MCU questions, let s discuss how to answer.

You have to write the program yourself, the first question is very easy to do, each is 256ms, and the 8-bit DA conversion is also 0-255,256 numbers, in fact, it is every 1 millisecond DA value minus 1.

The second question, 51 single-chip microcomputer I forgot t1 in the way of what timer, mainly look at the number of timing, counter, if 8 bits, the maximum value of the count is 2 8 = 256, 16 bits is 2 16 = 65536, the initial value is equal to the maximum value of the count minus the value to be counted, for example, with a 12-bit timing counter, the crystal oscillator is 12m, and then 12 dividend, then an instruction cycle is 1us, to time 10ms, then the count value is 10ms 1us=10000

Initial value = 2 12 (counter maximum) - value to count = 65535 - 10000 = 55535Because it is added from zero, it is necessary to reduce it by 1, that is, the maximum value of 8 bits is the maximum value of 65535

This means that if you want to time 10ms, 12m12 crossover frequency will cost 1us every time the technology is used, then 10,000 counts are 10ms

Because the maximum value of the counter is 65535, then you have to add exactly 10,000 from 55,535 to 65,535, and it takes 10 ms to add each number, and 10,000 is 10ms

When it is added to a value greater than the maximum value of the counter, the corresponding timer overflow flag and interrupt request flag are set, and then timed by interrupt or query.

I don't want to say anything more about the third question, it's too simple, it's OK to check the table and send the number!

This is not difficult!!

In the first question, a DAC chip is used, and the specified analog voltage will be output if the specified data is inputted.

The second question, the initial value calculation, is the most basic.

The third question, the eight data lines of the single-chip microcomputer are connected to the digital tube section, and then there are four strobe lines to strobe the digital tube. Strobe the first tube, then the second, all the way to the fourth.

Since it's a discussion, you have to say something first...

Idea: The timer timing enters the interrupt once every 50ms, and the flag bit in the interrupt function is added by 1, and the actual time is 50ms 20 1s when the interrupt enters 20 times. So whenever 1s p1 1 flips the level, it also achieves the effect of high level time 1s and low level time 1s, the total time of high and low level is period 2s, so this waveform is a square wave with a period of 2s.

The program is as follows, and I have tried to write the comments in as much detail as possible:

Question 2 is incomplete. For a crystal oscillator of around 12MHz, the maximum overflow time of Timerx can be reached.

Question 3a, question 4a

Question 5 lacks the same premise as Question 2. Assuming a 12MHz crystal oscillator, B is chosen.

Question 6: C, Question 7, C, B

Question 8C, Question 9D