EI331 Signals and Systems Lecture 8 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 8 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University March 21, 2019

Contents 1. Causal LTI Systems Described by Difference Equations 1.1 Linear Constant-coefficient Difference Equations 1.2 Iterative Method 1.3 Homogeneous and Particular Solutions 1.4 Zero-input and Zero-state Responses 1.5 Systems of First-order Difference Equations 2. Block Diagram Representations of First-order Systems Described by Differential/Difference Equations 3. Singularity Functions 1/38

Linear Constant-coefficient Difference Equations Example. Balance of bank account y [ n ] = ( 1 + r ) y [ n − 1 ] + x [ n ] r interest rate Example. Discretization of differential equation ⇒ y ( nT ) − y (( n − 1 ) T ) y ′ ( t ) = x ( t ) = ≈ x ( nT ) T Let x [ n ] = x ( nT ) , y [ n ] = y ( nT ) . Discretized equation y [ n ] = y [ n − 1 ] + Tx [ n ] (Euler’s method) Example. Exponential smoothing y [ n ] = ( 1 − α ) y [ n − 1 ] + α x [ n ] , α ∈ ( 0 , 1 ) smoothing factor 2/38

Linear Constant-coefficient Difference Equations System described by linear constant-coefficient difference equation N M � � a k y [ n − k ] = b k x [ n − k ] k = 0 k = − M 1 where a 0 � = 0 , a N � = 0 • also called recursive equation or recursion • N : order of difference equation • focus on M 1 = 0 for causal systems • input-output relation specified implicitly • solve difference equation for explicit input-output relation • difference equation alone does not uniquely determine T • need auxiliary conditions, typically initial conditions 3/38

Linear Constant-coefficient Differential Equations Initial value problem (IVP) N � Ly = f , where L = a k τ k k = 0 with initial conditions y [ k ] = y k , k = n 0 − 1 , n 0 − 2 , . . . , n 0 − N • N -th order difference equation needs N initial conditions • may use any N consecutive values as “initial” values • often n 0 = 0 • typically f [ n ] = 0 for n < n 0 4/38

Initial Rest • If input x [ n ] = 0 for n < n 0 , output y [ n ] = 0 for n < n 0 ◮ output zero until changed by input ◮ equivalent to causality for linear systems • Adapt initial time n 0 to input x : if x becomes nonzero at n 0 , use y [ n 0 − k ] = 0 for k = 1 , 2 , . . . , N , i.e. solve � Ly = f y [ k ] = 0 , k = n 0 − 1 , n 0 − 2 , . . . , n 0 − N • Linear constant-coefficient difference equation with initial rest condition specifies causal and LTI system for right-sided inputs 5/38

Iterative Method Iteratively compute y [ n ] from y [ n − 1 ] , . . . , y [ n − N ] and x � M � N y [ n ] = 1 � � b k x [ n − k ] − a k y [ n − k ] a 0 k = 0 k = 1 Special case N = 0 M � b k � � y [ n ] = x [ n − k ] a 0 k = 0 • explicit function of present and past input values • nonrecursive equation, no need for auxiliary conditions • causal LTI system with finite impulse response (FIR) M � b k � � h [ n ] = δ [ n − k ] a 0 k = 0 6/38

Iterative Method Example. Consider exponential smoothing y [ n ] − ay [ n − 1 ] = x [ n ] with y [ − 1 ] = y − 1 . • For n ≥ 0 , go forward in time y [ 0 ] = ay [ − 1 ] + x [ 0 ] = ay − 1 + x [ 0 ] y [ 1 ] = ay [ 0 ] + x [ 1 ] = a 2 y − 1 + ax [ 0 ] + x [ 1 ] y [ 2 ] = ay [ 1 ] + x [ 2 ] = a 3 y − 1 + a 2 x [ 0 ] + ax [ 1 ] + x [ 2 ] . . . n � y [ n ] = ay [ n − 1 ] + x [ n ] = a n + 1 y − 1 + a n − k x [ k ] k = 0 7/38

