Hi All, This is a theoritical type question. We know that, under all conditions there is an ambient thermal noise present and its value is -174+10log(B) dBm. Where B is bandwidth in Hz. For any signal to be received the input power of signal should be higher than this limit. Now if you consider the case of wideband radios for example CDMA-SS or UWB signals, each symbol is coded over multiple chips(pulses). Hence for each chip to be correctly detected at the receiver, the received power of each chip should be greater than -174+10log(Bc) (where Bc is the bandwidth of chip). Is my understanding correct ? Now since Bc is much higher number than B of narrowband radios, is it safe to assume that sensitivity of Wideband receivers is much lower than that of narrow band receivers ? For a given pathloss does this then mean CDMA transmitters trasmit at much high power than GSM transmitters? regards ashu
CDMA receiver sensitivity
Started by ●June 12, 2009
Reply by ●June 12, 20092009-06-12
On Jun 12, 9:44�am, ashu <ashutosh.ghildi...@gmail.com> wrote:> Hi All, > > This is a theoritical type question. > > We know that, under all conditions there is an ambient thermal noise > present and its value is �-174+10log(B) dBm. Where B is bandwidth in > Hz. For any signal to be received the input power of signal should be > higher than this limit. > > Now if you consider the case of wideband radios for example CDMA-SS or > UWB signals, each symbol is coded over multiple chips(pulses). Hence > for each chip to be correctly detected at the receiver, the received > power of each chip should be greater than -174+10log(Bc) (where Bc is > the bandwidth of chip). > > Is my understanding correct ? > > Now since Bc is much higher number than B of narrowband radios, is it > safe to assume that sensitivity of Wideband receivers is much lower > than that of narrow band receivers ? > > For a given pathloss does this then mean CDMA transmitters trasmit at > much high power than GSM transmitters? > > regards > ashuYou forgot to include the noise figure of the receiver, which makes the noise higher. Also, the signal should exceed the noise by more than 0 dB. John
Reply by ●June 12, 20092009-06-12
On Jun 12, 4:48�pm, John <sampson...@gmail.com> wrote:> On Jun 12, 9:44�am, ashu <ashutosh.ghildi...@gmail.com> wrote: > > > > > Hi All, > > > This is a theoritical type question. > > > We know that, under all conditions there is an ambient thermal noise > > present and its value is �-174+10log(B) dBm. Where B is bandwidth in > > Hz. For any signal to be received the input power of signal should be > > higher than this limit. > > > Now if you consider the case of wideband radios for example CDMA-SS or > > UWB signals, each symbol is coded over multiple chips(pulses). Hence > > for each chip to be correctly detected at the receiver, the received > > power of each chip should be greater than -174+10log(Bc) (where Bc is > > the bandwidth of chip). > > > Is my understanding correct ? > > > Now since Bc is much higher number than B of narrowband radios, is itThanks john yes, NF will come to play, but the minimal signal strength (assuming 0 NF) lets say will be this ... right ?> > safe to assume that sensitivity of Wideband receivers is much lower > > than that of narrow band receivers ? > > > For a given pathloss does this then mean CDMA transmitters trasmit at > > much high power than GSM transmitters? > > > regards > > ashu > > You forgot to include the noise figure of the receiver, which makes > the noise higher. Also, the signal should exceed the noise by more > than 0 dB. > > John
Reply by ●June 12, 20092009-06-12
On Fri, 12 Jun 2009 06:44:39 -0700, ashu wrote:> Hi All, > > This is a theoritical type question. > > We know that, under all conditions there is an ambient thermal noise > present and its value is -174+10log(B) dBm. Where B is bandwidth in Hz. > For any signal to be received the input power of signal should be higher > than this limit. > > Now if you consider the case of wideband radios for example CDMA-SS or > UWB signals, each symbol is coded over multiple chips(pulses). Hence for > each chip to be correctly detected at the receiver, the received power > of each chip should be greater than -174+10log(Bc) (where Bc is the > bandwidth of chip). > > Is my understanding correct ? > > Now since Bc is much higher number than B of narrowband radios, is it > safe to assume that