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TI公司DSP的VP口捕获高清连续视频流或图片的驱动程序源码!
TI公司DSP的VP口捕获高清连续视频流或图片的驱动程序源码!-TI
- 2022-03-18 23:59:24下载
- 积分:1
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开发环境:ccs。基于dsp的急救车与交通灯实验。(汇编语言)...
开发环境:ccs。基于dsp的急救车与交通灯实验。(汇编语言)-Development Environment: ccs. Dsp-based emergency vehicles and traffic lights test. (Assembly language)
- 2022-01-25 18:00:01下载
- 积分:1
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dsp642 for bootloader based on BIOS
dsp642 for bootloader based on BIOS-dsp642 for Bootloader based on the BIOS
- 2022-03-22 03:29:24下载
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2000系列,TMS320LF2407开发板原理图
2000系列,TMS320LF2407开发板原理图-2000 series, TMS320LF2407 development board schematics
- 2022-10-10 16:50:03下载
- 积分:1
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DSP的下程序,实现定点运算,希望互相交流
DSP的下程序,实现定点运算,希望互相交流-DSP next procedure, to achieve fixed-point computing, hoping to exchange
- 2022-03-23 05:59:34下载
- 积分:1
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合众达公司VPM642版的平台音频测试程序
合众达公司VPM642版的平台音频测试程序-Tatsu VPM642 united company a platform for the audio version of test procedures
- 2022-04-19 16:28:04下载
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这是测试SEED
这是测试SEED-VPM642系统中网络的接口的测试的程序。-This is a test SEED-VPM642 system network interface, the test procedure.
- 2023-08-01 14:00:03下载
- 积分:1
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用C语言进行DSP软件设计的优化考虑.rar
用C语言进行DSP软件设计的优化考虑.rar-Using C language optimized DSP software design considerations. Rar
- 2022-12-25 12:20:03下载
- 积分:1
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基于TI的DSP的G726_CODEC
基于TI的DSP的G726_CODEC-based on TI"s DSP G726_CODEC
- 2022-08-02 14:45:06下载
- 积分:1
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1、(1)32bit乘法的指令解释
(2)volume1的load.asm基础上实现一个
16bit数组的乘法累加的函数,并进行...
1、(1)32bit乘法的指令解释
(2)volume1的load.asm基础上实现一个
16bit数组的乘法累加的函数,并进行 -o2 / -o3 / 手工优化
2、c环境
C调用汇编函数,汇编函数调用c函数
addarr3(int * arr1, int * arr2, int * arr3, int * arr4, n)
//汇编函数,3个数组的对应位置相加,结果放在arr4[n]中,
汇编函数调用C的子函数,它把arr1和arr2相加放到arr3[n]中;
addarr2(int * arr1, int * arr2, int * arr3, n)-1, (1) 32bit multiplication instructions explained (2) the load.asm volume1 on the basis of a a 16bit multiplication cumulative array of functions, and-o2 /-o3/2 manual optimization, c environment compilation called C function, the compilation function call c function addarr3 (int* arr1. int* arr2, arr3 int*, int* arr4, n)// compilation function, 3 arrays corresponding location together, the results on arr4 [n], the compilation of the C function call function, arr1 put it together and put arr2 arr3 [n]; addarr2 (int* arr1, int* arr2, int* arr3, n)
- 2022-07-08 14:51:46下载
- 积分:1