# I A pulse of frequencies that start in phase - pulse period?

1. Nov 27, 2017

### Natalie Johnson

Hi,

I start at 100 MHz have 10 carrier frequencies all starting in phase at t=0 and the carrier frequencies have 25 kHz spacing. So my frequency range is 100-100.25 MHz, with 10 single frequencies equally spaced in this range.

They come in phase every 4x10-5 seconds. The equation Frequency = 1 / Time_period works for this..... with frequency being the frequency of the carrier spacing

The time period between all carriers coming back in phase (after going out of phase at t=0 or at any time after they realign) is directly related to the spacing of the carrier frequencies using this equation. I am struggling to see how this is related but it works. Perhaps I am looking for an elegant explanation of this

Thanks

Last edited: Nov 27, 2017
2. Nov 27, 2017

### .Scott

The 100MHz doesn't matter.
If you have only one 25KHz signal, it will be at 0 phase once every cycle. That is once every 40usec, the period.
With two, you still have both of them matching up at phase=0 every 40usec. The 1st one going through 1 cycle each period, the second going through 2 such cycles.
That pattern continues up to ten.

3. Nov 27, 2017

### tech99

In effect you have a 25kHz source and you are just looking at one wavelength of it.
This principle applies in other fields. It is possible to find the length of a cable by looking at the repetitive ripple of a VSWR sweep for instance.
It is also possible to ascertain the frequency of a microwave receiver by receiving harmonics of a low frequency carrier and finding the spacing between adjacent low frequency pips.From memory, F = F1^2 / (F1-F2).

4. Nov 27, 2017

### Natalie Johnson

It's 10 single frequencies, equally spaced every 25khz from a starting frequency of 100MHz to a finish frequency 100.250. Even though all the single frequencies have different time periods, my script shows they all come back in phase every 40 micro seconds.

Also it doesn't matter how many frequencies I add (with increasing the bandwidth past 100.250 to accomaddate more single frequencies), as long as the spacing is 25khz ... it's always repeating every 40 ms...

If each of these frequencies placed in th 250khz bandwidth are different frequencies, they all have different periods so how would they all come back in phase in the manner you suggest? It's clearly happening but I'm finding it difficult visualising.

How is it a 25khz source if it's a new frequency spaced between, it's the spacing not a repeating single frequency?

Last edited: Nov 27, 2017
5. Nov 27, 2017

### Tom.G

6. Nov 28, 2017

### .Scott

I'll take another shot at it.

Let's say we have two sine wave functions, one $A(t)$ at frequency $a$ cycles per second, the other $B(t)$ at frequency $b$ cycles per second. And we will assume that they start in phase at time $t_0$.
Those functions would be:
$A(t) = sin(2\pi a(t-t_0))$
$B(t) = sin(2\pi b(t-t_0))$

But we are interested in phase. So here are the formulas for phase:
$phase(A(t)) = 2\pi a(t-t_0)$
$phase(B(t)) = 2\pi b(t-t_0)$

And we are specifically interested in when these phases match:
$P_{diff} = phase(A(t)) - phaseB(t)) =$ a multiple of $2\pi$
$P_{diff} = ( 2\pi a(t-t_0) ) - ( 2\pi b(t-t_0) ) =$ a multiple of $2\pi$
$P_{diff} /(2\pi)= ( a(t-t_0) ) - ( b(t-t_0) ) =$ is an integer
$P_{diff} /(2\pi)= ( (a-b)(t-t_0) ) =$ is an integer
This will happen when
$( (a-b)(t-t_0) ) = 0, 1, 2, 3, ...$
$t-t_0 = 0/(a-b), 1/(a-b), 2/(a-b), 3/(a-b), ...$

So what happens is that at $t=t_0$, $P_{diff}$ is also zero.
As t increases, the phase difference $P_{diff}$ also increases. The first time they will be in phase is when the phase difference is $2\pi$ ($360^{\circ}$).
That will only happen when $t-t_0 = 1/(a-b)$. That's the first time they match up. The seconds time will be another $1/(a-b)$ seconds, and so on.

In your case, the frequency difference $(a-b)$ is 25KHz, so the period when they repeat will be 1/25KHz = 40usec. So:
at $t_0+0$usec, the phase difference expressed in full cycles ($2\pi$) would be $0(a-b) = 0\times25KHz = 0$ cycles ($0^{\circ}$).
at $t_0+5$usec: $5usec(a-b) = 5usec\times25KHz = 0.125$ cycles ($45^{\circ}$).
at $t_0+10$usec: $10usec(a-b) = 10usec\times25KHz = 0.25$ cycles ($90^{\circ}$).
at $t_0+20$usec: $20usec(a-b) = 20usec\times25KHz = 0.5$ cycles ($180^{\circ}$).
at $t_0+40$usec: $40usec(a-b) = 40usec\times25KHz = 0.5$ cycles ($360^{\circ}$).

That should show you why this works with only two. Now look at the case with 10.
The difference between the 1st and 2nd is 25KHz, so they will resync every 40usec.
The difference between the 1st and 3rd is 50KHz, so they will resync every 20usec.
The difference between the 1st and 4th is 75KHz, so they will resync every 40usec/3.
...
The difference between the 1st and 10th is 225KHz, so they will resync every 40usec/9.

So some of the pairs will sync more often that once every 40usec. But they will all be in phase when the first two signals are in phase.