MASSACHUSETTS INSTITUTE OF TECHNOLOGYDepartment of Electrical Engineering and Computer Science6.003: Signals and Systems — Spring 2004Tutorial 3Monday, February 23 and Tuesday, February 24, 2004Announcements• Problem set 3 is due this Friday.Today’s Agenda• Choosing a Basis• Eigenfunctions of LTI Systems• CT Fourier Series– Dirichlet conditions– Properties of Fourier series• DT Fourier Series• Parseval’s Relation411 Choosing a BasisSo far, we have thought of signals as functions of time: x(t)andx[n]. However, it is sometimes moreconvenient to express signals in terms of another basis. Let’s define a (discrete) basis to be a set of signalsindexed by k called {φk(t)} in CT and {φk[n]} in DT so that there exists coefficients aksuch that we canexpress a large class of signals in terms of the basis signals as superposition sums of the basis signals:x(t)=kakφk(t)x[n]=kakφk[n].Then, given a basis, the coefficients akare just as good at specifying the signal as the original x(t)andx[n]. In DT, it is easy to identify a “default” basis. Since:x[n]=+∞k=−∞x[k]δ[n − k],we see that we can choose the basis signals to be φk[n]=δ[n − k], so that ak= x[k].We do this all the time in other areas. The coefficients akare boldfaced for emphasis.• Whole numbers - Represent a general number by their decimal coefficients so that the basis is {10k}:32618 = (3 ×104)+(2 × 103)+(6 × 102)+(1 × 101)+(8 × 100).Now, let’s change to the {16k} basis and write 32618 in base 16:32618 = (7 × 163)+(15 × 162)+(6 × 161)+(10 × 160).We see that if we know the basis, the set of numbers {3, 2, 6, 1, 8} and {7, 15, 6, 10} both identify thenumber 32618.• Vec tors - Represent any vector by orthogonal components:v =(3, −1, 2) = 3 × (1, 0, 0) + −1 × (0, 1, 0) + 2 × (0, 0, 1).Let’s rotate our coordinate system so that we can express v in terms of new basis vectors:v =(3, −1, 2) = 3 × (1, 0, 0) +−√32− 1 ×0,√32, −12 +√3 −12×0,12,√32 .• Currency - Represent an amount by more elementary denominations:$34.92 = 1 × (20 dollar bill) + 1 × (10 dollar bill) + 4 × (1 dollar bill)+3 × (quarter) + 1 × (dime) + 1 × (nickel) + 2 × (penny)42Of course, we can break down $32.41 several ways, and can even write it terms of foreign currency.• Taylor series (more specifically, the MacLaurin series) - Represent a CT signal f (t) with continuousderivatives as a polynomial; the basis is {tk}:f(t)=f(0) × t0+(f (0) × t)+f (0)2× t2+f (0)3!× t3+ ···=∞k=0f(k)(0)k!tkWhat’s the most appropriate basis for representing signals that are the inputs and outputs of LTI systems?To answer this question, we need to determine the eigenfunctions of LTI systems.2 Eigenfunctions of LTI SystemsWe saw how to determine the output y(t)(y[n]) of any CT (DT) LTI system given the input x(t)(x[n]) andthe impulse response h(t)(h[n]) through the use of the convolution integral (sum). In general, the outputlooked nothing like the input and the calculation was rather tedious. Convolution came from expressingsignals as superpositions of shifted scaled impulses. Perhaps another set of basis signals would provide moreinsight on LTI system behavior.We showed in lecture that a certain set of input signals, namely complex exponentials of the form x(t)=est(x[n]=zn), are eigenfunctions of LTI systems, i.e. the corresponding outputs are simply scaled versions ofinputs of this form, and this scaling factor is the eigenvalue. We showed that the outputs of CT and DTLTI systems in response to x(t)=estand x[n]=znare y(t)=H(s)est= H(s)x(t)andy[n]=H(z)zn=H(z)x[n], respectively, where the eigenvalues H(s)andH(z) associated with the given eigenfunctions are:H(s)=+∞−∞h(τ)e−sτdτH(z)=+∞k=−∞h[k]z−kwhere h(t)andh[n] are the impulse responses of the systems.x(t)=esty(t)=H(s)estCT LTIh(t)x[n]=znx(t)=H(z)estDT LTIh[n]Because of the superposition property of LTI systems, this suggests another way of writing signals. Insteadof expressing the input signal as the linear combination of scaled and shifted impulses, we can also express theinput as the linear combination of complex exponentials. Then, the output is the same linear combinationof the exponentials scaled by the appropriate eigenvalue. So, if the inputs are:43x(t)=kakesktx[n]=kakznkthen the outputs are:y(t)=kakH(sk)eskty[n]=kakH(zk)znkIf you are familiar with linear algebra from 18.03 or 18.06, you’ve seen the concepts of eigenvectors andeigenvalues of matrices. This is exactly the same idea. We think of a matrix A as a linear transformation(system) from one (input) vector x into another (output) vector y:y = Ax.Let vkbe the eigenvectors of A with corresponding eigenvalues λkso that when A is applied to aneigenvector, the result is the same eigenvector scaled by the eigenvalue:λkvk= AvkThis is powerful, because if we express an arbitrary input vector x as the superposition sum of eigenvectors:x =kakvk,then we can express the output vector as the same linear combination with each component scaled bythe corresponding eigenvalue:y = Ax= Akakvk =kakAvk−→ y =kakλkvk,Thus, for the linear transformation A, {vk} is the most natural basis for representing vectors.We are now see a need to express signals in terms of complex exponentials. But how do we do so? Thisquestion motivates us to spend this tutorial studying the Fourier series, which expresses periodic signals inthat form.443 CT Fourier SeriesLet us begin by representing periodic CT signals as the linear combination of complex exponentials usingthe Fourier series representation. We will then do the same with periodic DT signals. Later, we will do thesame with aperiod CT and DT signals using the Fourier transform.Sayx(t) is periodic with fundamentalperiod T and fundamental frequency ω0=2π/T. Then, we can write x(t) as the sum of the harmonicallyrelated components φk(t)=ejkω0t(all with period T ) and the coefficients are unique. They are related by:CT Fourier Series:x(t)=+∞k=−∞akejkω0t(Synthesis equation)ak=1TTx(t)e−jkω0tdt (Analysis equation)The coefficients akare known as Fourier series coefficients. We can extract akby exploiting the orthogo-nality of the φk(t) components, that is:1TTφ∗m(t)φk(t)dt = δ[m − k].As discussed in the prevous section on choosing a basis, we are simply using a different
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