sensitivity of Wideband receivers is much lower than > that of narrow band receivers ? > > For a given pathloss does this then mean CDMA transmitters trasmit at > much high power than GSM transmitters?No, your understanding is not correct. The important way that it is not correct is in assuming that each chip has to be correctly received for CDMA (or any other spread-spectrum technique) to work. The CDMA receive process works like a bandpass filter in that many small bits of input are assembled together into a slowly-varying signal -- it's just that what's assembled together doesn't happen to be narrow band. To use the noise power in your calculations you have to take the bandwidth of the signal _after_ despreading. The unimportant (at least for this application) way that your understanding is incorrect is in taking the -178dBm/Hz number as gospel. Yes, that's the correct number when all the antenna sees is black-body radiation at 30 degrees centigrade. No, that's not the correct number if you have a narrow beam antenna pointed at the sun, the moon, or at a dark patch of the sky. So for cell phone applications you can take the effective temperature of the surroundings to be 30C, and your assumption is correct. But it's not universally correct. -- http://www.wescottdesign.com
Reply by ●June 12, 20092009-06-12
Hi Tim, thanks, not considering the temprature issue to be important. (lets say its 30C) I think we still have to receieve all the chips correctly, only then orthogonal PN sequence will despread the signal properly else the correlation will give another noise like signal.. isnt it ? that is the way we maintain multi user possibility in CDMA. Another thing is for each individual chip to be received correctly, the chip power after BPF must be greater than -174+10log(Bc) where Bc is the bandwidth of the chip. (which is very large). Essentially what I want to say is that the chip power must be greater than the noise power in the chip bandwidth(or the bandwidth after band pass filter) for the chip to be received correctly. regards ashu On Jun 12, 5:13�pm, Tim Wescott <t...@seemywebsite.com> wrote:> On Fri, 12 Jun 2009 06:44:39 -0700, ashu wrote: > > Hi All, > > > This is a theoritical type question. > > > We know that, under all conditions there is an ambient thermal noise > > present and its value is �-174+10log(B) dBm. Where B is bandwidth in Hz. > > For any signal to be received the input power of signal should be higher > > than this limit. > > > Now if you consider the case of wideband radios for example CDMA-SS or > > UWB signals, each symbol is coded over multiple chips(pulses). Hence for > > each chip to be correctly detected at the receiver, the received power > > of each chip should be greater than -174+10log(Bc) (where Bc is the > > bandwidth of chip). > > > Is my understanding correct ? > > > Now since Bc is much higher number than B of narrowband radios, is it > > safe to assume that sensitivity of Wideband receivers is much lower than > > that of narrow band receivers ? > > > For a given pathloss does this then mean CDMA transmitters trasmit at > > much high power than GSM transmitters? > > No, your understanding is not correct. > > The important way that it is not correct is in assuming that each chip > has to be correctly received for CDMA (or any other spread-spectrum > technique) to work. �The CDMA receive process works like a bandpass > filter in that many small bits of input are assembled together into a > slowly-varying signal -- it's just that what's assembled together doesn't > happen to be narrow band. �To use the noise power in your calculations > you have to take the bandwidth of the signal _after_ despreading. > > The unimportant (at least for this application) way that your > understanding is incorrect is in taking the -178dBm/Hz number as gospel. � > Yes, that's the correct number when all the antenna sees is black-body > radiation at 30 degrees centigrade. �No, that's not the correct number if > you have a narrow beam antenna pointed at the sun, the moon, or at a dark > patch of the sky. �So for cell phone applications you can take the > effective temperature of the surroundings to be 30C, and your assumption > is correct. �But it's not universally correct. > > --http://www.wescottdesign.com
Reply by ●June 12, 20092009-06-12
