Project 5: Audio Compression

Aidan Curran

Background
List of Code used
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5

Background

Reality Check 11 Chapter 11 Numerical Analysis Problems 1-5

List of Code used

Initial Code from textbook: Simple Audio Codec

Code for part 1: Audio Compression Adapted from Simple Codec

Code for part 2: Audio Compression for chords

Code for part 2: Compression Portion for chords

Code for part 3: Compression using a windowing function

Code for part 3: Compression for chords w/ a windowing function

Code for part 3: Compression Portion for chords 2

Code for part 5: Audio Compression for incoming audio files

Problem 1

Investigate the ability of MDCT to represent puretones. Begin with $b = 4$ bits per window of size $n = 32$ . Pick a tone of frequency between 100 Hz and 1000 Hz, and calculate the difference (as RMSE) between the original signal and the signal after encoding/decoding. You should cut the original signal to $xshort = x(n+1: end-n);$ for comparison with the output signal, since the latter lacks n entries at the left and right ends. Plot a short section of the original and decoded signal.

Here is the code used for this part: Audio Comp 1

     The frequency I picked for this part is 256 Hz

     out=audioComp1(cos((1:2^(12))*2*pi*(FREQUENCY)/2^(13)));

out=audioComp1(cos((1:2^(12))*2*pi*256/2^(13)));

RMSE =
    0.0091

Initial Audio

Compressed Audio

Problem 2

Build chords and evaluate the RMSE as in Step 1. Simple intervals can be constructed by a simple addition of multiple pure tones. Rational ratios of frequencies with low numerators and denominators are pleasing to the ear: A 2 : 1 ratio of frequencies gives an octave, 1.25 : 1 ratio gives a third, a 1.5 : 1 gives a fifth, and so forth. How does the RMSE depend on the number of bits used in the coder?

Here is the code used for this part: Chords

Here is the code used for this part: Audio Comp 2

chords(256,2,1)

RMSE =
    0.0092

Initial Audio

Compressed Audio

chords(256,1.25,1)

RMSE =
    0.0109

Initial Audio

Compressed Audio

chords(256,1.5,1)

RMSE =
    0.0069

Initial Audio

Compressed Audio

     With a higher bit quantization, the RMSE is gets smaller and smaller.
     It also decreases the number of higher and lower frequency bits being
     removed or adjusted.

Problem 3

A "windowing function" is often used to reduce codec error, due to the fact that the function being represented is not periodic over the window, but is being represented by periodic functions. The windowing function scales the input signal $x$ smoothly to zero at each end of the window, partially mitigating this problem. A common choice is to replace $x_{j}$ with $x_{j}h_{j}$ , where $h_{j}=\sqrt{2}sin\frac{(j-\frac{1}{2})\pi}{2n}$ for a length 2n window, where $j=1,...,2n$ . To undo the windowing function, multiply the inverse MDCT output $w$ componentwise by the same $hj$ . This results in multiplying $w_{2}$ componentwise by the second half of the $h_{j},j=n+1,...,2n$ , and $w_{3}$ by the first half $h_{j},j=1,...,n$ before combining into the decoded signal. Compare RMSE, plots, and audible sound as in Steps 1 and 2.

Here is the code used for this part: Audio Comp 3

Here is the code used for this part: Chords 2

Here is the code used for this part: Audio Comp 4

out=audioComp3(cos((1:2^(12))*2*pi*256/2^(13)));

RMSE =
    0.0011

Initial Audio

Compressed Audio

chords2(256,2,1)

RMSE =
    0.0037

Initial Audio

Compressed Audio

chords2(256,1.25,1)

RMSE =
    0.0048

Initial Audio

Compressed Audio

chords2(256,1.5,1)

RMSE =
    0.0016

Initial Audio

Compressed Audio

Problem 4

Explain the method for undoing the windowing that is suggested in Step 3. In other words, assume that if $Z_{1}$ and $Z_{2}$ are each multiplied componentwise by the entire windowing function $h$ , and $NMZ_{1}$ and $NMZ_{2}$ in equation (11.38) are each multiplied

$h_{j} = \sqrt{2}sin\frac{(j-\frac{1}{2})\pi}{2n}$

$Z_{1}$ = [x1h1 ; x2h2 ; x3h3 ; x4h4]

$Z_{2}$ = [x3h1 ; x4h2 ; x5h3 ; x6h4]

          =
         [ x1h1 - Rx2h2;
           -Rx1h1 + x2h2;
           x3h3 + Rx4h4;
           Rx3h3 + x4h4]

          =
         [ x3h1 - Rx4h2;
           -Rx3h1 + x4h2;
           x5h3 + Rx6h4;
           Rx5h3 + x6h4]

[x3; x4]= 1/2 ( $NMZ_{1}$ )n,...,2n-1 + 1/2 ( $NMZ_{2}$ )0,...,n-1

x_{3}

$\frac{1}{2}x_{3}[h_{3}^2 + h_{1}^2]$

$\frac{1}{2} x_{3} \sqrt{2}^2 sin^2(\frac{1-\frac{1}{2}}{2n})+\sqrt{2}^2cos^2(\frac{1-\frac{1}{2}}{2n})$

$\frac{1}{2}x_{3}2 = x_{3}$

x_{4}

$\frac{1}{2}x_{4}[h_{4}^2 + h_{1}^2]$

$\frac{1}{2} x_{4} \sqrt{2}^2 sin^2(\frac{1-\frac{1}{2}}{2n})+\sqrt{2}^2cos^2(\frac{1-\frac{1}{2}}{2n})$

$\frac{1}{2}x_{4}2 = x_{4}$

Problem 5

Import a .wav file with the MATLAB audioread command, or download an audio file of your choice. (Alternatively, load handel can be used. If you download a stereo file, you will need to work with each channel separately.) Reproduce the file (or a segment of it) using various values of b and with and without windowing. Compute RMSE for your choices of parameters and exhibit the results using the sound command.

Here is the code used for this part: Audio Comp 5

The song that I chose to demonstrate compression is the Tristram Theme from Diablo Soundtrack.

This is what the original song sounds like: