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UNM CS 530 - The Fourier Transform

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The Fourier Transform• Introduction• Orthonormal bases for Rn– Inner product– Length– Orthogonality– Change of basis– Matrix transpose• Complex vectors• Orthonormal bases for Cn– Inner product– Hermitian transpose• Orthonormal bases for 2π periodic functions– Shah basis– Harmonic signal basis– Fourier series• Fourier transformOrthonormal bases for RnLet u = [u1,u2]Tand v = [v1,v2]Tbe vectors inR2. We define the inner product of u and v tobehu,vi = u1v1+ u2v2.We can use the inner product to define notionsof length and angle. The length of u is given bythe square root of the inner product of u withitself:|u| = hu,ui12=qu21+ u22.The angle between u and v can also be definedin terms of inner product:hu,vi = |u||v|cosθwhereθ = cos−1hu,vi|u||v|.OrthogonalityAn important special case occurs whenhu,vi = |u||v|cosθ = 0.When cosθ equals zero, θ = π/2 = 90◦.Orthonormal bases for RnAny n orthogonal vectors which are of unit lengthhui,uji =1 if i = j0 otherwise.form an orthonormal basis for Rn. Any vectorin Rncan be expressed as a weighted sum of u1,u2, u3,...,un:v = w1u1+ w2u2+ w3u3+ ... + wnun.• Question How do we find w1, w2, w3,...,wn?• Answer Using inner product.ExampleConsider two orthonormal bases. The first basisis defined by the vectors u1=10and u2=01. It is easy to verify that these two vectorsform an orthonormal basis:10,01= 1·0+ 0·1 = 010,10= 1·1+ 0·0 = 101,01= 0·0+ 1·1 = 1.Example (contd.)The second, by the vectors u01=cosθsinθandu02=−sinθcosθ. It is also easy to verify thatthese two vectors form an orthonormal basis:cosθsinθ,−sinθcosθ= −cosθsinθ+ cosθsinθ = 0cosθsinθ,cosθsinθ= cos2θ+ sin2θ = 1−sinθcosθ,−sinθcosθ= cos2θ+ sin2θ = 1.Example (contd.)Let the coefficients of v in the first basis be w1and w2:v = w110+ w201.What are the coefficients of v in the second ba-sis? Stated differently, what values of w01andw02satisfy:v = w01cosθsinθ+ w02−sinθcosθ?Figure 1: Change of basis.Example (contd.)To find w01and w02, we use inner product:w01=cosθsinθ,w1w2w02=−sinθcosθ,w1w2.Example (contd.)The above can be written more economically inmatrix notation:w01w02=cosθ sinθ−sinθ cosθw1w2w0= Aw.If the rows of A are orthonormal, then A isan orthonormal matrix. Multiplying by an or-thonormal matrix effects a change of basis. Achange of basis between two orthonormal basesis a rotation.Matrix transposeIf A rotates w by θA =cosθ sinθ−sinθ cosθthen A−1= ATrotates w0by −θAT=cosθ −sinθsinθ cosθ.In other words, ATundoes the action of A, i.e.,they are inverses:AAT=hcos2θ+ sin2θ cosθsinθ−sinθcosθcosθsinθ−sinθcosθ cos2θ+ sin2θi=1 00 1.For orthonormal matrices, multiplying by thetranspose undoes the change of basis.Complex vectors in C2v = [a1eiθ1,a2eiθ2]Tis a vector in C2.• Question Can we define length and angle inC2just like in R2?• Answer Yes, but we need to redefine innerproduct:hu,vi = u∗1v1+ u∗2v2.Note that this reduces to the inner product forR2when u and v are real. The norm of a com-plex vector is the square root of the sum of thesquares of the amplitudes. For example, forv ∈ C2:|u| = hu,ui12=pu∗1u1+ u∗2u2.Orthonormal bases for Cn• Question How about orthonormal bases forCn, do they exist?• Answer Yes. If hui,uji = 0 when i 6= j andhui,uji = 1 when i = j, then the uiform anorthonormal basis for Cn.• Question Do complex orthonormal matricesexist?• Answer Yes, except they are called unitarymatrices and (A∗)Tundoes the action of A.That isA(A∗)T= Iwhere (A∗)T= AHis the Hermitian trans-pose of A.The space of 2π periodic functionsA function, f, is 2π periodic iff f(t) = f(t +2π). We can think of two complex 2π periodicfunctions, e.g., f and g, as infinite dimensionalcomplex vectors. Length, angle, orthogonal-ity, and rotation (i.e., change of basis) still havemeaning. All that is required is that we gener-alize the definition of inner product:hf,gi =Zπ−πf∗(t)g(t)dt.The length (i.e., the norm) of a function is:|f| = hf, fi12=rZπ−πf∗(t) f(t)dt.Two functions, f and g, are orthogonal whenhf,gi =Zπ−πf∗(t)g(t)dt = 0.Scaling Property of the ImpulseThe area of an impulse scales just like the areaof a pulse, i.e., contracting an impulse by a fac-tor of a changes its area by a factor of1|a|:Z∞−∞Π(at)dt =1|a|= limε→0Zε−εδ(at)dt.It follows that:limε→0Zε−ε|a|δ(at)dt = limε→0Zε−εδ(t)dt = 1.Since the impulse is defined by the above inte-gral property, we conclude that:|a|δ(at) = δ(t).Shah basisThe Shah function is a train of impulses:III(t) =∞∑n=−∞δ(t −n).We can use the scaling property of the impulseto define a 2π periodic Shah function:12πIIIt2π=12π∞∑n=−∞δt2π−n=2π2π∞∑n=−∞δ2πt2π−n=∞∑n=−∞δ(t −2πn).Shah basis (contd.)Consider the infinite set of 2π periodic Shahfunctions,12πIIIt−τ2π, for −π ≤ τ < π. Because12πIIIt−τ2π= δ(t −τ) for −π ≤t ≤ π it followsthat12πIIIt −τ12π,12πIIIt −τ22π=Zπ−πδ(t −τ1)δ(t −τ2)dtequals 0 when τ16= τ2and equalsRπ−πδ(t−τ1)dt =1 when τ1= τ2. It follows that the infinite setof 2π periodic Shah functions,12πIIIt−τ2π, for−π ≤ τ < π form an orthonormal basis for thespace of 2π periodic functions.. . .. . .1. . . . . .Figure 2: Making a 2π periodic Shah function.Shah basis (contd.)• Question How do we find the coefficients,w(τ), representing f(t) in the Shah basis?How do we find w(τ) such thatf(t) =12πZπ−πw(τ)IIIt −τ2πdτ?• Answer Take inner products of f with theinfinite set of 2π periodic Shah functions:w(τ) =12πIIIt −τ2π, f(t).Shah basis (contd.)Because12πIIIt−τ2π= δ(t −τ) for −π ≤t ≤π itfollows thatw(τ) =12πZπ−πf(t)IIIt −τ2πdt=Zπ−πf(t)δ(t −τ)dtwhich by the sifting property of the impulse isjust:w(τ) = f(τ).We see that the coefficients of f in the Shahbasis are just f itself!Harmonic signal basis• Question How long is a harmonic signal?• Answer The length of a harmonic signal is|ejω t| = hejω t,ejω ti12=Zπ−πe−jω tejω


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UNM CS 530 - The Fourier Transform

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