On Jun 12, 12:02 pm, ashu <ashutosh.ghildi...@gmail.com> wrote:> Hi Tim, > thanks, not considering the temprature issue to be important. (lets > say its 30C) > > I think we still have to receieve all the chips correctly, only then > orthogonal PN sequence will despread the signal properly else the > correlation will give another noise like signal.. isnt it ? that is > the way we maintain multi user possibility in CDMA. > > Another thing is for each individual chip to be received correctly, > the chip power after BPF must be greater than -174+10log(Bc) > where Bc is the bandwidth of the chip. (which is very large). > > Essentially what I want to say is that the chip power must be greater > than the noise power in the chip bandwidth(or the bandwidth after band > pass filter) for the chip to be received correctly. >The chip power need not be greater than the noise power. In fact, in practice, the chip SNR may be as low as (say) -20 dB and the decoding can still be successful. Think of it as each chip carrying the same bit with a multiplicative factor determined by the spreading sequence. So, if you have 128 chips corresponding to the same bit, they can all be combined together to form the estimate of the transmitted bit, so each individual chip need not be sent at a high power.> regards > ashu > > On Jun 12, 5:13 pm, Tim Wescott <t...@seemywebsite.com> wrote: > > > On Fri, 12 Jun 2009 06:44:39 -0700, ashu wrote: > > > Hi All, > > > > This is a theoritical type question. > > > > We know that, under all conditions there is an ambient thermal noise > > > present and its value is -174+10log(B) dBm. Where B is bandwidth in Hz. > > > For any signal to be received the input power of signal should be higher > > > than this limit. > > > > Now if you consider the case of wideband radios for example CDMA-SS or > > > UWB signals, each symbol is coded over multiple chips(pulses). Hence for > > > each chip to be correctly detected at the receiver, the received power > > > of each chip should be greater than -174+10log(Bc) (where Bc is the > > > bandwidth of chip). > > > > Is my understanding correct ? > > > > Now since Bc is much higher number than B of narrowband radios, is it > > > safe to assume that sensitivity of Wideband receivers is much lower than > > > that of narrow band receivers ? > > > > For a given pathloss does this then mean CDMA transmitters trasmit at > > > much high power than GSM transmitters? > > > No, your understanding is not correct. > > > The important way that it is not correct is in assuming that each chip > > has to be correctly received for CDMA (or any other spread-spectrum > > technique) to work. The CDMA receive process works like a bandpass > > filter in that many small bits of input are assembled together into a > > slowly-varying signal -- it's just that what's assembled together doesn't > > happen to be narrow band. To use the noise power in your calculations > > you have to take the bandwidth of the signal _after_ despreading. > > > The unimportant (at least for this application) way that your > > understanding is incorrect is in taking the -178dBm/Hz number as gospel. > > Yes, that's the correct number when all the antenna sees is black-body > > radiation at 30 degrees centigrade. No, that's not the correct number if > > you have a narrow beam antenna pointed at the sun, the moon, or at a dark > > patch of the sky. So for cell phone applications you can take the > > effective temperature of the surroundings to be 30C, and your assumption > > is correct. But it's not universally correct. > > > --http://www.wescottdesign.com
Reply by ●June 12, 20092009-06-12
A question for you. If you have two spreading sequences, one length 4 and the other length 256, which performs better (and why)? Regards ashu wrote:> Hi Tim, > thanks, not considering the temprature issue to be important. (lets > say its 30C) > > I think we still have to receieve all the chips correctly, only then > orthogonal PN sequence will despread the signal properly else the > correlation will give another noise like signal.. isnt it ? that is > the way we maintain multi user possibility in CDMA. > > Another thing is for each individual chip to be received correctly, > the chip power after BPF must be greater than -174+10log(Bc) > where Bc is the bandwidth of the chip. (which is very large). > > Essentially what I want to say is that the chip power must be greater > than the noise power in the chip bandwidth(or the bandwidth after band > pass filter) for the chip to be received correctly. > > regards > ashu > > On Jun 12, 5:13 pm, Tim Wescott <t...