Section 3: Summary. Section 4. Evaluating Convolution Integrals. Convolution integral (memorise this):

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1 Section 3: Summary Convolution integral (memorise this): f(t) input g(t) impulse response y(t) output y(t) g(t τ) f(τ) dτ Way to find the output of a linear system, described by a differential equation, for an arbitrary input: Find general solution to equation for input. Set boundary conditions y() ẏ() to get the step response. Differentiate to get the impulse response. Use convolution integral together with the impulse response to find the output for any desired input. Section 4 Evaluating Convolution Integrals A way of rearranging the convolution integral is described and illustrated. The differences between convolution in time and space are discussed and the concept of causality is introduced. The section ends with an eample of spatial convolution. 4 42

2 Convolution Summary Differential Equation ay + by + cy + d f(t) solve ay + by + cy + d with boundary conditions y() and y() Step response Splitting up Integrals Suppose we have a function: a, t < f(t) b, < t < k c, k < t and we want to evaluate the integral f(τ) dτ, we can split it up as follows: differentiate Any Input: f(t) Impulse response: g(t) Corresponding convolution Output: y(t) a dτ, t < a dτ + b dτ, < t < k a dτ + k b dτ + k c dτ, k < t y(t) g(t τ) f(τ) dτ 43 44

3 Eample Find the impulse response of d 2 y + y f(t) dt2 hence find the output for (i) input f(t) t, t > and (ii) input f(t) H(t) H(t ).. Find the General Solution with f(t) Complimentary function is y A cos(3t)+b sin(3t) Particular integral is y General solution is y + Acos(3t) + B sin(3t) 2. Set boundary conditions y() ẏ() to get the step response. + A 3B A and B Thus the Step Response is y ( cos(3t)) Differentiate the step response to get the impulse response. g(t) dy dt 3 sin(3t) 4. Use the convolution integral to find the output for the required input. For part (i) the required input is a ramp starting at the origin: f(t) t when t > and f(t) otherwise. y(t) y(t) g(t τ)f(τ) dτ sin(3(t τ)) τ dτ 3 t sin(3t) t 46

4 Case (a): t < g(t τ) dτ so y(t) for all t <. For part (ii) the required input is a pulse of unit height and unit duration: f(t) H(t) H(t ). y(t) g(t τ)f(τ)dτ g(t τ) dτ, t < Case (b): < t < y(t) g(t τ) dτ ( cos(3t)) sin(3(t τ)) dτ 3 g(t τ) dτ + g(t τ) dτ, < t < Case (c): < t g(t τ) dτ + g(t τ) dτ + g(t τ) dτ, t > y(t) g(t τ) dτ sin(3(t τ)) dτ 3 [ ] cos(3(t τ)) {cos(3(t )) cos(3t)} 47 48

5 Part (ii) Another Way The input for part (ii) is composed of two step functions. We can therefore calculate the output using the step response, r(t) ( cos(3t)). Input H(t) H(t ) Output r(t) r(t ) Alternate Convolution Integral The normal convolution integral y(t) g(t τ) f(τ) dτ can be inconvenient to compute when we have a complicated epression for g(t). Hence, for t >, y(t) ( cos(3t)) ( cos(3(t ))) We would therefore like to derive an alternative version of the convolution integral that has a term of the form g(τ) rather than g(t τ) as this will be easier to calculate in cases where g is a complicated epression. {cos(3(t )) cos(3t)} 4 5

6 Arguments of f and g Eample Substitute u t τ in the convolution formula. Assume all signals are zero for t < and set the lower integral limit to zero. We have du dτ, g(t τ)f(τ)dτ g(u)f(t u)du t g(u)f(t u)du As u is the variable of integration, we can call it anything, as it disappears when the integration has been evaluated. We therefore choose to rename u as τ. Hence: g(t τ)f(τ)dτ g(τ)f(t τ)dτ So it does not matter which way round you get the arguments to the functions in the convolutions integral, provided both functions are zero for t <. Consider a linear system with impulse response { 3t g(t) 2 4t + 7, t >, otherwise Find the output for the input f(t) t, (t ) and f(t), (t < ). Note that everything is zero for t <, so we can use y(t) f(t τ)g(τ) dτ. y(t) f(t τ)g(τ) dτ (t τ) (3τ2 4τ + 7) dτ t4 4 2t t

7 Spatial Convolution Systems with time-varying input & output. Causal: no output before the input that causes it. g(t), t < Systems with input, output a function of position. An input can affect the output on either side. g() can be non-zero for any. Consider a one-dimensional strip of a material that is known to deform linearly according to g() cosh() when subject to a unit force at. g() Spatial Convolution Eample Calculate the deformation of a strip of material with spatial impulse response as described on the previous page in response to a uniform f() 2 load of f(). applied from to 2. y() g( τ)f(τ) dτ cosh( τ) dτ 2 + cosh( τ) dτ + 2 cosh( τ) dτ 2 cosh( τ) dτ 2 { arctan ( e 2 ) arctan ( e )} This is a spatial impulse response

8 L Variable Impulse Response y a L y y a Consider a taut string suspended between two points a distance L apart. It is subject to a uniform loading of K per unit length which results in a small displacement. y a If we knew the deformation caused by a point load, we could integrate in a style similar to the convolution integral to find the shape under the distributed load. If it was possible to have a spatial impulse response g() then we could say y() L g( τ)f(τ) dτ 55 But a normal impulse response is not possible because the shape of g changes depending on the position of the point load along the string. We have a function g(, a) where the point load is at position a. The function g that gives the displacement under a point load depends on both the position of the load, a, and the position at which you want to know the displacement,. 56

9 If we can find this g(, a) we can work out the complete displacement under the continuous load K using L L y() g(, a)f(a) da g(, a)k da y a L r r 2 T F Segment Segment 2 To find g(, a) we first work out the maimum displacement, d, for a point load, F, at position a. T 2 d Now resolve vertically: T(sin(r ) + sin(r 2 )) Again approimate: cos(r ) cos(r 2 ) This gives us: T(tan(r ) + tan(r 2 )) ( d a + d L a ) T Ld a(l a) T a(l a) so d TL This enables us to write down equations for the two straight segments of the g(, a) function. Resolve horisontally: T cos(r ) T 2 cos(r 2 ) Use the approimation: cos(r ) cos(r 2 ) This gives us: T T 2. Call this tension T. Segment : < a Segment 2: > a g(, a) g(, a) ( a ) d ( L L a (L a) ) d TL a(l ) TL 57 58

10 Finally, we work out the shape of a string of length L with tension T under a uniform load of K per unit length. Section 4: Summary If f(t) g(t) for all t < then y() L g(, a)f(a) da L g(, a)k da L g seg 2 K da + g seg K da a(l )K L (L a)k da + da TL TL ( ) K (L ) 2T g(t τ)f(τ)dτ g(τ)f(t τ)dτ Systems for which g(t) for all t < are called causal systems. Systems with time-varying inputs and outputs are causal. Systems that have inputs and outputs that vary as a function of spatial location can have g() for any. 5 6