EI331 Signals and Systems Lecture 5 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University March 12, 2019
Contents 1. Linearity 2. DT Linear Time-invariant Systems 2.1 Impulse Response 2.2 Convolution 2.3 Properties of Convolution 3. CT Linear Time-invariant Systems 3.1 Impulse Response 3.2 Convolution 3.3 Properties of Convolution 1/44
Linearity System is linear if it has superposition property T ( a 1 x 1 + a 2 x 2 ) = a 1 T ( x 1 ) + a 2 T ( x 2 ) or, equivalently, if it is additive and homogeneous, 1. additivity T ( x 1 + x 2 ) = T ( x 1 ) + T ( x 2 ) 2. homogeneity T ( ax ) = aT ( x ) Example. y ( t ) = tx ( t ) is linear ( y = ax for a ( t ) = t ) Example. y ( t ) = x 2 ( t ) is nonlinear ( y = x 2 ) Example. y ( t ) = sin( x ( t )) is nonlinear ( y = sin x ) Example. y ( t ) = x (sin t ) is linear ( y = x ◦ sin ) 2/44
Linearity Why care about linear systems? 1. accurate models for many systems ◮ resistor, capacity, Newton’s law, etc 2. mathematical tractability, many powerful tools 3. linearization of nonlinear systems ◮ “small signal” perturbation around “operating point” ⇒ ∆ y ( t ) ≈ f ′ ( x 0 ( t ))∆ x ( t ) y ( t ) = f ( x ( t )) = where ∆ y ( t ) = y ( t ) − f ( x 0 ( t )) , ∆ x ( t ) = x ( t ) − x 0 ( t ) ◮ provides insights for behavior of nonlinear system 3/44
Linearity General superposition property �� � � = a k T ( x k ) T a k x k k k 1. finitely many terms: by induction 2. infinitely many terms: need continuity property, i.e. � � T k →∞ x k lim = lim k →∞ T ( x k ) in some sense. Will (implicitly) assume continuity in this course; can check for concrete systems. 4/44
Linearity Zero-in zero-out property For linear system T ( 0 ) = 0 where 0 is zero signal, i.e. x ( t ) = 0 , ∀ t or x [ n ] = 0 , ∀ n T ( 0 ) called zero-input response of system Proof. Use homogeneity. Example. y ( t ) = 2 x ( t ) + 1 is nonlinear (!) T ( 0 ) = 1 T is incrementally linear if ˜ T ( x ) = T ( x ) − T ( 0 ) is linear 5/44
Interconnections of Linear Systems Basic system interconnections preserve linearity T 2 ◦ T 1 y x T 1 T 2 T 1 y T 1 + T 2 x + T 2 y x + T 1 ( I − T 1 ◦ T 2 ) − 1 ◦ T 1 T 2 provided I − T 1 ◦ T 2 is invertible 6/44
Contents 1. Linearity 2. DT Linear Time-invariant Systems 2.1 Impulse Response 2.2 Convolution 2.3 Properties of Convolution 3. CT Linear Time-invariant Systems 3.1 Impulse Response 3.2 Convolution 3.3 Properties of Convolution 7/44
Representation of DT Signals by Impulses Recall DT unit impulse � n = 0 1 , δ [ n ] = n � = 0 0 , C Z = { x [ n ] ∈ C : n ∈ Z } , space of doubly infinite sequences of complex numbers 1. linear space analogous to C n 2. { τ k δ : k ∈ Z } is a basis of C Z ∞ ∞ x = � x [ k ] τ k δ, x [ n ] = � x [ k ] δ [ n − k ] , ∀ n ∈ Z or k = −∞ k = −∞ Above called sifting property of DT unit impulse 8/44
