MAS160/510 Supplemental NotesBasis functions and transformsWhile it’s typical to concentrate on representing signals as weighted sumsof complex exponential functions, here we’re going to examine the moregeneral case of linear combinations of any basis functions.Let’s assume we have a set of N basis functions (where N may or maynot be finite) φ0(t), φ1(t), ...φN−1(t). We will represent a function x(t):x(t) ≈ ˆx(t) =N−1Xi=0aiφi(t). (1)Note that for the moment we use the approximation symbol ≈ ratherthan the equal symbol because we haven’t spelled out the conditions underwhich the sum is exact rather than an approximation.For the moment, assume the basis functions φiare real-valued. They aresaid to be orthogonal over some interval (t1≤ t ≤ t2) if for all jZt2t1φi(t)φj(t) =(0, i 6= jλj, i = j. (2)For complex-valued basis sets this c omplicates slightly toZt2t1φi(t)φ∗j(t) =(0, i 6= jλj, i = j. (3)If λj= 1 for all j, then the basis functions are said to be orthonormal.We can use the definition of orthogonality in (3) above to determine thetransform coefficients. To compute akfor a particular basis function φkwemultiply both sides of (1) by φ∗kand integrate:Zt2t1φ∗kˆx(t)dt =Zt2t1φ∗k"N−1Xi=0aiφi(t)#dt. (4)Now we reverse the order of summation and integration:Zt2t1φ∗kˆx(t)dt =N−1Xi=0aiZt2t1φ∗kφi(t)dt (5)And by (3) we know that the right side is equal to akλk, since all other termsof the summation are zero. So the coefficient for the kth basis function issimplyak=1λkZt2t1φ∗kˆx(t)dt. (6)1In the case where the basis functions are complex exponentials (which areorthonormal), this becomesak=1t2− t1Zt2t1e−jkω0tˆx(t)dt. (7)A common measure of the degree to which the sum of weighted basisfunctions approximates the desired function x(t) is the mean squared error,or MSE:MSE =1t2− t1Zt2t1"x(t) −N−1Xi=0aiφi(t)#2dt. (8)Now if for all functions x(t) in some class the MSE goes to zero in the limit:limN→∞1t2− t1Zt2t1"x(t) −N−1Xi=0aiφi(t)#2dt = 0 (9)we say that the series approximation converges in the mean to x(t), and thatthe set of {φi(t)} is complete for all the functions x(t) over that interval.Walsh FunctionsAn interesting set of orthonormal basis functions is the Walsh functions.These are defined only on the range (0 ≤ t ≤ 1).There are a number of conventions for ordering the Walsh basis set. Allgive the same set of functions, but they generate them in differing orders.Here we will be using sequency order, where the index k gives the numberof zero-crossings in the interval.The first sixteen Walsh functions are shown in sequency order in Figure1. Note that for even k the ends match in sign, while for odd k they areopposite. We can generate these by a recursion formula. First we define thezeroth-order basis function:φ0(t) =n1, 0 ≤ t ≤ 1. (10)Now, if there isn’t a zero-crossing at 1/2 we can add one simply by invertingthe second half of the interval. So for odd k,φk(t) =(φk−1(t), 0 ≤ t <12−φk−1(t),12< t ≤ 1. (11)2For even k, we can repeat a squeezed version of φk/2. But if k/2 is odd, weneed to invert the second copy so the ends match up, or else there would bean extra, unwanted cross ing in the middle:φk(t) =(φk/2(2t), 0 ≤ t <12(−1)k/2φk/2(2t − 1),12< t ≤ 1. (12)Example: Represent the functionx(t) = sin(πt) (13)with Walsh functions over the interval (0 ≤ t ≤ 1). Find the first fourcoefficients.To find the coefficient for each basis function φk(t) we solve (6):ak=Z10φk(t)sin(πt)dt. (14)For k = 0, this is simplya0=Z10(1) sin(πt)dt = −1πcos(πt)10= 0.637. (15)Similarly,a1=Z1/20(1) sin(πt)dt +Z11/2(−1) sin(πt)dt = 0 (16)a2=Z1/40(1) sin(πt)dt +Z3/41/4(−1) sin(πt)dt +Z13/4(1) sin(πt)dt = −0.264(17)a3=Z1/40(1) sin(πt)dt+Z1/21/4(−1) sin(πt)dt+Z3/41/2(1) sin(πt)dt+Z13/4(−1) sin(πt)dt = 0.(18)30 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 105101520253035404550Figure 1: The first 16 Walsh basis functions on the interval from 0 to