@seemywebsite.com> wrote: >> On Fri, 12 Jun 2009 06:44:39 -0700, ashu wrote: >>> Hi All, >>> This is a theoritical type question. >>> We know that, under all conditions there is an ambient thermal noise >>> present and its value is -174+10log(B) dBm. Where B is bandwidth in Hz. >>> For any signal to be received the input power of signal should be higher >>> than this limit. >>> Now if you consider the case of wideband radios for example CDMA-SS or >>> UWB signals, each symbol is coded over multiple chips(pulses). Hence for >>> each chip to be correctly detected at the receiver, the received power >>> of each chip should be greater than -174+10log(Bc) (where Bc is the >>> bandwidth of chip). >>> Is my understanding correct ? >>> Now since Bc is much higher number than B of narrowband radios, is it >>> safe to assume that sensitivity of Wideband receivers is much lower than >>> that of narrow band receivers ? >>> For a given pathloss does this then mean CDMA transmitters trasmit at >>> much high power than GSM transmitters? >> No, your understanding is not correct. >> >> The important way that it is not correct is in assuming that each chip >> has to be correctly received for CDMA (or any other spread-spectrum >> technique) to work. The CDMA receive process works like a bandpass >> filter in that many small bits of input are assembled together into a >> slowly-varying signal -- it's just that what's assembled together doesn't >> happen to be narrow band. To use the noise power in your calculations >> you have to take the bandwidth of the signal _after_ despreading. >> >> The unimportant (at least for this application) way that your >> understanding is incorrect is in taking the -178dBm/Hz number as gospel. >> Yes, that's the correct number when all the antenna sees is black-body >> radiation at 30 degrees centigrade. No, that's not the correct number if >> you have a narrow beam antenna pointed at the sun, the moon, or at a dark >> patch of the sky. So for cell phone applications you can take the >> effective temperature of the surroundings to be 30C, and your assumption >> is correct. But it's not universally correct. >> >> --http://www.wescottdesign.com >
Reply by ●June 12, 20092009-06-12
On Jun 12, 7:44�pm, Karl Molnar <karl.mol...@ericsson.com> wrote:> A question for you. > > If you have two spreading sequences, one length 4 and the other length > 256, which performs better (and why)? > > Regards > > ashu wrote: > > Hi Tim, > > thanks, not considering the temprature issue to be important. (lets > > say its 30C) > > > I think we still have to receieve all the chips correctly, only then > > orthogonal PN sequence will despread the signal properly else the > > correlation will give another noise like signal.. isnt it ? �that is > > the way we maintain multi user possibility in CDMA. > > > Another thing is for each individual chip to be received correctly, > > the chip power after BPF must be greater than -174+10log(Bc) > > where Bc is the bandwidth of the chip. (which is very large). > > > Essentially what I want to say is that the chip power must be greater > > than the noise power in the chip bandwidth(or the bandwidth after band > > pass filter) for the chip to be received correctly. > > > regards > > ashu > > > On Jun 12, 5:13 pm, Tim Wescott <t...@seemywebsite.com> wrote: > >> On Fri, 12 Jun 2009 06:44:39 -0700, ashu wrote: > >>> Hi All, > >>> This is a theoritical type question. > >>> We know that, under all conditions there is an ambient thermal noise > >>> present and its value is �-174+10log(B) dBm. Where B is bandwidth in Hz. > >>> For any signal to be received the input power of signal should be higher > >>> than this limit. > >>> Now if you consider the case of wideband radios for example CDMA-SS or > >>> UWB signals, each symbol is coded over multiple chips(pulses). Hence for > >>> each chip to be correctly detected at the receiver, the received power > >>> of each chip should be greater than -174+10log(Bc) (where Bc is the > >>> bandwidth of chip). > >>> Is my understanding correct ? > >>> Now since Bc is much higher number than B of narrowband radios, is it > >>> safe to assume that sensitivity of Wideband receivers is much lower than > >>> that of narrow band receivers ? > >>> For a given pathloss does this then mean CDMA transmitters