Topic 3 The δ-function & convolution. Impulse response & Transfer function In this lecture we will described the mathematic operation of the convolution of two continuous functions. As the name suggests, two functions are blended or folded together. We will then discuss the impulse response of a system, and show how it is related to the transfer function of the system. First though we will define a special function called the δ-function or unit impulse. It is, like the Heaviside step function u(t), a generalized function or \distribution" and is best defined by considering another function in conjunction with it. 3.1 The δ-function Consider a function 1=w 0 < t < w g(t) = 0 otherwise One thing of note about g(t) is that Z w g(t)dt = 1: 0− The lower limit 0− is a infinitesimally small amount less than zero. Now, suppose that the width w gets very small, indeed as small at 0+, an number an infinitesimal amount bigger than zero. At that point, g(t) has become like the δ function, a very thin, very high spike at zero, such that Z 1 Z 0+ δ(t)dt = δ(t)dt = 1 −∞ 0− 1 3/2 g(t) δ (t) 1/ w 0 w 0 (a) (b) Figure 3.1: As w becomes very small the function g(t) turns into a δ-function δ(t) indicated by the arrowed spike. In some sense it is akin to the derivative of the Heaviside unit step Z t δ(t)dt = u(t) : −∞ More formally the delta function is defined in association with any arbitrary function f (t), as The delta function ... Z 1 f (t)δ(t)dt = f (0) : −∞ Picking out values of a function in this way is called sifting of f (t) by δ(t). We can also see that Z 1 f (t)δ(t − τ)dt = f (τ) ; −∞ a result that we will return to. δ (t−τ ) A δ (t−τ ) δ f(t) (t) f(t) t t 0 0 τ 0 τ (a) (b) (c) Figure 3.2: (a,b) Sifting. (c) With an amplitude A. Although the δ-function is infinitely high, very often you will see a described as the unit δ-function, or see a δ-function spike with an amplitude A by it. This is to R 1 R 1 denote a delta-function where −∞ δ(t)dt = 1 or −∞ Aδ(t)dt = A. 3/3 3.2 Properties of the δ-function Fourier transform of the delta function: FT [δ(t)] = 1 Proof: Use the definition of the δ-function and sift the function f (t) = e−i!t: Z 1 δ(t)e−i!tdt = e−i!0 = 1 : −∞ Symmetry: The δ-function has even symmetry. δ(t) = δ(−t) Parameter Scaling: 1 δ(at) = δ(t) jaj R 1 Proof: To prove this return to the fundamental definition, −∞ f (t)δ(t)dt = f (0) If a ≥ 0, substitute (at) for t (no swap in limits) Z 1 Z 1 f (at)δ(at)d(at) = a f (at)δ(at)dt = f (0) −∞ −∞ Z 1 But f (at)δ(t)dt = f (0) ) a δ(at) = δ(t): −∞ Now if a < 0, substitute (at) for t (but need to swap limits as a negative) Z −∞ Z 1 f (at)δ(at)d(at) = − a f (at)δ(at)dt = f (0) 1 −∞ Z 1 But f (at)δ(t)dt = f (0) ) −a δ(at) = δ(t): −∞ So jaj δ(at) = δ(t) covers both cases, and the stated definition follows immedi- ately. 3/4 3.3 Fourier Transforms that involve the δ-function Fourier Transform of ei!0t Z 1 i!0t i(!0−!)t FT e = e dt = 2πδ(! − !0) : −∞ Fourier Transform of 1 FT [1] = 2πδ(!) : You could obtain this either by putting !0 = 0 just above, or by using the dual property, FT [1] = 2πδ(−!), then the even symmetry property δ(−!) = δ(!). Fourier Transform of cos !0t 1 FT [cos ! t] = FT ei!0t + e−i!0t = π (δ(! − ! ) + δ(! + ! )) : 0 2 0 0 Fourier Transform of sin !0t 1 FT [cos ! t] = FT ei!0t − e−i!0t = −iπ (δ(! − ! ) − δ(! + ! )) : 0 2i 0 0 Fourier Transform of Complex Fourier Series | yes, this can be useful! " n=1 # n=1 X in!0t X FT Cne = 2π Cnδ(! − n!0) : n=−∞ n=−∞ 3.4 Convolution We turn now to a very important technique is signal analysis and processing. The convolution of two functions