Hello welcome to another module in this massive

open online course on the principles of MIMI, CDMA, OFDM wireless communication systems.

So, in the previous lecture we were looking at CDMA and how to generate the code sequences

for CDMA which stands for code division for multiple access and we have said that the

CDMA code sequences are generated as PN sequences. So, for CDMA when we look at CDMA the code

sequence of CDMA is generated as a PN sequence which stands for a Pseudo Noise. The sequence

it looks like a random series of 1Õs and plus 1Õs and minus 1Õs, this is generated

as a PN sequence or a Pseudo Noise sequence. PN sequence means a Pseudo Noise sequence

and this is generated using an LFSR structure which stands for a linear feedback shift register

structure. So, the PN sequences these are generated using

the LFSR. Which is also known as the linear feedback shift register structure. We had seen an example

of an LFSR structure for d equal to 4 registers, we said that this is a maximal length linear

feedback shift register and goes through all the possible states except the all 0 states

therefore, it goes through 2 to the power of d minus 1 or 2 to the power of 4 minus

1 that is 16 minus 1 or 15 states. And the PN sequence that is generated by this

LFSR that is given as, we had also derived that in the last lecture the PN sequence from

d equal to 4 linear feedback shift register that is given as well 1 1 1 1 0 0 0 1 0 0

1 1 0 1 0. So, this is the PN sequence that is generated by the LFSR and the length of

this PN sequence is equal to 15. So, this is the PN sequence that is generated by the

linear feedback shift register and the length of this PN sequence that is equal to 15, we

have 15 symbols in this PN sequence. Now, therefore, now what we are going to do

we have to convert this into a sequence of minus 1Õs and plus 1Õs. So, what we are going to do we are going to

digitally modulate this PN sequence by mapping by modulating, we are going to use BPSK modulation,

we are going to map the one information symbol 1 goes to a minus 1 this is the standard modulation.

And information symbol 0 goes to plus 1. So, we are going to map 1 the information symbol

1 to a minus 1 to a voltage level this is the typical convention. The information symbol

1 is mapped to the voltage level minus 1, the information symbol 0 is map to the voltage

level plus 1 therefore, now my PN sequence becomes. So, my earlier PN sequence let me

write that or let me write it over here, let me write that again my PN sequence information

sequence is 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0. Now, after modulation this becomes remember

1 is mapped to a minus 1, so this becomes minus 1, minus 1, minus 1, minus 1, 1, 1,

1, minus 1, 1, 1, minus 1, minus 1, 1, minus 1, minus 1. So, after modulation this PN sequence

becomes this is the PN sequence after modulation and this

sequence is given as minus 1, minus 1, minus 1, minus 1, 1, 1, 1, minus 1, 1, 1, minus

1, minus 1, 1, minus 1, minus 1 and the sequence is again of length 15 which remains unchanged.

So, this the modulated PN sequence. So, this is the PN sequence, this is the Pseudo Noise

sequence which looks like a random sequence or noise like sequence of plus 1Õs and minus

1Õs which is used as a code in the CDMA system. The code by which the user symbol is multiplied.

Now, to understand properties of CDMA better let us examine the properties of this PN sequence.

So, let us start examining the properties. Let us examine the properties of this PN sequence.

Now, let me write this PN sequence this PN sequence is minus 1, minus 1, minus 1, minus

1, 1, 1, 1, minus 1, 1, 1, minus 1, minus 1, 1, minus 1, 1., this is the sequence as

we all ready seen this is the sequence of length 15.

Now, let us calculate the number of plus 1Õs and minus 1Õs in this sequence, so we have.

So, if we want to calculate the number of plus minus 1Õs, 1st we have 4 minus 1Õs

here, 1 minus 1 here, 2 minus 1Õs here, 1 minus 1 here. So, number of minus 1Õs is

4 plus or 5 plus 2 7 plus 1 8. So, the total number of minus 1Õs in the system is equal to 8.

