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TLC2543是TI公司的12位串行模數(shù)轉(zhuǎn)換器，使用開關(guān)電容逐次逼近技術(shù)完成A/D轉(zhuǎn)換過程。由于是串行輸入結(jié)構(gòu)，能夠節(jié)省51系列單片機(jī)I/O資源；且價(jià)格適中，分辨率較高，因此在儀器儀表中有較為廣泛的應(yīng)用。
TLC2543的特點(diǎn)
（1）12位分辯率A/D轉(zhuǎn)換器；
（2）在工作溫度范圍內(nèi)10μs轉(zhuǎn)換時(shí)間；
（3）11個(gè)模擬輸入通道；
（4）3路內(nèi)置自測(cè)試方式；
（5）采樣率為66kbps；
（6）線性誤差±1LSBmax；
（7）有轉(zhuǎn)換結(jié)束輸出EOC；
（8）具有單、雙極性輸出；
（9）可編程的MSB或LSB前導(dǎo)；
（10）可編程輸出數(shù)據(jù)長(zhǎng)度。
TLC2543的引腳排列及說明
TLC2543有兩種封裝形式：DB、DW或N封裝以及FN封裝，這兩種封裝的引腳排列如圖1，引腳說明見表1
TLC2543電路圖和程序欣賞
#include<reg52.h>
#include<intrins.h>
#define uchar unsigned char
#define uint unsigned int
sbit clock=P1^0; sbit d_in=P1^1;
sbit d_out=P1^2;
sbit _cs=P1^3;
uchar a1,b1,c1,d1;
float sum,sum1;
double sum_final1;
double sum_final;
uchar duan[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f};
uchar wei[]={0xf7,0xfb,0xfd,0xfe};
void delay(unsigned char b) //50us
{
unsigned char a;
for(;b>0;b--)
for(a=22;a>0;a--);
}
void display(uchar a,uchar b,uchar c,uchar d)
{
P0=duan[a]|0x80;
P2=wei[0];
delay(5);
P2=0xff;
P0=duan[b];
P2=wei[1];
delay(5);
P2=0xff;
P0=duan[c];
P2=wei[2];
delay(5);
P2=0xff;
P0=duan[d];
P2=wei[3];
delay(5);
P2=0xff;
}
uint read(uchar port)
{
uchar i,al=0,ah=0;
unsigned long ad;
clock=0;
_cs=0;
port<<=4;
for(i=0;i<4;i++)
{
d_in=port&0x80;
clock=1;
clock=0;
port<<=1;
}
d_in=0;
for(i=0;i<8;i++)
{
clock=1;
clock=0;
}
_cs=1;
delay(5);
_cs=0;
for(i=0;i<4;i++)
{
clock=1;
ah<<=1;
if(d_out)ah|=0x01;
clock=0;
}
for(i=0;i<8;i++)
{
clock=1;
al<<=1;
if(d_out) al|=0x01;
clock=0;
}
_cs=1;
ad=(uint)ah;
ad<<=8;
ad|=al;
return(ad);
}
void main()
{
uchar j;
sum=0;sum1=0;
sum_final=0;
sum_final1=0;
while(1)
{
for(j=0;j<128;j++)
{
sum1+=read(1);
display(a1,b1,c1,d1);
}
sum=sum1/128;
sum1=0;
sum_final1=(sum/4095)*5;
sum_final=sum_final1*1000;
a1=(int)sum_final/1000;
b1=(int)sum_final%1000/100;
c1=(int)sum_final%1000%100/10;
d1=(int)sum_final%10;
display(a1,b1,c1,d1);
}
}
標(biāo)簽：
2543
TLC
上傳時(shí)間：
2013-11-19
上傳用戶：shen1230
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摘 要:用一種新的思路和方法,先計(jì)算低通、再計(jì)算高通濾波器的有關(guān)參數(shù),然后組合成帶通濾波器.關(guān)鍵詞:濾波器;參數(shù);新思路中圖分類號(hào): TN713. 5 文獻(xiàn)識(shí)別碼:B 文章編號(hào):1008 - 1666 (1999) 04 - 0089 - 03A New Consideration of the Band Filter’s CalculationGuo Wencheng( S hao Yang B usiness and Technology school , S haoyang , Hunan ,422000 )Abstract :This essay deals with a new method of calculating the band filters - first calculatingthe relevant parameters of low - pass filters ,then calculating the ones of high - pass filters.Key words :filter ; parameters ;new considercation八十年代后,信息產(chǎn)業(yè)得到了迅猛發(fā)展. 帶通濾波器在微波通信、廣播電視和精密儀器設(shè)備中得到了廣泛應(yīng)用. 帶通濾波器性能的優(yōu)劣,對(duì)提高接收機(jī)信噪比,防止鄰近信道干擾,提高設(shè)備的技術(shù)指標(biāo),有著十分重要的意義.我在長(zhǎng)期的教學(xué)實(shí)踐中,用切比雪夫型方法設(shè)計(jì)、計(jì)算出寬帶濾波器集中參數(shù)元件的數(shù)據(jù). 該濾波器可運(yùn)用在檢測(cè)微波頻率的儀器和其他設(shè)備中. 再將其思路和計(jì)算方法介紹給大家,供參考.
