EI331 Signals and Systems Lecture 6 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 6 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University March 14, 2019

Contents 1. CT Linear Time-invariant Systems 1.1 Impulse Response 1.2 Convolution 1.3 Properties of Convolution 2. Properties of LTI Systems 3. Causal LTI Systems Described by Differential Equations 1/35

Representation of CT Signals by Impulses Sifting property of CT unit impulse � x ( t ) = x ( a ) δ ( t − a ) da R Interpreted as limit as ∆ → 0 of ∞ � ˆ x ∆ ( t ) = x ( k ∆) p ∆ ( t − k ∆)∆ k = −∞ where x ( t ) p ∆ ( t ) = 1 ∆[ u ( t ) − u ( t − ∆)] t − ∆ ∆ 2 ∆ k ∆ 0 2/35

CT Linear Systems Response of linear system � � ∞ � ˆ y ∆ = T (ˆ x ∆ ) = T x ( k ∆) τ k ∆ p ∆ ∆ k = −∞ ∞ ∞ � � x ( k ∆)ˆ = x ( k ∆) T ( τ k ∆ p ∆ )∆ = h k ∆ ∆ k = −∞ k = −∞ where ˆ h k ∆ = T ( τ k ∆ p ∆ ) is response to shifted pulse τ k ∆ p ∆ . In the limit ∆ → 0 , • ˆ x ∆ → x and ˆ y ∆ → y = T ( x ) • for k ∆ → a , have τ k ∆ p ∆ → δ a , expect ˆ h k ∆ → h a = T ( δ a ) � � y = x ( a ) h a da , or y ( t ) = x ( a ) h a ( t ) da R R 3/35

CT Linear Time-invariant (LTI) Systems Unit impulse response 1 h = h 0 = T ( δ ) time invariance = ⇒ h a = T ( δ a ) = τ a ( T ( δ )) = τ a h Response of LTI system – Convolution integral � y ( t ) = x ( τ ) h ( t − τ ) d τ, ∀ t ∈ R R LTI system is fully characterized by unit impulse response! � Conversely, given h , system T ( x )( t ) � x ( τ ) h ( t − τ ) d τ is LTI R 1 For proof of existence, see Theorem 2 of VI.3 in Kˆ osaku Yosida. 4/35

Impulse Responses of Simple LTI Systems Identity h ( t ) = δ ( t ) Scaler multiplication h ( t ) = K δ ( t ) Time shift h ( t ) = δ a ( t ) � δ ( t − a ) Integrator � t h ( t ) = δ ( τ ) d τ = u ( t ) −∞ Differentiator h ( t ) = δ ′ ( t ) (to be defined) 5/35

Convolution � ( x 1 ∗ x 2 )( t ) = x 1 ( τ ) x 2 ( t − τ ) d τ, ∀ t ∈ R R Not always defined for arbitrary x 1 and x 2 Example. For x 1 ( t ) = u ( t ) = x 2 ( − t ) , integral divergent for all t . Sufficient conditions for absolute convergence 1. Either x 1 or x 2 has compact support supp x = { t : x ( t ) � = 0 } , i.e. x 1 or x 2 vanishes outside finite interval. 2. x 1 , x 2 both right-sided (or left-sided), i.e. x i ( t ) = 0 for t ≤ t i (or t ≥ t i ), ∀ i = ⇒ x 1 ∗ x 2 also right-sided (or left-sided) 6/35

Convolution Sufficient conditions for absolute convergence (cont’d) 3. One of x 1 and x 2 has finite L 1 norm and the other finite L p norm for 1 ≤ p ≤ ∞ , where L p norm 2  �� � 1 / p  | x ( t ) | p dt , if 1 ≤ p < ∞  � x � p � R  sup | x ( t ) | , if p = ∞ .  t ∈ R If � x 1 � 1 < ∞ , then � x 1 ∗ x 2 � p ≤ � x 1 � 1 · � x 2 � p . 4. � x 1 � p < ∞ and � x 2 � q < ∞ for 1 ≤ p , q ≤ ∞ and p − 1 + q − 1 = 1 . In this case, � x 1 ∗ x 2 � ∞ ≤ � x 1 � p · � x 2 � q . 2 More precisely, � x � ∞ = sup { B ≥ 0 : | x ( t ) | ≤ B for almost every t } 7/35

