Post by thread:Re: [PICLIST] period=1/f calculation

Irfan T = 1 /f so if the frecuency range is 30kHz-50Khz then time periody range is T = (1/30)ms-(1/50)ms or T= (1000/30)us-(1000/50)us so you need divided 1000/# counts for get a time periode in us Francisco Armenta "Muhammad Irfan(TECHNOMAN)" wrote: > > I need to calculate time period of a frequency in the range of 30kHz-50kHz. using PIC microcontroller. > I have sampled frequency for 0.1 Sec. and have counts in register. How can I divide 1 by these count? > Can any one help me please. > Irfan > > -- > http://www.piclist.com hint: PICList Posts must start with ONE topic: > [PIC]: PIC only [EE]: engineering [OT]: off topic [AD]: advertisements -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]: PIC only [EE]: engineering [OT]: off topic [AD]: advertisements

"Muhammad Irfan(TECHNOMAN)" wrote: > > I need to calculate time period of a frequency in the range of 30kHz-50kHz. using PIC microcontroller. > I have sampled frequency for 0.1 Sec. and have counts in register. How can I divide 1 by these count? > Can any one help me please. > Irfan > > -- > http://www.piclist.com hint: PICList Posts must start with ONE topic: > [PIC]: PIC only [EE]: engineering [OT]: off topic [AD]: advertisements -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]: PIC only [EE]: engineering [OT]: off topic [AD]: advertisements

OK, let's say that you have 250 counts accumulated over .1 seconds. If you multiply that by 10 you get 2,500 counts, which is the frequency. 1/2,500 = .0004 Note that 10,000/2,500 = 4 If we shift the decimal point four places to the left we get .0004, which is the answer we are looking for. Now, the count you have is in binary, not decimal, so we apply the same trick, but from a binary perspective. We will multiply the 1 by some convenient binary power such as 128 256 512 1024 2048, etc. Since your maximum stored count is going to be 5,000 this corresponds to a 1/x value of .0002 In reality, since the sample gate time was .1 seconds, this corresponds to an ACTUAL period of .00002 which is merely a matter of shifting the decimal point. So here's one way to do it: 1) Load the Numerator with 10,000 expressed in binary/hex. 2) Divide the Numerator by the Denominator (x) using binary/hex division. 3) Convert the answer into Decimal form. 4) Pad to the left with zeros. 5) Place the decimal point at the proper location. 6) Display decimal result. For example, if the frequency is 8.100 kHz, then the accumulator will have a count of 810. (Note the loss of the last digit). 10,000/810=12 (Note the loss of all but the integer part of the answer). .00012 is the displayed decimal answer, arrived at by placing the decimal point 5 places to the left. If you have a divide routine that can handle a 24 bit numerator, then you could use 10,000,000 as the numerator. That would yield a final answer of .00012345 You should realize that the last digit (5) COULD be in error due to the .1 second sampling period, even though in this particular example it happens to be right on. You will get a frequency of 8.100 kHz even if the actual frequency is 8.109 kHz. While the error is small at high frequencies, it is large at low frequencies. For example, from 10 Hz to 19.9 Hz the perceived frquency will be 10 Hz. When the frequency is low it is better to measure the period and then calculate the frequency. Since p=1/f and f=1/p you can use the same mathematical trick presented here to determine frequency from the period. Fr. Tom McGahee {Original Message removed}

Muhammad, you will have to use a divide routine. You don't mention which pic you are using, but a look at the application notes will sort you out. You could even customise one of the divide routines to become a dedicated inverting routine. I would use fixed point math with a 32/16 bit or 16/8 bit unsigned divide (you don't mention how bits store your count). A better solution may be to use the capture module and measure the period of the signal directly, possibly averaging that over say 64 or 128 readings. See AN526 and/or AN544 for lots of unsigned divide routines. Roland {Original Message removed}

part 1 3378 bytes content-type:text/plain; charset=us-ascii (decoded 7bit) Here is absolutely crazy code that may help. I never wrote something like this :) Don't know how to explain its operation, but what it does is approximate calculation of 32768/x using linear interpolation between points where x=1,2,4,8,16,32,64,128,256. The routine can be extended for other number of bits in x. I attached MATLAB functions that allow to compare the approximated and ideal functions. ;test cblock x, yh, yl endc clrf x ; movlw 129 ; movwf x again call Reciprocal8 nop incf x, f goto again ;**************************************************** ; ; Approximate calculation of 32768/x ; Input: x ; Output: y = 256 * yh + yl ~= 32768/x ; ; To calculate reciprocal, linear interpolation ; is used between points, where x=1,2,4,8,16,32,64,128 ; ; Size: 19 instructions ; Execution speed(worst case including return): ; 6+4*8-1+4+4+2=47 ; ; 4 Aug 2000 by Nikolai Golovchenko ;**************************************************** Reciprocal8 movf x, w skpnz retlw 1 ;error - division by zero movwf yh clrf yl setc Reciprocal8a rrf yl, f rlf yh, f skpc goto Reciprocal8a clrc rrf yh, w clrf yh subwf yh, f Reciprocal8b rrf yh, f rrf yl, f skpc goto Reciprocal8b retlw 0 ;ok ;**************************************************** Hope it helps, Nikolai On Thursday, August 03, 2000 Roland Andrag wrote: {Quote hidden}> Muhammad, you will have to use a divide routine. You don't mention which pic > you are using, but a look at the application notes will sort you out. You > could even customise one of the divide routines to become a dedicated > inverting routine. I would use fixed point math with a 32/16 bit or 16/8 > bit unsigned divide (you don't mention how bits store your count). > A better solution may be to use the capture module and measure the period of > the signal directly, possibly averaging that over say 64 or 128 readings. > See AN526 and/or AN544 for lots of unsigned divide routines. > Roland > {Original Message removed} part 2 365 bytes content-type:application/octet-stream; name="recipr_plot.m" (decode) part 3 720 bytes content-type:application/octet-stream; name="recipr.m" (decode) part 4 129 bytes -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

--- "Muhammad Irfan(TECHNOMAN)" <spam_OUTtechnokhTakeThisOuTCYBER.NET.PK> wrote: > I need to calculate time period of a frequency in > the range of 30kHz-50kHz. using PIC microcontroller. > I have sampled frequency for 0.1 Sec. and have > counts in register. How can I divide 1 by these > count? > Can any one help me please. > Irfan > > -- > http://www.piclist.com hint: PICList Posts must > start with ONE topic: > [PIC]: PIC only [EE]: engineering [OT]: off topic > [AD]: advertisements You can do this by using fixed point mathematic routines AN617 from Microchip's Application Notes. According to this what accuracy you want you should select one of them (24/16 bits for example etc.). Good look. Emil __________________________________________________ Do You Yahoo!? Kick off your party with Yahoo! Invites. http://invites.yahoo.com/ -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.