Hello, welcome to another module in this massive
open online course on the Principles of the
CDMA, MIMO, OFDM by Wireless Communication
System. So, in the previous module we have
looked at SVD or Singular Value Decomposition
based MIMO wireless transmission, we have
looked at how the Singular Value decomposed,
that first what is the singular value decomposition?
How it can be applied in the context of MIMO
transmission? That is the post processing
at the receiver, the pre coding at the transmitter,
the special multiplexing or the decoupled
special channels and finally, the Optimal
Power Allocation across the various sub channels
to maximize the some rate or achieve the Shannon
capacity of the MIMO wireless channel.
Now, to better understand all these aspects,
let us look at a simple example which will
summarize these aspects related to SVD or
singular Value Decomposition based MIMO channel.
So, today we are going to look at SVD Based
MIMO transmission, but we are going to consider
a simple example to help understand these
aspects comprehensively.
So, let us say we are with 3 cross 3 MIMO
channel, that is the MIMO channel matrix is
H which is given as 2, minus 6 0 3 4 0 0 0
2. So, this is the 3 cross 3 MIMO channel,
which means that basically my r is equal to
3, my t is equal to 3 that is number of receive
antennas is equal to 3, number of transmit
antennas is equal to 3, my channel matrix
h is this 3 cross 3 matrix. Now let us denotes
these columns by C 1 bar, C 2 bar, C3 bar.
Now, you can see that in this MIMO channel
matrix, in this 3 cross 3 MIMO channel matrix
which has 3 columns, C 1 bar, C 2 bar, C 3
bar, these 3 columns are orthogonal. You can
see that, if I take any 2 columns out of these
3 columns, I will have C 1 bar hermitian,
C i bar hermitian. So, if I take any 2 columns,
I will have C i bar hermitian, C j bar equal
to 0. On this matrix the columns are orthogonal,
for instance let us look at C 1 bar and C
2 bar, I have C 1 bar that is the first column
equals 2 3 0, C 2 bar that is a second column
equals minus 6 4 0.
So, if I look at C 1 bar hermitian C 2 bar
that is equal to the row 2 3 0 times, the
column vector minus 6 4 0 and this is equal
to 2 into minus 6 minus 12, plus 3 into 4
12, plus 0 into 0 0, which is equal to 0.
So, C 1 bar and C 2 bar are orthogonal. Similarly
you can verify this for the other column,
that is you can see C 2 bar and C 3 bar are
orthogonal, that is C 2 bar hermitian 3 bar
is 0, similarly C 1 bar, hermitian C 3 bar
is 0, that is C 1 bar and C 3 bar orthogonal.
So, all the columns of this 3 cross 3 channel
matrix are orthogonal. How do we do the singular
value decomposition? I am going to illustrate
now how to perform the singular value decomposition.
Let us look at, so the SVD can be found as
follows, let us go back to our channel matrix
which is 2 minus 6 0 3 4 0 0 0 2.
So, let us look at this channel matrix, now
you can see I can 
normalize each column that is I can make each
column as unit norm therefore, I will have
H equals the first column divided by the norm
of the first column, the norm of the first
column is square root of 2 square, plus 3
square, plus 0 square, that is square root
of 4 plus 9 that is 13, divide by the square
root of 13. So, I have 2 divided by 13, 3
divided by 13, 0 divided by 13 which is 0.
Similarly the second column is minus 6 4 0,
the norm of this second column is square root
of 6 square, plus 4 square, plus 0 square,
that is square root of 36 plus 16 that is
square root of 52. So, I can divide by square
root of 52, to normalize this column or make
this column unit norm.
So, I will have minus 6 divided by square
root of 52, 4 divided by square root of 52,
0 divided by square root to 52 is 0 and the
third column has only one element 2. So, the
norm of this is square root of 2 square, which
is square root of 4 which is 2. So, I have
2 divided by 2 which is 1 and now since we
are dividing by the norm I have to multiply
each column by the corresponding norm, so
I have square root of 13, square root of 52,
and 2 , 2. So, this is the diagonal, this
is the diagonal matrix.
