what exactly norm(x) function do?

Good Afternoon,

I have to convert a part of my matlab code into C++.

I would be great if you can explain me what exactly norm(x) do on a vector of complex number so that I can write a C++ code for it.

I just asked a similar question on stack overflow but could not get much response.

https://stackoverflow.com/questions/73807235/nomalization-of-complex-number-in-c?noredirect=1#comment130326770_73807235

Thank you in advance.

[Adam Danz]

The doc page for norm gives you the forumula :)

See more info section.

[Manu Chaudhary]

Thank you Adam.

[Steven Lord]

Are you performing the conversion to C++ manually or using MATLAB Coder? According to the Extended Capabilities section on its documentation page you can automatically generate C or C++ code from the norm function using MATLAB Coder.

[Manu Chaudhary]

Hi Steven,

Thank you updating me regarding the MATLAB coder. I will try to use this feature of MATLAB.

[Manu Chaudhary]

The formula of norm(x) is defined as above on your website. Just for confirmation, |vk| means a^2 + b^2?

Answer 1

[James Tursa]

The equivalent function to norm( ) for a complex column vector x is

sqrt(x' * x)

where x' is the complex conjugate transpose. Essentially sqrt(dot(x,x)). Note that dot( ) does the complex conjugate part of the calculation automatically.

E.g.,

x = [1;2;3;4] + [5;6;7;8]*1i

x = 
   1.0000 + 5.0000i
   2.0000 + 6.0000i
   3.0000 + 7.0000i
   4.0000 + 8.0000i

norm(x)

ans = 14.2829

sqrt(x'*x)

ans = 14.2829

sqrt(dot(x,x))

ans = 14.2829

So in C/C++ just run a loop to sum up the individual element-by-element multiplies for the dot product result. For an element of the form a+b*i you will be summing a^2+b^2 and then taking the sqrt( ) of the final sum.

[Manu Chaudhary]

int sum=0;
for(int i=0;i < data_size; i++){
    // real_number and img_number are the vectors
    
    sum= pow(real_number[i],2)+ pow(img_number[i],2);  
    }
    
    float normalized_value= sqrt(sum);
    

I think, this is the algorithm for norm(x) in c/c++ ?

[James Tursa]

You need to do

sum = sum + pow(etc) + pow(etc);

Also I would advise doing all of the arithmetic in double.

[Manu Chaudhary]

Thank you James. I think now the conversion of norm x in c++ is correct.

int sum=0;
for(int i=0;i < data_size_val; i++){
    sum= sum +pow(real_number[i],2)+ pow(img_number[i],2);  // Need some correction Here
    }
    
    float normalized_value= sqrt(sum);
    std::cout << "The normalized value" << normalized_value << std::endl;

[James Tursa]

Yes this algorithm is correct, but I still advise changing your code to do the arithmetic in double to avoid potential pitfalls where the int calculations might overflow.

Answer 2

[Matt J]

Assuming you only care about p=2, If you already have code that computes norm(x) for real x, you can extend it to complex x via,

norm(x) = norm( [norm(real(x)), norm(imag(x)) ] )

which is easily verified below,

x=complex(rand(5,1), rand(5,1));

norm(x)

ans = 1.2548

norm( [norm(real(x)), norm(imag(x)) ] )

ans = 1.2548

Alternatively, if you have an implementation of abs, you could do norm(abs(x),p) which will work for any p-norm:

p=3;
norm(x,p)

ans = 1.0042

norm(abs(x),p)

ans = 1.0042