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Hi, I am kind of new to this ADSP.... I am working on the ADSP BF533 EZ-kit lite example "Audio talk through" ..... In the C code they have set the SLEN$,since the codec(AD1836)sends 24 bits.... but then the DMA transfer is32 bit.... i don't understand this... Also... in the chapter Sport,while talking adt this SLEN..tells that the DMA can dump 32 bit to the TX buffer of the SPORT of which,the SLEN (in this case 24bits) bits are transfered thor' the sport... If it does like that,when r the remaining (32-24=8 bits) 8 bits transfered thro' the sport? Please i am in a pathetic stage.. now at my work.. so kindly help me to understand this....i will be very grateful... thanking u all in advance and eagerly waiting for the early reply from the expertiese..... ________________________________________________________________________ Yahoo! India Matrimony: Find your life partner online Go to: http://yahoo.shaadi.com/india-matrimony |
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adt Sport and DMA
Started by ●October 25, 2004
Reply by ●October 25, 20042004-10-25
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Someone more knowledgeable should correct any errors you see here. Even though the data from the codec is 24-bits, data inside the DSP is transferred in 32-bit words. The SLEN parameter indicates how many clocks are needed to transfer one codec word, which is then placed in 32-bit memory. At 08:42 AM 10/25/2004, shwetha shwetha wrote: >Hi, > >I am kind of new to this ADSP.... > >I am working on the ADSP BF533 EZ-kit lite example >"Audio talk through" ..... > >In the C code they have set the SLEN$,since the >codec(AD1836)sends 24 bits.... > >but then the DMA transfer is32 bit.... > >i don't understand this... > >Also... in the chapter Sport,while talking adt this >SLEN..tells that the DMA can dump 32 bit to the TX >buffer of the SPORT of which,the SLEN (in this case >24bits) bits are transfered thor' the sport... > >If it does like that,when r the remaining (32-24=8 >bits) 8 bits transfered thro' the sport? > >Please i am in a pathetic stage.. now at my work.. so >kindly help me to understand this....i will be very >grateful... > >thanking u all in advance and eagerly waiting for the >early reply from the expertiese..... > >________________________________________________________________________ >Yahoo! India Matrimony: Find your life partner online >Go to: http://yahoo.shaadi.com/india-matrimony >_____________________________________ >Note: If you do a simple "reply" with your email client, only the author >of this message will receive your answer. You need to do a "reply all" if >you want your answer to be distributed to the entire group. > >_____________________________________ >About this discussion group: > >To Join: Send an email to > >To Post: Send an email to > >To Leave: Send an email to > >Archives: http://groups.yahoo.com/group/adsp > >Other Groups: http://www.dsprelated.com/groups.php3 > >Yahoo! Groups Links > > Steve Holle Link Communications, Inc. 1035 Cerise Rd. Billings, MT 59101 |
Reply by ●October 26, 20042004-10-26
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Hi, I have been looking into making sound with my BF-533 EZ Kit lite this week-end, so i read about SPORT, DMA, AD1836 ( and I2S ) and practiced a little. Maybe i can share what i have understood so far. Anyone, pardon me and please correct me if i am wrong here :-) I'll talk about the I2S example. Both the AD1836 codec and the SPORT were configured to deal with 24 bits word size, and here we know each of those 24bits words actually represent one sample. When SPORT receives a sample from the codec, only 24 bits are exchanged through the SPORT. Once a 24 bits sample has been received, it is placed in the SPORT receive FIFO. Note that, because the "8-deep" FIFO lines are 16bits wide and that our word (sample) size is 24bits, one sample will take 2 lines ( so our SPORT FIFO is actually "4-deep" here). In other words, inside the FIFO, one sample will occupy 32bits although only 24 are really meaningful to us. The 8 extra bits were not transmitted through the SPORT, it is just some extra room in the FIFO line that we don't use. One of the states of the receive FIFO can look like this : FIFO line #7 - [15:0] sample 3 Lowest 16bits FIFO line #6 - [15:8] 8 unused bits [7:0] sample 3 Highest 8bits FIFO line #5 - [15:0] sample 2 Lowest 16bits FIFO line #4 - [15:8] 8 unused bits [7:0] sample 2 Highest 8bits FIFO line #3 - [15:0] sample 1 Lowest 16bits FIFO line #2 - [15:8] 8 unused bits [7:0] sample 1 Highest 8bits FIFO line #1 - [15:0] sample 0 Lowest 16bits FIFO line #0 - [15:8] 8 unused bits [7:0] sample 0 Highest 8bits If the "secondary side" bit would not been set ( in fact, it is, in the I2S example ), then sample 0 is left channel sample #1, sample 1 is right channel sample #1, sample 2 is left channel sample #2, sample 3 is right channel sample #2, sample 4 is left channel sample #3, sample 5 is right channel sample #3 .... and so on... Then comes the DMA. It can be configured to do transfers of 8, 16 or 32 bits items ( not 24 bits ... ). In the examples, it is configured to do 32 bits word exchanges between the memory and the SPORT, that is, with the SPORT's receive/transmit FIFOs. The 8 extra bits are actually part of each 32bits exchange. We store one sample as a 32bits integer in view of being processed, but we have to keep in mind that only the bits [SLEN:0] actually represent the sample. I hope that answer your question :) May i know which example you are talking about exactly ? the I2S one or the TDM one ? I haven't looked into the TDM one, but for the I2S one, there are more things going on mainly because SPORT is configured with the secondary side bit activated for both receive and transmit. That means SPORT transmit and received to/from both its primary and secondary data lines at the same time (transmists to the AD1836's DSDATA1 & DSDATA2 lines and receives from ASDATA1 & ASDATA2 lines), meaning that we actually deal with 4 channels ( the 2 stereo ADCs and 2 first of the 3 stereo DACs ). The receive FIFO state described above would change. sample 0 is StereoADC1 Left channel sample #1 sample 1 is StereoADC2 Left channel sample #1 sample 2 is StereoADC1 Right channel sample #1 sample 3 is StereoADC2 Right channel sample #1 sample 4 is StereoADC1 Left channel sample #2 sample 5 is StereoADC2 Left channel sample #2 sample 6 is StereoADC1 Right channel sample #2 sample 7 is StereoADC2 Right channel sample #2 ... and so on... Cheers, Laurent ----- Original Message ----- From: "Steve Holle" <> To: "shwetha shwetha" <>; <> Sent: Monday, October 25, 2004 12:06 PM Subject: Re: [adsp] adt Sport and DMA > > > Someone more knowledgeable should correct any errors you see here. Even > though the data from the codec is 24-bits, data inside the DSP is > transferred in 32-bit words. The SLEN parameter indicates how many clocks > are needed to transfer one codec word, which is then placed in 32-bit memory. > > At 08:42 AM 10/25/2004, shwetha shwetha wrote: > >Hi, > > > >I am kind of new to this ADSP.... > > > >I am working on the ADSP BF533 EZ-kit lite example > >"Audio talk through" ..... > > > >In the C code they have set the SLEN$,since the > >codec(AD1836)sends 24 bits.... > > > >but then the DMA transfer is32 bit.... > > > >i don't understand this... > > > >Also... in the chapter Sport,while talking adt this > >SLEN..tells that the DMA can dump 32 bit to the TX > >buffer of the SPORT of which,the SLEN (in this case > >24bits) bits are transfered thor' the sport... > > > >If it does like that,when r the remaining (32-24=8 > >bits) 8 bits transfered thro' the sport? > > > >Please i am in a pathetic stage.. now at my work.. so > >kindly help me to understand this....i will be very > >grateful... > > > >thanking u all in advance and eagerly waiting for the > >early reply from the expertiese..... > > > > > > > >________________________________________________________________________ > >Yahoo! India Matrimony: Find your life partner online > >Go to: http://yahoo.shaadi.com/india-matrimony > > > > > > > > > > > >_____________________________________ > >Note: If you do a simple "reply" with your email client, only the author > >of this message will receive your answer. You need to do a "reply all" if > >you want your answer to be distributed to the entire group. > > > >_____________________________________ > >About this discussion group: > > > >To Join: Send an email to > > > >To Post: Send an email to > > > >To Leave: Send an email to > > > >Archives: http://groups.yahoo.com/group/adsp > > > >Other Groups: http://www.dsprelated.com/groups.php3 > > > >Yahoo! Groups Links > > > > > > > > > > Steve Holle > Link Communications, Inc. > 1035 Cerise Rd. > Billings, MT 59101 > > > _____________________________________ > Note: If you do a simple "reply" with your email client, only the author of this message will receive your answer. You need to do a "reply all" if you want your answer to be distributed to the entire group. > > _____________________________________ > About this discussion group: > > To Join: Send an email to > > To Post: Send an email to > > To Leave: Send an email to > > Archives: http://groups.yahoo.com/group/adsp > > Other Groups: http://www.dsprelated.com/groups.php3 > > Yahoo! Groups Links > |
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Reply by ●October 27, 20042004-10-27
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Hi, I want to compute LOG in base 2 in few assembly instructions. In first approximation, log2 = a*mant*mant + b*mant + c + exp2
Is someone who has an idea ?
Thanks in advance.
