Differential Equations. Solving for Impulse Response. Linear systems are often described using differential equations.

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1 Differenial Equaions Linear sysems are ofen described using differenial equaions. For example: d 2 y d 2 + 5dy + 6y f() d where f() is he inpu o he sysem and y() is he oupu. We know how o solve for y given a specific inpu f. We now cover an alernaive approach: Differenial Equaion solve Solving for Impulse Response We canno solve for he impulse response direcly so we solve for he sep response and hen differeniae i o ge he impulse response. Differenial Equaion solve Sep response differeniae Impulse response Any inpu Impulse response convoluion Corresponding Oupu Any inpu convoluion Corresponding Oupu 17 18

2 Moivaion: Convoluion Solving for Sep Response If we know he response of a linear sysem o a sep inpu, we can calculae he impulse response and hence we can find he response o any inpu by convoluion. If we wan o find he sep response of dy d + 5y f() where f is he inpu and y is he oupu. I would be nice if we could pu f() H() and solve. Unforunaely we don know of a way o do his direcly. So we Suppose we wan o know how a car s suspension responds o los of differen ypes of road surface. We measure how he suspension responds o a sep inpu (or calculae he sep response from a heoreical model of he sysem). We can hen find he impulse response and use convoluion o find he car s behaviour for any road surface profile se f() 1, and solve for jus 0 2. se he boundary condiion y(0) 0 (also ẏ(0) 0 for second order equaions) o imply ha f() was zero for all < 0. We hus have a complee soluion because y 0 for < 0, and we have found y for all 0. 20

3 Boundary Condiion Jusificaion Prove ha y 0 a 0 by conradicion. We know ha y() 0 for all < 0. Therefore he only way for y o equal somehing oher han zero a 0 is if here is a sep disconinuiy in y a 0. Assume ha y has a sep of heigh h a 0. If y has a sep disconinuiy a 0 hen dy d mus have a dela funcion a 0. So we have: f() is a sep funcion so f() 1 for all. y h a 0. dy a 0. d Which violaes he original equaion a 0. dy f() 5y d As he RHS is finie bu he LHS is infinie. Therefore y mus be coninuous a 0, and we can use he iniial condiion y(0) Sep Response Example Sep 1: se f() 1, and solve for jus 0. dy d + 5y 1 Complimenary funcion: ẏ + 5y 0 y Ae 5 Paricular Inegral: ry y λ (a cons) y 1 5 General Soluion: y Ae Sep 2: se he boundary condiion y 0 a 0 y(0) 0 A A 1 5 So sep response is y() 1 5 ( 1 e 5 ) for 0. 22

4 Find he Impulse Response Sep Impulse Response Impulse Response g() inegrae differeniae Sep Response Sep response is y() 1 5 ( 1 e 5 ) for 0. Impulse response g() is given by: 0, < 0 g() [ d 1 ( 1 e 5 )] e 5, 0 d 5 23 d 2 y d dy d + 12y f() 1. Find he General Soluion wih f() 1 Complimenary funcion is y Ae 12 + Be Paricular inegral is y 1 12 General soluion is y Ae 12 + Be 2. Se boundary condiions y(0) ẏ(0) 0 o ge he sep response A + B 0 12A B 0 A and B 1 11 Thus Sep Response is y e e Differeniae he sep response o ge he impulse response. g() dy d e e 12 11, ( > 0) 24

5 Using he Impulse Response If we have a sysem inpu composed of impulses, f() 3δ( 1) + 4δ( 2) we can find he corresponding sysem oupu using superposiion. y() 3g( 1) + 4g( 2) 3 e ( 1) e 12( 1) e ( 2) e 12( 2) 11 More General Inpu Suppose our inpu is composed of los of dela funcions: f() n p n δ( q n ) Then he corresponding sysem oupu will be y() n p n g( q n ) 25 26

6 Secion 2: Summary Differenial Equaion ay + by + cy + d f() solve ay + by + cy + d 1 wih boundary condiions y(0) 0 and y(0) 0 Secion 3 Convoluion Sep response differeniae In his secion we derive he convoluion inegral and show is use in some examples. Impulse response Any inpu convoluion Corresponding Oupu 27 28

7 Convoluion Our goal is o calculae he oupu, y() of a linear sysem using he inpu, f(), and he impulse response of he sysem, g(). An impulse a ime 0 produces he impulse response. δ() Linear Sysem g() An impulse delayed o ime τ produces a delayed impulse response saring a ime τ. A scaled impulse a ime 0 produces a scaled impulse response. k δ() Linear Sysem k g() An impulse ha has been scaled by k and delayed o ime τ produces an impulse response scaled by k and saring a ime τ. k δ( τ ) Linear Sysem k g( τ ) δ( τ ) Linear Sysem g( τ ) τ τ τ τ 29 30

