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);
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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

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.

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

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

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

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.

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.