Now, let us calculate the number of plus 1 symbol in this CDMA in this PN sequence. I

want to calculate what is the total number of plus 1Õs, so total number of plus 1Õs

says we have 3 plus 1Õs over here, 2 plus 1Õs over here, another 1 over here. So, this

3 plus to 5 plus on 6 plus 1, 7, so total number of plus 1 is equal to 7. Now, if you

see we observe something very interesting the total number of plus 1 out of these 15

symbols, there is 15 plus 1Õs and minus 1Õs we have number of minus 1Õs is equal to 8

and number of plus 1Õs is equal to 7 therefore, what we have is a roughly the number of plus

1Õs and number of minus 1Õs is equal. In fact, the precise relation is that the

number of minus 1Õs is exactly 1 more than the number of plus 1Õs. So, what we observe is that the number of

minus 1Õs is equal to basically 1 plus the

number of plus 1Õs the number of minus 1 is 8, which is 1 plus the number of plus 1Õs

that is 1 plus 7. So, what you can see also is that the number

of minus 1Õs and number of plus 1Õs are approximately equal, that is out of a length

of 15 total length of 15 we have 8 minus 1Õs and 7 plus 1Õs. So, what we observe is that

the number of minus 1Õs and number of plus 1Õs is roughly equal and this is known as

the balance property. This property is known as the balance property. So, what we have

is number of minus 1Õs is approximately equal to

number of plus 1Õs and this is known as the balance property. That is approximately if

I take a length n PN sequence, approximately half of the symbols in the PN sequence are

minus 1 and approximately half of the symbols in the PN sequence are plus 1. So, approximately

half and half are plus 1 and minus 1 each this is known as the balance property. And

more precisely speaking the number of minus 1 symbol is exactly 1 more than the number

of plus 1 symbol in the PN sequence. This is known as the balance property of the PN

sequence. So, this is property number 1, let us now

look at property another interesting property let us now look at property number 2. For

this PN sequence and for this let us look at the runs what we call as the run, what

is the run? A run in a PN sequence is a string of a run in a PN sequence is a run is the

string of continuous values. So, let us look again at our PN sequence our

PN sequence what is our PN sequence? Our PN sequence is well minus 1, minus 1, minus 1,

minus 1, 1, 1, 1, minus 1, 1, 1, minus 1, minus 1, 1, minus 1, 1. So, this is my length

15 PN sequence. Now, let us look at the runs, that is then

a strings continuous strings of continuous values I have 1 string of 4 minus 1Õs here,

so this is my 1st run of length equal to 4; I have another string of length 3 1Õs, so

this is of n is equal to 3; I have another string of length 1. So, I have n is equal

to 1 I have another string of 1Õs of length is equal to 2, I have another string of minus

1Õs of length is equal to 2 then the rest of the string I have 3 strings of 1 minus

1 1 each of these is of length L is equal to 1. So, I have the 1st string which is of

4 minus 1Õs. So, the length is equal to 4, I have the 2nd string of length equal to 3

that is 3 1Õs, 3 consecutive plus 1 symbols, I have 1 string of length is equal to 1 which

is a minus 1, 1 string of a length equal to 2 of plus 1 symbols 1 string of length equal

to 2 of minus 1 symbols again 3 strings of length 1 each 1 minus 1 and 1.

So, the total number of strings, if you look at the total number of runs or strings, total

number of runs that is the number of strings is equal to 1 2 plus 1 2 3 4 5 6 7 8. So,

the total number of runs in this PN sequence that is total number of strings of continuous

values in this PN sequence is equal to 8. Now, out of these total number of runs, I

have 4 runs are of length 1, out of these total number of runs 8, I have 4 runs of length

1. So, the length of the run is basically the length of the string is the length of

the run and we have 4 strings of length 1 which means we have 4 runs of length 1 and

the total number of runs is 8 and now look at this 4 is basically half of 8. So, basically

which means that half of the 8 runs are of length, 1.

Now, let us look at the number of runs of length 2, we have 2 runs of length 2 which

means one-fourth of 8 , 8 is the total number of runs one-fourth of 8 is 2. So, one-fourth of 8 which is basically one-fourth

of total runs are of length 2. So, what we are observing is half of the runs are of length

1, one-fourth of the runs are of length 2 and interestingly if you look at this you

will further realize that you have 1 run of length 3. So, I have 1 run of length 3, which

is one-eighth of the total number of runs. So, one-eighth of 8 is 1.

