[SOLVED] [moved] Help me re-work this simple equation & solving a language barrier.

[David_]

Hello.

Math has never been my strong side but I am trying to learn things I with certainty should have learned a long time ago, but I can't research the subject online since I don't know the English terminology for what am looking for.

I have these two equation:

fo = (FTW)(fs)/2³² with 0 ≦ FTW ≦ 2³¹
fo = fs*(1-(FTW/2³²)) with 2³¹ < FTW < 2³² - 1

This comes from a DDS chip datasheet and
fo = output frequency
fs = system clock frequency
FTW = Frequency Tuning Word.

I want to re-write the equations so that I can use it to find out what FTW I need to gain a specific fo, and I think it should be a simple matter of re-arranging the equations, that is simple if you do know how this sort of thing is done(which I don't).

Could someone show me how(to re-arrange the equations give the FTW needed for a specific fo) and tell me what category of mathematics I need to learn about, this is just one of many cases where I am hindered by my lack of mathematical knowledge and I experience that in many cases application notes give you all you need in the form of a starting or original equation but that is nothing if I can't re-work it.

I want to use this to write a C function that will be used as:
FTW = Hz_out(200000); // tuning word needed for a 200kHz output

I also fail to understand what the conditions for when which equation should be used actually says, this is regarding a DDS with a 32-bit frequency tuning word and I do get that 0 ≦ FTW ≦ 2³¹ means 0 is less than or equal to FTW while FTW is less than or equal to 2³¹(2 147 483 648)
and that 2³¹ < FTW < 2³² - 1 means when 2³¹(2 147 483 648) is less than FTW while FTW is less than 2³² - 1(4 294 967 295), but why would the equations be different for when the tuning word is between 0 & 2³¹ and when the tuning word is between 2³¹ & 2³² - 1?
This might be a little to DDS specific for this thread but if anyone reading this knows why please enlighten me.

I would appreciate any help very much.
Regards

 

[srizbf]

fo = (FTW)(fs)/2³² with 0 ≦ FTW ≦ 2³¹
for the above:
FTW is;
FTW=(fo*2^32)/fs

-----------
fo = fs*(1-(FTW/2³²)) with 2³¹ < FTW < 2³² -
1

for this , FTW is:
FTW=(1-(fo/fs))*2^32

 

[Dan Mills]

fo = FTW * Fs / 2³²
Multiply both sides by 2³² and cancel:
2³² *fo = FTW * Fs
Divide both sides by Fs and cancel:
2³² * fo / Fs = FTW

FTW = (2³² / Fs) * fo (How I would probably implement in fixed point C, it makes the first part a constant.

It is just high school algebraic manipulation, any math book aimed at 14 - 16 year olds will cover this stuff.

The reason for the two equations is that at Fs/2 you reach the limit of the sampling theorem so the signal in the first Nyquest zone becomes the alias hence the different equation.

I would note that you can reach any frequency up to (But not including) Fs/2 just using the first set of maths, so the second one is basically a curiosity unless you need the thing to tune the other way (Which actually can be useful).

Regards, Dan.

 

[David_]

Thank you.
I don't know why I didn't think of it but I do have all my math books from my time in school, somewhere...
Good tip, thanks.