Calculation of Convolution 1. Plot both x 1 and x 2 as functions of τ , i.e. x 1 ( τ ) , x 2 ( τ ) 2. Reverse x 2 ( τ ) to obtain x 2 ( − τ ) 3. Given t , shift x 2 ( − τ ) by t to obtain x 2 ( t − τ ) 4. Multiply x 1 ( τ ) and x 2 ( t − τ ) pointwise to obtain g t ( τ ) = x 1 ( τ ) x 2 ( t − τ ) 5. Integrate g t over τ to obtain ( x 1 ∗ x 2 )( t ) , i.e. � ( x 1 ∗ x 2 )( t ) = g t ( τ ) d τ R 6. Repeat 1-5 for each t 8/35

Convolution Example. Let x ( t ) = e − at u ( t ) and h ( t ) = u ( t ) with a > 0 . x ( τ ) = e − a τ u ( τ ) For t < 0 , τ 0 ( x ∗ h )( t ) = 0 h ( τ ) = u ( τ ) τ For t ≥ 0 , 0 � t h ( − τ ) e − a τ d τ = 1 − e − at ( x ∗ h )( t ) = τ a 0 0 h ( t − τ ) Thus τ t ( < 0 ) 0 � 1 − e − at � h ( t − τ ) ( x ∗ h )( t ) = u ( t ) a τ 0 t ( > 0 ) 9/35

Convolution Example. Let x ( t ) = e − at u ( t ) and h ( t ) = u ( t ) with a > 0 . � ( x ∗ h )( t ) = x ( τ ) h ( t − τ ) d τ x ( t ) R � e − a τ u ( τ ) u ( t − τ ) d τ = t 0 R � e − a τ d τ = h ( t ) 0 ≤ τ ≤ t � t e − a τ d τ = u ( t ) t 0 0 � 1 − e − at � = u ( t ) a 1 ( x ∗ h )( t ) a Also true for a < 0 t 0 10/35

Convolution Example. Compute x ∗ x , where x ( t ) = u ( t + T ) − u ( t − T ) . . x ( − τ ) 1 x ( τ )  0 , t < − 2 T    τ − T  t + 2 T , − 2 T ≤ t < 0 0 T ( x ∗ x )( t ) = 2 T − t , 0 ≤ t < 2 T x ( t − τ )     0 , t ≥ 2 T τ t − T t + T x ( t − τ ) τ t − T t + T ( x ∗ x )( t ) x ( t − τ ) 2 T τ t − T t + T x ( t − τ ) t − 2 T 0 2 T τ t − T t + T 11/35

Convolution x ( τ ) 1 τ Example. Let 0 T 2 T � h ( t − τ ) 1 , 0 < t < T x ( t ) = τ t − 2 T t 0 0 , otherwise 2 T � t , 0 ≤ t ≤ 2 T τ h ( t ) = t 0 t − 2 T 0 , otherwise 2 T Five cases τ t 0 t − 2 T 1. t < 0 2 T t − 2 T 2. 0 ≤ t ≤ T τ t 0 3. T < t ≤ 2 T 2 T 4. 2 T < t ≤ 3 T τ 5. t > 3 T 0 t − 2 T t 12/35

Identity Element Recall sampling property of δ � x ( t ) = x ( τ ) δ ( t − τ ) d τ, ∀ t ∈ R R δ identity element for convolution x = x ∗ δ = δ ∗ x x x x = x ∗ δ δ x = δ ∗ x x x δ 13/35

Properties of Convolution Commutativity x 1 ∗ x 2 = x 2 ∗ x 1 Bilinearity �� � �� � � � ∗ = a i b j ( x 1 i ∗ x 2 j ) a i x 1 i b j x 2 j i j i j Associativity x 1 ∗ x 2 ∗ x 3 = ( x 1 ∗ x 2 ) ∗ x 3 = x 1 ∗ ( x 2 ∗ x 3 ) Time shift ( τ a x 1 ) ∗ ( τ b x 2 ) = τ a + b ( x 1 ∗ x 2 ) 14/35

Associative Law x 1 ∗ ( x 2 ∗ x 3 ) = ( x 1 ∗ x 2 ) ∗ x 3 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 15/35

