The Dirac delta function There is a function called the pulse: 0 if |t| > 1 Π(t)= 2 1 otherwise. Note that the area of the pulse is one. The Dirac delta function (a.k.a. the impulse) can be de- fined using the pulse as follows: 1 t δ(t)= lim Π . ε ε ε −→0 The impulse can be thought of as the limit of a pulse as its width goes to zero and its area is normalized to one. Properties of the Dirac delta function The Dirac delta function obeys the following two properties: • integral property ε lim δ(t)dt = 1 ε−→0 Z−ε • sifting property ∞ f (t)= f (τ)δ(t − τ)dτ. Z−∞ Impulse response function In the continuum, the output of a linear shift invariant system is given by the convolution in- tegral: ∞ y(t)= x(τ)h(t − τ)dτ. Z−∞ Since functions remain unchanged by convolu- tion with the impulse: ∞ f (t)= f (τ)δ(t − τ)dτ Z−∞ we say that the impulse is the identity function of linear shift invariant operators. System identification Let’s say we have an unknown linear shift in- variant system, i.e., a black box, H : H x(t) −→ y(t) where ∞ y(t)= x(τ)h(t − τ)dτ. Z−∞ • Question How do we find the function, h, which characterizes the linear shift invariant system? • Answer Feed it an impulse and see what comes out: H δ(t) −→ ? System identification (contd.) By commutativity and the sifting property we see that: ∞ δ(τ)h(t − τ)dτ = Z ∞ −∞ h(τ)δ(t − τ)dτ = h(t). Z−∞ It follows that: H δ(t) −→ h(t). For this reason, h is called the impulse response function. The impulse response is the first of two ways to characterize a linear shift invariant system. Impulse Response of Shift Operator To identify the impulse response function of the shift operator, S∆, we apply the shift operator to an impulse and see what comes out: S∆ δ(.) −→ δ((.) − ∆). We conclude that δ((.) − ∆) is the impulse re- sponse of the shift operator. It follows that to apply S∆ to a function f , we can convolve f with δ((.) − ∆): ∞ f∆(t) = f (τ)δ((t − τ) − ∆)dτ Z ∞ −∞ = f (τ)δ(t − ∆ − τ)dτ Z−∞ = f (t − ∆). Impulse Response of ∇ Operator To identify the impulse response function of the differentiation operator, ∇, we apply the differ- entiation operator to an impulse and see what comes out: ∇ δ(.) −→ δ0(.). We conclude that δ0(.), the derivative of an im- pulse, is the impulse response of the differenta- tion operator. It follows that to apply ∇ to a function f , we can convolve f with δ0(.): ∞ f 0(t)= f (τ)δ0(t − τ)dτ. Z−∞ Harmonic signals A harmonic signal, exp( j2πst), is a complex function of a real variable, t. The real part is a cosine: π Re(e j2 st)= cos(2πst) and the imaginary part is a sine: π Im(e j2 st)= sin(2πst). The transfer function Let xs(t) and ys(t) be the input and output func- tions of a linear shift invariant system, H : H xs(t) −→ ys(t). We observe that the output function, ys(t), can be written as a product of the input function xs(t) and a function H(s,t) defined as follows: y (t) H(s,t)= s . xs(t) Consequently, H xs(t) −→ H(s,t)xs(t). Since the above holds for any input function, xs(t), it holds when xs(t)= exp( j2πst): π H π e j2 st −→ H(s,t)e j2 st. 