Now, if you look at this first matrix call
this U, now you can see that U hermition U
will be identity, because it has now orthogonal
columns and we have normalized each column
to have unit norm therefore, U hermition U
will be identity. However, if you look at
this matrix the singular values that is square
root of 13, square root of 52 and 2 are not
ordered because square root of 13 is less
than square root of 52, I have to arrange
the singular values in a decreasing order
therefore; I need square root of 50. So, I
need first singular value a square root of
52, the second singular value a square root
of 13, the third singular value of as 2 therefore,
now we have to order the singular values.
So, here the problem is that the singular
values are not ordered, the singular values
are not in a decreasing order, therefore I
can re write this now as, so let me write
this once more.
I have 2 divided By square root of 13, 3 divided
By square root of 13, 0, minus 6 divided By
square root of 52, 4 divided By square root
of 52 0 0 0 1 times, I have square root of
32 0 0 0 square root of 52, 0 0 0 2 what I
am going to do now, I am to switch the first
2 columns of this first matrix U. So, that
will give you minus 6 by square root of 52,
4 divided by square root of 52, 0, 2 divided
by square root of 13, 3 divided by square
root of 13, 0, 0 0 1.
Now, since I am switching the first 2 columns
of the first matrix, I have to switch the
first 2 rows of the second matrix, so I will
have 0 square root of 52, 0 square root of
13, 0 0 0 0 2 so this is what I have. Now
I want to again shift the columns to make
this second matrix a the diagonal matrix,
I want to swap columns one and two, which
means I can right this as the following thing
minus 6 divided by square root of 52, 4 divided
by square root of 52, 0, 2 divided by square
root of 13, 3 divided By square root of 13,
0 0 0 1 times.
Now, let me make this a diagonal matrix square
root of 52, square root of 13, 2 0 0 0 0 0
0 and now I can multiply this by the matrix
which is 0 1 0 0, 1 0 0, 0 0 1 and now you
can see this was swap the columns of the diagonal
matrix. So, now you can see I have my matrix
U, I have my diagonal matrix sigma, which
has the non negative singular values in decreasing
orders, and I have my matrix V hermitian and
now if you can verify this matrixes U, Sigma
and V satisfy the properties.
The matrixes U sigma v, I have hermition,
U is identity, I have V hermition V equals
V V hermitian equals identity, I have sigma
as the diagonal matrix square root of 52,
square root of 13, 2.
So, sigma is a diagonal matrix with singular
value sigma 1 equals square root of 52, sigma
2 equals square root of 13, sigma 3 equals
2, therefore all the singular values sigma1,
sigma 2, sigma 3 are greater than 0. So, the
singular values are positive, further we have
sigma 1, greater than sigma 2, greater than
sigma 3, which means the singular values are
decreasing orders.
So, we have the diagonal matrix sigma, the
3 singular value square roots of 52, square
root of 13, and 2 which are nonnegative. In
fact, they are positive, at these singular
values are in decreasing order, because square
root of 52 is greater than square root of
13, is greater than 2. So, this and together
with the properties that U hermition U is
identity, V hermition V equals V V hermitian
equals identity, this forms a valid singular
value decomposition for the matrix H, that
is the 3 cross 3 channel matrix H, that has
been described in the beginning of this SVD
Based MIMO transmission example. So, now,
let us proceed with the post processing and
preprocessing. So, now, we have the matrixes
U sigma V hermitian.
As we said at receiver we perform Y tilde
equals U hermitian y bar, which is basically
I have the structure of matrix U. So, U hermitian
is minus 6 by square root of 52, 4 divided
by square root of 52, 0, 2 divided by square
root of 13, 3 divided by square root of 13,
0, and 0 0 1 times, y 1, y 2, y3 this is basically
my y bar, this is basically my U hermitian
and this is my y tilde, which is y 1 tilde,
y 2 tilde, y 3 tilde, which is basically my
3 cross one vector y tilde. So, at the receiver
I am generating y tilde as U hermitian times
y bar, and since we know the matrix U, which
follows from the SVD of the channel matrix
H, I can perform U hermitian y bar to generate
the vector y tilde.
