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Several additional questions about SA option in PNA-X

Question asked by Macree on Dec 26, 2016
Latest reply on Jan 16, 2017 by Dr_joel

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I want to say thank you to Dr.Joel for previous answers.You wrote so much to explain these questions to me.But I still have questions that I can't figure out.The first thing is how to determine sample points according to window shape.When we do FFT,there is a equation:N=Fs/RBW,which is written on a DSP book I read.But the sample points changes when I change window shape.Does it mean the frequency resolution will change after windowing?What's the relation between window and sample points.Is there a function to calculate the sample points according to sample rate,window shape and RBW?The second question is,the "number of LOs" seems not equal to span*N/68MHz when I use it.I tested some frequency span and image reject selection and made a list.How does this happen?

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Macree Jan 10, 2017 1:01 AM

Another question is how dose the system change VBW in PNA-X.This SA function is based on FFT,and the traditional SA is based on sweep,so I think there should be some difference when we change VBW.Is there a digital low pass filter which determine the VBW?

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Dr_joel

@ Macree on Jan 16, 2017 10:06 AM

In all SA that use FFT, including PXA, UXA, MXA and PNA-X, the VBW function is emulated by taking repeated FFTs and averaging the result. For a RBW/VBW ratio of 10, there are 5 FFTs averaged before the result is processed. The FFT averaging occurs before image rejection and so it helps to stabilize the result for signals near the noise floor.

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Dr_joel

Jan 16, 2017 10:03 AM

On the LO's, thanks for the table. In a previous note I mentioned "it's complicated". In fact, for each setting such as normal, then number of LO's is either N or N+1, depending on complicated things, so on average there are (2N+1)/2 LOs. So in your table, for 1 GHz, it would be 1000*4.5/68=66 which is close to your measured value of 65. If the span is small, we might need 1 or 2 extra LO's so the scaling is slightly off. If you have more detailed questions on this please email me and I can follow up.

As to window and resolution. The window increases the number of FFT points, but does not change the resolution bandwidth at all. Rather, the extra sample points are used to lower the side lobes. It is true the DFT frequency bin is smaller but in fact the resolution (that is the ability to separate signals) is the same, or to put it another way the equivalent bandwidth of the filter. More DFT resolution does not necessarily mean more signal resolution. It depends upon the filter shape. The Gaussian filter, with 90 dB sidelobes is about 1.87 longer than the no window case. The flattop filter, which has a flatter inband response and a sharper skirt, but still the same 3 dB noise BW, has about 2.4 more taps.

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Macree Jan 10, 2017 1:01 AM
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Another question is how dose the system change VBW in PNA-X.This SA function is based on FFT,and the traditional SA is based on sweep,so I think there should be some difference when we change VBW.Is there a digital low pass filter which determine the VBW?
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- Dr_joel @ Macree on Jan 16, 2017 10:06 AM
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  In all SA that use FFT, including PXA, UXA, MXA and PNA-X, the VBW function is emulated by taking repeated FFTs and averaging the result. For a RBW/VBW ratio of 10, there are 5 FFTs averaged before the result is processed. The FFT averaging occurs before image rejection and so it helps to stabilize the result for signals near the noise floor.
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Dr_joel Jan 16, 2017 10:03 AM
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On the LO's, thanks for the table. In a previous note I mentioned "it's complicated". In fact, for each setting such as normal, then number of LO's is either N or N+1, depending on complicated things, so on average there are (2N+1)/2 LOs. So in your table, for 1 GHz, it would be 1000*4.5/68=66 which is close to your measured value of 65. If the span is small, we might need 1 or 2 extra LO's so the scaling is slightly off. If you have more detailed questions on this please email me and I can follow up.

As to window and resolution. The window increases the number of FFT points, but does not change the resolution bandwidth at all. Rather, the extra sample points are used to lower the side lobes. It is true the DFT frequency bin is smaller but in fact the resolution (that is the ability to separate signals) is the same, or to put it another way the equivalent bandwidth of the filter. More DFT resolution does not necessarily mean more signal resolution. It depends upon the filter shape. The Gaussian filter, with 90 dB sidelobes is about 1.87 longer than the no window case. The flattop filter, which has a flatter inband response and a sharper skirt, but still the same 3 dB noise BW, has about 2.4 more taps.
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Several additional questions about SA option in PNA-X

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