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vivkr

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chopping question (time vs. frequency domain)
Jul 16^th, 2007, 3:24am

Hi All,

I have a basic question regarding chopping, which is concerned not with the effect on the 1/f noise (which is obvious), but on
the signal itself.

Consider the following situation, where the input is a(t), chopping signal x(t), and one can approximate x(t) with
a (4/pi)*sin(wchop*t)

On chopping, the baseband signal a(t) is modulated to wchop. Let this signal be b(t).

Now, on demodulation, it would appear that b(t) is demodulated down to baseband giving back a(t), but
also that there is a component of b(t) which goes up to 2*wchop.

In time domain, one can easily visualize the chopping process to see that there ought to be no component
at 2*wchop. What is wrong with my reasoning in the frequency domain? Could someone suggest an explanation?
Obviously, the time domain analysis is correct, and one can see it in any example of chopping, and in transient simulations.

Thanks
Vivek

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Frank Wiedmann

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Re: chopping question (time vs. frequency domain)
Reply #1 - Jul 16^th, 2007, 4:51am

I believe that there are two things which make chopping a very special case of mixing:

The modulating signal is not sinusoidal but rectangular with no dc component.
The phase of the modulating and the demodulating signal is exactly the same.

If you take these points into account, you will probably find that the component at twice the modulating frequency comes out equal to zero.

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HdrChopper

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Re: chopping question (time vs. frequency domain)
Reply #2 - Jul 16^th, 2007, 5:19am

Hi vivek,

As Frank suggested, having a rectangular modulating signal it is a different story since it generates components at the odd harmonics of the clock frequency. The modulated baseband signal power content will be located at such frequencies before being demodulated. Once it is demodulated those odd harmonics become again even harmonics (including baseband), while the offset or low frequency noise being modulated in the same process are now located at the odd harmonics of the clock frequency.

The chopping technique, in order to get rid of such offset components and reduce 1/f noise contribution to your system make sense only if there is some sort of post-chopper filtering means. In this way the components resulting from the offset and low frequency noise modulation (shifted to the odd harmonics of the chopping clock frequency) are attenuated or even cancelled out, depeding on your filtering. This filter will also help remove the even harmonics resulting from the chopping operation and associated to the baseband signal modulation process.
I guess when you talked about the transient simulation - where you cannot see any 2nd clock harmonic component on your signal - you might have had some kind of filtering. That should explain it.

tosei

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vivkr

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Re: chopping question (time vs. frequency domain)
Reply #3 - Jul 16^th, 2007, 7:33am

Hi Frank, Tosei,

Many thanks for your responses.

I was assuming a sinusoidal model for the chopping signal just to make things simpler. If you look at my post, I have taken the fundamental component.
The remaining terms go down as 1/3, 1/5 etc. and I dropped them for ease. However, even when I write out the equations on paper assuming the identical phase,
the 2*wc term persists.

In simulation, I did not use any filtering, but still the 2*wc term is not there. Conceptually, I do not expect it to be there if I visualize chopping in the time
domain.

Any more hints as to what I may be doing wrong?

Thanks
Vivek

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HdrChopper

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Re: chopping question (time vs. frequency domain)
Reply #4 - Jul 16^th, 2007, 1:22pm

Hi Vivek,

When you considered the 2*wchop component resulting from the demodulation process of a(t), did you also consider the phase contribution of both harmonics at 2*wchop and -2wchop (bilateral spectral density)?. I'm not sure but these two should cancel out yield now harmonic components at 2*wchop.

tosei

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HdrChopper Community Fellow Offline Posts: 493	Re: chopping question (time vs. frequency domain) Reply #5 - Jul 16^th, 2007, 1:23pm I meant yielding NO harmonic components at 2*wchop....
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vivkr

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Re: chopping question (time vs. frequency domain)
Reply #6 - Jul 16^th, 2007, 11:05pm

Hi Tosei,

Perhaps, you can see where I am making an error. I repeat the calculation in time-domain as it is easier to type it in here:

Chopping signal x(t) ~ (4/pi)*sin(wchop*t) (Dropping 3rd, 5th and so harmonics, which have amplitude (4/pi)*(1/Nharm) )

Input signal a(t)

After chopping up: b(t) = a(t)*x(t) + n(t), where n(t) is the noise term from 1/f and other noise sources

After chopping down: c(t) = b(t)*x(t) = a(t)*x(t)*x(t) + n(t)*x(t)

The second term shows that noise is modulated up to the chopping frequency.

