I need to know how to read frequency - frequency

I got port that is on electronic card (Atmel card with 8051 micro controller) . This port named p3_0. The port get '1' or '0' logic.
1 0 1 0
''''''' ''''''''
' ' ' '
' ''''''' '''''' And so on....
I need to use C for 8051 (that is actually original C) and write a code that will know what the frequency is. finally I need the frequency in array of char cause I need to print it to LCD that only know to print char array[];. I use "itoa" function that convert the int number, the 'count' to char array [];. the frequency can be between 16Hz to 90Hz.
This is what I did :
void main(void)
{
init_lcd()///Print command
;print_lcd(1,"The Project"); //Print command
;print_lcd(2,"is starting:"); //Print command
for(i=0; i<20; i++){delay(1000);delay(1000);delay(1000);}
for(i=0; i<20; i++){delay(1000);delay(1000);delay(1000);}
while(1)
{
count = 0;
while(P3_0 == 1)
{
count ++;
init_lcd() //Print command
;print_lcd(1,"Counting..."); //Print command
for(i=0; i<1; i++){delay(1000);delay(1000);delay(1000);}//delay...
}
init_lcd()///Print command
;print_lcd(1,"Done Counting!"); //Print command
for(i=0; i<1; i++){delay(1000);delay(1000);delay(1000);}
;print_lcd(2,itoa(count)); //Print command
for(i=0; i<20; i++){delay(1000);delay(1000);delay(1000);}
for(i=0; i<20; i++){delay(1000);delay(1000);delay(1000);}
}
}
The idea is I check the port and I count every time its '1'. Finally I will decide of a time that every '1' counted and (doing the calculation of what the frequency is)
THE QUESTION IS: HOW CAN I KNOW EXACTLY HOW MUCH 1 IS ON THE FREQUENCY, HOW CAN I?
Also there is thing with the dalay when counting that "take from the frequency".
How can I solve that?
thanks for all who read enter code herethis!!
I really appreciate any help!,
THANKS ALL!

Their is a logic to measure frequency.
8051 has 2 timers/counters, configure one as timer and another as counter and give your frequency input to the counter pin and then start both counter and timer same time( anyhow their'll be 1 machine cycle delay in activating timers but you can compensate it by calculations), before this load counter with suitable value so that your module can measure low frequency. When counter overflows stop timer and and by knowing no. of counts and time elapsed you can calculate frequency.
Hope this logic works.

