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Differential Equations Solving for Impulse Response Linear systems are often described using differential equations. For example: We cannot solve for the impulse response directly so d2y dy we solve for the step response and then differentiate + 5 + 6y = f (t) dt2 dt it to get the impulse response. where f(t) is the input to the system and y(t) is the output. Differential solve Equation We know how to solve for y given a speciﬁc input f . Step response We now cover an alternative approach: differentiate Differential solve Equation Impulse response Impulse response Corresponding Any input convolution Output Corresponding Any input convolution Output 17 18 Motivation: Convolution Solving for Step Response If we know the response of a linear system to a step If we want to ﬁnd the step response of input, we can calculate the impulse response and hence we can ﬁnd the response to any input by convolution. dy + 5y = f (t) dt where f is the input and y is the output. It would be nice if we could put f (t) = H(t) and solve. Unfortu- nately we don’t know of a way to do this directly. So we 1. set f (t) = 1, and solve for just t ≥ 0 Suppose we want to know how a car’s suspension re- sponds to lots of different types of road surface. ˙ 2. set the boundary condition y(0) = 0 (also y(0) = We measure how the suspension responds to a step input (or calculate the step response from a theoreti- 0 for second order equations) to imply that f (t) cal model of the system). was zero for all t < 0. We can then ﬁnd the impulse response and use con- volution to ﬁnd the car’s behaviour for any road sur- We thus have a complete solution because y = 0 for face proﬁle. t < 0, and we have found y for all t ≥ 0. 19 20 Boundary Condition Justiﬁcation Step Response Example Prove that y = 0 at t = 0 by contradiction. Step 1: set f (t) = 1, and solve for just t ≥ 0. We know that y(t) = 0 for all t < 0. Therefore the only way for y to equal something other than zero at t = 0 is if there is a step discontinuity in y at t = 0 . dy + 5y = 1 dt Assume that y has a step of height h at t = 0 . If y has a step discontinuity at t = 0 then dy must have dt a delta function at t = 0. Complimentary function: y + 5y = 0 ⇒ y = Ae−5t ˙ So we have: • f (t) is a step function so |f (t)| ≤ 1 for all t. 1 Particular Integral: try y = λ (a const) ⇒ y = 5 • |y| ≤ h at t = 0. 1 General Solution: y = Ae−5t + 5 dy • dt → ∞ at t = 0. Which violates the original equation at t = 0. Step 2: set the boundary condition y = 0 at t = 0 dy dt = f (t) − 5y 1 y(0) = 0 ⇒ A + 5 = 0 ⇒ A = − 1 5 As the RHS is ﬁnite but the LHS is inﬁnite. Therefore y must be continuous at t = 0, and we can use the So step response is y(t) = 1 1 − e−5t for t ≥ 0. 5 initial condition y(0) = 0. 21 22 Find the Impulse Response d2y dy + 13 + 12y = f (t) dt2 dt Step → Impulse Response 1. Find the General Solution with f (t) = 1 integrate Complimentary function is y = Ae−12t + Be−t Impulse Step Response Response 1 Particular integral is y = 12 differentiate g(t) 1 General solution is y = 12 + Ae−12t + Be−t 1 Step response is y(t) = 5 1 − e−5t for t ≥ 0. ˙ 2. Set boundary conditions y(0) = y(0) = 0 to get the step response. 1 Impulse response g(t) is given by: 12 + A + B = 0 −12A − B = 0 0, t < 0 1 1 ⇒ A = 132 and B = − 11 g(t) = 1 −12t −t d 1 Thus Step Response is y = 12 + e132 − e 1 − e−5t = e−5t, t ≥ 0 11 dt 5 3. Differentiate the step response to get the impulse response. dy e−t − e−12t g(t) = = , (t > 0) dt 11 23 24 Using the Impulse Response If we have a system input composed of impulses, More General Input f (t) = 3δ(t − 1) + 4δ(t − 2) Suppose our input is composed of lots of delta func- we can ﬁnd the corresponding system output using tions: superposition. f (t) = pn δ(t − qn ) n y(t) = 3g(t − 1) + 4g(t − 2) Then the corresponding system output will be e−(t−1) − e−12(t−1) = 3 y(t) = pn g(t − qn ) 11 n e−(t−2) − e−12(t−2) +4 11 25 26 Section 2: Summary Differential Equation ay + by + cy + d = f(t) Section 3 solve ay + by + cy + d = 1 with boundary conditions y(0) = 0 and y(0) = 0 Convolution Step response In this section we derive the convolution integral and show its use in some examples. differentiate Impulse response Corresponding Any input convolution Output 27 28 Convolution Our goal is to calculate the output, y(t) of a linear sys- tem using the input, f (t), and the impulse response A scaled impulse at time t = 0 produces a scaled of the system, g(t). impulse response. An impulse at time t = 0 produces the impulse re- k δ(t) Linear k g(t) sponse. System t t δ (t) Linear g(t) System An impulse that