EI331 Signals and Systems Lecture 7 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 7 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University March 19, 2019

Contents 1. Causal LTI Systems Described by Differential Equations 1.1 Initial Rest 1.2 Jump from 0 − to 0 + 1.3 Higher-order ODE 1.4 Systems of First-order ODEs 1.5 Second Recipe for Higher-order ODE 1/31

Initial Rest Often work with right-sided inputs, i.e. x ( t ) = 0 for t < t 0 • stimulus turned on at some point Initial rest condition • If input x ( t ) = 0 for t < t 0 , output y ( t ) = 0 for t < t 0 ◮ output zero until changed by input (cf. Newton’s law) ◮ equivalent to causality for linear systems • Adapt initial time t 0 to input x : if x becomes nonzero at t 0 , use y ( k ) ( t 0 ) = 0 for k = 0 , 1 , . . . , N − 1 , i.e. solve � L y y = f y ( k ) ( t 0 ) = 0 , k = 0 , 1 , . . . , N − 1 Linear constant-coefficient ODE with initial rest condition specifies causal and LTI system for right-sided inputs 2/31

Initial Rest Example. Newton’s second law mx ′′ ( t ) = f ( t ) Initial rest • stays at origin x = 0 • zero velocity v = 0 (at rest!) • stays so unless changed by external force (Newton’s first law) If force starts at t = 0 • x ( 0 ) = 0 , v ( 0 ) = x ′ ( 0 ) = 0 f If force starts on at t = 1 • x ( 1 ) = 0 , v ( 1 ) = x ′ ( 1 ) = 0 3/31

Initial Rest Example. RLC circuit C d 2 dt 2 v ( t ) + 1 dtv ( t ) + 1 d Lv ( t ) = d dti S ( t ) R Initial rest • no stored energy in L , C + i S ( t ) i R i L i C • zero voltage and current v ( t ) R R L C If source on at t = 0 • v ( 0 ) = 0 − • i C ( 0 ) = Cv ′ ( 0 ) = 0 If source on at t = 1 • v ( 1 ) = 0 • i C ( 1 ) = Cv ′ ( 1 ) = 0 4/31

Initial Rest General IVP with first-order ODE � y ′ ( t ) + ay ( t ) = f ( t ) y ( t 0 ) = y 0 Solution � t y 0 e − a ( t − t 0 ) f ( τ ) e − a ( t − τ ) d τ y ( t ) = + � �� � t 0 zero-input response � �� � zero-state response Initial rest: zero-input response always 0; take t 0 → −∞ � t � t f = f · τ t 0 u f ( τ ) e − a ( t − τ ) d τ f ( τ ) e − a ( t − τ ) d τ y ( t ) = = = = = ⇒ y ( t ) = u ( t − t 0 ) t 0 −∞ 5/31

Initial Rest Example. y ′ ( t ) + 2 y ( t ) = x ( t ) with initial rest condition and input x ( t ) = u ( t + 1 ) Response � t � t x ( τ ) e − 2 ( t − τ ) d τ = u ( τ + 1 ) e − 2 ( t − τ ) d τ y ( t ) = −∞ −∞ For t < − 1 � t 0 · e − 2 ( t − τ ) d τ = 0 y ( t ) = −∞ For t > − 1 � t e − 2 ( t − τ ) d τ = 1 2 ( 1 − e − 2 ( t + 1 ) ) y ( t ) = − 1 6/31

Non-initial Rest Example. y ′ ( t ) + 2 y ( t ) = x ( t ) with initial condition y ( 0 ) = 0 and input x ( t ) = u ( t + 1 ) Response � t � t x ( τ ) e − 2 ( t − τ ) d τ = u ( τ + 1 ) e − 2 ( t − τ ) d τ y ( t ) = 0 0 For t > − 1 � t e − 2 ( t − τ ) d τ = 1 2 ( 1 − e − 2 t ) y ( t ) = 0 For t < − 1 � − 1 e − 2 ( t − τ ) d τ = 1 2 ( e − 2 − 1 ) e − 2 t y ( t ) = 0 7/31

