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Simulators >> Circuit Simulators >> Noise in Complex filters
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Message started by Megh on Mar 29^th, 2010, 10:06pm

Title: Noise in Complex filters
Post by Megh on Mar 29^th, 2010, 10:06pm

Cadence recommends following way to do the noise simulation in Complex filters: -

According to principle of superposition noise can be calculated as follows:
1. Calculate output noise (V**2/Hz) when only portI is active:
- at Iout: Ioutn_I(f);
- at Qout: Qoutn_I(f).
2. Now calculate output noise (V**2/Hz) when only portQ is active:
- at Iout: Ioutn_Q(f);
- at Qout: Qoutn_Qf).

Now you can calculate the total otput noise (V**2/Hz) at both outputs:
Ioutn(f) = Ioutn_I(f) + Ioutn_Q(f);
Qoutn(f) = Qoutn_I(f) + Qoutn_Q(f);
______________________________________________________________________________
The complex BPF is made of 2 LPFs. Both of them have gain = 0dB. Gain is 0dB for complex out to complex input.
Now I see that input referred noise is double that of output referred noise even though gain is 0dB . I don't know why this is happening. Please help.
Thanks
Megh

Title: Re: Noise in Complex filters
Post by ray_wang on Apr 11^th, 2010, 6:40pm

Obviously cadence does not support complex filter simulation well. What I did is to plot output noise, plot gain(from I input to I output), both with two ports active. And VNIN=VNOUT/gain*sqrt(2). Here, sqrt(2) is the summed noise from IQ channels.

I don't use that "VNIN" direct plot. It is wrong. I guess it might be that spectre does not know how to calculate correctly the gain of a complex filter. That's why I said cadence does not support complex filter simulation well.

Title: Re: Noise in Complex filters
Post by Andrew Beckett on Apr 12^th, 2010, 9:48am

I'm probably misunderstanding something here. In order to calculate the noise at the output of a circuit, you don't need input ports to be "active" - so I'm not sure what you're actually doing here.

Please make clear:

a) what analysis you're running
b) what the circuit setup is in each.

Noise analyses compute the output noise, and then functions like input-referred noise are derived from that - they can only be referred back to a single input (what does it mean that have it referred to two inputs at different phases?).

Apologies if this is a dumb question, but if I don't understand what you're trying to do, then the chances are others don't either, and so you may be limiting your opportunity to get a useful answer.

Regards,

Andrew.

Title: Re: Noise in Complex filters
Post by ray_wang on Apr 12^th, 2010, 6:40pm

VNIN=VNOUT/gain. I think, for spectre output, VNOUT is correct and VNIN is incorrect because of incorrect gain computation (in this complex filter case), so I use AC magnitude as gain instead to calculate VNIN.
I think the proper AC magnitude for Iin or Qin is 1/sqrt(2), then AC magnitude for Iout is exactly the gain for IQ channel VNIN compuation.
If you use 1 as AC magnitude for Iin or Qin, AC magnitude for Iout is the gain we are normally referrring to from Iin to Iout, but VNOUT/Iout is just one channel VNIN. For IQ channels VNIN, we need to multiply VNOUT/Iout by sqrt(2)

Title: Re: Noise in Complex filters
Post by Megh on Apr 12^th, 2010, 10:22pm

Dear Ray/Andrew,
Thanks for replying. Sorry if I sounded confusing.
Actually my problem has been articulated correctly by Ray in his latest reply.
Here is the situation
I have created a complex filter using 2 low pass filters (active RC: Ideal components). Both low pass filters have gain of 0dB. I run a noise analysis with ac source at I input and zero at the Q input. I get output referred noise at I output. Now when I see input referred noise from spectre it is actually double that of output referred noise. The thing that is confusing me is why it is double even though my gain is 0dB.
@Ray
You are right that VNin using direct plot is wrong in Cadence.
What I did is to plot output noise, plot gain(from I input to I output), both with two ports active. And VNIN=VNOUT/gain*sqrt(2). Here, sqrt(2) is the summed noise from IQ channels.
You can specify only one input port in noise analysis. How did you plot output noise with both inputs?
Here is what I have understood from your reply (Please correct me if I am wrong).
1) Put AC source on both inputs I and Q (90 deg phase apart).
2) For noise analysis specify I input as the input source.
3) Calculate output referred noise at I output. (VNoutI)
4) Calculate gain Iout/Iin (Gain_I_I)
5) Calculate Total input referred noise for complex filter is given by (VNoutI/(Gain_I_I))*sqrt(2)

If this is correct can you please elaborate a bit on sqrt 2 part?

