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receiver response to a CW input for E8362B

Question asked by lzw on Mar 10, 2006
Latest reply on Mar 30, 2006 by Dr_joel

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Hi Gurus,

Can anyone explain a little bit about the receiver response to a CW input in the non-ratio measurement?

For example if we apply a -60 dBm 1.8 GHz CW at the port 2, and use receiver B to do the non ratio power measurement. It will give us two tones at the display. One is at 1.8 GHz and other is at the somewhere lower frequency end with offset (2nd LO+2nd IF+41.66...kHz).

From the block diagram, there is a a filter in front of the ADC. If we do the peak power reading of the two displayed tones, it power level is almost the same. When it seems there is no effect on the image tone from the 2nd mixer?

Thanks in advance!

Sincerely yours,

lzw

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dgun Mar 10, 2006 10:22 AM
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Hi Gurus,

Can anyone explain a little bit about the receiver response to a CW input in the non-ratio measurement?

For example if we apply a -60 dBm 1.8 GHz CW at the port 2, and use receiver B to do the non ratio power measurement. It will give us two tones at the display. One is at 1.8 GHz and other is at the somewhere lower frequency end with offset (2nd LO+2nd IF+41.66...kHz).

From the block diagram, there is a a filter in front of the ADC. If we do the peak power reading of the two displayed tones, it power level is almost the same. When it seems there is no effect on the image tone from the 2nd mixer?

"

I'm assuming that you have the PNA source turned off. If so, then there are a number of tones that you might see for a CW input from an external source. The external source will have harmonics, sidebands, and subharmonics. The PNA receiver will also have subharmonics at (1.8 GHz - (N-1)*IF)/N and the respective images at (1.8 GHz - (N+1)*IF)/N, where N is the positive integer subharmonic. Odd subharmonics are dominant. You usually won't see these because they will be outside the IF bandwidth of the points you measure. If you use an IF bandwidth greater than the point spacing, you will see them for sure. At frequencies below 50 MHz, the IF jumps around quite a bit, so the above equations usually won't work.

For high power levels that put the receiver into compression, you will also see tones at M *1.8 GHz, where M is an integer.

Of course, you shouldn't see any of this if you use the internal source and sweep it with the measurement.
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Dr_joel Mar 10, 2006 10:25 AM
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You are correct, the second mixer has no effect on the tones, because they are - at the same frequency!

Let me explain: The first mixer is not image free or image reject, so that a signal below or above the LO will convert with almost the same conversion efficientcy: Example: RF = 1 GHz fixed, LO sweeps from.9 to 1.1. When the LO is 8 MHz below the RF, an 8 MHz IF signal will show up at the input to the second converter, be converted, sampled by the ADC at 41 KHz, and be displayed. Then, when the LO sweeps to 8 MHz above the 1 GHz signal, another 8 MHz signal (same frequency as the first one) shows up in the first IF, is down-converted to the 41 kHz, sampled by the ADC and shows up again. The first converter might have a slight difference between the first and second signals (that is, the conversion loss may be different for LO=1-.008 and LO=1+.008 GHz), but not very much different.

So, there are not 2 signals, but only 1, at 1 GHz, that get's sampled twice, once with the LO 8 MHz below and once again when the LO is 8 MHz above.
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dgun Mar 10, 2006 5:18 PM
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"

The PNA receiver will also have subharmonics at (1.8 GHz - (N-1)*IF)/N and the respective images at (1.8 GHz - (N+1)*IF)/N, where N is the positive integer subharmonic.

For high power levels that put the receiver into compression, you will also see tones at M *1.8 GHz, where M is an integer.

"

I should clarify that these receiver tones are false responses, and not actual tones coming into or out of the test ports.
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lzw Mar 14, 2006 12:47 AM
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Thanks for your explanation.

I would like to ask another question: If I set the span=2MHz, and number of points for the linear sweep as 1001, so I have 2kHz/points. If I set the IF BW=10kHz, the LO sweep shall give at least 5 samples input to the ADC, or sampeld 5 times (I assume the LO sweeps through all the chosen points). Why only one peak is displayed for the CW input signal instead of four as I hade expected?

So the number of points set in sweep is not the same as the number of points of used, isn't it?