DT Linear Systems Response of linear system � ∞ � ∞ ∞ � � � y = T ( x ) = T x [ k ] τ k δ = x [ k ] T ( τ k δ ) = x [ k ] h k k = −∞ k = −∞ k = −∞ where h k = T ( τ k δ ) is system response to shifted impulse τ k δ . In general, h k for different k unrelated to each other. Recall in linear algebra, n n n � � � x = x k e k = ⇒ A x = x k A e k = x k f k k = 1 k = 1 k = 1 where f k = A e k is image of e k under linear transformation A . Just going from finite dimensions to infinite dimensions! 9/44
DT Linear Time-invariant (LTI) Systems Unit impulse (sample) response h = T ( δ ) time invariance = ⇒ h k = T ( τ k δ ) = τ k ( T ( δ )) = τ k h Response of LTI system – Convolution sum ∞ ∞ y = � x [ k ] τ k h , y [ n ] = � x [ k ] h [ n − k ] , ∀ n ∈ Z or k = −∞ k = −∞ LTI system is fully characterized by unit impulse response! ∞ Conversely, given h , system T ( x ) � � x [ k ] τ k h is LTI k = −∞ 10/44
DT LTI Systems 1 δ [ n ] h [ n ] 1 n n 0 0 1 2 2 . 5 2 2 x [ n ] 0 . 5 0 . 5 n n 0 1 0 1 2 3 x [ 0 ] δ [ n ] x [ 0 ] h [ n ] 0 . 5 0 . 5 n n 0 0 1 2 2 2 x [ 1 ] δ [ n − 1 ] x [ 1 ] h [ n − 1 ] n n 1 1 2 3 11/44
Impulse Responses of Simple LTI Systems Identity h [ n ] = δ [ n ] Scaler multiplication h [ n ] = K δ [ n ] Time shift h [ n ] = δ a [ n ] � δ [ n − a ] Accumulator n � h [ n ] = δ [ k ] = u [ n ] k = −∞ First difference h [ n ] = δ [ n ] − δ [ n − 1 ] 12/44
Convolution of Sequences ∞ ( x 1 ∗ x 2 )[ n ] = � x 1 [ k ] x 2 [ n − k ] , ∀ n ∈ Z k = −∞ Not always defined for arbitrary x 1 and x 2 Example. For x 1 [ n ] = u [ n ] = x 2 [ − n ] , sum divergent for all n . Sufficient conditions for absolute convergence 1. Either x 1 or x 2 has finite support supp x = { n : x [ n ] � = 0 } , i.e. x 1 [ n ] or x 2 [ n ] nonzero only for finitely many n . 2. x 1 , x 2 both right-sided (or left-sided), i.e. x i [ n ] = 0 for n ≤ n i (or n ≥ n i ), ∀ i = ⇒ x 1 ∗ x 2 also right-sided (or left-sided) 13/44
Convolution of Sequences 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 � 1 / p � ∞ � | x [ n ] | p if 1 ≤ p < ∞ , � x � p � n = −∞ sup | x [ n ] | , if p = ∞ . n ∈ Z 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 . 14/44
Calculation of Convolution 1. Plot both x 1 and x 2 as functions of k , i.e. x 1 [ k ] , x 2 [ k ] 2. Reverse x 2 to obtain Rx 2 , i.e. x 2 [ − k ] 3. Given n , shift Rx 2 by n to obtain τ n ( Rx 2 ) , i.e. x 2 [ n − k ] 4. Multiply x 1 and τ n ( Rx 2 ) component-wise to obtain g n = x 1 · τ n ( Rx 2 ) , i.e. g n [ k ] = x 1 [ k ] x 2 [ n − k ] 5. Sum g n over k to obtain ( x 1 ∗ x 2 )[ n ] , i.e. ∞ � ( x 1 ∗ x 2 )[ n ] = g n [ k ] k = −∞ 6. Repeat 1-5 for each n 15/44