Related

Parallel Dynamic Programming with CUDA

It is my first attempt to implement recursion with CUDA. The goal is to extract all the combinations from a set of chars "12345" using the power of CUDA to parallelize dynamically the task. Here is my kernel:
__device__ char route[31] = { "_________________________"};
__device__ char init[6] = { "12345" };
__global__ void Recursive(int depth) {
// up to depth 6
if (depth == 5) return;
// newroute = route - idx
int x = depth * 6;
printf("%s\n", route);
int o = 0;
int newlen = 0;
for (int i = 0; i<6; ++i)
{
if (i != threadIdx.x)
{
route[i+x-o] = init[i];
newlen++;
}
else
{
o = 1;
}
}
Recursive<<<1,newlen>>>(depth + 1);
}
__global__ void RecursiveCount() {
Recursive <<<1,5>>>(0);
}
The idea is to exclude 1 item (the item corresponding to the threadIdx) in each different thread. In each recursive call, using the variable depth, it works over a different base (variable x) on the route device variable.
I expect the kernel prompts something like:
2345_____________________
1345_____________________
1245_____________________
1234_____________________
2345_345_________________
2345_245_________________
2345_234_________________
2345_345__45_____________
2345_345__35_____________
2345_345__34_____________
..
2345_245__45_____________
..
But it prompts ...
·_____________
·_____________
·_____________
·_____________
·_____________
·2345
·2345
·2345
·2345
...
What I´m doing wrong?
What I´m doing wrong?
I may not articulate every problem with your code, but these items should get you a lot closer.
I recommend providing a complete example. In my view it is basically required by Stack Overflow, see item 1 here, note use of the word "must". Your example is missing any host code, including the original kernel call. It's only a few extra lines of code, why not include it? Sure, in this case, I can deduce what the call must have been, but why not just include it? Anyway, based on the output you indicated, it seems fairly evident the launch configuration of the host launch would have to be <<<1,1>>>.
This doesn't seem to be logical to me:
I expect the kernel prompts something like:
2345_____________________
The very first thing your kernel does is print out the route variable, before making any changes to it, so I would expect _____________________. However we can "fix" this by moving the printout to the end of the kernel.
You may be confused about what a __device__ variable is. It is a global variable, and there is only one copy of it. Therefore, when you modify it in your kernel code, every thread, in every kernel, is attempting to modify the same global variable, at the same time. That cannot possibly have orderly results, in any thread-parallel environment. I chose to "fix" this by making a local copy for each thread to work on.
You have an off-by-1 error, as well as an extent error in this loop:
for (int i = 0; i<6; ++i)
The off-by-1 error is due to the fact that you are iterating over 6 possible items (that is, i can reach a value of 5) but there are only 5 items in your init variable (the 6th item being a null terminator. The correct indexing starts out over 0-4 (with one of those being skipped). On subsequent iteration depths, its necessary to reduce this indexing extent by 1. Note that I've chosen to fix the first error here by increasing the length of init. There are other ways to fix, of course. My method inserts an extra _ between depths in the result.
You assume that at each iteration depth, the correct choice of items is the same, and in the same order, i.e. init. However this is not the case. At each depth, the choices of items must be selected not from the unchanging init variable, but from the choices passed from previous depth. Therefore we need a local, per-thread copy of init also.
A few other comments about CUDA Dynamic Parallelism (CDP). When passing pointers to data from one kernel scope to a child scope, local space pointers cannot be used. Therefore I allocate for the local copy of route from the heap, so it can be passed to child kernels. init can be deduced from route, so we can use an ordinary local variable for myinit.