Even I use the setting of 2kHz/points and the IF BW to 3 kHz, so there is frequency overlapp between adjacent points in the sweep. I have not found more than one peak on display either. I have changed the CW power level from -65 dBm to -30 dBm in 5 dB step. (Did I mess up your reply, Daniel?)

Any suggestion about how should I choose the number of points used in a sweep. If I use more points than needed in setup, will I get punishment in sweep time?

Thanks again for your clarification!

Regards

lzw
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dgun Mar 14, 2006 8:48 AM
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I would like to ask another question: If I set the span=2MHz, and number of points for the linear sweep as 1001, so I have 2kHz/points. If I set the IF BW=10kHz, the LO sweep shall give at least 5 samples input to the ADC, or sampeld 5 times (I assume the LO sweeps through all the chosen points). Why only one peak is displayed for the CW input signal instead of four as I hade expected?
"

Why would you expect four peaks? There will still be one peak when the IF filter is centered on the IF response of the input CW signal. The other points will be somewhat smaller the IF response of the CW signal will be in the skirt of the IF filter. Using your example, increase the number of points to 10001 and you will see the shape of the IF filter.

"
So the number of points set in sweep is not the same as the number of points of used, isn't it?

Even I use the setting of 2kHz/points and the IF BW to 3 kHz, so there is frequency overlapp between adjacent points in the sweep. I have not found more than one peak on display either. I have changed the CW power level from -65 dBm to -30 dBm in 5 dB step. (Did I mess up your reply, Daniel?)
"

For a 1.8 GHz center frequency and span of 2 MHz, you won't see the IF image. Depending on the PNA in question, the IF image will be at a frequency around 15 or 16 MHz lower than the 1.8 GHz signal.

"
Any suggestion about how should I choose the number of points used in a sweep. If I use more points than needed in setup, will I get punishment in sweep time?

"

What are you trying to measure?
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Dr_joel Mar 14, 2006 11:42 AM
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Thanks for your explanation.

I would like to ask another question: If I set the span=2MHz, and number of points for the linear sweep as 1001, so I have 2kHz/points. If I set the IF BW=10kHz, the LO sweep shall give at least 5 samples input to the ADC, or sampeld 5 times (I assume the LO sweeps through all the chosen points). Why only one peak is displayed for the CW input signal instead of four as I hade expected?
"

Depends on if you are in stepped or swept mode. You will ALWAYS be in stepped mode if the IF is 1 kHz or less in the PNA. You can choose stepped or swept under sweep setup menu. If you are stepped, the PNA steps to each frequency before the ADC readings are used. If you use swept, the ADC reading are used when you are close to the frequency. If you are in auto-sweeptime (set sweep time to 0, and it resets to the min vaule), the ADC is sampling is continous, if the PNA can sweep that fast. If the sweep time is longer than minimum, the ADC readings are only used for readings near the frequency, and thrown away in between. ADC sample rate is always the same, and the IF BW determines how many readings are used to compute the signal mag and phase.

"
So the number of points set in sweep is not the same as the number of points of used, isn't it?

Even I use the setting of 2kHz/points and the IF BW to 3 kHz, so there is frequency overlapp between adjacent points in the sweep. I have not found more than one peak on display either. I have changed the CW power level from -65 dBm to -30 dBm in 5 dB step. (Did I mess up your reply, Daniel?)

"
In this case, you should see the IF filter response.

"

Any suggestion about how should I choose the number of points used in a sweep. If I use more points than needed in setup, will I get punishment in sweep time?

"

If you are measuring a known signal, you can just measure 1 point at the value you want. If you want to measure several known frequencies, look to using the segment sweep. If you want to use the PNA as a spectrum analyzer to look for unknown signals, set the IF BW to the level you need for resolution, and using the span that you want, set the number of points by Span/IFBW. This will give you a guarantee that you will have a measurment at each frequency in your span. If your span is greater than 2 times the PNA first IF (1.04 in the RF PNA, 8.33 in the microwave PNA) then you will see the IF image.

Joel
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lzw Mar 15, 2006 1:47 AM
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Thanks Dr_joel for your excellent explanation!

What I want to do is to use the PNA as a spectrum analyzer and measure the channel power for the GSMK signal. The filter response from the CW measurement will be used in the channel power integration. This is why I am interested the response of the possible input to the ADC.