Iterative Method Example (cont’d). Consider exponential smoothing y [ n ] − ay [ n − 1 ] = x [ n ] with y [ − 1 ] = y − 1 . • For n ≥ 0 , go forward in time y [ n ] = ay [ n − 1 ] + x [ n ] ay [ n − 1 ] = a 2 y [ n − 2 ] + ax [ n − 1 ] . . . a n − 1 y [ 1 ] = a n y [ 0 ] + a n − 1 x [ 1 ] a n y [ 0 ] = a n + 1 y [ − 1 ] + a n x [ 0 ] n � Summing up all equations, y [ n ] = a n + 1 y − 1 + a n − k x [ k ] k = 0 8/38

Iterative Method Example (cont’d). Consider exponential smoothing y [ n ] − ay [ n − 1 ] = x [ n ] with y [ − 1 ] = y − 1 . • n < − 1 , go backward in time y [ − 2 ] = a − 1 y [ − 1 ] − a − 1 x [ − 1 ] = a − 1 y − 1 − a − 1 x [ − 1 ] − 1 � y [ − 3 ] = a − 1 y [ − 2 ] − a − 1 x [ − 2 ] = a − 2 y − 1 − a − 3 − k x [ k ] k = − 2 . . . − 1 � y [ n ] = a − 1 y [ n + 1 ] − a − 1 x [ n + 1 ] = a n + 1 y − 1 − a n − k x [ k ] k = n + 1 9/38

Iterative Method Example (cont’d). Consider difference equation y [ n ] − ay [ n − 1 ] = x [ n ] with y [ − 1 ] = y − 1 . • Response � � n − 1 � � a n + 1 y − 1 a n − k x [ k ] y [ n ] = + − � �� � k = 0 k = n + 1 zero-input response � �� � zero-state response where by convention � m 2 k = m 1 · = 0 if m 2 < m 1 • Initial rest: causal LTI system with impulse response h [ n ] = a n u [ n ] ◮ infinite impulse response (IIR) system 10/38

Decomposition of Solutions (1) Linear inhomogeneous difference equation N � Ly = f , where L = a k τ k k = 0 Particular solution Ly p = f Homogeneous solution (natural response) Ly h = 0 General solution y = y p + y h 11/38

Homogeneous Solution Characteristic equation N � a k α N − k = 0 k = 0 • Obtained from Ly h = 0 upon substitution of y h [ n ] = α n • r distinct characteristic roots α i of multiplicity m i , i = 1 , 2 , . . . , r ; note � r i = 1 m i = N • Homogeneous solution takes form r m i � � A ik n k − 1 α n y h [ n ] = i i = 1 k = 1 i.e. space of all homogeneous solutions has basis α n 1 , n α n 1 , . . . , n m 1 − 1 α n 1 ; . . . ; α n r , n α n r , . . . , n m 1 − 1 α n r . 12/38

Particular Solution • Look for forced response of same form as input f f y p p � B k n k n p , 1 not characteristic root k = 0 p � B k n m + k n p , 1 characteristic root of multiplicity m k = 0 α n , α not characteristic root B α n α n Bn m i α n i , α i characteristic root of multiplicity m i i 2 ( e j ω n + e − j ω n ) and sin( ω n ) = 1 2 j ( e j ω n − e − j ω n ) Note cos( ω n ) = 1 are special cases 13/38

Example Fibonacci sequence � y [ n ] − y [ n − 1 ] − y [ n − 2 ] = 0 y [ 1 ] = y [ 2 ] = 1 • Characteristic equation √ ⇒ α 1 , 2 = 1 ± 5 α 2 − α − 1 = 0 = 2 • Homogeneous solution y [ n ] = A 1 α n 1 + A 2 α n 2 • Determine A 1 , A 2 from initial conditions � � 1 A 1 = y [ 1 ] = A 1 α 1 + A 2 α 2 = 1 √ = 5 ⇒ y [ 2 ] = A 1 α 2 1 + A 2 α 2 A 2 = − 1 2 = 1 √ 5 14/38