You're going to quickly hit some dynamic parallelism (and perhaps memory) limits here if you continue this. I believe the total number of kernel launches for this is 5^5, which is 3125 (I'm doing this quickly, I may be mistaken). CDP has a pending launch limit of 2000 kernels by default. We're not hitting this here according to what I see, but you'll run into that sooner or later if you increase the depth or width of this operation. Furthermore, in-kernel allocations from the device heap are by default limited to 8KB. I don't seem to be hitting that limit, but probably I am, so my design should probably be modified to fix that.
Finally, in-kernel printf output is limited to the size of a particular buffer. If this technique is not already hitting that limit, it will soon if you increase the width or depth.
Here is a worked example, attempting to address the various items above. I'm not claiming it is defect free, but I think the output is closer to your expectations. Note that due to character limits on SO answers, I've truncated/excerpted some of the output.
$ cat t1639.cu
#include <stdio.h>
__device__ char route[31] = { "_________________________"};
__device__ char init[7] = { "12345_" };
__global__ void Recursive(int depth, const char *oroute) {
char *nroute = (char *)malloc(31);
char myinit[7];
if (depth == 0) memcpy(myinit, init, 6);
else memcpy(myinit, oroute+(depth-1)*6, 6);
myinit[6] = 0;
if (nroute == NULL) {printf("oops\n"); return;}
memcpy(nroute, oroute, 30);
nroute[30] = 0;
// up to depth 6
if (depth == 5) return;
// newroute = route - idx
int x = depth * 6;
//printf("%s\n", nroute);
int o = 0;
int newlen = 0;
for (int i = 0; i<(6-depth); ++i)
{
if (i != threadIdx.x)
{
nroute[i+x-o] = myinit[i];
newlen++;
}
else
{
o = 1;
}
}
printf("%s\n", nroute);
Recursive<<<1,newlen>>>(depth + 1, nroute);
}
__global__ void RecursiveCount() {
Recursive <<<1,5>>>(0, route);
}
int main(){
RecursiveCount<<<1,1>>>();
cudaDeviceSynchronize();
}
$ nvcc -o t1639 t1639.cu -rdc=true -lcudadevrt -arch=sm_70
$ cuda-memcheck ./t1639
========= CUDA-MEMCHECK
2345_____________________
1345_____________________
1245_____________________
1235_____________________
1234_____________________
2345__345________________
2345__245________________
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1245__145___45____4______
...
1235__1235__235___25_____
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1235__1235__123___12____1
1234__234________________
1234__134________________
1234__124________________
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1234__124___24___________
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1234__1234__124___24_____
1234__1234__124___14_____
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1234__1234__124___24____2
1234__1234__124___14____4
1234__1234__124___14____1
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1234__1234__123___23_____
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1234__1234__123___12____2
1234__1234__123___12____1
========= ERROR SUMMARY: 0 errors
$
The answer given by Robert Crovella is correct at the 5th point, the mistake was in the using of init in every recursive call, but I want to clarify something that can be useful for other beginners with CUDA.
I used this variable because when I tried to launch a child kernel passing a local variable I always got the exception: Error: a pointer to local memory cannot be passed to a launch as an argument.
As I´m C# expert developer I´m not used to using pointers (Ref does the low-level-work for that) so I thought there was no way to do it in CUDA/c programming.
As Robert shows in its code it is possible copying the pointer with memalloc for using it as a referable argument.
Here is a kernel simplified as an example of deep recursion.
__device__ char init[6] = { "12345" };
__global__ void Recursive(int depth, const char* route) {
// up to depth 6
if (depth == 5) return;
//declaration for a referable argument (point 6)
char* newroute = (char*)malloc(6);
memcpy(newroute, route, 5);
int o = 0;
int newlen = 0;
for (int i = 0; i < (6 - depth); ++i)
{
if (i != threadIdx.x)
{
newroute[i - o] = route[i];
newlen++;
}
else
{
o = 1;
}
}
printf("%s\n", newroute);
Recursive <<<1, newlen>>>(depth + 1, newroute);
}
__global__ void RecursiveCount() {
Recursive <<<1, 5>>>(0, init);
}
I don't add the main call because I´m using ManagedCUDA for C# but as Robert says it can be figured-out how the call RecursiveCount is.
About ending arrays of char with /0 ... sorry but I don't know exactly what is the benefit; this code works fine without them.

