 2 � � 4 a = X ( j ω ) e = 0 = 4 a 4 a d ω 11/30

Fourier Inversion for L 1 Signals Given Fourier transform, � ∞ x ( t ) e − j ω t dt X ( j ω ) = F { x } ( j ω ) = −∞ Is inverse Fourier transform well-defined? Is it equal to x ? � ∞ x ( t ) ? = F − 1 { X } ( t ) = 1 X ( j ω ) e j ω t d ω 2 π −∞ Theorem. If x ∈ L 1 ( R ) is continuous and X = F { x } ∈ L 1 ( R ) , then x = F − 1 { X } . • e.g. Two-sided decaying exponential, Gaussian But, for x ( t ) ∈ L 1 ( R ) , X ( j ω ) is not necessarily in L 1 ( R ) • e.g. one-sided decaying exponential, rectangular pulse • if X ∈ L 1 ( R ) , x must be continuous 12/30

Fourier Inversion for L 1 Signals Inverse FT typically interpreted as principal value, i.e. � ∞ � W 1 1 X ( j ω ) e j ω t d ω = lim X ( j ω ) e j ω t d ω 2 π 2 π W →∞ −∞ − W may converge without being absolutely convergent Theorem. If x ∈ L 1 ( R ) satisfies Dirichlet conditions on all finite intervals, then � W X ( j ω ) e j ω t d ω = x ( t + ) + x ( t − ) 1 lim pointwise 2 π 2 W →∞ − W NB. Gibbs phenomenon at discontinuity Often also need to interpret FT as principal value � ∞ � T x ( t ) e − j ω t dt = lim x ( t ) e − j ω t dt T →∞ −∞ − T 13/30

Example: Rectangular Pulse x ( t ) x ( t ) = u ( t + T ) − u ( t − T ) 1 � T e − j ω t dt = 2 sin( ω T ) X ( j ω ) = ω − T − T t T Inverse FT � ∞ X ( j ω ) sin( ω T ) d ω = x ( 0 ) = 1 2 T πω −∞ π As T → ∞ , T • frequency domain ω 0 sin( ω T ) lim = δ ( ω ) πω T →∞ sinc( θ ) � sin( πθ ) • time domain: x ( t ) → 1 , DC � πθ 14/30

Example: Ideal Lowpass Filter Frequency response H ( j ω ) 1 H ( j ω ) = u ( ω + ω c ) − u ( ω − ω c ) Impulse response ω ω c − ω c � ω c e j ω t d ω = sin( ω c t ) h ( t ) = 1 h ( t ) 2 π π t − ω c π ω c ω c π NB. h ( t ) / ∈ L 1 ( R ) but H ( j ω ) ∈ L 1 ( R ) t 0 As ω c → ∞ , • time domain ◮ h ( t ) → δ ( t ) , becomes identity system • frequency domain ◮ H ( j ω ) → 1 , passes all frequencies � 15/30

Duality X 1 ( j ω ) 2 T x 1 ( t ) F π 1 T ω − T t 0 T 0 x 2 ( t ) X 2 ( j ω ) F π 1 W π W ω t − W W 0 0 16/30

Contents 1. CT Fourier Transform 2. Fourier Transform of L 1 Signals 3. Fourier Transform of More General Functions 4. Fourier Transform of Periodic Signals 17/30

Fourier Transform of More General Functions If x n → x , define Fourier transform of x by X ( j ω ) = F { x } � lim n F { x n } i.e. � � x ( t ) e − j ω t dt � lim x n ( t ) e − j ω t dt X ( j ω ) = n R R If x ∈ L 1 ( R ) , above definition is consistent with old one In general, convergence interpreted in distributional sense, i.e. for nice test function φ � � �� � x n ( t ) e − j ω t dt X ( j ω ) φ ( ω ) d ω � lim φ ( ω ) d ω n R R R Interchanging order of integration leads to alternative definition �� � � � � φ ( ω ) e − j ω t d ω X ( j ω ) φ ( ω ) d ω = lim x n ( t ) dt = x ( t )Φ( jt ) dt n R R R R 18/30

Schwarz Space Space of test functions is so-called Schwarz space on R , denoted S = S ( R ) Function φ ∈ S if it is • infinitely differentiable: φ ( k ) exists for all k ∈ N • rapidly decreasing: | t ℓ φ ( k ) ( t ) | < ∞ , � φ � ℓ, k � sup ∀ ℓ, k ∈ N t ∈ R Example. Gaussian g ( t ) = e − at 2 ∈ S Note φ ∈ S = ⇒ φ ∈ L 1 , Fourier transform Φ well-defined. Theorem. If φ ∈ S , then Φ ∈ S Example. Gaussian G ( j ω ) = � π a e − ω 2 4 a ∈ S 19/30

Example: Unit Impulse δ Method 1. Recall for ideal lowpass filter, h W ( t ) = sin( Wt ) F ← − − → H W ( j ω ) = u ( ω + W ) − u ( ω − W ) π t Since h W → δ as W → ∞ F { δ } = lim W →∞ H W ( j ω ) = 1 Can also use Gaussian instead of sinc . Recall 1 √ π ae − t 2 → G a ( j ω ) = e − a ω 2 F g a ( t ) = ← − − a 4 Since g a → δ as a → 0 F { δ } = lim a → 0 G a ( j ω ) = 1 20/30