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Date: Tue, 31 Dec 2013 13:59:11 -0600
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Subject: Re: [geda-user] Help sending a sine wave to a speaker
From: Ozzy Lash <ozzy DOT lash AT gmail DOT com>
To: geda-user AT delorie DOT com
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On Tue, Dec 31, 2013 at 1:26 PM, Rob Butts <r DOT butts2 AT gmail DOT com> wrote:

> It has been a while since I've written major C code so I'm having a hard
> time coming up with the code.
> I have a simple design where I use a microchip pic16f1825 to read a byte
> from an eeprom via spi and send it to a dac and ultimately to a speaker.
> As a dac and speaker test I want to send a simple 1000Hz sine wave to the
> dac at an 8KHz rate.
>
> Can anyone check my thoughts below?  Here's my thinking...
> If I set my timer to interrupt at an 8KHz rate and if I want to play the
> tone for 10 seconds the for loop might look something like this:
>
> for(j=0; j < 80000; j++){
>      // j is the sample index
>     // straight sine function means one cycle per 2*pi samples
>     // sample(j) = sin(j)
>     // multiply by 2*pi now it is one cycle per sample
>     // sample(j) = sin((2 * pi) * j)
>     // multiply by 1,000 samples per second now it is cycles per second
>     // sample(j) = sin(1000 * 2 * pi * j)
>     // divide by 8000 samples per second now it is cycles per 8000 samples
>     // sample(j) = sin(1000 * 2 * pi * j / 8000)
>     // the 1000 cycles is the frequency so in those terms
>     //  sample(j) = sin(freq * 2 * pi * j / 8000)
>     // now if I want to normalize the amplitude to 0xff  the final value
> is
>     // sample(j) = 0xff * sin(freq * 2 * pi * j / 8000)
>     // so...    declaring sample as a float
>     sample = 0xff * sin(freq * 2 * pi * j / 8000);
>     while(!update_dac){    // where update_dac is set to 1 in the timer
> interrupt handler
>     }
>     update_ dac = 0;
>     send_dac((UINT8) sample);
> }
>
> When I execute the code and step through it the value for sample when j =
> 1 is 180.238where I expect 3.146
>
> Can someone see my error or what is going on?
>
> Thanks
>

If freq=1000 and j=1, then sample is 255*sin(pi/4).  sin(pi/4)=sqrt(2)/2,
so sample should be about 180

Did you do the sin on a calculator and have it in degrees mode instead of
radians?

Bill

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<div dir=3D"ltr"><div class=3D"gmail_extra"><div class=3D"gmail_quote">On T=
ue, Dec 31, 2013 at 1:26 PM, Rob Butts <span dir=3D"ltr">&lt;<a href=3D"mai=
lto:r DOT butts2 AT gmail DOT com" target=3D"_blank">r DOT butts2 AT gmail DOT com</a>&gt;</span>=
 wrote:<br>
<blockquote class=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1p=
x #ccc solid;padding-left:1ex"><div dir=3D"ltr"><div>It has been a while si=
nce I&#39;ve written major C code so I&#39;m having a hard time=20
coming up with the code. <br></div><div>I have a simple design where I use =
a microchip=A0pic16f1825 to=20
read a byte from an eeprom via spi and send it to a dac and ultimately to a=
=20
speaker.=A0 As a dac and speaker test I want to send a simple 1000Hz sine w=
ave to=20
the dac at an 8KHz rate.=A0<br>=A0=A0=A0 <br>Can anyone=A0check my thoughts=
 below?=A0 Here&#39;s my thinking... <br>If I=20
set my timer to interrupt at an 8KHz rate and if I want to play the tone fo=
r 10=20
seconds the for loop might look something like this: <br>=A0 <br>for(j=3D0;=
=A0j &lt;=20
80000; j++){ <br>=A0=A0=A0=A0 //=A0j is the sample index <br>=A0=A0=A0 // s=
traight sine function=20
means one cycle per 2*pi samples <br>=A0=A0=A0 //=A0sample(j) =3D sin(j) <b=
r>=A0=A0=A0 //=20
multiply by 2*pi now it is one cycle per sample <br>=A0=A0=A0 // sample(j) =
=3D sin((2 *=20
pi) * j) <br>=A0=A0=A0 // multiply by 1,000 samples per second now it is cy=
cles per=20
second <br>=A0=A0=A0 // sample(j) =3D sin(1000 * 2 * pi * j) <br>=A0=A0=A0 =
// divide by 8000=20
samples per second now it is cycles per 8000 samples <br>=A0=A0=A0 // sampl=
e(j) =3D=20
sin(1000 * 2 * pi * j / 8000) <br>=A0=A0=A0 // the 1000 cycles is the frequ=
ency so in=20
those terms <br>=A0=A0=A0 //=A0 sample(j) =3D sin(freq * 2 * pi * j / 8000)=
 <br>=A0=A0=A0 // now=20
if I want to normalize the amplitude to 0xff=A0 the final value is <br>=A0=
=A0=A0 //=20
sample(j) =3D 0xff * sin(freq * 2 * pi * j / 8000) <br>=A0=A0=A0 // so...=
=A0=A0=A0=A0declaring=20
sample as a float <br>=A0=A0=A0 sample =3D 0xff * sin(freq * 2 * pi * j / 8=
000); <br>=A0=A0=A0=20
while(!update_dac){=A0=A0=A0 // where update_dac is set to 1 in the timer i=
nterrupt=20
handler <br>=A0=A0=A0 } <br>=A0=A0=A0 update_ dac =3D 0; <br>=A0=A0=A0 send=
_dac((UINT8) sample);=20
<br>}=A0<br>=A0 <br>When I execute the code and step through it the value f=
or sample=A0when j =3D 1 is 180.238where I expect 3.146</div><div>=A0</div>=
<div>Can someone see my error or what is going on?</div><div>=A0</div><div>=
Thanks</div>

</div>
</blockquote></div><br></div><div class=3D"gmail_extra">If freq=3D1000 and =
j=3D1, then sample is 255*sin(pi/4).=A0 sin(pi/4)=3Dsqrt(2)/2, so sample sh=
ould be about 180<br><br></div><div class=3D"gmail_extra">Did you do the si=
n on a calculator and have it in degrees mode instead of radians?<br>
<br></div><div class=3D"gmail_extra">Bill<br></div></div>

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