trasmit at > >>> much high power than GSM transmitters? > >> No, your understanding is not correct. > > >> The important way that it is not correct is in assuming that each chip > >> has to be correctly received for CDMA (or any other spread-spectrum > >> technique) to work. �The CDMA receive process works like a bandpass > >> filter in that many small bits of input are assembled together into a > >> slowly-varying signal -- it's just that what's assembled together doesn't > >> happen to be narrow band. �To use the noise power in your calculations > >> you have to take the bandwidth of the signal _after_ despreading. > > >> The unimportant (at least for this application) way that your > >> understanding is incorrect is in taking the -178dBm/Hz number as gospel. � > >> Yes, that's the correct number when all the antenna sees is black-body > >> radiation at 30 degrees centigrade. �No, that's not the correct number if > >> you have a narrow beam antenna pointed at the sun, the moon, or at a dark > >> patch of the sky. �So for cell phone applications you can take the > >> effective temperature of the surroundings to be 30C, and your assumption > >> is correct. �But it's not universally correct. > > >> --http://www.wescottdesign.comHi Dilip and Karl, thanks! I would like a more intuitive way of looking at "processing gain". In my understanding processing gain comes into picture only after the 'chip' has been correctly detected. And for the chip to be correctly detected, dont you think chip power should be higher than noise power(or voltage). If a chip (a blip in the time) is buried deep into noise, how can the receiver 'sense' the event of chip arrival. Karl: Look at the question another way, if sending more chips can really improve the reception, then perhaps there is no limit to the lowest signal strength one can receive. All we need to do send in many many like zillions of chips and we detect -1000000dBm signal ? regards ashu
Reply by ●June 12, 20092009-06-12
On Jun 12, 3:06�pm, ashu <ashutosh.ghildi...@gmail.com> wrote:> On Jun 12, 7:44�pm, Karl Molnar <karl.mol...@ericsson.com> wrote: > > > > > A question for you. > > > If you have two spreading sequences, one length 4 and the other length > > 256, which performs better (and why)? > > > Regards > > > ashu wrote: > > > Hi Tim, > > > thanks, not considering the temprature issue to be important. (lets > > > say its 30C) > > > > I think we still have to receieve all the chips correctly, only then > > > orthogonal PN sequence will despread the signal properly else the > > > correlation will give another noise like signal.. isnt it ? �that is > > > the way we maintain multi user possibility in CDMA. > > > > Another thing is for each individual chip to be received correctly, > > > the chip power after BPF must be greater than -174+10log(Bc) > > > where Bc is the bandwidth of the chip. (which is very large). > > > > Essentially what I want to say is that the chip power must be greater > > > than the noise power in the chip bandwidth(or the bandwidth after band > > > pass filter) for the chip to be received correctly. > > > > regards > > > ashu > > > > On Jun 12, 5:13 pm, Tim Wescott <t...@seemywebsite.com> wrote: > > >> On Fri, 12 Jun 2009 06:44:39 -0700, ashu wrote: > > >>> Hi All, > > >>> This is a theoritical type question. > > >>> We know that, under all conditions there is an ambient thermal noise > > >>> present and its value is �-174+10log(B) dBm. Where B is bandwidth in Hz. > > >>> For any signal to be received the input power of signal should be higher > > >>> than this limit. > > >>> Now if you consider the case of wideband radios for example CDMA-SS or > > >>> UWB signals, each symbol is coded over multiple chips(pulses). Hence for > > >>> each chip to be correctly detected at the receiver, the received power > > >>> of each chip should be greater than -174+10log(Bc) (where Bc is the > > >>> bandwidth of chip). > > >>> Is my understanding correct ? > > >>> Now since Bc is much higher number than B of narrowband radios, is it > > >>> safe to assume that sensitivity of Wideband receivers is much lower than > > >>> that of narrow band receivers ? > > >>> For a given pathloss does this then mean CDMA transmitters trasmit at > > >>> much high power than GSM transmitters? > > >> No, your understanding is not correct. > > > >> The important way that it is not correct is in assuming that each