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 −0.2 −0.2 −0.4 −0.4 −0.6 −0.6 −0.8 −0.8 −1 −1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 (a) (b) 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 −0.2 −0.2 −0.4 −0.4 −0.6 −0.6 −0.8 −0.8 −1 −1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 (c) (d) 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 −0.2 −0.2 −0.4 −0.4 −0.6 −0.6 −0.8 −0.8 −1 −1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 (e) (f) Figure 1: Re(t) (solid) and Im(t) (dashed) for harmonic signals. (a) exp( j2πt). (b) exp(− j2πt). (c) exp( j2π3t). (d) exp(− j2π3t). (e) exp( j2π12t). (f) exp(− j2π12t). 1 1 0.5 0.5 0 0 imaginary imaginary −0.5 −0.5 −1 −1 −1 −1 −0.5 1 −0.5 1 0.8 0.8 0 0.6 0 0.6 0.5 0.4 0.5 0.4 0.2 0.2 1 1 real 0 real 0 time time (a) (b) 1 1 0.5 0.5 0 0 imaginary imaginary −0.5 −0.5 −1 −1 −1 −1 −0.5 1 −0.5 1 0.8 0.8 0 0.6 0 0.6 0.5 0.4 0.5 0.4 0.2 0.2 1 1 real 0 real 0 time time (c) (d) 1 1 0.5 0.5 0 0 imaginary imaginary −0.5 −0.5 −1 −1 −1 −1 −0.5 1 −0.5 1 0.8 0.8 0 0.6 0 0.6 0.5 0.4 0.5 0.4 0.2 0.2 1 1 real 0 real 0 time time (e) (f) Figure 2: Harmonic signals visualized as space curve, [Re(t),Im(t)]. (a) exp( j2πt). (b) exp(− j2πt). (c) exp( j2π3t). (d) exp(− j2π3t). (e) exp( j2π12t). (f) exp(− j2π12t). The transfer function (contd.) Now let’s shift the input: π τ H π τ e j2 s(t− ) −→ H(s,t − τ)e j2 s(t− ). As expected, the output is shifted by the same amount. But notice that π τ π π τ e j2 s(t− ) = e j2 ste− j2 s π τ π = e− j2 s e j2 st where e− j2πsτ is just a (complex) constant. The transfer function (contd.) Linearity tells us the effect of multiplying the input of a linear shift invariant system by a con- stant: H kxs(t) −→ kys(t) so that π τ π H π τ π e− j2 s e j2 st −→ e− j2 s H(s,t)e j2 st or π τ H π τ e j2 s(t− ) −→ H(s,t)e j2 s(t− ). We can only conclude that H(s,t − τ)= H(s,t). Observe that H(s,t) is independent of t! Eigenfunctions It follows that the effect of applying a linear shift invariant operator H to a harmonic signal π H π e j2 st −→ H(s)e j2 st is to multiply it by a complex constant H(s) de- pendent only on frequency s. This multiplica- tion can change the amplitude and phase of the harmonic signal, but not its frequency. Eigenfunctions (contd.) Written somewhat differently, the effect of a linear shift invariant operator H on a harmonic signal is: π π H(s)e j2 st = H {e j2 st} or ∞ π π τ H(s)e j2 st = e j2 s h(t − τ)dτ. Z−∞ Observe the similarity between the above and the familiar equation relating eigenvector xi and eigenvalue λi of matrix A: λixi = Axi or λ i (xi) j = ∑A jk (xi)k . k Because of this similarity, we say that harmonic signals are the eigenfunctions of linear shift in- variant systems. e j2πst is like an eigenvector and H(s) is like an eigenvalue. The transfer function (contd.) Next week, we will see that (almost) any func- tion f can be uniquely decomposed into a weighted sum of harmonic signals, i.e., eigenfunctions of H : ∞ π f (t)= F(s)e j2 stds. Z−∞ F is called the Fourier transform of f . The transfer function (contd.) For the moment, we won’t consider the prob- lem of how to compute F. We simply observe that in the basis of eigenfunctions of H , each component F(s) of the representation of f (t) is modulated by a complex constant, H(s): ∞ ∞ π H(s)F(s)e j2 stds = f (τ)h(t − τ)dτ. Z−∞ Z−∞ Like the impulse response function, h, the trans- fer function, H, completely specifies the behav- ior of the linear shift invariant operator H . The transfer function (contd.) • Question What is the relationship between the impulse response function, h, and the trans- fer function, H? • Answer H is the Fourier transform of h: ∞ π h(t)= H(s)e j2 stds. Z−∞.