how to understand a example of tigersharc 201 dsp (visual dsp++5.0)?

shedmaze Question

Dec 20, 2012 00:00

hello,

right now i am learning the tigersharc dsp by myself, and i have bought a tigher sharc EZ-kit lite board, totally i has bought 3 boards.

i read an example in the following path: \VisualDSP 5.0\TS\Examples\ADSP-TS201 EZ-KIT Lite\Power_On_Self_Test

in this project, it has a file audio_test.c, and in this file, the following codes confused me, can you help me?? thank you and thank you.

void audio_int( void )
{
int iChannelX_Left;
int iChannelX_Right;
float fChannelX;

iChannelX_Left = ReadDataLeft;
iChannelX_Right = ReadDataRight;

// shift 24-bit codec data left channel
iChannelX_Left <<= 8;
iChannelX_Left >>= 8; // what has it done?? i don't know. it seems there is no change for the iChannelX_Left

// sign extend values
if( iChannelX_Left & 0x800000 )
{
iChannelX_Left |= 0xFF000000; // this part is also, i don't know what are they doing ...?
}

// shift 24-bit codec data right channel
iChannelX_Right <<= 8;
iChannelX_Right >>= 8;

// sign extend values
if( iChannelX_Right & 0x800000 )
{
iChannelX_Right |= 0xFF000000;
}

ts201

power_on_self_test

example

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ADIApproved Reply

Dec 20, 2012 00:00

Hi,

The first instructions perform a bitwise left shift assignment, followed by a bitwise right shift assignment. The aim here is to clear the most significant 8 bits of the data (as the int holds 32-bits, but the codec data is 24-bit).

Then, the code checks whether the sign-bit of this 24-bit value is set (to indicate a negative value). So, a bitwise AND is done with hex value 0x800000 - in binary 10000000 00000000 00000000. If bit 23 (the sign bit of the 240bit value) is set, the number should be negative, so the sign bit in the 32-bit integer that holds the value needs to be set by OR-ing the value with 0xFF000000 (this actually sets all of the Most-Significant 8 bits to 1).

Hope that clarifies the purpose of that particular section: basically, clearing the 8 most significant bits, the setting the sign bit based on bit 23 of the 24-bit codec data.

Regards,
Craig.

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shedmaze Reply

Dec 20, 2012 00:00

hello, Craig,

thanks for your so quickly replying... right now, i am in Korean, the time is 3:55AM, i am still writing a report.

anyway, tomorrow is a holiday~~

for the first question, i am still confused, the ReadDataLeft is a 'long' integer, so it is already 64bits, then it was given to the iChannelX_Left, i think it is not necessary to use the shift moving,can you give me a example for that.

for the dsp communication with the equipment, i can't do it very well. so please be patient.

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shedmaze Reply

Dec 20, 2012 00:00

hello,

the example means that can you explain the process of the following codes, thanks.

// shift 24-bit codec data left channel

iChannelX_Left <<= 8;

iChannelX_Left >>= 8;

best wishes,

SHENG MING

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ADIApproved Reply

Dec 21, 2012 00:00

Hi,

While ReadDataLeft is a 64-bit long, iChannelX_Left is a 32-bit int. The data coming from the Codec is only 24-bits, so there are still 8 bits of 'garbage' in the 8 most significant bits that need to be removed before iChannelX_Left is processed.

For example, if the data coming in is 0x22222222, the left/right shift would set it to 0x00222222. There may, for example, be a value where bit 31 is set - indicating a negative number, while bit 23 is NOT set because the value from the codec was not negative. This needs to be shifted out and corrected.

Regards,

Craig.

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EZAdministrator Reply

Aug 2, 2018 00:00

This question has been assumed as answered either offline via email or with a multi-part answer. This question has now been closed out. If you have an inquiry related to this topic please post a new question in the applicable product forum.

Thank you,
EZ Admin

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