Comparison of Initial Conditions Example. y ′ ( t ) + 2 y ( t ) = u ( t + 1 ) y x y ( 0 ) = 0 1 1 2 t t − 1 0 − 1 0 y initial rest 1 2 1 2 ( 1 − e 2 ) t − 1 0 8/31

Jump from 0 − to 0 + Need more care for initial condition with singular input Example. What’s impulse response of causal LTI system described by y ′ ( t ) + 2 y ( t ) = x ( t ) with initial rest condition? Method 1. Solve � h ′ ( t ) + 2 h ( t ) = δ ( t ) h ( 0 ) = 0 • For t � = 0 , reduces to � h ′ ( t ) + 2 h ( t ) = 0 h ( 0 ) = 0 • General solution h ( t ) = Ae − 2 t for t � = 0 ⇒ h ( t ) = 0 for t � = 0 , something wrong � • h ( 0 ) = 0 = 9/31

Jump from 0 − to 0 + Need more care for initial condition with singular input Example. What’s impulse response of causal LTI system described by y ′ ( t ) + 2 y ( t ) = x ( t ) with initial rest condition? Method 2. Response � t x ( τ ) e − 2 ( t − τ ) d τ = ⇒ h ( t ) = e − 2 t u ( t ) y ( t ) = −∞ Observation: h discontinuous at t = 0 • h ( 0 − ) = 0 , h ( 0 + ) = 1 , due to singularity of δ at t = 0 • For t > 0 , � t � t δ ( τ ) e − 2 ( t − τ ) d τ = δ ( τ ) e − 2 ( t − τ ) d τ h ( t ) = 0 − −∞ 10/31

Jump from 0 − to 0 + IVP � � y ′ ( t ) + ay ( t ) = f ( t ) y ′ ( t ) + ay ( t ) = f ( t ) vs. y ( 0 − ) = y 0 y ( 0 + ) = y 0 Solution � t y ( t ) = y ( 0 − ) e − at + f ( τ ) e − a ( t − τ ) d τ 0 − vs. � t y ( t ) = y ( 0 + ) e − at + f ( τ ) e − a ( t − τ ) d τ 0 + Initial rest: use y ( 0 − ) = 0 � 0 + f ( τ ) e a τ d τ y ( 0 + ) = y ( 0 − ) + 0 − • if f ( τ ) has no singularity at τ = 0 , y ( 0 + ) = y ( 0 − ) = 0 • if f ( τ ) has singularity at τ = 0 , y ( 0 + ) may be different 11/31

Jump from 0 − to 0 + Example. Impulse response of y ′ ( t ) + 2 y ( t ) = x ( t ) revisited. � h ′ ( t ) + 2 h ( t ) = δ ( t ) h ( 0 − ) = 0 • For t � = 0 , reduces to � � h ′ ( t ) + 2 h ( t ) = 0 , t > 0 h ′ ( t ) + 2 h ( t ) = 0 , t < 0 h ( 0 − ) = 0 h ( 0 − ) = 0 • General solution h ( t ) = A + e − 2 t u ( t ) + A − e − 2 t u ( − t ) • A + = h ( 0 + ) , but used A + = A − = h ( 0 − ) = 0 in first try � � t h ( t ) = h ( 0 − ) e − at + δ ( τ ) e − a ( t − τ ) d τ 0 + 12/31

Recipe for IVP with First-order ODE IVP � y ′ ( t ) + ay ( t ) = f ( t ) y ( t 0 ) = y 0 Solution for all cases � t y ( t ) = y ( t 0 ) e − a ( t − t 0 ) + f ( τ ) e − a ( t − τ ) d τ t 0 • If t 0 means t 0 + or t 0 − , be consistent in all places! • Matters only if f has singularity at t 0 Initial rest � t f ( τ ) e − a ( t − τ ) d τ y ( t ) = −∞ 13/31