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Reply by ●October 27, 20042004-10-27
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// input value = v = 2^EXP * (1 + MANT / 2^23) // log2(v) = EXP + log2(1 + MANT / 2^23) = EXP + log2(1+B), log2(1+B) can be approximated by a polynomial // log2(v) = EXP + c1*B + c2*B^2 + c3*B^3 = EXP + B*(c1+B*(c2+B*c3)) // Accuracy: Maximum error = +/- 0.0047dB (0.078%) If you don't need error checking (v<0) then take off 2 instructions If you don't need B^3 term take off 2 more instructions You should be able to do it in 9 cycles with the above taken off. Alex Young DSP software Engineer Consultant for Philips Digital Systems Laboratories
I want to compute LOG in base 2 in few assembly instructions. In first approximation, log2 = a*mant*mant + b*mant + c + exp2
Is someone who has an idea ?
Thanks in advance.
Crz gratuitement votre Le nouveau Yahoo! Messenger est arriv! Douvrez toutes les nouveaut pour dialoguer instantanent avec vos amis. Thargez GRATUITEMENT ici ! _____________________________________ Note: If you do a simple "reply" with your email client, only the author of this message will receive your answer. You need to do a "reply all" if you want your answer to be distributed to the entire group. _____________________________________ About this discussion group: To Join: Send an email to a...@yahoogroups.com To Post: Send an email to a...@yahoogroups.com To Leave: Send an email to a...@yahoogroups.com Archives: http://groups.yahoo.com/group/adsp Other Groups: http://www.dsprelated.com/groups.php3 Yahoo! Groups Links
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Reply by ●October 27, 20042004-10-27
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I have tried, but I don't know how compute the MANT multiply by a
float because the instruction MANT is left-adjusted in the fixed-point field
??
How can I do in assembly language ??
R0 = 2.4;;
XR1 = MANT FR0;; /* extract the mantissa */
XR2 = LOGB FR0;; /* extract the exponent */ XFR3 = R1 * R10;; /* mant x c3 */ Thanks in advance a...@philips.com wrote:
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Reply by ●October 28, 20042004-10-28
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You will need the FLOAT instruction to convert from fixed to floating-point, possibly with some shifts to make everything line up correctly. --- adsp ts <> wrote: > I have tried, but I don't know how compute the MANT multiply by > a float because the instruction MANT is left-adjusted in the > fixed-point field ?? > How can I do in assembly language ?? > R0 = 2.4;; > XR1 = MANT FR0;; /* extract the mantissa */ > XR2 = LOGB FR0;; /* extract the exponent */ > XFR3 = R1 * R10;; /* mant x c3 */ > > Thanks in advance > wrote: > > // input value = v = 2^EXP * (1 + MANT / 2^23) > // log2(v) = EXP + log2(1 + MANT / 2^23) = EXP + log2(1+B), > log2(1+B) can be approximated by a polynomial > // log2(v) = EXP + c1*B + c2*B^2 + c3*B^3 = EXP + > B*(c1+B*(c2+B*c3)) > // Accuracy: Maximum error = +/- 0.0047dB (0.078%) > > If you don't need error checking (v<0) then take off 2 > instructions > If you don't need B^3 term take off 2 more instructions > > You should be able to do it in 9 cycles with the above taken > off. > > Alex Young > DSP software Engineer > Consultant for Philips Digital Systems Laboratories > > adsp ts <> > 27/10/04 11:15 > To: > cc: (bcc: Alex Young/LEU/PDSL/PHILIPS) > Subject: [adsp] ADSP-TS101 : compute LOG > Classification: > Hi, > I want to compute LOG in base 2 in few assembly instructions. > In first approximation, log2 = a*mant*mant + b*mant + c + exp2 > > Is someone who has an idea ? > > Thanks in advance. __________________________________ |
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Reply by ●October 29, 20042004-10-29
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I have found a solution :
Extract only the 23 bits of mantissa and after Mask with 0x3F800000
So exemple, for 2,4 = 0x4019999A
XR0 = 23;;
XR1 = 0x3F800000;;
XR2 = 0x4019999A;;
XR3 = FEXT R2 BY R0;; /* R2 = 0x0019999A */
XR4 = R2 OR R3;;
And you have extract the mantissa for the number in float : R4 = 1,2
After, it is only computation with float number.
You can't use the FLOAT instruction because the MANT instruction
extracts the mantissa what is left-adjusted in the fixed-point field.
Thanks for your ideas and your help.
Jon Harris <j...@yahoo.com> wrote:
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Reply by ●October 29, 20042004-10-29
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Hi, I use a ADSP-TS101 and I have a different program for each DSP of the card. So I have created four project but I don't know how I shall do to load the 4 program on the card ? Shall I create a new project with dependenties ???
Thanks
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Reply by ●October 29, 20042004-10-29
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On Fri, 29 Oct 2004, adsp ts wrote: > I use a ADSP-TS101 and I have a different program for each DSP of the >card. So I have created four project but I don't know how I shall do to >load the 4 program on the card ? > > Shall I create a new project with dependenties ??? How do you boot each processor? If one processor boots the other 3, then you only need one project with overlays. If each processor boots independently, you will need 4 projects. Patience, persistence, truth, Dr. mike |