Counting up the value and performing calculation using LCD Arduino

I am new to Arduino programming. My condition is I want to count up a value of i from 0 using while loop. At the same time, a calculation, b is needed to be performed continuously when counting up the value. However, the counting start from 90 instead of start from 0 when the calculation b is performing. Can anyone can help me on it? Thanks.
#include <LiquidCrystal.h>
LiquidCrystal lcd(12, 11, 5, 4, 3, 2);
void setup() {
lcd.begin(20, 4);
}
void loop() {
int i;
while (i>=0){
i++;
lcd.setCursor(1,1);
lcd.print("b");
lcd.setCursor(3,1);
lcd.print(i);
delay(1000);
int c;
c = i + 2;
lcd.setCursor(1,2);
lcd.print(c);
delay(1000);
}
}
First of all your code should not even compile as b is not declared.
You initialize i with 0. So while (i>0) is never the case.
If you fix both you end up in an infinite loop. Not sure if you want that as there is some code after the while loop.
So pick a limit for i and use that in a for loop.
If you want to use a while loop make sure your loop condition is true >=0
You also might want to add a delay as otherwise your values change way too fast for you to see.

How to Optimally Shift Large Arrays n Number of Incidences

I am creating my own version of a music visualizer that responds to the frequency of music; a common project. I am using 2 strips of Neopixels, each with 300 LEDs making a total of 600 LEDs.
I have written functions, shown below, that create the desired affect of having a pulse of light travel down the strips independently. However, when running in real time with music, the updates per second is too slow to create a nice pulse; it looks choppy.
I believe the problem is the number of operations that must be preformed when the function is called. For each call to the function, a 300 value array per strip must be shifted 5 indices and 5 new values added.
Here is an illustration of how the function currently works:
-Arbitrary numbers are used to fill the array
-A shift of 2 indices shown
-X represents an index with no value assigned
-N represents the new value added by the function
Initial array: [1][3][7][2][9]
Shifted array: [X][X][1][3][7]
New array: [N][N][1][3][7]
Here if my code. Function declarations below loop(). I am using random() to trigger a pulse for testing purposes; no other functions were included for brevity.
#include <FastLED.h>
// ========================= Define setup parameters =========================
#define NUM_LEDS1 300 // Number of LEDS in strip 1
#define NUM_LEDS2 300 // Number of LEDS in strip 1
#define STRIP1_PIN 6 // Pin number for strip 1
#define STRIP2_PIN 10 // Pin number for strip 2
#define s1Band 1 // String 1 band index
#define s2Band 5 // String 2 band index
#define numUpdate 5 // Number of LEDs that will be used for a single pulse
// Colors for strip 1: Band 2 (Index 1)
#define s1R 255
#define s1G 0
#define s1B 0
// Colors for strip 2: Band 6 (Index 5)
#define s2R 0
#define s2G 0
#define s2B 255
// Create the arrays of LEDs
CRGB strip1[NUM_LEDS1];
CRGB strip2[NUM_LEDS2];
void setup() {
FastLED.addLeds<NEOPIXEL, STRIP1_PIN>(strip1, NUM_LEDS1);
FastLED.addLeds<NEOPIXEL, STRIP2_PIN>(strip2, NUM_LEDS2);
FastLED.setBrightness(10);
FastLED.clear();
FastLED.show();
}
void loop() {
int num = random(0, 31);
// Pulse strip based on random number for testing
if (num == 5) {
pulseDownStrip1();
}
pulseBlack1();
}
// ======================= FUNCTION DECLARATIONS =======================
// Pulse a set of colored LEDs down the strip
void pulseDownStrip1() {
// Move all current LED states by n number of leds to be updated
for (int i = NUM_LEDS1 - 1; i >= 0; i--) {
strip1[i] = strip1[i - numUpdate];
}
// Add new LED values to the pulse
for (int j = 0; j < numUpdate; j++) {
strip1[j].setRGB(s1R, s1G, s1B);
}
FastLED.show();
}
// Pulse a set of black LEDs down the strip
void pulseBlack1(){
// Move all current LED states by n number of leds to be updated
for (int i = NUM_LEDS1 - 1; i >= 0; i--) {
strip1[i] = strip1[i - numUpdate];
}
// Add new LED values to the pulse
for (int j = 0; j < numUpdate; j++) {
strip1[j].setRGB(0, 0, 0);
}
FastLED.show();
}
I am looking for any suggestions regarding optimizing this operation. Through my research, copying the desired values to a new array rather than shifting the existing array seems to be a faster operation.
If you have any advice on optimizing this process, or alternate methods to produce the same animation, I would appreciate the help.
The secret is to not shift it. Shift where you start reading it instead. Keep track of a separate variable that keeps the start position and alter your reading through the array to start there, roll back over to zero when it gets to the array length, and stop one short of where it starts.
Google the term "circular buffer" Look at the Arduino HardwareSerial class for a decent implementation example.

