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| | Inquiry of noise figure - RF Cafe Forums |
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James Post subject: Inquiry of noise figure Unread postPosted: Sat Mar 12, 2005 10:36 am
Dear Friends,
I am trying to improve the noise figure my system.
There is like this.
a) Current BPF component : 5.15 ~ 5.825GHz (700MHz Bandwidth) b) Will be changed BPF component : 5.725 ~ 5.825GHz (100MHz Bandwidth)
If I changed the a) to b), noise figure will be improved up to 5dB. Is that right? Assuming same input signal comes.
I appreciate any help or comment or feedback,
Many thanks
:roll:
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Kirt Blattenberger Post subject: Here's a start Unread postPosted: Sat Mar 12, 2005 6:38 pm Offline Site Admin User avatar
Joined: Sun Aug 03, 2003 2:02 pm Posts: 308 Location: Erie, PA Greetings James:
Changing the filter bandwidth will not affect the noise figure calculation, only the insertion loss (gain) of the filter will. However, reducing the bandwidth by a factor of 7 as you propose will reduce the noise power by 10*log(7)=8.45 dB. That in turn will improve your signal-to-noise ratio by 8.45 dB, since the noise power is dependent on the system bandwidth. So, the minimum detectable signal will be 8.45 dB lower. That translates to a range increase factor of 10^(8.45/20)=2.65.
These calculations hold for ideal environment and assume no other factors in the system equation change. It also assumes that the relative bandwidth changes are the noise power bandwidths of the filters and not just the 3 dB bandwidths, and that the final bandwidth at the detector is set by your new filter. At 100 MHz, it likely is not.
The benefit you will most likely realize is a reduction in the out-of-channel interference both from direct signals and from intermodulation products created by those out-of-band signals that end up in-band.
For all the formulas you need to do the calculations yourself, please go to this page and click on the links of interest. http://www.rfcafe.com/references/electrical.htm
On this page is a simple online cascade calculator: http://www.rfcafe.com/references/spread ... alc-ss.htm
Also, you can download RF Workbench (shareware by RF Cafe) that includes calculations for determining filter noise bandwidth: http://www.rfcafe.com/business/software ... kbench.htm
- Kirt Blattenberger :smt024
Top Profile
Guest Post subject: Unread postPosted: Sat Mar 12, 2005 8:01 pm
Dear Kirt: Thank for your careful answer regarding my question. Let me confirm a few things as follows:
1. I think the sensitvity will be 8.45dB better if I change 5.15 ~ 5.825GHz (700MHz Bandwidth) to 5.725 ~ 5.825GHz (100MHz Bandwidth). Is that right? Is sensitivy related with system bandwidth?
2. Based upon your reply, could you explain more details regarding range increase means? Does it cover more distance?
So, the minimum detectable signal will be 8.45 dB lower. That translates to a range increase factor of 10^(8.45/20)=2.65.
Best Regards, James
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Kirt Blattenberger Post subject: Unread postPosted: Sat Mar 12, 2005 9:43 pm Offline Site Admin User avatar
Joined: Sun Aug 03, 2003 2:02 pm Posts: 308 Location: Erie, PA Greetings James:
1. Your sensitivity will be improved if the changed BW represents the final BW that your detector or smapling system will see. That is, if somewhere down the receive chain, there is, say, a 50 kHz bandwith filter prior to detection, then having reduced the other filter from 700 MHz down to 100 MHz will have no measureable effect. As mentioned, the benefit you might experience would be a reduction of interfering signals that get translated inband. That still counts as a signal-to-noise improvement, but in a different way.
2. Signal strength in power falls off at a rate of 20 log(2) dB, which is approximately 6.02 dB for every doubling of the distance. So, every 6.02 dB of improved sensitivity theoretically results in being able to receive a signal twice as far away. I say theoretically because in the real world, multipath and fading due to barriers can make the change something other than 6.02 dB.
- Kirt Blattenberger :smt024
Posted 11/12/2012
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