So, one-eighth of the total runs are of length 3. So, what we

are observing is something very interesting, we are observing that half of the runs are

of length 1, one-fourth of the runs are of length 2, one-eighth of the runs are of length

3 and, if you have a longer PN sequence we will have one-sixteenth of the runs of length

4, one more 32 of the runs of length 5. In general 1 over 2 to the power of n of the

total runs, of the total numbers of fraction 1 over 2 to the power of n of the total number

of runs is of length n. So, what we are observing is the 2nd interesting

property, this is the run length property out of total number of runs. 1 over 2 to the

power of n of total runs are of length n. So, what we are observing

is and this is known as the run length property of the CDMA sequences which says

that 1 by 2 to the power of n fraction of runs are of length n, which means half that

is 1 by 2 to the power of 1 or half of the total runs are of length 1, one-fourth of

the runs are of length 2, one-eighth of the runs are of length 3 and so on, this is 1

as the run length property of the PN CDMA, PN sequence. This is known as the run length

property and this another important property of a CDMA sequence, which basically is a fundamental

property it must to obey if it indeed has to look at like a noise like sequence of plus

and minus 1, that is the randomly generated sequence of plus 1Õs and minus 1Õs.

Now we want to look at another property that is property number 3 that is a 3rd property.

So, I want to look at another property, property number 3 of this PN sequence and this is the

most important property, this is known as the auto correlation property of the PN sequence. And auto correlation of the property and sequence

is defined as 1 over N summation over n c n c of n minus d where, this quantity d is

the shift, this d is equal to shift, this quantity n is equal

to the length of the PN sequence. The capital N is equal to length of

and this is the auto correlation, this is the definition of the auto correlation which

is basically 1 over N summation of c n multiplied by c n minus d.

So, what are we doing? We are taking each chip c n we are multiplying by the chip corresponding

to the shifted PN sequence. What is c n minus d? C n minus d is basically shifting the PN

sequence by the quantity d and therefore, we are shifting the PN sequence by this amount

d, we are doing in element wise multiplication of these PN sequence and the shifted PN sequence

by d, we are summing this element wise product and then we are dividing by capital N, which

is the length of the PN sequence this is defined as the this is known as the auto correlation

of the PN sequence, this is the auto correlation of the PN sequence.

Now, let us see what happens? Now, realize also that each c n, each element c n is either

equals to plus or minus 1, each c n is either equal to plus or minus 1. Now, let us see

what happens to this out of the behavior of this auto correlation function of this PN

sequences is a function of the shift d. Now, if d is equal to 0, if d is equal to

0 then my auto correlation is 1 over N summation n c n c n minus d, but d is equal to 0, so

auto correlation becomes 1 over N summation over n c n into d equal to 0. So, this is

simply c n into c n, but c n into c n is c n square which is summation over n c square

of N, but c n is plus or minus 1 which means c square of n is 1. So, this is simply 1 over

N summation over n of 1. So, I will write the reasons since c n equals plus or minus

1 c square n is equal to 1 summation 1 over N times 1, which is equal to 1 over N summation

1, N times is equal to N, so this is equal to 1. So, if d is equal to 0 the auto correlation

of this PN sequence is equal to 1. So, if d is equal to 0 that is the auto correlation

of the self correlation of this PN sequence is equal to 1. Now, let us examine what happens

if d is not equal to 0 for instance let us take a simple example. Let us consider d is equal to 2, let us see

what happens when d is equal to 2, let us circularly shift this PN sequence by 2 symbols

and let us see what the value of the auto correlation is going to be.

So, I am going to write this sequence. So, I have c of n which is 1, 1, 1, 1, minus 1,

minus 1, minus 1 or by PN sequence is basically c n is basically minus 1, minus 1, minus 1,

minus 1, 1, 1, 1, minus 1, 1, 1, minus 1, minus 1, 1, minus 1, 1.. Now, I am going to

circularly shift this by 2. So, I am taking this PN sequence and I am circularly shifting

this PN sequence by 2 symbols. So, I am considering a shift of d is equal to 2 therefore, this

will become minus 1, minus 1, minus 1, minus 1, 1, 1, 1, minus 1, 1, 1, minus 1, minus

1, 1, minus 1 and this last symbols will be will come to the front because this is a circular

shift. So, this is this, so I have the sequence I have the circularly shifted sequence.

