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| − | + | To hear the audio files produced click [[here_ECE301Fall2008mboutin]] | |
| − | == | + | |
| + | == Matlab Code == | ||
<pre> | <pre> | ||
| + | %{ | ||
| + | Homework #1 | ||
| + | Problem 1 | ||
| + | Jeremiah Wise | ||
| + | |||
| + | This Program plays the first portion of the corus "Hail Purdue". | ||
| + | It first plats at a norlal speed then plays 2 times that speed. | ||
| + | On the third trial in is played at the original speed in the next | ||
| + | octave higher. | ||
| + | %} | ||
| + | |||
| + | %Frequencies associated with each note | ||
| + | A = 220; | ||
| + | Bb = 233.082; | ||
| + | B = 246.942; | ||
| + | C = 261.626; | ||
| + | Db = 277.183; | ||
| + | D = 293.665; | ||
| + | Eb = 311.127; | ||
| + | E = 329.628; | ||
| + | F = 349.228; | ||
| + | Gb = 369.994; | ||
| + | G = 391.995; | ||
| + | Ab = 415.305; | ||
| + | |||
| + | %First few notes of the chorus | ||
| + | chorus = [A, B, Db, D, E, Gb, Gb, G, G, G, D, E, F, Gb]; | ||
| + | |||
| + | delta = 0.0005; | ||
| + | |||
| + | %Part a | ||
| + | endTime = 0.4; | ||
| + | t = [0 : delta : endTime]; | ||
| + | |||
| + | for i = 1 : length(chorus) | ||
| + | note = sin(2 * pi * t * chorus(i)); | ||
| + | sound(note, 1/delta); | ||
| + | end | ||
| + | %Part b | ||
| + | endTime = endTime / 2; | ||
| + | t = 0 : delta : endTime; | ||
| − | + | for i = 1 : length(chorus) | |
| + | note = sin(2 * pi * t * chorus(i)); | ||
| + | sound(note, 1/delta); | ||
| + | end | ||
| + | %Part c | ||
| + | endTime = 0.4; | ||
| + | t = 0 : delta : endTime; | ||
| + | for i = 1 : length(chorus) | ||
| + | note = sin(4 * pi * t * chorus(i)); | ||
| + | sound(note, 1/delta); | ||
| + | end | ||
</pre> | </pre> | ||
Latest revision as of 12:54, 4 September 2008
To hear the audio files produced click here_ECE301Fall2008mboutin
Matlab Code
%{
Homework #1
Problem 1
Jeremiah Wise
This Program plays the first portion of the corus "Hail Purdue".
It first plats at a norlal speed then plays 2 times that speed.
On the third trial in is played at the original speed in the next
octave higher.
%}
%Frequencies associated with each note
A = 220;
Bb = 233.082;
B = 246.942;
C = 261.626;
Db = 277.183;
D = 293.665;
Eb = 311.127;
E = 329.628;
F = 349.228;
Gb = 369.994;
G = 391.995;
Ab = 415.305;
%First few notes of the chorus
chorus = [A, B, Db, D, E, Gb, Gb, G, G, G, D, E, F, Gb];
delta = 0.0005;
%Part a
endTime = 0.4;
t = [0 : delta : endTime];
for i = 1 : length(chorus)
note = sin(2 * pi * t * chorus(i));
sound(note, 1/delta);
end
%Part b
endTime = endTime / 2;
t = 0 : delta : endTime;
for i = 1 : length(chorus)
note = sin(2 * pi * t * chorus(i));
sound(note, 1/delta);
end
%Part c
endTime = 0.4;
t = 0 : delta : endTime;
for i = 1 : length(chorus)
note = sin(4 * pi * t * chorus(i));
sound(note, 1/delta);
end
