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How to generate, print barcode using .NET, Java sdk library control with example project source code free download:
si generate, create code 39 extended none in software projects Visual Studio Development Language =. s2 . ij (5.14). The distance between two signal constellation points s i and sk is thus si sk = . (sij skj )2 =. (si (t) sk (t))2 dt,. (5.15). 2(t). 1(t). Figure 5.3: Signal S pace Representation where the second equality is obtained by writing s i (t) and sk (t) in their basis representation (5.3) and using the orthonormal properties of the basis functions.

Finally, the inner product < s i (t), sk (t) > between two real signals si (t) and sk (t) on the interval [0, T ] is. < si (t), sk (t) >= si (t)sk (t)dt. (5.16). Similarly, the inner product < si , sk > between two real vectors is < si , sk >= si sT = k si (t)sk (t)dt =< barcode code39 for None si (t), sk (t) >,. (5.17). where the equality b bar code 39 for None etween the vector inner product and the corresponding signal inner product follows from the basis representation of the signals (5.3) and the orthonormal property of the basis functions (5.5).

We say that two signals are orthogonal if their inner product is zero. Thus, by (5.5), the basis functions are orthogonal functions.

. 5.1.3 Receiver Structure and Suf cient Statistics Given the channel ou barcode code39 for None tput r(t) = si (t) + n(t), 0 t < T , we now investigate the receiver structure to determine which constellation point si or, equivalently, which message mi , was sent over the time interval [0, T ). A similar procedure is done for each time interval [kT, (k +1)T ). We would like to convert the received signal r(t) over each time interval into a vector, as it allows us to work in nite-dimensional vector space to estimate the transmitted signal.

However, this conversion should not compromise the estimation accuracy. For this conversion, consider the receiver structure shown in Figure 5.4, where.

sij = si (t) j (t)dt,. (5.18). and nj = n(t) j (t)dt. (5.19). We can rewrite r(t) as (sij + nj ) j (t) + nr (t) =. j=1 j=1 rj j (t) + nr (t),. (5.20). where rj = sij + nj Software bar code 39 and nr (t) = n(t) N nj j (t) denotes the remainder noise, which is the component j=1 of the noise orthogonal to the signal space. If we can show that the optimal detection of the transmitted signal constellation point si given received signal r(t) does not make use of the remainder noise n r (t), then the receiver can make its estimate m of the transmitted message m i as a function of r = (r1 , . .

. , rN ) alone. In other words, r = (r1 , .

. . , rN ) is a suf cient statistic for r(t) in the optimal detection of the transmitted messages.

. T ( )dt 0 (t ). s i1+n1=r 1 r(t)=s i(t)+n(t). Find i: r Z ^ m=mi T ( )dt 0 ( t). s iN +nN =r N Figure 5.4: Receiver Code 3/9 for None Structure for Signal Detection in AWGN. It is intuitively clear that the remainder noise n r (t) should not help in detecting the transmitted signal s i (t) since its projection onto the signal space is zero.

This is illustrated in Figure 5.5, where we assume the signal lies in a space spanned by the basis set ( 1 (t), 2 (t)) while the remainder noise lies in a space spanned by the basis function nr (t), which is orthogonal to 1 (t) and 2 (t). The vector space in the gure shows the projection of the received signal onto each of these basis functions.

Speci cally, the remainder noise in Figure 5.5 is represented by nr , where nr (t) = nr nr (t). The received signal is represented by r + n r .

From the gure it appears that projecting r + nr onto r will not compromise the detection of which constellation s i was transmitted, since nr lies in a space orthogonal to the space where s i lies. We now proceed to show mathematically why this intuition is correct..

r+n r r s3 s4 (t) n 2 (t) 1(t). Figure 5.5: Projecti Code 39 Full ASCII for None on of Received Signal onto Received Vector r. Let us rst examine the distribution of r.

Since n(t) is a Gaussian random process, if we condition on the transmitted signal si (t) then the channel output r(t) = si (t) + n(t) is also a Gaussian random process and 122. r = (r1 , . . .

, rN ) is a Gaussian random vector. Recall that r j = sij + nj . Thus, conditioned on a transmitted constellation si , we have that (5.

21) rj . si = E[rj si ] = E[sij + nj sij ] = sij since n(t) has zero mean, and rj si = E[rj rj si ]2 = E[sij + nj sij sij ]2 = E[n2 ]. j Moreover, Cov[rj rk si ] = E[(rj rj ) Software barcode 3/9 (rk rk ). si ] = E[nj nk ] = E
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