The first term is the signal a(t)*x(t)*x(t) = a(t)*(16/pi^2)*(sin(wchop*t))^2 = (8/pi^2)*a(t)*(1 - cos(2*wchop*t))

I see the second harmonic term even though I don't expect to see it. Could you point out my mistake?

Thanks
Vivek

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Frank Wiedmann Community Fellow Offline Posts: 678 Munich, Germany	Re: chopping question (time vs. frequency domain) Reply #7 - Jul 16^th, 2007, 11:40pm I think you need to take into account all combinations resulting in a contribution to the 2nd harmonic: 1st with 1st, 1st with 3rd, 3rd with 5th, 5th with 7th, 7th with 9th, and so on. When you sum them all up (taking the phase into account), the result should be zero.
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boe Community Fellow Offline Posts: 615	Re: chopping question (time vs. frequency domain) Reply #8 - Jul 17^th, 2007, 4:09am Vivek, The sin(2pif0t) has a frequency component at +f0 and one at -f0. So, when you shift your signal back to f=0 with one component, the other gives you a mirror at 2f0. BOE
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HdrChopper

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Re: chopping question (time vs. frequency domain)
Reply #9 - Jul 17^th, 2007, 10:40am

Hi Vivek,

I think the problem is related to your initial assumption. When simplifying the modulating signal x(t) to just the first harmonic, your are violating the principle of the chopper modulation in order to properly recover the modulated signal: the bandwidth of the system you are chopping should be > 5x larger than the chopping frequency. This is required in order to let the (actual) rectangular modulating signal to go through the system without being distorted significantly. This guarantees that a(t) is going to be properly recovered after demodulation.

By assuming your modulating signal x(t) is just the first harmonic, you already forcing your system to be band-limited way below the chopping frequency (like if you were filtering out all the harmonics of x(t) but the first one, which is what you are mathematically assuming).

To clarify this point and for the sake of simplicity, let�s assume a(t) is constant and equal to 1. After demodulation, you will get K*(1-cos(2*wchop*t)) as you correctly stated before. Certainly, you will have 2*wchop harmonic components in your recovered signal a(t)*x(t)^2. If you afterwards filter this signal (as in every chopped system where you need to filter out the harmonic components resulting from the chopper operation) you�ll find the recovered DC signal is attenuated with respect to the input signal level a(t). The reason is that your modulating signal, being a pure tone, acts like a �filtered rectangular modulating signal�, with the direct consequence of attenuation on the recovered signal (once filtered).

Summarizing, if the modulating signal is assumed to be x(t) = sin(wchop*t), it is correct to find 2nd order harmonic components. When you consider a rectangular modulating signal, the 2nd and higher order harmonics components cancel out when you also consider the phase contribution of each one as Frank suggested originally.

Hope this clarify things a little bit more.

Tosei.

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vivkr

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Re: chopping question (time vs. frequency domain)
Reply #10 - Jul 17^th, 2007, 11:50pm

Hi Frank, Tosei,

Many thanks for your suggestions. Indeed, you are absolutely right that one ought not to take just the fundamental component, and
while looking at it this way, I realized that it was very silly of me to even go to such trouble, because the solution is very simple:

final output c(t) = a(t)*x(t)*x(t) + n(t)*x(t), where a(t) is the original signal, x(t) is the chopping signal and n(t) is the noise added.

Of course, since x(t) assumes values +1 and -1, x(t)*x(t) = 1 for all t. Hence, the output is obviously c(t) = a(t) + n(t)*x(t)

Sometimes, the obvious is not the obvious, but luckily, there is the Designer's Guide Forum to help.

Regards
Vivek

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