Thanks a lot guys for your time.
Megh

Title: Re: Noise in Complex filters
Post by Andrew Beckett on Apr 13^th, 2010, 2:23am

Megh,

When you specify the input port for noise analysis, it does not affect the output noise in any way. It is only used to compute metrics such as input referred noise and noise figure, so that they can compute the gain between that input port or source and the output.

You can even set the input port/source to "none" and the output noise will remain the same.

In your case, because the input is not a single source, you cannot use this method to compute the input referred noise. So instead, you need to compute the output noise as normal (and it doesn't matter what you specify for the input port/source - you could set it to none), and then divide by your computed AC gain.

Note that specifying the AC magnitude on a source is not used in noise analysis - it is only used in AC analysis.

So, compute the AC gain by specifying an appropriate AC magnitude and phase on the I and Q sources, and then measure the signal level at the output - from this you can find the AC gain (sounds as if this is what you're already doing). Then knowing that, you can divide the output noise by this gain, to give you the input referred noise that you're after.

Regards,

Andrew.

Title: Re: Noise in Complex filters
Post by ray_wang on Apr 14^th, 2010, 11:25pm

Megh, I agree with what Andrew said, the input port you specify in noise analysis affects not VNOUT but VNIN.
I believe VNOUT is always there no matter what port you specify and the port you specify determines VNIN and NF, so you don't need to care about that.

If the input is Z=VI+jVQ, with added noise, the total input becomes Z+ZN=VI+VIN+jVQ+jVQN. If you take VI+jVQ as signal magnitude, the noise magnitude is sqrt(VIN^2+VQN^2)=sqrt(2)*VIN.

Regards
Ray

Title: Re: Noise in Complex filters
Post by Megh on Apr 14^th, 2010, 11:53pm

I completely agree.
So now just to summarize the procedure : -
1) calculate the VNout_I
2) Have input as VI+jVQ.
3) Calculate gain Gain_I_I= Vout_I/VI
4) Total input referred noise for I side VNI= VNout_I/Gain_I_I
5) Total input referred noise for complete Complex filter VN_complex=VNI * sqrt(2)

I hope now I have got it correct :) :). Please let me know if I have missed anything
Thanks a lot to both of you

Title: Re: Noise in Complex filters
Post by Megh on Apr 19^th, 2010, 2:49am

Dear Andrew/Ray,
Just want to add little more confusion here. :)
When a tool like Cadence reports a noise does it take in to account negative frequencies? Basically my question is does the tool assume the waveform as Double Side Band and report total noise as Noise = 2* SSB noise ?
Now if this is the case then we need to divide our answer for noise by 2 as this is a polyphase filter case which has only positive frequencies.
Please respond.
Thanks
Megh

Title: Re: Noise in Complex filters
Post by Andrew Beckett on Apr 19^th, 2010, 7:57am

Are you now asking about pnoise analysis rather than noise? With noise (rather than pnoise) there is no frequency translation; the input and output frequencies are the same. With pnoise, you get the noise contributions from all the specified sidebands (some of which may be negative frequencies - so you get both lower and upper sidebands).

Regards,

Andrew.

Title: Re: Noise in Complex filters
Post by Megh on Apr 19^th, 2010, 10:59pm

Andrew,
I am asking for noise simulation (not pnoise).
1) When we get the noise spectral density in real system is it combination of spectral densities at both (positive and negative) frequencies?
2) In real system frequency response is symmetric about DC unlike complex system.
So how does this affect the noise numbers reported by noise analysis.

Title: Re: Noise in Complex filters
Post by ray_wang on Apr 28^th, 2010, 12:58am

Megh,
I think the noise spectral density is the combination as you mentioned. Since noise is random, a sign indicating whether it is "positive" or "negative" is quite unnecessary. So all noise TF should be symmetrical with f=0. I don't think we need to divide by 2.
I especially compared the output noise of CPF with that of LPF. They're almost the same. So I think for noise, the CPF looks almost like two seperate LPFs, except for the "coupling things" to adjust fc.

Ray

Title: Re: Noise in Complex filters
Post by Geoffrey_Coram on Apr 30^th, 2010, 5:07am

If the noise from a single resistor is 4 k T R, then it's 2*SSB noise, and you would need to divide by 2 for your complex filter.

Title: Re: Noise in Complex filters
Post by salty on May 13^th, 2010, 10:27am

I have though about this as well for filters and mixers.
If you are receiving a SSB signal for a LIF rcvr, then using spectre to get the noise from a single real channel overestimates the noise by 3dB. Reason is that it folds in USB and LSB noise.
If you have a DSB signal in a ZIF rcvr, then the noise from a single real channel is correct for the whole quadrature channel.

Since adding complex signals, sqrt(I^2 + Q^2) increasese at the same rate as the noise: sqrt(NI^2 + NQ^2).

It seems to me that the sqrt(2) is more of a SSB vs DSB issue.