EI331 Signals and Systems Lecture 16 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 16 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University April 18, 2019

Contents 1. Parseval’s Identity 2. Convolution Property 3. Multiplication Property 4. Systems Described by Linear Constant-coefficient ODEs 1/44

Parseval’s Identity Theorem. If x ∈ L 2 ( R ) , X = F { x } , then 2 = 1 � | x ( t ) | 2 dt = 1 � � x � 2 2 π � X � 2 | X ( j ω ) | 2 d ω 2 , or 2 π R R Note ω is angular frequency and ω 2 π is frequency | X ( j ω ) | 2 d ω � � | x ( t ) | 2 dt = 2 π R R Interpretation: Energy conservation • | x ( t ) | 2 power, or energy per unit time (second) • | X ( j ω ) | 2 energy per unit frequency (Hertz) | X ( j ω ) | 2 called energy-density spectrum 2/44

Parseval’s Identity Theorem. If x , y ∈ L 2 ( R ) , X = F { x } , Y = F { y } , then � � � x , y � = 1 x ( t ) y ∗ ( t ) dt = 1 X ( j ω ) Y ∗ ( j ω ) d ω 2 π � X , Y � , or 2 π R R “Proof.” � 1 � ∗ � � � x ( t ) y ∗ ( t ) dt = Y ( j ω ) e j ω t d ω x ( t ) dt 2 π R R R � � = 1 Y ∗ ( j ω ) e − j ω t d ω dt x ( t ) 2 π R R �� � = 1 � x ( t ) e − j ω t dt Y ∗ ( j ω ) d ω 2 π R R � = 1 X ( j ω ) Y ∗ ( j ω ) d ω 2 π R 3/44

Parseval’s Identity Example. sin( Wt ) F ← − − → u ( ω + W ) − u ( ω − W ) π t By Parseval’s identity sin 2 ( Wt ) � dt = 1 � | u ( ω + W ) − u ( ω − W ) | 2 d ω π 2 t 2 2 π R R � W = 1 d ω 2 π − W = W π 4/44

Contents 1. Parseval’s Identity 2. Convolution Property 3. Multiplication Property 4. Systems Described by Linear Constant-coefficient ODEs 5/44

Convolution Property For LTI systems T with • impulse response h R h ( t ) e − j ω t dt = F { h } • frequency response H ( j ω ) = � e j ω t is eigenfunction associated with eigenvalue H ( j ω ) T ( e j ω t ) = H ( j ω ) e ω t Input x is linear superposition of e j ω t x ( t ) = 1 � X ( j ω ) e j ω t d ω 2 π R Output � 1 � � X ( j ω ) e j ω t d ω y ( t ) = ( x ∗ h )( t ) = T 2 π R = 1 � X ( j ω ) T ( e j ω t ) d ω = 1 � X ( j ω ) H ( j ω ) e j ω t d ω 2 π 2 π R R 6/44

Convolution Property F F { x ∗ y } = F { x } F { y } , or ( x ∗ y )( t ) ← − − → X ( j ω ) Y ( j ω ) convolution in time ⇐ ⇒ multiplication in frequency “Proof”. � � � ( x ∗ y )( t ) e − j ω t dt = x ( τ ) y ( t − τ ) e − j ω t d τ dt R R R � �� � y ( t − τ ) e − j ω t dt = x ( τ ) d τ R R � x ( τ ) Y ( j ω ) e − j ωτ d τ = R � x ( τ ) e − j ωτ d τ = Y ( j ω ) R = Y ( j ω ) X ( j ω ) 7/44

Example x 1 � 1 − 2 | t | � � � u ( t + T 2 ) − u ( t − T x ( t ) = 2 ) T O t − T T 2 2 x = x 1 ∗ x 1 x 1 � ω T � � T · 2 sin � 2 2 4 X 1 ( j ω ) = T ω 1 ( j ω ) = 8 sin 2 � ω T O t − T T � 4 4 X ( j ω ) = X 2 4 ω 2 T X 1 ( j ω ) X ( j ω ) � T T 2 2 ω O ω O 8/44

Frequency Response of LTI System LTI system T • fully characterized by impulse response h y = T ( x ) = h ∗ x • also fully characterized by frequency response H = F { h } , if H is well-defined ◮ BIBO stable system, h ∈ L 1 ( R ) ◮ other systems: identity h = δ , differentiator h = δ ′ , ... Typically, convolution property implies Y = F { y } = HX = F { h } F { x } Instead of computing x ∗ h , can do y = F − 1 ( F { h } F { x } ) 9/44