Convolution Example. Let x [ n ] = α n u [ n ] and h [ n ] = u [ n ] with α ∈ ( 0 , 1 ) . x [ k ] = α k u [ k ] For n < 0 , 0 k ( x ∗ h )[ n ] = 0 h [ k ] = u [ k ] For n ≥ 0 , 0 k h [ − k ] n α k = 1 − α n + 1 � ( x ∗ h )[ n ] = 0 k 1 − α k = 0 h [ n − k ] Thus n ( < 0 ) 0 k � 1 − α n + 1 � h [ n − k ] ( x ∗ h )[ n ] = u [ n ] 1 − α 0 n ( > 0 ) k 16/44
Convolution Example. Let x [ n ] = α n u [ n ] and h [ n ] = u [ n ] with α ∈ ( 0 , 1 ) . ∞ � ( x ∗ h )[ n ] = x [ k ] h [ n − k ] x [ n ] = α n u [ n ] k = −∞ ∞ n 0 � α k u [ k ] u [ n − k ] = h [ n ] = u [ n ] k = −∞ � α k = n 0 0 ≤ k ≤ n n � α k = u [ n ] 1 1 − α k = 0 � 1 − α n + 1 � = u [ n ] ( x ∗ h )[ n ] 1 − α n Also true for α > 1 0 17/44
Convolution x [ k ] Example. Let 0 4 k � h [ n − k ] 0 ≤ n ≤ 4 1 , x [ n ] = 0 , otherwise n 0 4 k n − 6 h [ n − k ] � α n , 0 ≤ n ≤ 6 h [ n ] = n 0 4 k otherwise n − 6 0 , h [ n − k ] Five cases n 0 4 k n − 6 1. n < 0 h [ n − k ] 2. 0 ≤ n ≤ 4 n 3. 4 < n ≤ 6 0 n − 6 4 k 4. 6 < n ≤ 10 h [ n − k ] 5. n > 10 n 0 4 k n − 6 18/44
Identity Element Recall sifting property of δ ∞ � x [ n ] = x [ k ] δ [ n − k ] , ∀ n ∈ Z k = −∞ δ identity element for convolution x = x ∗ δ = δ ∗ x x x x = x ∗ δ δ x = δ ∗ x x x δ 19/44
Commutative Law x 1 ∗ x 2 = x 2 ∗ x 1 Proof. Change variables by m = n − k , ∞ ∞ � � ( x 1 ∗ x 2 )[ n ] = x 1 [ k ] x 2 [ n − k ] = x 1 [ n − m ] x 2 [ m ] = ( x 2 ∗ x 1 )[ n ] m = −∞ k = −∞ y y = x ∗ h x h y y = h ∗ x x h 20/44
Commutative Law h [ k ] Example. Let 0 6 k � 0 ≤ n ≤ 4 1 , x [ n − k ] x [ n ] = 0 , otherwise n 0 6 k n − 4 x [ n − k ] � α n , 0 ≤ n ≤ 6 h [ n ] = n otherwise 0 6 k 0 , n − 4 x [ n − k ] Five cases 0 n − 4 n 6 k 1. n < 0 x [ n − k ] 2. 0 ≤ n ≤ 4 3. 4 < n ≤ 6 n 0 6 k n − 4 4. 6 < n ≤ 10 x [ n − k ] 5. n > 10 n 0 6 k n − 4 21/44
Distributive Law x 1 ∗ ( x 2 + x 3 ) = x 1 ∗ x 2 + x 1 ∗ x 3 Proof. Provided x 1 ∗ x 2 and x 1 ∗ x 3 well-defined, ∞ ∞ ∞ � � � x 1 [ k ]( x 2 [ n − k ]+ x 3 [ n − k ]) = x 1 [ k ] x 2 [ n − k ]+ x 1 [ k ] x 3 [ n − k ] k = −∞ k = −∞ k = −∞ h 1 y h = h 1 + h 2 x + h 2 h 22/44
Bilinearity Scalar multiplication x 1 ∗ ( Kx 2 ) = K ( x 1 ∗ x 2 ) ( x 1 , x 2 ) �→ x 1 ∗ x 2 bilinear, i.e. linear in both x 1 and x 2 �� � �� � � � ∗ = a i b j ( x 1 i ∗ x 2 j ) a i x 1 i b j x 2 j i j i j 23/44
Associative Law ( x 1 ∗ x 2 ) ∗ x 3 = x 1 ∗ ( x 2 ∗ x 3 ) “Proof”. Under appropriate conditions ∞ � (( x 1 ∗ x 2 ) ∗ x 3 )[ n ] = ( x 1 ∗ x 2 )[ n − m ] x 3 [ m ] m = −∞ � � ∞ ∞ � � = x 1 [ k ] x 2 [ n − m − k ] x 3 [ m ] m = −∞ k = −∞ ∞ ∞ � � = x 1 [ k ] x 2 [ n − k − m ] x 3 [ m ] k = −∞ m = −∞ ∞ � = x 1 [ k ]( x 2 ∗ x 3 )[ n − k ]) = ( x 1 ∗ ( x 2 ∗ x 3 ))[ n ] k = −∞ When associative law holds, write x 1 ∗ x 2 ∗ x 3 without ambiguity 24/44
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