f (t) and g(t) is denoted by f ∗ g. The convolution is defined by an integral over the dummy variable τ. The convolution integral. The value of f ∗ g at t is Z 1 (f ∗ g)(t) = f (τ)g(t − τ)dτ −∞ 3/5 The process is commutative, which means that Z 1 Z 1 (f ∗ g)(t) ≡ (g ∗ f )(t) or f (τ)g(t − τ)dτ ≡ f (t − τ)g(τ)dτ −∞ −∞ 3.4.1 | Example 1 It is easier to \see" what is going on when convolving a signal f with a function g of even or odd symmetry. However, to get into a strict routine, it is best to start with an example with no symmetry. [Q] Find and sketch the convolution of f (t) = u(t)e−at with g(t) = u(t)e−bt, where both a and b are positive. [A] Using the first form of the convolution integral, the \short" answer must be the unintelligible Z 1 f ∗ g = u(τ)e−aτ u(t − τ)e−b(t−τ)dτ : −∞ First, make sketches of the functions f (τ) and g(t −τ) as τ varies. Function f (τ) looks just like f (t) of course. But g(t − τ) is a reflected (\time reversed") and shifted version of g(t). (The reflection is easy enough. To check that the shift is correct, ask yourself \where does the function g(p) drop?" The answer is at p = 0. So g(t − τ) must drop when t − τ = 0, that is when τ = t.) f(τ ) g(τ ) g(−τ ) g(t−τ ) τ τ τ τ 0 0 t 0 ! ! 0 (a) (b) ! (c) ! (d) Figure 3.3: We now multiply the two functions, BUT we must worry about the fact that t is a variable. In this case there are two different regimes, one when t < 0 and the other when t ≥ 0. Figure 3.4 shows the result. So now to the integration. For t < 0, the function on the bottom left of Figure 3.4 is everywhere zero, and the result is zero. For t ≥ 0 Z 1 Z t e−bt u(τ)e−aτ u(t−τ)e−b(t−τ)dτ = e−bt e(b−a)τ dτ = e(b−a)t − 1 −∞ 0 b − a 3/6 g(t−τ ) f(τ ) g(t−τ ) f(τ ) τ τ t 0 0 t f(τ )g(t−τ ) f(τ )g(t−τ ) t <0 t >0 τ τ 0 t 0 t Figure 3.4: So e−at − e−bt =(b − a) for t ≥ 0 f ∗ g(t) = 0 for t < 0 It is important to realize that the function at the bottom right of Figure 3.4 is NOT the convolution. That is the function you are about to integrate over for a particular value of t. Figure 3.5 shows the t > 0 part of the convolution for b = 2 and a = 1. 0.25 (exp(-x)-exp(-2*x)) 0.2 0.15 0.1 0.05 0 0 1 2 3 4 5 6 Figure 3.5: f ∗ g(t) plotted for t > 0 when b = 2 and a = 1. 3/7 3.4.2 | Example 2 [Q] Derive an expression for the convolution of an arbitrary signal f (t) with the function g(t) shown in the figure. Determine the convolution when f (t) = A, a constant, and when f (t) = A + (B − A)u(t). [A] Follow the routine. Function f (τ) looks exactly like f (t), but g(t − τ) is reflected and shifted. Multiply and integrate over τ from −∞ to 1. Because g only has finite range, we can pinch in the limits of integration, and the convolution becomes Z t Z t+a f ∗ g = − f (τ)dτ + f (τ)dτ t−a t g(t) f(t) g(t− τ ) f(t) 1 a t 1 −a t t−a t τ −1 t t+a −1 Figure 3.6: When f (t) = A, a constant, it is obvious by inspection that the convolution is zero for all t. When f (t) = A + (B − A)u(t), we have to be more careful because there is a discontinuity in the function. From Figure 3.7(a): • The convolution is zero for all t < −a and all t > a (Diagram positions 1,2,5). • The maximum value is when t = 0 (Position 4). By inspection, or using the integrals above, (f ∗ g)(t = 0) = a(B − A). • For −a < t < 0, (Position 3) Z t Z 0 Z t+a f ∗g = − (A)dτ+ Adτ+ Bdτ = −aA+−tA+(t+a)B = (a+t)(B−A) t−a t 0 showing that the increase in correlation value is linear. Symmetry tells us that the decrease for 0 < t < a will also be linear. One can see from Figure 3.7(b) that this convolution provides a rudimentary de- tector of steps in the signal.