what exactly norm(x) function do?

2022 9 21
2 个回答
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更新时间：2022 9 22
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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.

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Adam Danz
Adam Danz 2022-9-21
移动：Adam Danz 2022-9-22
The doc page for norm gives you the forumula :)
See more info section.
Manu Chaudhary
Manu Chaudhary 2022-9-21
移动：Adam Danz 2022-9-22
Thank you Adam.
Steven Lord
Steven Lord 2022-9-22
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
Manu Chaudhary 2022-9-22
Hi Steven,
Thank you updating me regarding the MATLAB coder. I will try to use this feature of MATLAB.
Manu Chaudhary
Manu Chaudhary 2022-9-22
编辑：Manu Chaudhary 2022-9-22
The formula of norm(x) is defined as above on your website. Just for confirmation, |vk| means a^2 + b^2?

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 采纳的回答

James Tursa
James Tursa 2022-9-21
编辑：James Tursa 2022-9-21
Ran in:

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Ran in:
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.

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Manu Chaudhary
Manu Chaudhary 2022-9-22
编辑：Manu Chaudhary 2022-9-22
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
James Tursa 2022-9-22
编辑：James Tursa 2022-9-22
You need to do
sum = sum + pow(etc) + pow(etc);
Also I would advise doing all of the arithmetic in double.
Manu Chaudhary
Manu Chaudhary 2022-9-22
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
James Tursa 2022-9-22
编辑：James Tursa 2022-9-22
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.

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更多回答（1 个）

Matt J
Matt J 2022-9-22
编辑：Matt J 2022-9-22
Ran in:

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Ran in:
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

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