Calculating the Bytes Per Pixel and BitRate of a JPEG Image

Hello.

Graduated with an Electrical Engineering degree in 1999. Decided to purse a part-time MBA this year and was given an option to do an Engineering Elective with a partner University and completely getting my behind kicked as a lot has changed in the two decades since I graduated. Never had Matlab when I graduated and my focus back then was Electrical Motors, so this is all very very new to me. Everything I've learnt is from Google and the MatLab documentation (I travel a lot for work and therefore don't have much on-campus resources), so my apologies if I've even gotten the basics wrong.

For the first part of the assignment (four parts total), I have to "read" and image into Matlab, determine the size of the matrix, calculate the uncompressed image size, save it as a JPEG with varying quality factors, calculate compression ratios, etc. etc. for a total of 15 sub-tasks. I'm using the image file posted here: https://imgur.com/WsEAGHf

Does the code I have so far appear correct? I've tried to comment on what each function is attempting to accomplish.

%Reading image into a matrix called LENA in Matlab
LENA = imread('lena-256x256.jpg');

%Determining number of rows and columns of the matrix
[LENA_X, LENA_Y] = size(LENA);

%Calculating uncompressed image size assuming each pixel is stored in one byte
LENA_uncompressed_image_size = LENA_X * LENA_Y;

%Saving the image with different compression ratios of 95, 60, 30 and 5
imwrite(LENA, 'LENA_95.jpg', 'jpg', 'Quality', 95);
imwrite(LENA, 'LENA_60.jpg', 'jpg', 'Quality', 60);
imwrite(LENA, 'LENA_30.jpg', 'jpg', 'Quality', 30);
imwrite(LENA, 'LENA_5.jpg', 'jpg', 'Quality', 5);

Here is what is stumping me. I am supposed to calculate the bit-rate of each compressed file using the formula BitRate = (FileSize / PixelCount). I'm unable to figure out how to read the file size of each image into Matlab. Additionally, they have stated that the original uncompressed file has a bit-rate of 8. Using the formula they have provided, and just using the "Size" in Windows as the file size (45577 bytes) and the pixel count (256x256 = 65536) I have no idea how to get 8 bits-per-pixel.

Any help is appreciated.

Thanks, Steve.

[KALYAN ACHARJYA]

*Graduated with an Electrical Engineerin degree in 1999. Decided to purse a part MatLab documentation (I travel a lot for work and ..............................gotten the basics wrong.

Please note, nowadays you can learn easily,. all resources are available in the fingertip, use it and explore the learning.

Good Wishes!

Answer 1

[Image Analyst]

I wouldn't get hung up on the term "rate" - there is no time(s) involved here. I would just compute the compression ratio from the size in memory divided by the size on disk. Here's a start:

fileInfo = dir('*.jpg')
bytesOnDisk = [fileInfo.bytes]
% Let's list them all and also compute compression ratio.
for k = 1 : length(fileInfo)
  thisName = fullfile(fileInfo(k).folder, fileInfo(k).name);
  imageInfo = imfinfo(thisName);
  imageSize = imageInfo.Width * imageInfo.Height * imageInfo.BitDepth/8;
  fprintf('The size of %s in memory is %d bytes, the disk size is %d\n',...
    fileInfo(k).name, imageSize, imageInfo.FileSize);
  % Now compute compression ratio...
end

[whatchamacallit]

Thanks! I'm glad to see you read what I was asking. Following your steps (but ignoring the loop as that is way above my understanding for now), I came up with this code for calculating the compression ratio and bitrate. The assignment hints that we have to use the compression ratio and the uncompressed file bitrate, we have to calculate the bitrates of the compressed file sizes. So this is what I think they want:

%Reading image disk file sizes
LENA_Uncompressed_FileInfo = imfinfo('lena256.jpg');
LENA_95_FileInfo = imfinfo('LENA_95.jpg');
LENA_60_FileInfo = imfinfo('LENA_60.jpg');
LENA_30_FileInfo = imfinfo('LENA_30.jpg');
LENA_05_FileInfo = imfinfo('LENA_05.jpg');

%Computing Compression Ratios
LENA_95_Compression_Ratio = LENA_Uncompressed_FileInfo.FileSize / 
LENA_95_FileInfo.FileSize;
LENA_60_Compression_Ratio = LENA_Uncompressed_FileInfo.FileSize / 
LENA_60_FileInfo.FileSize;
LENA_30_Compression_Ratio = LENA_Uncompressed_FileInfo.FileSize / 
LENA_30_FileInfo.FileSize;
LENA_05_Compression_Ratio = LENA_Uncompressed_FileInfo.FileSize / 
LENA_05_FileInfo.FileSize;

%Computing bitrates using the compression ratio and the bitrate of the 
uncompressed file
bitrate_uncompressed_file = 8;
LENA_95_Bitrate = bitrate_uncompressed_file / 
LENA_95_Compression_Ratio;
LENA_60_Bitrate = bitrate_uncompressed_file / 
LENA_60_Compression_Ratio;
LENA_30_Bitrate = bitrate_uncompressed_file / 
LENA_30_Compression_Ratio;
LENA_05_Bitrate = bitrate_uncompressed_file / 
LENA_05_Compression_Ratio;

Is this how you would calculate the bitrates of the compressed files?

Thanks

[Guillaume]

Well according to your definition of a bit rate, it is FileInfo.FileSiize divided by the size of the image so

%FileInfo: the structure returned by imfinfo
bitrate = FileFinfo.FileSize / (FileInfo.Width * FileInfo.Height)

not at all what you've calculated.

Calling that a bit rate is very misleading but if that's what they ask...

You're no longer using size to find the size of the image, which is good. Note that your original:

someimage = imread('somefile');
[height, width] = size(someimage);

will only give you the correct width if the image matrix is 2D, which will never be the case for colour images.

[height, width, ~] = size(someimage);

is the correct to do it for both colour and greyscale images.

[whatchamacallit]

I totally agree. The reason why the formula they have provided doesn't work is because the original uncompressed filesize is 45577 bytes and the number of pixels is 65536, which gives us 5.56 bits per pixel, and not the 8. Hence I have no idea what they are doing, so I'm going with the second option which is using the compression ratio to figure out bit rate. I have no idea if this correct or completely wrong, but since I'm doing this remotely I have no one to really ask on campus.

And you correct btw, this is only grayscale, which is for the first part of this course. The image in question is here: https://imgur.com/WsEAGHf

Thanks.

[Image Analyst]

Yes, good to see that you're now using the file size from the directory into and not from the way you used size() originally like Guillaume said. It's a common beginner mistake, so much so that Steve Eddins wrote a blog on it: https://blogs.mathworks.com/steve/2011/03/22/too-much-information-about-the-size-function/

I also agree that bit rate is a bad term. Unless you know the time over which the image was transmitted, how can you know a rate? Maybe they mean bpp, bits per pixel, like if the original RGB image was 24 bits per pixel (8 bits for Red, 8 bits for green, and 8 bits for blue), and now it's 14.3 or whatever.

Answer 2

[KALYAN ACHARJYA]

Use dir() and check the file size in return
And find rows and column of the respective write image, row*com gives the total pixel counts

Answer 3

[Masoumeh Farhadi Nia]

Hello. I need to gain the bitrate for a bulk of video sequences comprssed utilizing different codecs and crfs. I just need to know the correct form of bitrate calculation utilizing videoinfo files in matlab. Is there an body can help?