Frequency Response of LTI System h ( t ) H ( j ω ) id δ ( t ) 1 e − j ω t 0 τ t 0 δ ( t − t 0 ) d δ ′ ( t ) j ω dt � t 1 u ( t ) j ω + πδ ( ω ) −∞ sin( ω c t ) ideal lowpass u ( ω + ω c ) − u ( ω − ω c ) π t 1 τ e − t /τ u ( t ) 1 1st order lowpass 1 + j τω 10/44

Examples Example. Differentiation property y = x ′ = x ∗ δ ′ Y ( j ω ) = X ( j ω ) F { δ ′ } = j ω X ( j ω ) Example. Integration property � t y ( t ) = x ( τ ) d τ = ( x ∗ u )( t ) −∞ � 1 � = 1 Y ( j ω ) = X ( j ω ) U ( j ω ) = X ( j ω ) j ω + πδ ( ω ) j ω X ( j ω )+ π X ( 0 ) δ ( ω ) 11/44

Example Unit ramp function u − 2 ( t ) = tu ( t ) = ( u ∗ u )( t ) Convolution property suggests ω 2 + 2 π F { u − 2 } ( j ω ) = U 2 ( j ω ) = − 1 j ω δ ( ω ) + π 2 δ ( ω ) δ ( ω ) But π j ω δ ( ω ) and δ ( ω ) δ ( ω ) not well-defined! Know F { u − 2 } = − 1 ω 2 + j πδ ′ ( ω ) Convolution property not applicable here! Rule of thumb. Applicable when formula is well-defined 12/44

Example Response of LTI system with impulse response h ( t ) = e − at u ( t ) to input x ( t ) = e − bt u ( t ) , a , b > 0 Method 1. Direct convolution y = x ∗ h Method 2. Solve following ODE with initial rest condition y ′ ( t ) + ay ( t ) = e − bt u ( t ) Method 3. Fourier transform. 1 1 1 H ( j ω ) = a + j ω, X ( j ω ) = b + j ω = ⇒ Y ( j ω ) = ( a + j ω )( b + j ω ) b − a ( e − at − e − bt ) u ( t ) 1 1 1 1 If a � = b , Y ( j ω ) = b − a ( a + j ω − b + j ω ) = ⇒ y ( t ) = � � If a = b , Y ( j ω ) = − d 1 ⇒ y ( t ) = − d da e − at u ( t ) = te − at u ( t ) = da a + j ω � � Can also use Y ( j ω ) = j d 1 and differentiation property d ω a + j ω 13/44

Example Response of LTI system with impulse response h ( t ) = e − at u ( t ) to input x ( t ) = cos( ω 0 t ) , a > 0 . Frequency response 1 1 a 2 + ω 2 e − j arctan ω H ( j ω ) = a + j ω = √ a Method 1. Use eigenfunction property x ( t ) = 1 2 e j ω 0 t + 1 2 e − j ω 0 t e j ω 0 t y ( t ) = 1 2 H ( j ω 0 ) e j ω 0 t + 1 2 H ( − j ω 0 ) e − j ω 0 t = Re a + j ω 0 = a cos( ω 0 t ) + ω 0 sin( ω 0 t ) cos( ω 0 t − arctan ω 0 1 = a ) a 2 + ω 2 a 2 + ω 2 � 0 0 14/44

Example Response of LTI system with impulse response h ( t ) = e − at u ( t ) to input x ( t ) = cos( ω 0 t ) . Frequency response 1 1 a 2 + ω 2 e − j arctan ω H ( j ω ) = a + j ω = √ a Method 2. Use Fourier transform of X X ( j ω ) = πδ ( ω − ω 0 ) + πδ ( ω + ω 0 ) Y ( j ω ) = π H ( j ω 0 ) δ ( ω − ω 0 ) + π H ( − j ω 0 ) δ ( ω + ω 0 ) y ( t ) = 1 2 H ( j ω 0 ) e j ω 0 t + 1 2 H ( − j ω 0 ) e − j ω 0 t 15/44