Now, I have to take the element wise multiplication and I have to take the sum which is going

to be minus 1 into minus 1 is plus 1 plus minus 1 into 1 is minus 1. So, minus 1 and

calculating this way plus 1 plus 1 minus 1 minus 1 plus 1 minus 1 plus 1 minus 1 minus

1 minus 1 minus 1 plus 1 plus 1. And now, if you take the sum of all these things you

will see I have 1 minus 1 is 0 plus 2 minus 2 is 0 1 minus 1 is 0 1 minus 1 is 0 minus

3 plus 2 at this sum is minus 1 the element wise sum is minus 1 and now, dividing by N

therefore, the auto correlation now, I am taking the element wise sum and now I have

to divide by N. So, therefore, the final auto correlation will be equal to minus 1 by N,

which in this case N is equal to 15. So, this is minus 1 by 15. and what we are observing

is something very interesting, we are observing that if the auto correlation, if d shift is

equal to 0 then the auto correlation is equal to 1 for any shift other than 0 the auto correlation

decreases drastically it is minus 1 divided by N. So, the auto correlation property can be summarized

as follows the auto correlation property of the PN sequence.

If d equal to 0 auto correlation is equal to, if d is not equal to 0 autocorrelation

is equal to minus 1 by n. So, for any shift non-zero, for is 0 shift the auto correlation

is 1 for a non-zero shift auto correlation is minus 1 over N. We had seen an example

of this by considering a shift of d is equal to 2. Therefore, what you can see something

very interesting as soon as the shift moves away from 0 the auto correlation decreases

drastically, it becomes minus 1 over N which means as N increases. Remember N is the length

of the spreading sequence of the PN sequence, if N is large then 1 over N is small which

means 1 over N becomes progressively is approximately equal to 0 for larger, which means the auto

correlation of this PN sequence for any shift other than the 0 shift is approximately equal

to 0 for a PN sequence which a large length N which is a very desirable property as we

are going to see shortly or in subsequent modules.

So, what happens is this N, this approximately equal to 0 for large N or the PN sequence

decreases as 1 over N with large N. Therefore I can plot this auto correlation,

The auto correlation function of the PN sequence it looks like this that is at 0 this is equal

to 1, but away from 0 what happens is this auto correlation it was sharply and this is

equal to minus 1 by N. So, the moment I have any other shift other than the 0 shift,

this auto correlation fall sharply. So, this is the auto correlation property of the PN

sequence this is the plot of the auto correlation. This is the plot of the auto correlation of

the PN sequence. So, for a 0 shift the auto correlation, so

this auto correlation equal to 1 for a 0 shift and for any non-zero shift, for any shift

other than 0, it falls to 1 by N which is approximately equal to 0 for

a large N. So, the auto correlation property of the PN sequence relates that the correlation

is equal to 1 or the self correlation is equal to 1 for the 0 shift, but for any non-zero

shift the auto correlation is minus 1 over N which is approximately equal to 0 for large

N and these are the salient properties of the PN sequences which are used in a CDMA

system. One is the balance property, two is the run length property and third the most

important property is the auto correlation property, which we are going to see is will

help us in extracting the multipath components in the wireless channel in a CDMA system and

that gives raise the multipath diversity and this is a very important property of the PN

sequences in a CDMA system. So, this module which deals with the properties

of the PN sequences we will stop here and continue with other aspects of the CDMA systems

in subsequent modules. Thank you very much.

@ 6:42 Last symbol of Modulated Sequence should be 1 as you are mapping 0 –> 1

Watch at 1.25x

Run properties at 13:06

auto correlation at 21:47

skip to 7:05

u are ultimate sir

good job

lot of errors here. For balance property the no. of 1s is always more than the 0s therefor if we have 4 shift reg. then the total sequence would be 2*m – 1 which would be 15. therefor the no. of 1s would be 8 and no. of zeros(or -1s) would be 7. only look for videos if you have doubt or your whole concept would be ruined.

Why 1logic is coded by -1V instead of +1V.