(5.164)
We know the following equations hold for the determinant and trace of the above matrix.
(5.165)
(5.166)
Since the determinant is zero the matrix is singular and hence cannot be inverted, implying the following
to hold true when defining the first eigenvalue, . Using the above equations we see the
first eigenvalue equals . Applying the ESD, we have the following:
(5.167)
The first eigenvalue can be used in the above equation to determine the corresponding eigen vector
given below
(5.168)
where the value of K was chosen to make the eigen vector orthonormal, .
Note that the solution that maximizes the power equation is the eigen vector that corresponds to
the largest eigenvalue. Using the ESD theorem we can reconstruct the channel matrix as follows:
(5.169)
Assuming , one can simply choose as the solution.
5.4 LINK BUDGET DISCUSSION
In this section, we will discuss an example of a link budget of a wireless communication system. The
link budget will show where the system power is distributed between the transmitter, channel, and
receiver. The link budget consists of the following parameters:
1. Base station transmit power
2. Base station cabling loss
3. Base station antenna gain
4. Path loss
5.
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