[time-nuts] 10 MHz Oscillator comparison part II

[Paramithiotti, Luciano Paolo S]

^Well, some time back I proposed a DMTD style front-end which mixes two unequal frequencies with a common oscillator into a common frequency. In a second ^stage it is mixed down by a second LO in more traditional DMTD style.

^LO1 = (f1 + f2)/2
^IF1 = abs(f1 - LO1) = abs(f2 - LO2) = abs(f1 - f2)/2
^LO2 = IF1 - IF2

^The IF1 filtering needs to filter out the difference frequency and supress the sum frequency. However, since both sides of the chain will have same ^frequencies after first mixer, correlation between the sides will create smaller response differences unless built very different. 
^The mixer oscillator contribution for LO1 and LO2 will correlate between the channels.

Magnus, can you write down a block diagram?

Luciano
IZ5JHJ

Luciano P. S. Paramithiotti



-----Original Message-----
From: time-nuts-bounces at febo.com [mailto:time-nuts-bounces at febo.com] On Behalf Of Magnus Danielson
Sent: giovedì 13 gennaio 2011 5.34
To: time-nuts at febo.com
Subject: Re: [time-nuts] 10 MHz Oscillator comparison part II

On 12/01/11 19:24, Bob Camp wrote:
> Hi
>
> If you have a random frequency like 7.352 MHz that neither divides or 
> multiplies to 10 MHz harmonic or sub harmonic, you can indeed mix the 
> signal to 10 MHz.
>
> If you do so, you will need to filter the outputs, since the mixing 
> spurs will mess up the input to the multiplier.
>
> If the generator you use for the mixing has more noise or jitter than 
> the sources, that noise is likely to de-correlate unless the chains 
> are absolutely identical. Since they multiply to two different 
> frequencies, they really can't be identical. Net result is your 
> measurement is messed up by the noise of the generator.

Well, some time back I proposed a DMTD style front-end which mixes two unequal frequencies with a common oscillator into a common frequency. In a second stage it is mixed down by a second LO in more traditional DMTD style.

LO1 = (f1 + f2)/2
IF1 = abs(f1 - LO1) = abs(f2 - LO2) = abs(f1 - f2)/2
LO2 = IF1 - IF2

The IF1 filtering needs to filter out the difference frequency and supress the sum frequency. However, since both sides of the chain will have same frequencies after first mixer, correlation between the sides will create smaller response differences unless built very different. 
The mixer oscillator contribution for LO1 and LO2 will correlate between the channels.

Cheers,
Magnus

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