Difference between revisions of "HW1.4 Satya Hegde ECE301Summer2009" - Rhea

Latest revision as of 20:01, 23 June 2009

%Satya Hegde %HW 1, Problem 4 Part (a)

%A has a frequency of 440 Hz. Using the table provided, note A has a %frquency of 5x/3, where x is the frequency of middle C. %Thus 5x/3 = 440, yielding %x = 264Hz

x = 264;

del = 0.0001;

t = 0:del:1;  %definition of time vector

%Definition of Notes based on frequencies

C = sin(2*pi*x.*t);

E = sin(2*pi*(5*x/4).*t);

G = sin(2*pi*(3*x/2).*t);

%Lower notes each have a frequency equalling half their higher-octave %counterpart (ex.: G = 3x/2, so Gl = 3x/4)

Gl = sin(2*pi*(3*x/4).*t);

El = sin(2*pi*(5*x/8).*t);

Al = sin(2*pi*(5*x/6).*t);

Als = sin(2*pi*233.08.*t);  %Given

Bl = sin(2*pi*(15*x/16).*t);

%Playing of Mario Brothers

sound(E,1/del)

sound(E,1/del)

sound(E,1/del)

sound(G,1/del)

sound(C,1/del)

sound(E,1/del)

sound(G,1/del)

sound(C,1/del)

sound(Gl,1/del)

sound(El,1/del)

sound(Al,1/del)

sound(Bl,1/del)

sound(Als,1/del)

sound(Al,1/del)