How to implement the given equation in MATLAB?

I am studying the following equation:

here is vector of QPSK symbols, belongs to , is a random complex number, is vector and n ranges from 1 to N. Also note that I have all these values.

My query is how to implement this equation (1) in MATLAB.

Any help in this regard will be highly appreciated.

[chaaru datta]

Any help pls .... I am trying it but not getting it correctly...

[Sam Chak]

Hi @chaaru datta

It is important to understand that is NOT an equation. It is more like a problem to be solved.

Mathematically, it implies the value of x for which attains its minimum.

In your case, you should find the solution that is in the set of values .

I understand that this kind of problem can be difficult to understand for people who are not familiar with the math notations. I took a long time to get familiar with them because I don't have a Math Degree and they are often found in academic journal papers involving mathematical optimization problems.

Would suggest you to type out the values of all paramaters.

Not many people understand that QPSK stands for Quadrature Phase Shift Keying.

[chaaru datta]

Thank you so much sir for your reply..... I am sharing the values of all parameters by assuming n to be ranging from 1 to 8.

.

Answer 1

[Sam Chak]

Thanks @chaaru datta. But the data you shared are non-copyable because they are images.

You can click the MATLAB code button and dump the data there.

Anyhow, you can try this simple method to find the value that gives the minimum. I copied only 3 sets of data pairs only, but you should get the idea how to do it.

Psi = -0.06-1.16i;
ym  = [-0.41+0.41i 0.54-0.58i -0.62-0.71i]
sm  = [0.70+0.70i -0.70-0.70i -0.70+0.70i]
fcn = (abs(Psi*ym - sm)).^2
[fmin, idx] = min(fcn)
sm(idx)

[chaaru datta]

Thanks sir for your efforts and assistance....now by running your sample code, I found that -0.70 + 0.70i gives the minimum.

"So it means now will be 1x8 vector containing -0.70 + 0.70i inside it on all 8 different indexes."

Is my understanding mentioned in double quotes correct sir....

[chaaru datta]

@Sam Chak Can you please confirm it sir....

[Sam Chak]

Hi @chaaru datta, I didn't count how many data you have because they are images. Have requested you supply them, but you didn't. I just took the first 3 data points and demonstrated how to find the minimum. If you have 8 data points, there you go...

[chaaru datta]

Thanks a lot sir for resolving my query....

Here again I am submitting the data....

Psi = -0.06-1.16i;

ym = [-0.41+0.41i, 0.54-0.58i, -0.62-0.71i, 0.64+0.56i, -0.70+0.75i, 0.68-0.60i, -0.54-0.55i, 0.43+0.79i];

sm = [0.70+0.70i, -0.70-0.70i, -0.70+0.70i, 0.70-0.70i,0.70+0.70i, -0.70-0.70i, -0.70+0.70i, 0.70-0.70i];

Thanks once again for your help sir...

[Sam Chak]

Please note that sm has some repeated values.

Psi = -0.06-1.16i;
ym = [-0.41+0.41i, 0.54-0.58i, -0.62-0.71i, 0.64+0.56i, -0.70+0.75i, 0.68-0.60i, -0.54-0.55i, 0.43+0.79i];
sm = [0.70+0.70i, -0.70-0.70i, -0.70+0.70i, 0.70-0.70i, 0.70+0.70i, -0.70-0.70i, -0.70+0.70i, 0.70-0.70i];
fcn = (abs(Psi*ym - sm)).^2
[fmin, idx] = min(fcn)
sm(idx)

[chaaru datta]

Yes sir ...."sm" has some repeated values because its a QPSK signal and hence first four values are unique and after that same 4 values will be repeated.