If else function....how to use return

#include <stdio.h>
#include <cs50.h>
int GetPositiveInt();
int main (void)
{
int min; /*variable to hold minutes*/
printf ("How many minutes does it take you to use a shower?");
scanf ("%d", &min);
int numbtl = min * 12; /*computes number of bottles*/
if (min > 0)
{
printf ("Taking a shower you use %d bottles of water", numbtl);
}
else
{
printf ("Please enter the positive number: \n");
scanf ("%d", &min);}
return min;
}
}
I've written this program but I've got some bug in else place.
Here is the text which I get trying to execute this program.
~/workspace/pset1/ $ make water
clang -ggdb3 -O0 -std=c11 -Wall -Werror -Wshadow water.c -lcs50 -lm -o water
~/workspace/pset1/ $ ./water
How many minutes does it take you to use a shower?0
Please enter the positive number:
10
~/workspace/pset1/ $
Let me explin what I mean by this code and its execution. This code is used to compute how many bottles of water you use while taking a shower. 1 minute of shower equals 12 bottles.
If you enter the positive integer (for example, 10) the result will be like this:Taking a shower you use 120 bottles of water. BUT if you enter 0 or negative integer the program will ask you to enter the positive integer. And here is when the problem occurs. After entering the positive integer I get the result of 0 bottles.
Lets break down your program and see what it is doing; first
int GetPositiveInt();
Is a forward declaration for a function that is never used or defined, you can remove it entirely.
int main (void)
{
int min; /*variable to hold minutes*/
printf ("How many minutes does it take you to use a shower?");
scanf ("%d", &min);
Here we define the main function (the entry point to a c program) and declare a variable of type int called min. Then print out a line asking for user input and read in their response storing it in the min variable.
int numbtl = min * 12; /*computes number of bottles*/
Here you process min by multiplying it by 12 and storing the result in numbtl.
if (min > 0)
{
printf ("Taking a shower you use %d bottles of water", numbtl);
}
Here you check if min is valid, if it is you print the response. Here you can see the success path of your program is correct, the problem is what happens when min is not greater than 0. (Note there was a minor formatting error in this next bit that I corrected by what I assume you meant - note that this is why correct indenting and formatting is important).
else
{
printf ("Please enter the positive number: \n");
scanf ("%d", &min);
}
If min is not valid (ie less than or equal to 0) then you ask for another input and store it in min.
return min;
}
Lastly you return whatever is stored in min. Note you never actually do anything with this second value, except return it.
Side note: The return value on main is used as the exit status of your application. You can see the exit status of the last command in bash with echo $?.
$ ./water
How many minutes does it take you to use a shower?
2
Taking a shower you use 24 bottles of water
$ echo $?
2
This is probably not what you want. The exit status is normally 0 to indicate success and a positive number otherwise. This last return you will likely want return 0 to indicate your program has run successfully.
Now for the logic of your program, what it looks like you are trying to do is obtain some input from the user; validate it (obtaining new input if its invalid); then process it. So the first thing to do is move your processing to the end of your program
int main (void)
{
int min; /*variable to hold minutes*/
...
int numbtl = min * 12; /*computes number of bottles*/
printf ("Taking a shower you use %d bottles of water", numbtl);
return 0;
}
Now we just need to acquire and validate the user input; a typical algorithm to do this is:
prompt for user input
while (input is not valid) {
reprompt for user input
}
process user input
Converting this to c your applications ends up with;
int main (void)
{
int min; /*variable to hold minutes*/
printf ("How many minutes does it take you to use a shower? ");
scanf ("%d", &min);
while (min <= 0)
{
printf ("Please enter the positive number: ");
scanf ("%d", &min);
}
int numbtl = min * 12; /*computes number of bottles*/
printf ("Taking a shower you use %d bottles of water", numbtl);
return 0;
}
This will continue to ask the user for a positive number until they either enter one or hit crtl+c to kill the command. Once a valid number has been obtained it will processes it and print out the result.
#include <stdio.h>
#include <string.h>
int main (void)
{
long min;
char msg[255];
strcpy(msg,"How many minutes does it take you to use a shower?\n");
do {
printf ("%s", msg);
scanf ("%lu", &min);
strcpy(msg , "Please enter the positive number: \n");
}while( min <= 0 );
long numbtl = min * 12; /*computes number of bottles*/
printf ("Taking a shower you use %lu bottles of water", numbtl);
return 0;
}
You can also use a goto label. Here I used a goto label named start. If a number less than zero is entered then the program goes to beginning of the program. Hope, It helps.
#include<stdio.h>
int main (void)
{
int min; /*variable to hold minutes*/
start :
printf ("How many minutes does it take you to use a shower?\nEnter here : ");
scanf ("%d", &min);
if (min <= 0)
{
printf("Enter a number grater than zero\n\n");
goto start;
}
int numbtl = min * 12; /*computes number of bottles*/
printf ("Taking a shower you use %d bottles of water", numbtl);
return 0;
}

How to run two loops at the same time?

I have been developing a very simple text game using Objective C and Xcode. It is almost done but I am having a problem, the scanf method stops the loop and asks for user input while I need the computer to be running the rest of the loop, the solution I came up with was running two while loops at the same time, one being the logic loop and another being a loop for user input.
I have been doing my research and it looks like using threads are the way to go, I just have not found a tutorial that will break it down for a n00b in Objective C (I am decent in java, I just have never worked with threads). If anybody could explain them or link me to a very broken down tutorial that would be great. Or if anybody has another idea I am open to anything else.
Necessary Code (The scanf I am having a problem with has asterisks on the line):
while(running != 0)
{
if(gameState == 1)
{
if(timeToGenerateNum == true)
{
while(randNumber < 10000000)
{
randNumber = arc4random() % 100000000;
}
NSLog(#"%i", randNumber);
timeToGenerateNum = false;
}
else
{
while(time <= 2500)
{
NSLog(#"Testing");
time++;
******************scanf("%i", &userNum);************************
if(userNum == randNumber)
{
score += time;
time = 0;
timeToGenerateNum = true;
}
}
NSLog(#"Game Over! Your score was %i!", score);
running = 0;
}
}
else if(gameState == 2)
{
NSLog(#"To play, simply type in the number that appears on the screen.");
NSLog(#"But be careful, you only have a short amount of time before GAME OVER!");
NSLog(#"The quicker you type in the number the more score you get!");
NSLog(#"Are you ready to start, if so type '1' and press enter!");
scanf("%i", &gameState);
}
}
You're going to have to learn a bit about BSD (Unix, Linux) input/output to pull this off: replace your call to scanf with a non-blocking function you write to acquire input from the user's keyboard.
This function should immediately return whatever the user typed, or immediately return with a zero character count if she didn't type anything.
Read up on the select(2) system call, and keep in mind that keyboard input (standard input) is the first file descriptor, file descriptor zero.