PSNR of image using Matlab

Question

I want to find out PSNR and SNR values of two images A and B of same dimension in Matlab.I used the following code

[peaksnr, snr] = psnr(A,B)

and getting an error

Undefined function 'psnr' for input arguments of type 'uint8'.

Then I converted both the images into double as follows

A = double(A);
B = double(B);

and again PSNR is calculated.Now I'm getting an error

Undefined function 'psnr' for input arguments of type 'double'.

Why this is happening?Please help me to troubleshoot the error.

What version of Matlab do you use? `psnr` seems to a be a recent addition — Luis Mendo, Oct 06 '14 at 08:52
Personally, I think this is a better way to compute PSNR for images: http://stackoverflow.com/a/16265510/3250829 — rayryeng, Oct 23 '14 at 20:15

score 4 · Accepted Answer · answered Oct 06 '14 at 09:14

4

         function PSNR = psnr(distImg, origImg)

            origImg = double(origImg);
            distImg = double(distImg);

            [M N] = size(origImg);
            error = origImg - distImg;
            MSE = sum(sum(error .* error)) / (M * N);

            if(MSE > 0)
            PSNR = 10*log(255*255/MSE) / log(10);
            else
            PSNR = 99;
            end 

        end

answered Oct 06 '14 at 09:14

omid

400
3
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Why do you assign a `PSNR` of 99 if the MSE is negative? By definition, the mean **squared** error will never be negative. Also the `log(10)` division is incorrect. Consult the definition of the PSNR here: http://en.wikipedia.org/wiki/Peak_signal-to-noise_ratio – rayryeng Oct 23 '14 at 20:13
Hi in matlab when error occurs MSE is negative thus PSNR=99 is maximum. Also base of log function in matlab is 2 and this formula is correct.Because log(a)/log(b)=log(a) in base b. – omid Nov 16 '14 at 08:51
Ah yes, ok. What you should do is use `log10`. It avoids having to divide in order to get it into the right base. However, the MSE will **NEVER** be negative. Sum of squared errors was created to circumvent any negative deviation, so that I don't agree with. The arbitrary 99 is not correct either IMHO. I looked at the PSNR source in MATLAB, and there is no -99 assignment anywhere. – rayryeng Nov 16 '14 at 16:58
BTW the base of the log function is **e**, not 2. `log2` is the function that is base 2. Please read up about MATLAB internals before answering any other questions. – rayryeng Nov 17 '14 at 15:03

score 0 · Answer 2 · edited Apr 06 '16 at 14:22

0

psnr = 10*log10(255/sqrt(mean((image1(:) - image2(:)).^2)))

edited Apr 06 '16 at 14:22

CubeJockey

2,209
8
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answered Apr 06 '16 at 12:35

Marco

1
1

score 0 · Answer 3 · answered May 13 '16 at 09:02

There is some mistakes with above code where it should consider log10 in Matlab. So the correct code (even for color image) is as below:

function PSNR = MyPsnr(distImg, origImg)
        origImg = double(origImg);
        distImg = double(distImg);

        [M N P] = size(origImg);
        error = origImg - distImg;
        MSE = sum(sum(sum(error.^2))) / (M * N * P);

        if(MSE > 0)
            PSNR = 20*log10(max(max(max(origImg))))-10*log10(MSE);
        else
            PSNR = 99;
        end

what's differnce of log10(2)/log10(8) with log2(2)/log2(8) – omid Nov 25 '16 at 07:27 — omid, Nov 25 '16 at 07:27

PSNR of image using Matlab

3 Answers3