Higher-order ODE N d k � Ly = a k dt k y = f , ( a N � = 0 ) k = 0 General solution y = + y h y p ���� ���� homogeneous solution particular solution Characteristic equation N � a k λ k = 0 k = 0 � d � k dt → λ ; note d k • LHS obtained from L by substituion d dt k = dt • N (complex) roots by Fundamental Theorem of Algebra (root of multiplicity k counted as k roots) 14/31

Higher-order ODE Homogeneous solution • r distinct characteristic roots λ 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 t k − 1 e λ i t y h ( t ) = i = 1 k = 1 i.e. space of all homogeneous solutions has basis e λ 1 t , te λ 1 t , . . . , t m 1 − 1 e λ 1 t ; . . . ; e λ r t , te λ r t , . . . , t m r − 1 e λ r t . • When a k ∈ R , ∀ k , complex roots σ ± j ω appear in pairs ◮ in calculus, used e σ t cos( ω t ) and e σ t sin( ω t ) ◮ here, use e ( σ + j ω ) t and e ( σ − j ω ) t ◮ equivalent by Euler’s formula 15/31

Higher-order ODE Particular solution • Look for forced response of same form as input f f y p p � t p , 0 not characteristic root B k t k k = 0 p � B k t m + k t p , 0 characteristic root of multiplicity m k = 0 e at , a not characteristic root Be at e λ i t , λ i characteristic root of multiplicity m i Bt m i e λ i t 2 ( e j ω t + e − j ω t ) and sin( ω t ) = 1 2 j ( e j ω t − e − j ω t ) are Note cos( ω t ) = 1 special cases 16/31

IVP with Second-order ODE Example. Second-order system y ′′ + 3 y ′ + 2 y = x at initial rest. Let x ( t ) = e − t u ( t ) . • Characteristic equation λ 2 + 3 λ + 2 = 0 = ⇒ λ 1 = − 1 , λ 2 = − 2 • Homogeneous solution y h ( t ) = A 1 e − t + A 2 e − 2 t • For t > 0 , particular solution y p ( t ) = Bte − t p ( t ) + 2 y p ( t ) = Be − t = x ( t ) = e − t = y ′′ p ( t ) + 3 y ′ ⇒ B = 1 • General solution y ( t ) = te − t + A 1 e − t + A 2 e − 2 t ⇒ y ( t ) = te − t + e − 2 t − e − t • Initial rest y ( 0 ) = y ′ ( 0 ) = 0 = 17/31

Systems of First-order ODEs Consider N -th order ODE with a N = 1 (WLOG) y ( N ) + a N − 1 y ( N − 1 ) + · · · + a 1 y ′ + a 0 y = f (1) Let Y k = y ( k ) , k = 0 , 1 , . . . , N − 1 • Y ′ k = Y k + 1 for k = 0 , 1 , . . . , N − 2 N − 1 = y ( N ) = f − � N − 1 k = 0 a k y ( k ) = f − � N − 1 • Y ′ k = 0 a k Y k (1) equivalent to Y ′ = AY + bf where       . . . Y 0 0 1 0 0 0 0 . . .  Y 1   0 0 1 0 0   0        . . . . . . . ...  .   . . . . .   .  Y = , A = , b = . . . . . . .             . . . Y N − 2 0 0 0 0 1 0       Y N − 1 − a 0 − a 1 − a 2 − a 3 . . . − a N − 1 1 18/31

Systems of First-order ODEs Initial value problem (IVP) � y ( N ) + a N − 1 y ( N − 1 ) + · · · + a 1 y ′ + a 0 y = f (2) y ( k ) ( t 0 ) = y k , k = 0 , 1 , . . . , N − 1 equivalent to � Y ′ = AY + bf (3) Y ( t 0 ) = Y 0 where Y 0 = ( y 0 , y 1 , . . . , y N − 1 ) T . Solution to (3) � t e A ( t − t 0 ) Y 0 f ( τ ) e A ( t − τ ) bd τ Y ( t ) = + � �� � t 0 zero-input response � �� � zero-state response ∞ matrix exponential e At � 2 ( At ) 2 + . . . � ( At ) n = I + At + 1 n ! n = 0 19/31