chip > > >> has to be correctly received for CDMA (or any other spread-spectrum > > >> technique) to work. �The CDMA receive process works like a bandpass > > >> filter in that many small bits of input are assembled together into a > > >> slowly-varying signal -- it's just that what's assembled together doesn't > > >> happen to be narrow band. �To use the noise power in your calculations > > >> you have to take the bandwidth of the signal _after_ despreading. > > > >> The unimportant (at least for this application) way that your > > >> understanding is incorrect is in taking the -178dBm/Hz number as gospel. � > > >> Yes, that's the correct number when all the antenna sees is black-body > > >> radiation at 30 degrees centigrade. �No, that's not the correct number if > > >> you have a narrow beam antenna pointed at the sun, the moon, or at a dark > > >> patch of the sky. �So for cell phone applications you can take the > > >> effective temperature of the surroundings to be 30C, and your assumption > > >> is correct. �But it's not universally correct. > > > >> --http://www.wescottdesign.com > > Hi Dilip and Karl, > > thanks! I would like a more intuitive way of looking at "processing > gain". > > In my understanding processing gain comes into picture only after the > 'chip' has been correctly detected. > And for the chip to be correctly detected, dont you think chip power > should be higher than noise power(or voltage). > If a chip (a blip in the time) is buried deep into noise, how can the > receiver 'sense' the event of chip arrival. > > Karl: Look at the question another way, if sending more chips can > really improve the reception, then perhaps there is no limit to the > lowest signal strength one can receive. All we need to do send in many > many like zillions of chips and we detect -1000000dBm signal ? > > regards > ashuDSSS is not some magic technique that allows you to receive a signal of arbitrarily low power just by increasing the length of the spreading code. Given an information bit rate, if you make the spreading code longer, each chip gets narrower. Narrower chips means that the resulting signal is spread over a larger bandwidth. Spreading the same signal power (you don't change the total signal power by spreading it) over a larger bandwidth results in a lower SNR, since there is more noise power in the larger frequency band. Therefore, by increasing chip rate, you're actually going to receive your signal at a lower SNR at the receiver. However, the processing gain inherent in the despreading process allows you to undo this "power dilution" by sucking all of the signal power from the wide swath of bandwidth and compressing it into a smaller region, commensurate with the information bitrate. This is done optimally by cross-correlating the spreading sequence and the received signal; since the received noise is uncorrelated with the spreading sequence, the signal of interest is enhanced much more than the noise is. This gives you the processing gain. It's important to note, though, that you can't pull out any more signal power at the receiver than you transmitted at the other end. So in an ideal world, the best that a spreader/despreader combination can do is give you perfect reversibility (and thus act as if they aren't there at all). It's not a magic bullet that allows you to just spread the signal out further to improve reception. The key statistic is always Eb/No; how much energy did you receive per information bit? That's not to say that DSSS isn't useful; it has advantages that would drive you to use the technique when appropriate. The wide bandwidth of the waveform makes the despreader's output less susceptible to multipath interference (you get nice sharp peaks in the despreader output if your code is a good one), you can use the spreading process to "push your signal toward the noise floor" and make it more inconspicuous, or you could use sets of carefully-selected spreading codes for multiple access (i.e. CDMA). Jason
Reply by ●June 12, 20092009-06-12