Associative Law Example. x 1 ( t ) = 1 , x 2 ( t ) = u ( t ) , x 3 ( t ) = δ ′ ( t ) (defined later) 1. ( x 2 ∗ x 3 )( t ) = δ ( t ) , so x 1 ∗ ( x 2 ∗ x 3 ) = 1 2. x 1 ∗ x 2 and ( x 1 ∗ x 2 ) ∗ x 3 undefined! 3. x 1 ∗ x 3 = 0 , so ( x 1 ∗ x 3 ) ∗ x 2 = 0 Sufficient conditions for associative law 1. At least two of x 1 , x 2 and x 3 have compact supports 3 2. x 1 , x 2 , x 3 all right-sided (or left-sided), = ⇒ x 1 ∗ x 2 ∗ x 3 also right-sided (or left-sided) 3. One signal (say x 3 ) has finite L p norm for 1 ≤ p ≤ ∞ and others finite L 1 norm. � x 1 ∗ x 2 ∗ x 3 � p ≤ � x 1 � 1 · � x 2 � 1 · � x 3 � p 3 δ a and its derivatives (to be defined) have support { a } . 16/35

Contents 1. CT Linear Time-invariant Systems 1.1 Impulse Response 1.2 Convolution 1.3 Properties of Convolution 2. Properties of LTI Systems 3. Causal LTI Systems Described by Differential Equations 17/35

Memory For LTI systems ∞ � y [ n ] = ( x ∗ h )[ n ] = x [ k ] h [ n − k ] , ∀ n ∈ Z k = −∞ � y ( t ) = ( x ∗ h )( t ) = x ( τ ) h ( t − τ ) d τ, ∀ t ∈ R R memroyless ⇐ ⇒ h = K δ All LTI systems except for scalar multiplication have memory 18/35

Invertibility y x x T 1 T I y x x h 1 h δ Impulse responses of a system and its inverse satisfy h ∗ h 1 = δ Necessary but not sufficient (requires associativity) • e.g. first difference h = δ − τ 1 δ , accumulator h 1 = u 19/35

Causality For LTI systems ∞ n � � y [ n ] = ( x ∗ h )[ n ] = x [ k ] h [ n − k ] = x [ k ] h [ n − k ] k = −∞ k = −∞ causal ⇐ ⇒ h [ n ] = 0 for all n < 0 � t � ∞ y ( t ) = ( x ∗ h )( t ) = x ( τ ) h ( t − τ ) d τ = x ( τ ) h ( t − τ ) d τ −∞ −∞ causal ⇐ ⇒ h ( t ) = 0 for all t < 0 20/35

Stability Recall BIBO stability: � x � ∞ < ∞ = ⇒ � T ( x ) � ∞ < ∞ For LTI systems BIBO stable ⇐ ⇒ � h � 1 < ∞ Proof. Sufficiency. Assume � h � 1 < ∞ . Recall � x ∗ h � ∞ ≤ � x � ∞ � h � 1 . Thus � x � ∞ < ∞ = ⇒ � x ∗ h � ∞ < ∞ . Necessity. Assume BIBO stability. Let x = R (¯ h / | h | ) , where R is time reversal and ¯ h is complex conjugate of h 4 . Note � x � ∞ = 1 . By stability, � x ∗ h � ∞ < ∞ . Note � h � 1 is value of x ∗ h at time zero. Thus � h � 1 ≤ � x ∗ h � ∞ < ∞ . 4 when h takes zero value, use convention 0 / 0 = 0 . 21/35

Unit Step Response Unit step response of LTI systems s � T ( u ) = u ∗ h DT LTI n � s [ n ] = h [ k ] −∞ h [ n ] = s [ n ] − s [ n − 1 ] CT LTI � t s ( t ) = h ( τ ) d τ −∞ h ( t ) = s ′ ( t ) 22/35

Contents 1. CT Linear Time-invariant Systems 1.1 Impulse Response 1.2 Convolution 1.3 Properties of Convolution 2. Properties of LTI Systems 3. Causal LTI Systems Described by Differential Equations 23/35

Linear Constant-coefficient Differential Equations Characteristics of R , L , C i R ( t ) = 1 Rv ( t ) + i S ( t ) i R i L i C � t i L ( t ) = 1 v ( τ ) d τ L v ( t ) R L C −∞ i C ( t ) = C d dtv ( t ) − Kirchhoff’s current law i R ( t ) + i L ( t ) + i C ( t ) = i S ( t ) Second order ordinary differential equation (ODE) C d 2 dt 2 v ( t ) + 1 dtv ( t ) + 1 d Lv ( t ) = d dti S ( t ) R 24/35