標(biāo)簽：
帶通濾波器設(shè)計(jì)
計(jì)算
上傳時(shí)間：
2014-12-28
上傳用戶：Yukiseop
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JAVA課程設(shè)計(jì)實(shí)例。多個(gè)大型系統(tǒng)源代碼，如考試系統(tǒng)單機(jī)版、C\S版、B\S版。
標(biāo)簽：
JAVA
設(shè)計(jì)實(shí)例
大型
上傳時(shí)間：
2014-01-20
上傳用戶：zuozuo1215
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RSA算法 :首先, 找出三個(gè)數(shù), p, q, r, 其中 p, q 是兩個(gè)相異的質(zhì)數(shù), r 是與 (p-1)(q-1) 互質(zhì)的數(shù)...... p, q, r 這三個(gè)數(shù)便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 這個(gè) m 一定存在, 因?yàn)?r 與 (p-1)(q-1) 互質(zhì), 用輾轉(zhuǎn)相除法就可以得到了..... 再來, 計(jì)算 n = pq....... m, n 這兩個(gè)數(shù)便是 public_key ，編碼過程是, 若資料為 a, 將其看成是一個(gè)大整數(shù), 假設(shè) a < n.... 如果 a >= n 的話, 就將 a 表成 s 進(jìn)位 (s
標(biāo)簽：
person_key
RSA
算法
上傳時(shí)間：
2013-12-14
上傳用戶：zhuyibin
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USB Manager(usbmgr) 0.4.8 Shuu Yamaguchi <shuu@wondernetworkresources.com> Special Helper: Philipp Thomas When USB devices connect to or disconnect from a USB hub, the usbmgr works as the following according to configuration. a) It loads and unloads files Linux kernel modules. b) It execute file to setup USB devices.
標(biāo)簽：
wondernetworkresources
Yamaguchi
Manager
Special
上傳時(shí)間：
2014-01-27
上傳用戶：zhaiyanzhong
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prolog 找路例子程序：
=== === === === === ===
Part 1-Adding connections
Part 2-Simple Path
example
| ?- path1(a,b,P,T).
will produce the response:
T = 15
P = [a,b] ?
Part 3 - Non-repeating path
As an example, the query:
?- path2(a,h,P,T).
will succeed and may produce the bindings:
P = [a,depot,b,d,e,f,h]
T = 155
Part 4 - Generating a path below a cost threshold
As an example, the query:
?- path_below_cost(a,[a,b,c,d,e,f,g,h],RS,300).
returns:
RS = [a,b,depot,c,d,e,g,f,h] ?
RS = [a,c,depot,b,d,e,g,f,h] ?
no
==================================
標(biāo)簽：
Part
connections
example
prolog
上傳時(shí)間：
2015-04-24
上傳用戶：ljt101007
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《大型數(shù)據(jù)庫(kù)技術(shù):Oracle 9i高級(jí)程序設(shè)計(jì)教程》隨書光盤。
本書結(jié)合大量的實(shí)例，介紹如何利用Oracle來管理和維護(hù)數(shù)據(jù)，以及使用JSP、PHP和C＃開發(fā)C/S模式和B/S模式網(wǎng)絡(luò)數(shù)據(jù)庫(kù)應(yīng)用程序。全書分為管理篇、開發(fā)篇和實(shí)踐篇3個(gè)部分。管理篇和開發(fā)篇Oracle數(shù)據(jù)庫(kù)的管理和開發(fā)技術(shù)；實(shí)踐篇完整地介紹使用JSP、PHP和C#開發(fā)的基于Oracle數(shù)據(jù)庫(kù)項(xiàng)目的實(shí)例。這些實(shí)例既可以作為獨(dú)立的系統(tǒng)運(yùn)行，也可以取其中的一部分作為應(yīng)用軟件的一個(gè)模塊，具有很強(qiáng)的實(shí)用性。本書所附光盤內(nèi)容為本書的所涉及的源代碼。
本書適合Oracle數(shù)據(jù)庫(kù)管理員和應(yīng)用程序開發(fā)人員，以及對(duì)數(shù)據(jù)庫(kù)技術(shù)感興趣的讀者閱讀。
標(biāo)簽：
Oracle
大型
光盤
數(shù)據(jù)庫(kù)技術(shù)
上傳時(shí)間：
2015-05-06
上傳用戶：xmsmh
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數(shù)字運(yùn)算，判斷一個(gè)數(shù)是否接近素?cái)?shù)
A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in base b is a Niven number if the sum of its digits divides its value.
Given b (2 <= b <= 10) and a number in base b, determine whether it is a Niven number or not.
Input
Each line of input contains the base b, followed by a string of digits representing a positive integer in that base. There are no leading zeroes. The input is terminated by a line consisting of 0 alone.
Output
For each case, print "yes" on a line if the given number is a Niven number, and "no" otherwise.
Sample Input
10 111
2 110
10 123
6 1000
8 2314
0
Sample Output
yes
yes
no
yes
no
標(biāo)簽：
數(shù)字
運(yùn)算
上傳時(shí)間：
2015-05-21
上傳用戶：daguda
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LCD-7279的經(jīng)過調(diào)試多次已經(jīng)能夠?qū)崿F(xiàn)在指定的位置顯示特定的數(shù)據(jù)。
程序lcd1.c實(shí)現(xiàn)基本的功能，即：鍵盤輸入0、1、2、3、4、5、6、7、8、9、a\b\c\d\e\f 在液晶的指定位置顯示實(shí)現(xiàn)滿屏或半屏顯示點(diǎn)陣和字符，調(diào)入一幅圖畫的代碼進(jìn)行顯示；
標(biāo)簽：
7279
LCD
lcd
調(diào)試
上傳時(shí)間：
2014-01-14
上傳用戶：hgy9473
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We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.
標(biāo)簽：
represented
integers
group
items
上傳時(shí)間：
2016-01-17
上傳用戶：jeffery
