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CX CXY CY X CY in .NET Integration UPCA in .NET CX CXY CY X CY

CX CXY CY X CY using visual studio .net toembed upc symbol on asp.net web,windows application Barcode FAQs (9.10). Notes is equal to CX + CX UPC-A for .NET Y + CY X + CY ,. which, upon noting that CXY is a scalar and therefore equal to its transpose, simpli es to CX + 2 CY X + CY . On the other hand, if we put w := + j , and use (9.7), then (see Problem 22) wH Kw = (CX +CY ) + (CX +CY ) + 2 (CY X CXY ) . Clearly, if CX = CY and CXY = CY X , (9.

11) then (9.10) is equal to wH Kw/2. Conversely, if (9.

10) is equal to wH Kw/2 for all w = + j , then (9.11) holds (Problem 29). We say that a complex Gaussian random vector Z = X + jY is circularly symmetric or proper if (9.

11) holds. If Z is circularly symmetric and zero mean, then its characteristic function is E[e j( X+ Y ) ] = e w. H Kw/4. w = + j . (9.12). The density c orresponding to (9.9) is (assuming zero means) exp 1 x y 2 CX CXY CY X CY (2 )n det ..

fXY (x, y) = where (9.13). :=. CX CXY CY X CY It is shown i UPC Code for .NET n Problem 30 that under the assumption of circular symmetry (9.11), e z K z , fXY (x, y) = n det K.

H 1. z = x + jy,. (9.14). and that K is invertible if and only if is invertible. Notes 9.1: Introduc .NET UPC Code tion Note 1.

We show that if X has the density in (9.2), then its characteristic function is e j m C /2 . Write E[e j X ] = =.

e j x f (x) dx e j x exp[ 1 (x m) C 1 (x m)] 2 dx. (2 )n/2 detC Gaussian random vectors Now make the multivariate change of variable y = C 1/2 (x m), or equivalently, x = C1/2 y + m. Then dx = . detC1/2 . dy = detC dy (see Note 2), and E[e j X ] = e j (C 1/2 y+m). e y y/2 d etC dy n/2 detC (2 ). = e j m = e j m Put t := C1/2 so that e j(C e j(C 1/2 ). e y y/2 dy (2 )n/2 n y2 ] i=1 i dy. (2 )n/2. 1/2 ). 1 y exp[ 2. (C1/2 ) y = Then E[e j X ] = e j m ti yi . e yi /2 e jti yi dy 2 IRn i=1 n = e j m e yi /2 e jti yi dyi . 2 Since the int GS1 - 12 for .NET egral in parentheses is of the form of the characteristic function of a univariate N(0, 1) random variable, E[e j X ] = e j m e ti /2. = e e = e j m e C /2 = e j m C /2 . 9.4: Density function Note 2.

Recall that an n n matrix C is symmetric if it is equal to its transpose; i.e., C = C .

It is positive de nite if a Ca > 0 for all a = 0. We show that the determinant of a positive-de nite matrix is positive. A trivial modi cation of the derivation shows that the determinant of a positive-semide nite matrix is nonnegative.

At the end of the note, we also de ne the square root of a positive-semide nite matrix. We start with the well-known fact that a symmetric matrix can be diagonalized [30]; i.e.

, there is an n n matrix P such that P P = PP = I and such that P CP is a diagonal matrix, say 1 .. P CP = = .

. i=1 j m t t/2. 0 . . . Problems Next, from P visual .net UPC-A Supplement 5 CP = , we can easily obtain C = P P . Since the determinant of a product of matrices is the product of their determinants, detC = det P det det P .

Since the determinants are numbers, they can be multiplied in any order. Thus, detC = det det P det P = det det(P P) = det det I = det = 1 n . Rewrite P CP = as CP = P .

Then it is easy to see that the columns of P are eigenvectors of C; i.e., if P has columns p1 , .

. . , pn , then Cpi = i pi .

Next, since P P = I, each pi satis es pi pi = 1. Since C is positive de nite, 0 < piCpi = pi ( i pi ) = i pi pi = i . Thus, each eigenvalue i > 0, and it follows that detC = 1 n > 0.

Because positive-semide nite matrices are diagonalizable with nonnegative eigenvalues, it is easy to de ne their square root by C := P P , where 1 .. := .

. n. . . Thu s, det C = 1 n = detC. Furthermore, from the de nition of C, it is clear that it is positive semide nite and satis es C C = C. We also point out that since C = then C 1 = P 1 P , where 1 is diagonal with diagonal P P , if C is positive de nite, entries 1/ i ; hence, C 1 = ( C ) 1 .

Finally, note that 1 CC C = (P P )(P 1 P )(P P ) = I..
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