Now, let us look at the preprocessing at the
transmitter, at the transmitter remember I
perform x bar equals, V x tilde, that is this
is also termed as remember we also term this
as 
pre coding therefore, I have x 1, x 2, x 3
is equal to V which is 0 1 0, 1 0 0, 0 0 1
times x 1 tilde, x 2 tilde, x 3 tilde. So,
this is basically my x bar, this is basically
my matrix V, this is basically my vector x
tilde and this is the preprocessing at the
transmitter, this step is also known as “Transmit
Preprocessing or Pre Coding”.
So, this step is also known as the Transmit
Pre processing step in SVD Based MIMO transmission
or basically pre coding. Now after this what
I am got at the receiver as we already seen
this in the previous module, I am going to
have y tilde equals sigma times x tilde plus
w tilde.
So, what I am going to have at the receiver
is I am going to have y tilde, equals sigma
x tilde, plus w tilde. Now y tilde equals
y 1 tilde, y 2 tilde ,y 3 tilde which is equal
to sigma that is square root of 52, square
root of 13, 2 0 0 0 0 0 0 times x 1 tilde,
x 2 tilde, x 3 tilde, plus w 1 tilde, w 2
tilde, w 3 tilde and therefore, now if you
can look at this since sigma is diagonal across
each sub channel I have a 3 decoupled channels
in this MIMO wireless system, namely I can
now separate these as y 1 tilde equals square
root of 52, times x one tilde, plus w 1 tilde,
y 2 tilde equals to square root of 13, x 2
tilde plus w 2, tilde y 3 tilde square equals
2, times x 3 tilde plus w 3 tilde.
So, what do I have as a result I have 3 Decoupled
channels and this is basically 
this we said Spatial Multiplexing, Why is
this Spatial Multiplexing? Because now I am
able to transmit 3 symbols, what are the 3
symbols? x 1 tilde, x 2 tilde, x 3 tilde on
the same frequency, at the same time. Because
I am using the spatial dimension in the 3
cross MIMO channel I am using the 3 transmit
and 3 receive antenna to simultaneously multiplex
3 symbols x 1 tilde, x 2 tilde, x 3 tilde
across the MIMO wireless channel, this is
also known as “Spatial Multiplexing”,
that is what we have seen in the previous
module this is known as Spatial Multiplex.
Now, let us look at Optimal Power Allocation,
in this Optimal Power Allocation that is if
the transmit power is P, Optimal Power Allocation
for what? Optimal Power Allocation to maximize
some rate that is basically, which implies
maximizing the some rate, this automatically
implies to achieve capacity or Shannon 
capacity I have P1 equals 1 over lambda, minus
sigma square divided by sigma 1square, subscript
plus which is 1 over lambda, minus sigma square
divided by sigma 1 square equals 52, remember
the plus sign indicates the quantity, if the
quantity inside the bracket is greater than
0 and if is equal to 0.
If the quantity inside the brackets is less
than or equal to 0 and similarly P 2 equals
1 by lambda minus sigma square by sigma 2
square superscript, plus which is 1 over lambda,
minus sigma square divided by 13, plus P 3
equals 1 over lambda minus sigma square divided
by sigma 3 square, plus which is 1 over lambda,
minus sigma square divided by 4 plus.
Now, to complete this example let us consider
a 
noise power sigma square equals 0 dB, which
implies 10 log 10 sigma square equals 0, which
implies sigma square equals 10 to the power
of 0, divided by 10 which is equal to 1. So,
the noise power is 1, let us consider P equals
the total power, that the total transmit power
b equal to 3 dB, which means 10 log to the
base 10, P equals 3, which implies P equals
10 to the power 3 by 10 which is equal to
10 to the power of point 3, which is approximately
2 as we have seen several times 3 dB equals
2.
