Part IA Engineering Contents & Examples Questions Mathematics Section 1 Linear System & Impulse Response Lent Term Paper 8: questions 1–3. Section 2 Differential Equations to Describe Linear Systems Convolution Paper 8: questions 4 and 5. Fourier Series Section 3 Convolution Paper 8: questions 6 and 7. Probability Section 4 Evaluating Convolution Integrals Paper 8: question 8. Section 5 Fourier Series Richard Prager Paper 9: questions 1 and 2. January 2009 1 2 Section 1 Contents & Examples Qs contd. Linear System & Impulse Response Section 6 General Fourier Series Paper 9: question 3 and first part of 4. We motivate the study of linear time-invariant systems. Section 7 Convergence & Half Range Serieses Paper 9: end of Q4 and questions 5–7 The principle of superposition is explained. Section 8 Complex Fourier Series Step functions and delta functions are introduced, to- Paper 9: question 8. gether with their corresponding responses. Section 9 Probability Paper 10. Examples are given to illustrate the use of the step response with superposition. The sifting theorem is stated and illustrated with some examples. 3 4 Linear Systems Motivation f(t)Linear y(t) System Many engineering problems concern linear systems. ForcesSystem Strains 1. Linear time-invariant systems satisfy the principle VoltagesSystem Currents of superposition. Pressure System Density If input f1(t) output y1(t) and input f (t) → output y (t) 2 → 2 Heat flow System Temperature then input αf (t)+ βf (t) output αy (t)+ βy (t) 1 2 → 1 2 Power System Kinetic Energy where α and β are any constants. 2. Sine waves have the special property that a sine In a linear system the output is computed as some wave at the input leads to a (possibly different) sine linear combination of the inputs (including inputs from wave at the output. the past, if we are considering a system with a time- varying input and output). Sine wave with Sine wave Linear amplitude or System phase changed 5 6 Superposition Example Step Function Linear System H(t) r(t) H(t) t H(t−1) r(t−1) 1 1 0, t< 0 H(t)= −H(t−1) −r(t−1) ( 1, t> 0 H(t) r(t) Linear H(t) − H(t−1) r(t) − r(t−1) System step response 1 1 7 8 Calculation of Superposition Dirac Delta Function Find the output of a linear system f(t) with step response f(t) 0, t< 0 2 height = 1/w r(t)= 1 e 5t, t 0 ( − − ≥ when the input is the pulse f(t). 1 t −w/2 w/2 t Case (a): When 0 t< 1 the input is the same as a As w 0 the pulse f(t) becomes narrower and ≤ → scaled step function so the output y(t) is given by taller. In the limit as w 0 the pulse f(t) becomes a → delta function: ( ). y(t) = 2 1 e 5t , 0 t< 1 δ t − − ≤ The delta function is a spike with unit area. It goes Case (b): When 1 t ≤ bang when its argument is zero. f(t) = 2H(t) 2H(t 1) − − δ(t) = 0 except at t = 0 therefore the output y(t) is given by b δ(t)dt = 1 provided a< 0 and b> 0 Za y(t) = 2r(t) 2r(t 1) − 5t − 5(t 1) = 2 1 e− 1+ e− − δ − − (t)Linear impulse response = 2 e5 1 e 5t , 1 t − − ≤ System 9 10 Integrating the Delta Function Impulse Response From the previous page Impulse integrate Step b Response Response δ(t) dt = 1 provided a< 0 and b> 0 differentiate Za g(t) r(t) thus T 0, T< 0 δ(t) dt = Find the output, g(t) of a linear system with step re- ( 1, T> 0 5t Z−∞ sponse r(t) = 1 e− when the input is the delta = H(T ) − function δ(t). The integral of a delta function is a step function. Conversely, the derivative of a step function is a delta function. r(t) = 1 e 5t − − dr 5t Delta integrate Step g(t) = = 5e− Function Function dt differentiate 5t δ (t) H(t) So the impulse response of the system is 5e− . 11 12 Sifting Theorem Sifting Examples δ(t−b) f(t) t t a b c b π cos(2t) δ(t) dt = cos(0) = 1 π Z− δ(t−b) f(t) π π cos(2t) δ t dt = cos(π)= 1 Z π − 2 − t − a b c 0 π cos(2t) δ t dt = 0 π − 2 Z− π π π tδ t + dt = π 2 −2 c Z− δ(t b)dt = 1 provided a < b and c > b π π Za − tδ t + dt = 0 0 2 c Z δ(t b)f(t)dt = f(b) provided a < b and c > b Za − 13 14 Section 2 Section 1: Summary Differential Equations to Superposition: If input f (t) output y (t) 1 → 1 and input f2(t) output y2(t) then input αf (t)+ βf (t) → output αy (t)+ βy (t) Describe Linear Systems 1 2 → 1 2 where α and β are any constants. Sifting: We motivate the convolution integral, which will be c presented in section 3, using an example of a car go- ( ) ( ) = ( ) provided and δ t b f t dt f b a < b c > b ing up a step. Za − Step function and step response. A technique is described for solving a linear differen- tial equation to obtain the step response of the sys- Impulse function and impulse response. tem. We set the input to 1, and solve with initial con- ditions y =y ˙ = 0 for t = 0. The impulse response Finding the system response to a pulse by combining can then be obtained by differentiating the step re- scaled and delayed step responses using superposi- sponse. tion. The utility of this technique, when used together with convolution, is outlined. 15 16 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 + 6 = ( ) 2 y f t dt 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.