C51 MCU C language questions 3 and 4 255

Asynchronous serial communication means that both parties use a character (including specific additional bits) as the unit of data transmission, and the sender does not necessarily transmit the characters at intervals. Synchronous serial communication is the one that allows a sequence of characters to be sent consecutively, each with the same number of data bits and no start and stop bits. Asynchronous serial communication means that the time interval between two bytes in the communication is not fixed, while the time interval between two adjacent bits in the same byte is fixed.

Synchronous serial communication means that the time interval of each byte is equal during the communication process, and the time interval of bits per byte is also fixed. The first bit of the asynchronous communication data frame is the start bit, which is in the logical "1" state when there is no data transmission on the communication line. When the transmitting device wants to send a character data, it first emits a logical "0" signal, and this logic low is the starting bit.

The start bit is transmitted to the receiving device through the communication line, and when the receiving device detects this logic low, it begins to prepare to receive the data bit signal.

It depends on how you program, and it's not very good to review, first of all.

1) To count the external signal level duration, you need to count the pulse signal internally, so c t=0;

2) To start the timing counter with an external signal, make gate=1;

3) So from the options given, only the 80 and 09 symbols are required;

80 is the start timer 1, mode 0, 13-bit timing counter (8192)09 is the start timer 0; Mode 1, 16-bit timing counter (65536).

When c t=0 is the timing function: add 1 counter to count pulse f, and add 1 to the counter for each pulse until the timer tfx overflows;

c t=1 is the counting function: add 1 counter to count the external signal pulses from the input pins t0 ( and t1 (, each pulse comes, the counter adds 1 until the timer tfx overflows;

The microcontroller supports C51 programming for assembly programming.

1. Assembly programming, traditional assembly**. Disadvantages: Readability is not strong, advantages: **Streamlined 2, C51 programming, using special C language programming. Disadvantages: **Not streamlined, advantages: strong readability.

Compiler software such as Keilc and IAR with support for both C51 and assembly**.

It's too big a topic. If you really want to learn single-chip microcomputer, buy a development board handle, more than 100.

Write the navigation beacon light control program, the system crystal oscillator frequency unit is 6MHz, the navigation beacon light is required to be extinguished during the day, and lit intermittently at night, with a time interval of 2 seconds, that is, 2 seconds of brightness, 2 seconds of darkness, and the cycle is carried out in a cycle.

Clause. First, you need to design a clock first.

Clause. Second, the time is not clear, day or night, this statement is not accurate.

The following figure shows the interface circuit of a three-position LED 8-segment common anode static display of an AT89C51 application system, 1) What are the advantages and disadvantages of the static display circuit compared with the display circuit of dynamic scanning?

Let's read a book. 2) Write out the segment code showing the characters "x", "y", "z", note that the lowest digit of the segment code is "a" segment, and the highest digit of the segment code is "dp" segment y:91h.

x, z: cannot be displayed.

It is known that the addresses of the PA port, PB port, and PC port of 8255A are FF7CH, FF7DH, and FF7EH respectively, and these three ports have been initialized by the initialization program to mode 0 output, please write the program segment that makes the three-position LED common anode display display from left to right.

Which one of the monitors is mounted on the left?

This topic is too vague.

Suppose the left side is the PA port and the right side is the PC port.

mov dptr, #0ff7ch

mov a, #30h ;3.The segment code of MOVX @dptr, A

inc dptr

mov a, #80h ;8 Segment code MOVX @dptr, A

inc dptr

mov a, #92h ;5 Segment code movx @dptr, aend

Now who is still doing you, this is still using old antiques to occupy a bunch of mouth lines to do the bus 8255 question. Let you, the teacher who wrote the question, update it first.

8255 Amount, I haven't used it for a long, long time, I don't know if the information is still tender and can't be found.

The colon is missing after the subroutines L9 and L3.