his project was built and tested with WinAVR-20060125. Make sure the MCU target define in the Makefiles corresponds to the AVR you are using!! To build the code, just install WinAVR and run "make" from the console in echomaster and echoslave subdirs. "make program" will program the device if you have a AVRISP attached. Remember to set the AVR device to at least 8MHz. The AVR may use the programmable clock from MC1319x, just remember to check if the MC1319x and SPI communication is working FIRST! Otherwise you wont get any clock signal to the AVR and then you can t program it or reset the fuses! The MC1319x has default clock output of 32kHz so you will have to set your programmer to a very low frequency (<=32kHz/4) to be able to program it while it is running on that!
標簽： the 20060125 project WinAVR
上傳時間： 2014-10-10
上傳用戶：yan2267246
Test program to loop on Successive Approximation A-to-D conversion. Allows digital codes and resulting DAC output to be viewed on scope.
標簽： Approximation Successive conversion program
上傳時間： 2015-08-07
上傳用戶：頂得柱
8點基二fft Fast Fourier Transform (FFT). Using 8 points, takes about 1.2 ms to execute one FFT. Sets up General Purpose Timer 1 to generate events at 10 kHz. Will produce 10 kHz output on T1PWM and T1PWM pins.
標簽： FFT Transform Fourier execute
上傳時間： 2013-12-18
上傳用戶：Yukiseop
book:simulation and softeware radiao for mibole code:psk-based digital transmission schemes ofdm transmission technology cdma transmission technology multiple access protocals cellular telecommunication systems
標簽： transmission simulation psk-based softeware
上傳時間： 2015-08-11
上傳用戶：003030
The book "The Finite Difference Time Domain for Electromagnetics" by Karl Kunz and Raymond Luebbers, CRC Press, 1993, contains an FDTD code and output files in Appendix B. The same code and output files are contained in this directory.
標簽： Electromagnetics Difference The Luebbers
上傳時間： 2013-12-29
上傳用戶：chongcongying
EXAMPLE SOURCE CODE FOR IMPLIB FILTER This filter accepts input through the standard input stream, convertsit and outputs it to the standard output am. The streams are linkedthrough pipes, such that the input stream is the output from the import librarian being invoked, and the output stream is connected to the message window of the IDE, ie.
標簽： input standard EXAMPLE accepts
上傳時間： 2014-11-18
上傳用戶：siguazgb
EXAMPLE SOURCE CODE FOR TASM FILTER his filter accepts input through the standard input stream, converts it and outputs it to the standard output stream. The streams are linked through pipes, such that the input stream is the output from the assembler being invoked, and the output stream is connected to the message window of the IDE, ie.
標簽： input standard EXAMPLE accepts
上傳時間： 2014-01-13
上傳用戶：小碼農lz
KVM (for Kernel-based Virtual Machine) is a full virtualization solution for Linux on x86 hardware. It consists of a loadable kernel module (kvm.ko) and a userspace component. Using KVM, one can run multiple virtual machines running unmodified Linux or Windows images. Each virtual machine has private virtualized hardware: a network card, disk, graphics adapter, etc. The kernel component of KVM is included in mainline Linux, and will appear in Linux 2.6.20. KVM is open source software.
標簽： virtualization Kernel-based for hardware
上傳時間： 2015-08-20
上傳用戶：lijianyu172
/* * EULER S ALGORITHM 5.1 * * TO APPROXIMATE THE SOLUTION OF THE INITIAL VALUE PROBLEM: * Y = F(T,Y), A<=T<=B, Y(A) = ALPHA, * AT N+1 EQUALLY SPACED POINTS IN THE INTERVAL [A,B]. * * INPUT: ENDPOINTS A,B INITIAL CONDITION ALPHA INTEGER N. * * OUTPUT: APPROXIMATION W TO Y AT THE (N+1) VALUES OF T. */
標簽： APPROXIMATE ALGORITHM THE SOLUTION
上傳時間： 2015-08-20
上傳用戶：zhangliming420
ACM試題An Easy Problem Description As we known, data stored in the computers is in binary form. The problem we discuss now is about the positive integers and its binary form. Given a positive integer I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of 1 s in whose binary form is the same as that in the binary form of I. For example, if "78" is given, we can write out its binary form, "1001110". This binary form has 4 1 s. The minimum integer, which is greater than "1001110" and also contains 4 1 s, is "1010011", i.e. "83", so you should output "83".
標簽： Description computers Problem binary
上傳時間： 2013-12-11
上傳用戶：libenshu01
