Hi~
I am now using cadence to design mixer. Spectrum is analysed by setting PSS, but only positive spurs(magnitude and phase information) are shown in results.

First question is whether negative frequency can be shown with PSS analyse?

Then,I check the bandwidth of the down-conversion mixer by utilizing PAC Analyse. I set my mixer with a 13GHz LO frequency and a RF frequency range from 1GHz to 30GHz. After that, the mixer would translate the frequency range to -12GHz to 17GHz. So, I ploted the voltage gain from the RF port(1GHz-12.99GHz&13.01GHz-30GHz) to the IF port(-12GHz--10MHz&10MHz-17GHz) with the help of PAC Analyse.The Figure of 'bandwidth' is shown below.

Here is my second question. Is this method used to test the bandwidth of mixer convincing?

On the other hand,'negative bandwidth'and'positive bandwidth'in this figure are asymmetric. In my opinion, positive bandwidth should equals negative bandwidth in spectrum, since we have cosωt=1/2(exp(jωt)+exp(-jωt)) and sinωt=j/2(-exp(jωt)+exp(-jωt)).

My third question is whether this ‘negative and positive bandwidth’ from PAC analyse is the same conception as what we normally think about negative bandwidth and positive bandwidth?

Hoping for ur xplanation pls.

SpectreRF has no difficulty working with negative frequency. By default it generally converts negative frequency signals into positive frequency signals before displaying them using the complex conjugate. You can control whether it performs this mapping, and whether to use input or the output frequency as the X-axis using the freqaxis setting.

I have no idea what you are talking about when you talk about positive and negative bandwidth.

-Ken

Hi,


short answers, 1) yes, 2) no.

The thing is that bandwidth is a measure of how much of the frequency spectrum a signal occupies. Basically I think Ken was implying that it is a scalar quantity, so you're using the wrong terminology. I simply assumed that you were talking about the spectrum of a baseband signal.

The signal is contained in two channels, the in-phase channel, I, and the quadrature-phase channel, Q. For transmission, the I channel modulates one phase of the RF carrier, and the Q channel modulates a 90 degree phase shifted version of the carrier, and the two are summed together. For the purposes of mathematical manipulation, I believe that you can represent the two channels as real and imaginary parts.

Again, "bandwidth" is not transformed, the signal itself can be represented as a magnitude, A(f), and a phase, θ(f). The I-channel by itself is symmetrical both in magnitude and phase about zero frequency (phase might have rotational symmetry, I can't remember and I'm lazy to check). The Q channel by itself is also symmetric both in magnitude and phase. The combination I+jQ is NOT symmetric either in magnitude or phase. However, there is no part of the physical design where the signal I+jQ can be represented by a single voltage. It is just a mathematical representation and not physical. I and Q are not combined until they modulate a carrier. The modulated carrier is a real, physical signal, and it's spectrum is symmetric both in magnitude and phase.


best regards,
Aaron