f(x) f( x) f a (x) = } form an orthonormal basis. Similarly, for the space A, { sin(nx)

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1 ECE 8-6 Homework 1 Solutions 1. a) It is known that the symmetric (even) part of a signal can be obtained as: f(x) + f( x) f s (x) = (1) Similarly, the antisymmetric (odd) part of a signal can be written as: f a (x) = f(x) f( x) Therefore, f(x) = f s (x) + f a (x). b) For the space S, { cos(nx) } form an orthonormal basis. Similarly, for the space A, { sin(nx) } form an orthonormal basis. It is straightforward to show that both of these sets are orhonormal since 1 π π cos(nx)cos(mx)dx = δ nm and 1 π π sin(nx)sin(mx)dx = δ nm. In order to establish that they form bases for the given spaces, we need to show that they span the space. By Fourier series theory, we can show that any function, f(x) L [ π, π] can be written as a combination of { cos(nx) }, { sin(nx) () }. For f(x) S, the expansion coefficients for { sin(nx) } are equal to zero. Therefore, S is spanned by { cos(nx) }. A similar argument can be made for the subspace, A. c)projection onto V 1 is v 1 = 1 ( π π x dx)+( 1 π π π x sin(x)dx) sin(x) π + ( 1 π π π x cos(x)dx) cos(x) π = π 4cos(x). Similarly, the orthogonal pro- jection to V can be found as v = v 1 + ( 1 π π π x sin(x)dx) sin(x) 4cos(x) + cos(x). A similar calcu- ( 1 π π π x cos(x)dx) cos(x) π = π lation yields the projection onto V as v = π 4 9 π + 4cos(x) + cos(x) cos(x).these projections are plotted in Figure 1. d) The inner products are.49, 1.45 and 1.85, respectively for the different subspaces. The reason why the inner products are not exactly equal to zero is that a continuous time inner product is approximated by a discrete summation in MATLAB.. a) a) We need to show that {1, t, t,...} is a linearly independent set. We will first show that any subset of these vectors are linearly independent. Therefore, N i= α i t i = for all t iff α i =. Writing the Nth 1

2 1 8 Original signal Projection onto V1 Projection onto V Projection onto V Figure 1: The signal and its projections derivatives and computing α i s iteratively yield: α N (N)(N 1)(N )... = α N = α N 1 (N 1)(N ) α N (N)(N 1)...t = α N 1 = () b) Gram-Schmidt procedure can be computed with the following MAT- LAB algorithm: function z=gramschmidt(x,n,m); for i=1:n; s(i,:)=x(i,:); end; e(1)=s(1,:)*conj(s(1,:). ); phi(1,:)=s(1,:)/sqrt(e(1)); for i=:n; th(i,:)=zeros(1,m); for r=i-1:-1:1; th(i,:)=th(i,:)+(s(i,:)*conj(phi(r,:). ))*phi(r,:); end;

3 th(i,:)=s(i,:)-th(i,:); e(i)=th(i,:)*conj(th(i,:). ); phi(i,:)=th(i,:)/sqrt(e(i)); end; z=phi(1:n,:); b)for the polynomial set {1, t, t, t,...}, the corresponding orthonormal set is known as the Legendre polynomials. The first five functions in the orthonormal set is shown in Figure Time Figure : The first five orthonormal polynomials. a) Since {x 1, x,...} is an orthonormal basis, y 1 and y can be written as y 1 = i α i x i and y = j β j x j. Using these representations for y 1 and y, we can write the inner product as: < y 1, y > = < α i x i, β j x j > i j = α i βj < x i, x j > i j = α i βj δ ij i j = i α i β j

4 = i < y 1, x i >< y, x i > (4) where we used the definition of orthonormal basis and the expansion coefficients to simplify the equality. 1 b)using the result from part (a) and the fact that e jωt is an orthonormal basis and < f, 1 e jωt >= F(ω), we get Parseval s equality. c)using the equality proven in part (a), < h T (t nt), h T (t pt) > = 1 T 1 [ π/t,π/t] (ω)e j(n p)tω dω = T π/t e j(n p)tω dω π/t = Tδ(n p) (5) Therefore, h T (t nt) are an orthogonal set of functions. By Shannon s sampling theorem, if f(t) is bandlimited in [ π/t, π/t], then f(t) = n f(nt) sin(π(t nt)/t). π(t nt)/t This shows that for any f(t) U,f(t) = n a(n)h T (t nt). Therefore, h T (t nt) form a basis for U.The coefficients in the expansion can be obtained through inner product of f(t) with h T (t nt) since h T (t nt) form an orthogonal basis.therefore, < f(t), h T (t nt) > = 1 F(ω)H(ω)e jntω dω π/t Therefore, f(nt) = 1 T < f(t), h T(t nt) >. = T F(ω)e jntω dω π/t = Tf(nT). (6) 4. a) The Fourier series coefficients will be significant for rad/sec, 4 rad/sec, 8 rad/sec and 4 rad/sec. Note the signal does not have to be necessarily periodic to define Fourier series coefficients. As long as the signal is defined in a finite interval, we can consider that interval as one period. 4

5 .9 Magnitude of Fourier Series Coefficients k Figure : The magnitude of the Fourier series coefficients Actual Signal Approximation Time Figure 4: The magnitude of the Fourier series coefficients b) The most significant coefficients are at k = 1,, 19, 1 corresponding to the frequencies identified in part (a). c) x=:.1:pi; 5

6 f = exp( (x. )/1).*(cos(*x)+*sin(4*x)+.4*cos(*x).*cos(4*x)); for k=1:5; coef(k)=(1/length(x))*sum(f.*exp(-j**k*x)); end dc=(1/length(x))*sum(f); fpart=zeros(size(f))+dc; for n=1:6; fpart=fpart+coef(n)*exp(j**n*x)+conj(coef(n))*exp(-j**n*x); end 6