# Exercises d'algebre 1re annee by Calvo, Doyen, et al.

By Calvo, Doyen, et al.

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49). This yields ˆ f n = (ψ n , f ) . 50) If a signal can be represented by a ﬁnite set of linearly independent functions, one can always construct the corresponding reciprocal basis and use it to represent the signal. Thus any given signal has two alternative representations, one in terms of the direct and other in terms of the reciprocal basis functions. 51) n=1 the expansion in terms of the reciprocal basis functions will be represented by N f˜n ψ n (t) . 52) n=1 The two sets of coeﬃcients are related by the Gram matrix.

Since the ranks of C and A are identical, σ 22 = 0 so that the TLS solution is also the same. A diﬀerent situation arises when the rank of C is N + 1 so that σ 22 = 0. Then y is not in the subspace spanned by the columns of A and Ax = y has no solutions. However, as long as there is a nonzero projection of the vector y into the range of A there exists a nontrivial LMS solution. The TLS solution also exists and is distinct from the LMS solution. Clearly, the two solutions must approach each other when σ 22 → 0.

N only a small number of which are linearly independent. This situation is found, for example, in digital communications where it is frequently advantageous to use modulation formats involving a set of linearly dependent waveforms. , by using symmetry arguments). 150) provides an alternative and systematic approach. 152). Upon taking account of the orthogN ∗ vnk = δ jk we obtain the following identity onality n=1 vnj N N φn (t) φ∗n (t ) = n=1 R σ 2j uj (t) u∗j (t ) . 159) j=1 Multiplying both sides by f ∗ (t) and f (t ) and integrating with respect to both variables give N 2 R n=1 2 σ 2j |(f, uj )| .