# Thread: Number Theory for Electronic Signal Frequency Synthesis

1. I am new to this forum. I am an electrical engineer designing frequency synthesizers for electronic test and measurement equipment.

I have a design problem and I think that number theory could help me solve it. I'm not a mathematician, so I will state the problem the best I can.

Definitions:
F. Any number between 4000 and 8000
P1 = 50, or other predetermined constant.
P2. Unknown

When F mod P1 = 0, is there a value P2 such that F mod P2 >= C?

C is a fixed number, for example 10.

Thank you.

2. Originally Posted by Garan
I am new to this forum. I am an electrical engineer designing frequency synthesizers for electronic test and measurement equipment.

I have a design problem and I think that number theory could help me solve it. I'm not a mathematician, so I will state the problem the best I can.

Definitions:
F. Any number between 4000 and 8000
P1 = 50, or other predetermined constant.
P2. Unknown

When F mod P1 = 0, is there a value P2 such that F mod P2 >= C?

C is a fixed number, for example 10.

Thank you.
Hi Garan! Welcome to MHB!

Just to clarify, suppose we pick P2=8001. Doesn't that already satisfy your requirement?

Originally Posted by I like Serena
Hi Garan! Welcome to MHB!

Just to clarify, suppose we pick P2=8001. Doesn't that already satisfy your requirement?
P2 is restricted to something similar to P1.

P1 and P2 represent reference frequencies generated by two different crystal oscillators. For various design reasons, it would be best if P2 is "similar" to P1. P2 preferably should be an integer, but can be a rational number.

For example if P1 is 50 and P2 were 55, then this is realizable in the hardware.
But for certain values of F between 4000 and 8000, F mod P1 = 0 and F mod P2 will also get close to 0. I would like to be able to find a value of P2 that will keep F mod P2 > some constant C, (10 for example) whenever F mod P1 = 0.

4. Originally Posted by Garan
P2 is restricted to something similar to P1.

P1 and P2 represent reference frequencies generated by two different crystal oscillators. For various design reasons, it would be best if P2 is "similar" to P1. P2 preferably should be an integer, but can be a rational number.

For example if P1 is 50 and P2 were 55, then this is realizable in the hardware.
But for certain values of F between 4000 and 8000, F mod P1 = 0 and F mod P2 will also get close to 0. I would like to be able to find a value of P2 that will keep F mod P2 > some constant C, (10 for example) whenever F mod P1 = 0.
Well... suppose we pick F, P1, and C, then we can pick a rational P2:
$$P_2 = P_1\left(1+\frac C F\right)$$
so we'll have:
$$F \bmod P_2 = C$$

Alternatively, we can iterate $P_2$ one frequency at a time until we find one that gives a sufficient modulo.

Originally Posted by I like Serena
Well... suppose we pick F, P1, and C, then we can pick a rational P2:
$$P_2 = P_1\left(1+\frac C F\right)$$
so we'll have:
$$F \bmod P_2 = C$$

Alternatively, we can iterate $P_2$ one frequency at a time until we find one that gives a sufficient modulo.
Thank you. I will use Matlab to do the brute-force search based on your suggestions.

I was hoping for that there was an elegant closed-form solution from number theory

6. Originally Posted by Garan
Thank you. I will use Matlab to do the brute-force search based on your suggestions.

I was hoping for that there was an elegant closed-form solution from number theory
Well... suppose we tweak your problem statement a bit.
Let's make it: given $F \bmod P_1 = 0$, find $P_2$ such that $F \bmod P_2 = C$, where $C = \left\lfloor\frac{P_1}{2}\right\rfloor$.
Then we have $F = C + kP_2$ for some integer $k$.
That means $kP_2 = F - C$.
So if we factor $F-C$ in any way, we find $k$ and $P_2$.

Unfortunately we don't have any real conditions on $P_2$.
Rather, it seems that we're searching for a practical way to find some $P_2$ that does what we want within certain practical boundaries.
That's not really a number theory problem, that's more like asking for a practical down-to-earth engineering solution.