The reason I hade expected "four peaks" was that I thought the IF filter should be the near rectangular type, so the trace data should have something close to 4-5 frequency points with similar amplitude input to ADC. (May I ask why IF can only be set to ... 2KHz, 3kHz, 5kHz, 7kHz, 10kHz ... fixed frequency?)

One thing I would like to report is that when I save trace to file, *.prn. The frequency is also same frequency for all the list. Save to CITI file looks ok, not for SnP file either for the non-ratio measurement. How shall get over this?

The firmware version is A.04.83 on my PNA.

Thanks again!

Sincerely yours,
lzw
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dgun Mar 21, 2006 3:39 PM
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What I want to do is to use the PNA as a spectrum analyzer and measure the channel power for the GSMK signal. The filter response from the CW measurement will be used in the channel power integration. This is why I am interested the response of the possible input to the ADC.

The reason I hade expected "four peaks" was that I thought the IF filter should be the near rectangular type, so the trace data should have something close to 4-5 frequency points with similar amplitude input to ADC. (May I ask why IF can only be set to ... 2KHz, 3kHz, 5kHz, 7kHz, 10kHz ... fixed frequency?)
"

See below to learn all about the IF filter:

http://www.agilent.com/find/pulsedRF

Technical Presentation: Pulsed-RF S-Parameter Measurements Using a Vector Network Analyzer. pages 26-34
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lzw Mar 23, 2006 5:26 AM
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Thanks Daniel,

I have a question in another topic.

Is there any plan to implement the variable smoothing ratio for each segment with a segment sweep in the firmware?

Best regards

lzw
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dgun Mar 24, 2006 3:32 PM
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Is there any plan to implement the variable smoothing ratio for each segment with a segment sweep in the firmware?
"

You should ask this in a new topic so there is a greater chance someone who knows the answer will see it.
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Dr_joel Mar 25, 2006 9:52 PM
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Thanks Daniel,

I have a question in another topic.

Is there any plan to implement the variable smoothing ratio for each segment with a segment sweep in the firmware?

Best regards

lzw
"

Short answer: Nope.

Long answer: This gets really tough as there is no control of the minimum number of points in each segment. So, even if we had an entry in the segment table for different smoothing per segment, it get's problematic how to actually make it work. For example, would you want the smoothing window to restart with each segment, or take data from previous segments to get enough points to smooth the current segment.

Now, if you really wanted different smoothing per frequency band, you might consider using more channels, instead of segments. Each channel (you can have up to 32), could have it's own smoothing function. For the most part, this would not be a lot slower than doing segment sweeps.

Finally, please let us in on why you need such functionality. What are you trying to measure, and how are you using smoothing. I ask this as smoothing is sometimes mis-understood and used in ways that elliminate or distort the true response of a DUT.

Best regards, Joel
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lzw Mar 29, 2006 11:47 PM
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Hi Dr Joel,
Actually we are using the segment sweep to meaasure the highly rejective cavity filter. We use segment sweep to reduce the sweep time for the whole frequency band to be measured. Each segment has different power and IF BW settings, of course also the number of points. At the somwhere of 15-20 MHz away from the passband edge, the filter attantuation has a steep slope (> 100 dB/octave) and high antentuation (-60 dB~- -80 dB) compared to passband. To reduce the noise floor for segement where the attentuation is more 90 dB, I would like to have a smoothing ratio around 20%, but for the place mentioned early I have a problem by choosing the number of the points. If a reasonable response is expected, I must have many points than I really wanted. Otherwise I have to turn off the smoothing on the trace.

Hope this is what you would like to kinow.

Best regards

lzw
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Dr_joel Mar 30, 2006 9:24 AM
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Hi lzw,
Thanks for the information. In general, we would say that it's a bit unusual for using smoothing to reduce noise in the stop band of a filter. The reason for this is that for some filters, there can be a "spurious passband" that can be quite narrow, and as such, the smoothing would reduce the height of the peak, and show a filter which passes isolation when in fact it does not. If you need to reduce noise, another idea is to reverse the directional coupler on port 2 of the VNA, which is possible if you have the configurable test set. In this way, you improve your noise floor by about 14 dB. Now, in the passband, the max power of the VNA will be too high for the receiver, so it is necessary to reduce the power from the source, but in the stop band, you can use high power again.

Joel
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receiver response to a CW input for E8362B

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