Subspace-based Estimators - Spectral Analysis: Parametric and Non-Parametric Digital Methods - page 205

eigenvalue, then it immediately comes esp { V b } ⊂ esp { A } ⊥ . Inversely, any vector
of esp { A } ⊥ is an eigenvector of x Γ associated with the null eigenvalue. Hence
esp { V b } = esp { A } ⊥ . This implies that the other eigenvectors of x Γ associated
with strictly positive eigenvalues, given esp { V s }, are in esp { A }. Then we have
esp { V s } ⊂ esp { A }. Now, x Γ being hermitian, we know that there is an
orthonormal basis of eigenvectors, which implies that the dimension of esp { V s } is
P. Hence esp { V s } = esp { A }.

In the presence of a white noise, the eigenvalues of:

H2

ΓΓ = AA + σ I

xx ss

are those of A Γ ss A H increased by σ 2 and the eigenvectors are unchanged, hence the
results of Theorem 8.1.

We indicate by signal subspace the subspace esp { V s } spanned by the
eigenvectors associated with the biggest P eigenvalues of Γ xx , and by noise subspace
the subspace esp { V b } spanned by the M - P eigenvectors associated with the
smallest eigenvalue σ 2 . Thus, it immediately results that the complex sinusoid
vectors a ( f i ) are orthogonal to the noise subspace, given:

H

()

Va f == 0, i 1, … , P [8.16]

b i

Thus, the signal vectors corresponding to the searched frequencies are the
vectors a ( f ) of the form given in [8.9] which are orthogonal to V b .

A number P of independent vectors of the form of [8.9] should exist in a
subspace of dimension P of M so that the resolution of the non-linear system
[8.16] gives the searched frequencies in a unique way. This is true provided that
P + 1 < M.

THEOREM 8.2. If P + l < M, any system of P vectors of the form a ( f ) given in [8.9],
for different values of f, forms a free family of

M .

DEMONSTRATION. In M , an infinity of vectors of the form a ( f ) [8.9] exist
when f scans the set of real numbers. However, any family of I < M vectors
{ a ( f 1 ) , …, a ( f I )}, where the parameters f i are different, is free because it is easy to
verify that the matrix formed by these vectors is of the Vandermonde type. Thus, in
a subspace of dimension I < M, only I vectors of this type exist. Actually, if another
one existed, noted { a ( f I+ 1 )} with f I+ 1 ≠ f i for i = 1, …, I in this subspace of dimension
I , it would be written as a linear combination of these I vectors and the system { a ( f 1 )
, …, a ( f I+ 1 )} would not be free, which is impossible if I + 1 < M. Therefore, in