Therefore we must have P 1, plus P 2, plus
P 3 equals 2, which means if I now substitute
for P 1, P 2, P 3 I have 1 minus 1 over lambda,
minus one by 52 remember P 1 equals 1over
lambda, minus sigma square by 52, but sigma
square is 1, plus 1 over lambda minus sigma
square by 13 by sigma square is 1, plus 1
over lambda minus sigma square by 4, but sigma
square equals 1 equals the total power which
is 2, and now I can solve for 1 over lambda
equals 2, plus 1 over 52, plus 1 over 13,
plus 1 over 14, divided By 3 equals 0.7821.
So, what I am doing I am finding the power,
each of the power Optimal Power Allocation,
the sum of the power must be equal to the
total transmit power that is 3 dB or 2, from
this equation what I am finding this quantity
1 over lambda.
One over lambda where lambda is the Lagrange
multiplier of the optimization problem, remember
that. So, I found out 1 over lambda and 1
over lambda is 0.7821, I do not need to find
lambda because really all I need is 1 over
lambda to find each of the powers. Now substituting
this 1 over lambda, I can find each of the
powers so power 1.
So, if I look at this. So, power 1 is now
10 log to the base 10, 1 over lambda minus
sigma square divided By 52, equals 10 log
to the base 10, 0.7821 minus 1 over 52, which
is equal to minus 1.1755 dB, similarly P 2
equals 10 log to the base 10, 0.7821 minus
1 over 13 equals, minus 1.517 dB, P 3 equals,
10 log 10 0.7821 minus 1 over 4, equals minus
2.74 dB.
So, what we have done is we have found P 1
we have found the expression for P 1, we have
found the power P 2 we have found the power
P 3 at this are the powers, this is what are
this P 1, P 2, P 3 these 
are the powers for Optimal Power Allocation,
what we have said is these are the power is
that maximize the some rate of the MIMO wireless
channel or basically achieve the Shannon capacity
of the MIMO wireless channel and if you observe
closely you will also observe something interesting
that the power 1 is minus 1.17 dB, power 2
is minus 1.51 dB, power 3 is minus 2.74 dB.
You will observe that the power is decreasing,
Power allocated decreases as the gain sigma
i square of the, you will observe interestingly
that is, for various channels i, the channel
i which have a higher gain, that is higher
value of sigma i square have a higher power,
the channel which have a lower value of sigma
i square, that is lower gain, have lower power.
So, the power allocated in the Optimal Power
Allocation decreases as the gain of the channel
progressively decreases, which means very
interestingly you are allocating lesser power
to poorer channel and your allocating more
power to stronger channel and that helps you
optimize the power allocation for MIMO wireless
system towards maximizing the transmitted
rate between the transmitter and receiver,
we also said this algorithm known as the “Water
Filling Power Allocation”.
The Water Filling Power Allocation basically
states that, in a Weasel where the weasel
height is high, the water level or the height
of the water level is the basically the net
water level, which is basically level of water
minus as the height of the weasel is lower.
That is the depth of the water is greater,
where the depth of the water level is. So,
this is water filling allocation, it basically
translates into the fact that in the MIMO
power allocation more power is allocated to
the stronger channels, less power is allocated
to the weaker channel.
Therefore in this module, we are comprehensively
seen of an example of SVD Based MIMO transmission,
in which we have considered a simple 3 cross
3 MIMO wireless channel, we have looked at
the a Singular Value Decomposition of this
MIMO wireless channel, the post processing
at the transmitter, the post processing at
the receiver, the preprocessing or pre coding
at the transmitter, the diagonalization into
the decoupled MIMO channels, spatial multiplexing
and then finally, Optimal Power Allocation.
So, we will conclude this module here.
Thank you very much.