On Jun 12, 3:06�pm, ashu <ashutosh.ghildi...@gmail.com> wrote:> On Jun 12, 7:44�pm, Karl Molnar <karl.mol...@ericsson.com> wrote: > > > > > > > A question for you. > > > If you have two spreading sequences, one length 4 and the other length > > 256, which performs better (and why)? > > > Regards > > > ashu wrote: > > > Hi Tim, > > > thanks, not considering the temprature issue to be important. (lets > > > say its 30C) > > > > I think we still have to receieve all the chips correctly, only then > > > orthogonal PN sequence will despread the signal properly else the > > > correlation will give another noise like signal.. isnt it ? �that is > > > the way we maintain multi user possibility in CDMA. > > > > Another thing is for each individual chip to be received correctly, > > > the chip power after BPF must be greater than -174+10log(Bc) > > > where Bc is the bandwidth of the chip. (which is very large). > > > > Essentially what I want to say is that the chip power must be greater > > > than the noise power in the chip bandwidth(or the bandwidth after band > > > pass filter) for the chip to be received correctly. > > > > regards > > > ashu > > > > On Jun 12, 5:13 pm, Tim Wescott <t...@seemywebsite.com> wrote: > > >> On Fri, 12 Jun 2009 06:44:39 -0700, ashu wrote: > > >>> Hi All, > > >>> This is a theoritical type question. > > >>> We know that, under all conditions there is an ambient thermal noise > > >>> present and its value is �-174+10log(B) dBm. Where B is bandwidth in Hz. > > >>> For any signal to be received the input power of signal should be higher > > >>> than this limit. > > >>> Now if you consider the case of wideband radios for example CDMA-SS or > > >>> UWB signals, each symbol is coded over multiple chips(pulses). Hence for > > >>> each chip to be correctly detected at the receiver, the received power > > >>> of each chip should be greater than -174+10log(Bc) (where Bc is the > > >>> bandwidth of chip). > > >>> Is my understanding correct ? > > >>> Now since Bc is much higher number than B of narrowband radios, is it > > >>> safe to assume that sensitivity of Wideband receivers is much lower than > > >>> that of narrow band receivers ? > > >>> For a given pathloss does this then mean CDMA transmitters trasmit at > > >>> much high power than GSM transmitters? > > >> No, your understanding is not correct. > > > >> The important way that it is not correct is in assuming that each chip > > >> has to be correctly received for CDMA (or any other spread-spectrum > > >> technique) to work. �The CDMA receive process works like a bandpass > > >> filter in that many small bits of input are assembled together into a > > >> slowly-varying signal -- it's just that what's assembled together doesn't > > >> happen to be narrow band. �To use the noise power in your calculations > > >> you have to take the bandwidth of the signal _after_ despreading. > > > >> The unimportant (at least for this application) way that your > > >> understanding is incorrect is in taking the -178dBm/Hz number as gospel. � > > >> Yes, that's the correct number when all the antenna sees is black-body > > >> radiation at 30 degrees centigrade. �No, that's not the correct number if > > >> you have a narrow beam antenna pointed at the sun, the moon, or at a dark > > >> patch of the sky. �So for cell phone applications you can take the > > >> effective temperature of the surroundings to be 30C, and your assumption > > >> is correct. �But it's not universally correct. > > > >> --http://www.wescottdesign.com > > Hi Dilip and Karl, > > thanks! I would like a more intuitive way of looking at "processing > gain". > > In my understanding processing gain comes into picture only after the > 'chip' has been correctly detected. > And for the chip to be correctly detected, dont you think chip power > should be higher than noise power(or voltage). > If a chip (a blip in the time) is buried deep into noise, how can the > receiver 'sense' the event of chip arrival. > > Karl: Look at the question another way, if sending more chips can > really improve the reception, then perhaps there is no limit to the > lowest signal strength one can receive. All we need to do send in many > many like zillions of chips and we detect -1000000dBm signal ? > > regards > ashu- Hide quoted text - > > - Show quoted text -Hello Ashu, Think of the coding gain as arising from a sort of integration (part of a correlation) where the noise tends to average out. So yes the signal power can be lower than the noise power. As to why you can't average forever - one reason is the averaged noise power will decrease by the square root of the size of the averaging interval. I.e., average 100 times longer will reduce your noise by about 10 times. Clay Look up the coding gain used by the GPS system!