Example Solve difference equation y [ n ] + 2 y [ n − 1 ] = x [ n ] − x [ n − 1 ] with input x [ n ] = n 2 and initial condition y [ − 1 ] = − 1 • Characteristic equation α + 2 = 0 = ⇒ α = − 2 • Homogeneous solution y h [ n ] = A ( − 2 ) n • Particular solution ◮ find f [ n ] = x [ n ] − x [ n − 1 ] = n 2 − ( n − 1 ) 2 = 2 n − 1 ◮ look for solution of form y p [ n ] = B 1 n + B 0 y p [ n ] + 2 y p [ n − 1 ] = 3 B 1 n + 3 B 0 − 2 B 1 = f [ n ] = 2 n − 1 ◮ compare coefficients � � B 1 = 2 3 B 1 = 2 3 = ⇒ B 0 = 1 3 B 0 − 2 B 1 = − 1 9 15/38

Example (cont’d) Solve difference equation y [ n ] + 2 y [ n − 1 ] = x [ n ] − x [ n − 1 ] with input x [ n ] = n 2 and initial condition y [ − 1 ] = − 1 • General solution y [ n ] = 2 3 n + 1 9 + A ( − 2 ) n • Determine A from initial condition y [ − 1 ] = 2 3 ( − 1 ) + 1 ⇒ A = 8 9 + A ( − 2 ) − 1 = − 1 = 9 • Complete solution 2 3 n + 1 8 9 ( − 2 ) n y [ n ] = + 9 � �� � � �� � forced response natural response 16/38

Decomposition of Solutions (2) IVP � Ly = f y [ k ] = y k , k = n 0 − 1 , n 0 − 2 , . . . , n 0 − N Zero-input response: linear in initial state � Ly zi = 0 y zi [ k ] = y k , k = n 0 − 1 , n 0 − 2 , . . . , n 0 − N Zero-state response: linear in input � Ly zs = f y zs [ k ] = 0 , k = n 0 − 1 , n 0 − 2 , . . . , n 0 − N Complete solution y = y zi + y zs 17/38

Zero-state Response � Ly zs = f y zs [ k ] = 0 , k = n 0 − 1 , n 0 − 2 , . . . , n 0 − N Focus on case where f [ n ] = 0 for n < n 0 = ⇒ initial rest Causal and LTI system with output n � y zs [ n ] = ( f ∗ h )[ n ] = u [ n − n 0 ] f [ k ] h [ n − k ] k = n 0 where h is impulse response of Ly = x , i.e. � Lh = δ, n ≥ 0 h [ k ] = 0 , k = − 1 , − 2 , . . . , − N or equivalently � Lh = 0 , n ≥ 1 h [ k ] = 0 , k = − 1 , − 2 , . . . , 1 − N ; h [ 0 ] = 1 / a 0 18/38

Example Solve difference equation y [ n ] + 2 y [ n − 1 ] = x [ n ] − x [ n − 1 ] with input x [ n ] = n 2 u [ n ] and initial condition y [ − 1 ] = − 1 • Homogeneous solution y h [ n ] = A ( − 2 ) n • Zero-input response, i.e. homogeneous solution with y zi [ − 1 ] = − 1 A ( − 2 ) − 1 = − 1 = ⇒ y zi = 2 ( − 2 ) n ⇒ A = 2 = • Impulse response of y [ n ] + 2 y [ n − 1 ] = x [ n ] , i.e. solution of following IVP � h [ n ] + 2 h [ n − 1 ] = δ [ n ] h [ − 1 ] = 0 19/38

Example (cont’d) Solve difference equation y [ n ] + 2 y [ n − 1 ] = x [ n ] − x [ n − 1 ] with input x [ n ] = n 2 u [ n ] and initial condition y [ − 1 ] = − 1 • Homogeneous solution y h [ n ] = A ( − 2 ) n • Impulse response � h [ n ] + 2 h [ n − 1 ] = δ [ n ] , n ≥ 0 h [ − 1 ] = 0 ◮ by iterative method h [ 0 ] = δ [ 0 ] − 2 h [ − 1 ] = 1 ◮ now solve � h [ n ] + 2 h [ n − 1 ] = 0 , n ≥ 1 h [ 0 ] = 1 ◮ h [ 0 ] = A = 1 = ⇒ h [ n ] = ( − 2 ) n u [ n ] 20/38