The Dirac delta functionThere is a function called the pulse:Π(t) =0 if |t| >121 otherwise.Note that the area of the pulse is one. The Diracdelta function (a.k.a. the impulse) can be de-fined using the pulse as follows:δ(t) = limε−→01εΠtε.The impulse can be thought of as the limit ofa pulse as its width goes to zero and its area isnormalized to one.Properties of the Dirac delta functionThe Dirac delta function obeys the followingtwo properties:• integral propertylimε−→0Zε−εδ(t)dt = 1• sifting propertyf(t) =Z∞−∞f(τ)δ(t − τ)dτ.Impulse response functionIn the continuum, the output of a linear shiftinvariant system is given by the convolution in-tegral:y(t) =Z∞−∞x(τ)h(t − τ)dτ.Since functions remain unchanged by convolu-tion with the impulse:f(t) =Z∞−∞f(τ)δ(t − τ)dτwe say that the impulse is the identity functionof linear shift invariant operators.System identificationLet’s say we have an unknown linear shift in-variant system, i.e., a black box, H :x(t)H−→ y(t)wherey(t) =Z∞−∞x(τ)h(t − τ)dτ.• Question How do we find the function, h,which characterizes the linear shift invariantsystem?• Answer Feed it an impulse and see what comesout:δ(t)H−→ ?System identification (contd.)By commutativity and the sifting property wesee that:Z∞−∞δ(τ)h(t − τ)dτ =Z∞−∞h(τ)δ(t − τ)dτ = h(t).It follows that:δ(t)H−→ h(t).For this reason, h is called the impulse responsefunction. The impulse response is the first oftwo ways to characterize a linear shift invariantsystem.Impulse Response of Shift OperatorTo identify the impulse response function of theshift operator, S∆, we apply the shift operator toan impulse and see what comes out:δ(.)S∆−→ δ((.) − ∆).We conclude that δ((.) − ∆) is the impulse re-sponse of the shift operator. It follows that toapply S∆to a function f, we can convolve fwith δ((.) − ∆):f∆(t) =Z∞−∞f(τ)δ((t − τ) − ∆)dτ=Z∞−∞f(τ)δ(t − ∆− τ)dτ= f(t − ∆).Impulse Response of ∇ OperatorTo identify the impulse response function of thedifferentiation operator, ∇, we apply the differ-entiation operator to an impulse and see whatcomes 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 afunction f, we can convolve f with δ0(.):f0(t) =Z∞−∞f(τ)δ0(t − τ)dτ.Harmonic signalsA harmonic signal, exp( j2πst), is a complexfunction of a real variable, t. The real part isa cosine:Re(ej2πst) = cos(2πst)and the imaginary part is a sine:Im(ej2πst) = sin(2πst).The transfer functionLet xs(t) and ys(t) be the input and output func-tions of a linear shift invariant system, H :xs(t)H−→ ys(t).We observe that the output function, ys(t), canbe written as a product of the input functionxs(t) and a function H(s,t) defined as follows:H(s,t) =ys(t)xs(t).Consequently,xs(t)H−→ H(s,t)xs(t).Since the above holds for any input function,xs(t), it holds when xs(t) = exp( j2πst):ej2πstH−→ H(s,t)ej2πst.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1−0.8−0.6−0.4−0.200.20.40.60.810 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1−0.8−0.6−0.4−0.200.20.40.60.81(a) (b)0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1−0.8−0.6−0.4−0.200.20.40.60.810 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1−0.8−0.6−0.4−0.200.20.40.60.81(c) (d)0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1−0.8−0.6−0.4−0.200.20.40.60.810 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1−0.8−0.6−0.4−0.200.20.40.60.81(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).00.20.40.60.81−1−0.500.51−1−0.500.51timerealimaginary00.20.40.60.81−1−0.500.51−1−0.500.51timerealimaginary(a) (b)00.20.40.60.81−1−0.500.51−1−0.500.51timerealimaginary00.20.40.60.81−1−0.500.51−1−0.500.51timerealimaginary(c) (d)00.20.40.60.81−1−0.500.51−1−0.500.51timerealimaginary00.20.40.60.81−1−0.500.51−1−0.500.51timerealimaginary(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:ej2πs(t−τ)H−→ H(s,t − τ)ej2πs(t−τ).As expected, the output is shifted by the sameamount. But notice thatej2πs(t−τ)= ej2πste− j2πsτ= e− j2πsτej2πstwhere e− j2πsτis just a (complex) constant.The transfer function (contd.)Linearity tells us the effect of multiplying theinput of a linear shift invariant system by a con-stant:kxs(t)H−→ kys(t)so thate− j2πsτej2πstH−→ e− j2πsτH(s,t)ej2πstorej2πs(t−τ)H−→ H(s,t)ej2πs(t−τ).We can only conclude thatH(s,t − τ) = H(s,t).Observe that H(s,t) is independent of t!EigenfunctionsIt follows that the effect of applying a linearshift invariant operator H to a harmonic signalej2πstH−→ H(s)ej2πstis 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 theharmonic signal, but not its frequency.Eigenfunctions (contd.)Written somewhat differently, the effect of alinear shift invariant operator H on a harmonicsignal is:H(s)ej2πst= H {ej2πst}orH(s)ej2πst=Z∞−∞ej2πsτh(t − τ)dτ.Observe the similarity between the above andthe familiar equation relating eigenvector xiandeigenvalue λiof matrix A:λixi= Axiorλi(xi)j=∑kAjk(xi)k.Because of this similarity, we say that harmonicsignals are the eigenfunctions of linear shift in-variant systems. ej2πstis like an eigenvector andH(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 weightedsum of harmonic signals, i.e., eigenfunctions ofH :f(t) =Z∞−∞F(s)ej2πstds.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 observethat in the basis of eigenfunctions of H , eachcomponent F(s) of the representation of f(t) ismodulated by a complex constant, H(s):Z∞−∞H(s)F(s)ej2πstds =Z∞−∞f(τ)h(t − τ)dτ.Like the impulse response function, h, the trans-fer function, H, completely specifies