has been scaled by k and delayed to time t = τ produces an impulse response scaled by t t k and starting at time τ . An impulse delayed to time t = τ produces a delayed k δ(t−τ ) Linear k g(t−τ ) impulse response starting at time τ . System δ(t−τ ) Linear g(t−τ ) τ t τ t System τ t τ t 29 30 Consider the input, f (t) to be made up of a sequence of strips of width ∆τ . Each of these strips is similar to a delta function and thus leads to a system out- put of an appropriately scaled and delayed impulse response. f(t) δ (t−τ ) f(τ ) ∆ τ t leads to response y(t) = g(t − τ )f (τ )dτ g(t−τ ) f(τ ) ∆τ −∞ • Treat t as a constant when evaluating the integral. ∆τ The integration variable is τ . τ t The response of the system, y(t) is thus the sum of • t is time as it relates to the output of the system these delayed, scaled impulse responses. (Provided y(t). g(t) = 0 for t < 0.) y(t) ≈ g(t − τ )f (τ )∆τ • τ is time as it relates to the input of the system All f (τ ). slices Let the width of the slices tend to zero. The sum turns into an integral called the convolution integral. t y(t) = g(t − τ )f (τ )dτ −∞ 31 32 Convolution Example 2 Convolution Example 1 For the same system (g(t) = e−5t , t ≥ 0), ﬁnd the output for input Consider a system with impulse response 0 ,t<0 f(t) g(t) = 0, t < 0 e−5t , t ≥ 0 f (t) = v, 0 < t < k v 0, t > k t Find the output for input f (t) = H(t) (step function). k Using the convolution integral, the answer is given by t y(t) = g(t − τ )f (τ )dτ t −∞ y(t) = g(t − τ )f (τ )dτ t −∞ = e−5(t−τ )H(τ )dτ −∞ t g(t − τ ) × 0 dτ, −∞ t<0 t e−5(t−τ )dτ = 0 0 −∞ g(t − τ ) × 0 dτ 1 −5(t−τ ) t t + 0 g(t − τ ) v dτ, 0<t<k = e 5 0 = 1 0 = 1 − e−5t −∞ g(t − τ ) × 0 dτ 5 k + 0 g(t − τ ) v dτ t + k g(t − τ ) × 0 dτ, t > k 33 34 Case (a): t < 0 t −∞ g(t − τ ) × 0 dτ = 0 so y(t) = 0 for all t < 0. Convolution Example 3 Case (b): 0 < t < k t t y(t) = g(t − τ ) v dτ = e−5(t−τ ) v dτ For the same system (g(t) = e−5t , t ≥ 0), ﬁnd the 0 0 v −5(t−τ ) t output for input = e 5 0 v 0, t<0 f(t) = 1 − e−5t f (t) = 5 sin(ωt), t > 0 t Case (c): t > k k k Using the convolution integral, the answer is given by −5(t−τ ) y(t) = g(t − τ ) v dτ = e v dτ 0 0 v −5(t−τ ) k t = e y(t) = g(t − τ )f (τ )dτ 5 0 −∞ v 5k = e − 1 e−5t t g(t − τ ) × 0 dτ, 5 −∞ t<0 = 0 −∞ g(t − τ ) × 0 dτ y(t) t + 0 g(t − τ ) sin(ωτ ) dτ, 0 < t (a) (b) k (c) t 35 36 Convolution Summary Differential Equation Case (a): t < 0 ay + by + cy + d = f(t) t −∞ g(t − τ ) × 0 dτ = 0 so y(t) = 0 for all t < 0. solve ay + by + cy + d = 1 Case (b): 0 < t with boundary conditions t y(0) = 0 and y(0) = 0 y(t) = g(t − τ ) sin(ωτ ) dτ 0 t = e−5(t−τ ) sin(ωτ ) dτ 0 Step response t = Im e−5(t−τ )eiωτ dτ 0 t differentiate e(5+iω)τ = Im e−5t 5 + iω 0 Impulse response: g(t) eiωt − e−5t = Im 5 + iω Any Corresponding convolution 5 sin(ωt) − ω cos(ωt) + ωe−5t Input: f(t) Output: y(t) = 25 + ω 2 t y(t) = g(t − τ ) f (τ ) dτ −∞ 37 38 Complete Example Find the impulse response of 3. Differentiate the step response to get the impulse d2y dy response. + 3 + 2y = f (t) dt2 dt dy hence ﬁnd the output when the input f (t) = H(t)e−t . g(t) = = e−t − e−2t dt 1. Find the General Solution with f (t) = 1 4. Use the convolution integral to ﬁnd the output for Complimentary function is y = Ae−t + Be−2t the required input. Particular integral is y = 1 2 The required input is f (t) = e−t , t > 0. 1 General solution is y = 2 + Ae−t + Be−2t t y(t) = g(t − τ )f (τ )dτ −∞ t 2. Set boundary conditions y(0) = y(0) = 0 to get ˙ = e−(t−τ ) − e−2(t−τ ) e−τ dτ 0 the step response. t = e−t − eτ −2t dτ 1 0 2+A+B =0 t −A − 2B = 0 = τ e−t − eτ −2t 0 ⇒ A = −1 and B = 1 2 = (t − 1) e−t + e−2t −2t Thus Step Response is y = 1 − e−t + e 2 2 39 40 Section 3: Summary Convolution integral (memorise this): Section 4 f (t) = input g(t) = impulse response Evaluating Convolution Integrals y(t) = output t y(t) = g(t − τ ) f (τ ) dτ −∞ A way of rearranging the convolution integral is de- Way to ﬁnd the output of a linear system, described scribed and illustrated. by a differential equation, for an arbitrary input: The differences between convolution in time and space • Find general solution to equation for input = 1. are discussed and the concept of causality is intro- • Set boundary conditions y(0) = y(0) = 0 to get ˙ duced. the step response. The section ends with an example of spatial convolu- • Differentiate to get the impulse response. tion. • Use convolution integral together with the impulse response to ﬁnd the output for any desired input. 41 42

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posted: | 7/10/2011 |

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