8 Consider he inpu, f() o be made up of a sequence of srips of widh τ. Each of hese srips is similar o a dela funcion and hus leads o a sysem oupu of an appropriaely scaled and delayed impulse response. f() δ( τ ) f( τ) τ leads o response g( τ ) f( τ ) τ y() g( τ)f(τ)dτ τ τ The response of he sysem, y() is hus he sum of hese delayed, scaled impulse responses. (Provided g() 0 for < 0.) y() g( τ)f(τ) τ All slices Trea as a consan when evaluaing he inegral. The inegraion variable is τ. is ime as i relaes o he oupu of he sysem y(). τ is ime as i relaes o he inpu of he sysem f(τ). Le he widh of he slices end o zero. The sum urns ino an inegral called he convoluion inegral. y() g( τ)f(τ)dτ 31 32

9 Convoluion Example 1 Consider a sysem wih impulse response { 0, < 0 g() e 5, 0 Find he oupu for inpu f() H() (sep funcion). Convoluion Example 2 For he same sysem (g() e 5, 0), find he oupu for inpu 0, < 0 f() v, 0 < < k 0, > k v f() k Using he convoluion inegral, he answer is given by y() 1 5 g( τ)f(τ)dτ e 5( τ) H(τ)dτ 0 e 5( τ) dτ [ ] 1 5 e 5( τ) 0 ( 1 e 5 ) y() g( τ)f(τ)dτ g( τ) 0 dτ, < 0 0 g( τ) 0 dτ + 0 g( τ) v dτ, 0 < < k 0 g( τ) 0 dτ + k 0 g( τ) v dτ + k g( τ) 0 dτ, > k 33 34

10 Case (a): < 0 g( τ) 0 dτ 0 so y() 0 for all < 0. Case (b): 0 < < k y() g( τ) v dτ 0 0 e 5( τ) v dτ Case (c): > k [ e 5( τ) ] v 5 v ( 1 e 5 ) 5 k k y() g( τ) v dτ 0 0 e 5( τ) v dτ y() 0 [ e 5( τ) ] k v 5 0 v ( e 5k 1 ) e 5 5 Convoluion Example 3 For he same sysem (g() e 5, 0), find he oupu for inpu f() { 0, < 0 sin(ω), > 0 f() Using he convoluion inegral, he answer is given by y() g( τ)f(τ)dτ g( τ) 0 dτ, < 0 0 g( τ) 0 dτ + 0 g( τ) sin(ωτ) dτ, 0 < (a) (b) k (c) 35 36

11 Convoluion Summary Case (a): < 0 Differenial Equaion ay + by + cy + d f() g( τ) 0 dτ 0 so y() 0 for all < 0. Case (b): 0 < y() g( τ)sin(ωτ) dτ 0 0 e 5( τ) sin(ωτ) dτ { } Im 0 e 5( τ) e iωτ dτ Im e 5 e(5+iω)τ 5 + iω 0 { e iω e 5 } Im 5 + iω 5sin(ω) ω cos(ω) + ωe ω 2 Any Inpu: f() solve ay + by + cy + d 1 wih boundary condiions y(0) 0 and y(0) 0 Sep response differeniae Impulse response: g() convoluion Corresponding Oupu: y() y() g( τ) f(τ) dτ 38 37

12 Complee Example Find he impulse response of d 2 y d 2 + 3dy d + 2y f() hence find he oupu when he inpu f() H()e. 3. Differeniae he sep response o ge he impulse response. g() dy d e e 2 1. Find he General Soluion wih f() 1 Complimenary funcion is y Ae + Be 2 Paricular inegral is y 1 2 General soluion is y Ae + Be 2 2. Se boundary condiions y(0) ẏ(0) 0 o ge he sep response A + B 0 A 2B 0 A 1 and B 1 2 Thus Sep Response is y 1 2 e + e Use he convoluion inegral o find he oupu for he required inpu. The required inpu is f() e, > 0. y() g( τ)f(τ)dτ 0 ( e ( τ) e 2( τ) ) e τ dτ 0 e e τ 2 dτ [ τe e τ 2] 0 ( 1) e + e 2 40

13 Secion 3: Summary Convoluion inegral (memorise his): f() inpu g() impulse response y() oupu y() g( τ) f(τ) dτ Way o find he oupu of a linear sysem, described by a differenial equaion, for an arbirary inpu: Find general soluion o equaion for inpu 1. Se boundary condiions y(0) ẏ(0) 0 o ge he sep response. Differeniae o ge he impulse response. Use convoluion inegral ogeher wih he impulse response o find he oupu for any desired inpu. Secion 4 Evaluaing Convoluion Inegrals A way of rearranging he convoluion inegral is described and illusraed. The differences beween convoluion in ime and space are discussed and he concep of causaliy is inroduced. The secion ends wih an example of spaial convoluion

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