 # Continuous-Time Convolution EE 313 Linear Systems and Signals Fall 2005 Initial conversion of content to PowerPoint by Dr. Wade C. Schwartzkopf Prof. Brian.

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Continuous-Time Convolution EE 313 Linear Systems and Signals Fall 2005 Initial conversion of content to PowerPoint by Dr. Wade C. Schwartzkopf Prof. Brian L. Evans Dept. of Electrical and Computer Engineering The University of Texas at Austin

4 - 2 Useful Functions Unit gate function (a.k.a. unit pulse function) What does rect(x / a) look like? Unit triangle function 0 1 1/2-1/2 x rect(x) 0 1 1/2-1/2 x (x)(x)

4 - 3 Unit Impulse (Functional) Mathematical idealism for an instantaneous event Dirac delta as generalized function (a.k.a. functional) Unit area: Sifting provided g(t) is defined at t = 0 Scaling: Note that  (0) is undefined  t  t

4 - 4 Unit Impulse (Functional) Generalized sifting Assuming that a > 0 By convention, plot Dirac delta as arrow at origin Undefined amplitude at origin Denote area at origin as (area) Height of arrow is irrelevant Direction of arrow indicates sign of area With  (t) = 0 for t  0, it is tempting to think f(t)  (t) = f(0)  (t) f(t)  (t-T) = f(T)  (t-T) t (1) 0 Simplify unit impulse under integration only

4 - 5 Unit Impulse (Functional) We can simplify  (t) under integration What about? Answer: 0 What about? By substitution of variables, Other examples What about at origin?

4 - 6 Unit Impulse (Functional) Relationship between unit impulse and unit step What happens at the origin for u(t)? u(0 - ) = 0 and u(0 + ) = 1, but u(0) can take any value Common values for u(0) are {0, ½, 1} u(0) = ½ is used in impulse invariance filter design: L. B. Jackson, “A correction to impulse invariance,” IEEE Signal Processing Letters, vol. 7, no. 10, Oct. 2000, pp. 273-275.

4 - 7 LTI Systems Recall for LTI system with input f and output y –Homogeneity: c f(t)  c y(t) –Time-invariance: c f(t-T)  c y(t-T) –Adding additivity: If a signal ( F(t) ) can be expressed as a sum of shifted ( t - T i ) and weighted ( c i ) copies of a simpler signal ( f(t) ), we easily find a system’s output ( Y(t) ) to that signal if we only know system’s output ( y(t) ) to that simpler signal A common choice for f(t) is the impulse

4 - 8 Impulse Response Impulse response of a system is response of the system to an input that is a unit impulse (i.e., a Dirac delta functional in continuous time) Linear constant coefficient differential equation When initial conditions are zero, this differential equation is LTI and system has impulse response b 0 is coefficient (could be 0) of b 0 D N f(t) on right-hand side N is highest order of derivative in differential equation

4 - 9 Impulse Response In following plug, where did b n come from? In solving these differential equations for t  0, Funny things happen to y’(t) and y”(t) In differential equations class, solved for m(t) –Likely ignored  (t) and  ’(t) terms –Solution for m(t) is really valid for t  0 +

4 - 10 t F(t)F(t)  t=n  System Response Signals as sum of impulses But we know how to calculate the impulse response ( h(t) ) of a system expressed as a differential equation Therefore, we know how to calculate the system output for any input, F(t)

4 - 11 Convolution Integral Commonly used in engineering, science, math Convolution properties –Commutative: f 1 (t) * f 2 (t) = f 2 (t) * f 1 (t) –Distributive: f 1 (t) * [f 2 (t) + f 3 (t)] = f 1 (t) * f 2 (t) + f 1 (t) * f 3 (t) –Associative: f 1 (t) * [f 2 (t) * f 3 (t)] = [f 1 (t) * f 2 (t)] * f 3 (t) –Shift: If f 1 (t) * f 2 (t) = c(t), then f 1 (t) * f 2 (t - T) = f 1 (t - T) * f 2 (t) = c(t - T). –Convolution with impulse, f(t) *  (t) = f(t) –Convolution with shifted impulse, f(t) *  (t-T) = f(t-T) important later in modulation

4 - 12 Graphical Convolution Methods From the convolution integral, convolution is equivalent to –Rotating one of the functions about the y axis –Shifting it by t –Multiplying this flipped, shifted function with the other function –Calculating the area under this product –Assigning this value to f 1 (t) * f 2 (t) at t

4 - 13 3  2 f()f() 2 -2 + t2 + t g(t-  ) * 2 2 t f(t)f(t) -22 3 t g(t)g(t) Graphical Convolution Example Convolve the following two functions: Replace t with  in f(t) and g(t) Choose to flip and slide g(  ) since it is simpler and symmetric Functions overlap like this:

4 - 14 3  2 f()f() 2 -2 + t2 + t g(t-  ) 3  2 f()f() 2 -2 + t2 + t g(t-  ) Graphical Convolution Example Convolution can be divided into 5 parts I. t < -2 Two functions do not overlap Area under the product of the functions is zero II. -2  t < 0 Part of g(t) overlaps part of f(t) Area under the product of the functions is

4 - 15 Graphical Convolution Example III. 0  t < 2 Here, g(t) completely overlaps f(t) Area under the product is just IV. 2  t < 4 Part of g(t) and f(t) overlap Calculated similarly to -2  t < 0 V. t  4 g(t) and f(t) do not overlap Area under their product is zero 3  2 f()f() 2 -2 + t2 + t g(t-  ) 3  2 f()f() 2 -2 + t2 + t g(t-  )

4 - 16 Graphical Convolution Example Result of convolution (5 intervals of interest): t y(t)y(t) 024-2 6

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