Example Response of ideal lowpass filter with impulse response h ( t ) to input x ( t ) . h ( t ) = sin( ω c t ) x ( t ) = sin( ω i t ) , π t π t Fourier transforms X ( j ω ) = u ( ω + ω i ) − u ( ω − ω i ) H ( j ω ) = u ( ω + ω c ) − u ( ω − ω c ) Use Y ( j ω ) = X ( j ω ) H ( j ω ) , � � X ( j ω ) , if ω i ≤ ω c x ( t ) , if ω i ≤ ω c Y ( j ω ) = = ⇒ y ( t ) = H ( j ω ) , if ω i > ω c h ( t ) , if ω i > ω c NB. Convolution of two sinc is another sinc 16/44

Example Convolution of Gaussians is another Gaussian − ( t − µ i ) 2 � � 1 x i ( t ) = exp � 2 σ 2 2 πσ 2 i i Fourier transform (complex conjugate of characteristic function) � − i µ i ω − σ 2 � 2 ω 2 i X i ( j ω ) = exp For y = x 1 ∗ x 2 , − i ( µ 1 + µ 2 ) ω − σ 2 1 + σ 2 � � 2 ω 2 Y ( j ω ) = X 1 ( j ω ) X 2 ( j ω ) = exp 2 − ( t − µ 1 − µ 2 ) 2 � � 1 y ( t ) = exp � 2 ( σ 2 1 + σ 2 2 ) 2 π ( σ 2 1 + σ 2 2 ) 17/44

System Connections LTI systems in series connection y = ( x ∗ h 1 ) ∗ h 2 = x ∗ ( h 1 ∗ h 2 ) = ( x ∗ h 2 ) ∗ h 1 y h = h 1 ∗ h 2 x h 1 h 2 h commutative y h = h 2 ∗ h 1 x h 2 h 1 h Order of processing usually not important for LTI systems 18/44

System Connections LTI systems in series connection Y = ( XH 1 ) H 2 = X ( H 1 H 2 ) = ( XH 2 ) H 1 H = H 1 H 2 H 1 ( j ω ) H 2 ( j ω ) y x H ( j ω ) commutative y H = H 2 H 1 x H 2 ( j ω ) H 1 ( j ω ) H ( j ω ) Order of processing usually not important for LTI systems 19/44

System Connections LTI systems in systems in parallel connection h 1 y h = h 1 + h 2 x + h 2 h H 1 ( j ω ) y H = H 1 + H 2 x + H 2 ( j ω ) H ( j ω ) 20/44

System Connections LTI systems in systems in feedback connection + + y x h 1 - y = h 1 ∗ ( x − h 2 ∗ y ) h 2 h = ? h + + y Y = H 1 X − H 1 H 2 Y H 1 ( j ω ) x - H 2 ( j ω ) H = Y H 1 X = 1 + H 1 H 2 H ( j ω ) 21/44

Contents 1. Parseval’s Identity 2. Convolution Property 3. Multiplication Property 4. Systems Described by Linear Constant-coefficient ODEs 22/44

Multiplication Property Dual of convolution property F { xy } = 1 → 1 � F 2 π F { x }∗ F { y } , or x ( t ) y ( t ) ← − − X ( j θ ) Y ( j ( ω − θ )) d θ 2 π R multiplication in time ⇐ ⇒ convolution in frequency 1 � Proof. Let Z ( j ω ) = R X ( j θ ) Y ( j ( ω − θ )) d θ 2 π � � �� � 1 Z ( j ω ) e j ω t d ω = 1 1 e j ω t d ω X ( j θ ) Y ( j ( ω − θ )) d θ 2 π 2 π 2 π R R R �� � = 1 � X ( j θ ) 1 Y ( j ( ω − θ )) e j ω t d ω d θ 2 π 2 π R R = 1 � X ( j θ ) y ( t ) e j θ t d θ = x ( t ) y ( t ) 2 π R 23/44

Example: Modulation X ( j ω ) Baseband signal x ( t ) A Carrier ω ω 1 − ω 1 0 p ( t ) = cos( ω c t ) P ( j ω ) = πδ ( ω − ω c ) + πδ ( ω + ω c ) P ( j ω ) π π Modulated signal ω ω c y ( t ) = x ( t ) p ( t ) − ω c 0 Y ( j ω ) = 1 2 X ( j ( ω − ω c ))+ 1 2 X ( j ( ω + ω c )) Y ( j ω ) A / 2 ω ω c − ω c 0 − ω c − ω 1 − ω c + ω 1 ω c − ω 1 ω c + ω 1 24/44