All, question for you. I thought I had this sorted, but I think I'm missing a key point here... Okay, UMTS right. We have a chip rate of 3.84Mchip/sec. Now, the bandwidth of our system is calculated by: W = Fc/2 (1+a), where Fc = 3.84M and a = 0.22 (filter roll-off). W = 2.34MHz for UMTS at baseband Now, question time. To simulate this using complex baseband representation, I construct the signal as follows For a spreading factor (SF) = 4 we have room for 3.84M/4 = 960k sym/sec. Therefore, I can construct a complex vector (960k elements long) and then repeat each symbol by the SF and multiply by my spreading sequence. So we would have something like Tx Signal = 960k symbols * SF elements Spread Seq = 3.84M elements long Spread Signal = element by element multiplication of the Tx signal and the Spread seq The result of this is then fed into my RRC filter prior to transmission, so the sequence looks like this for, say, QPSK, where Fc = chip rate 3.84M, SF = 4 QPSK (960K signal at rate Fc/SF) -> Spread (signal at rate Fc) -> Upsample x 2 -> RRC (signal at rate 2Fc) -> TX So, question, the bandwidth of my spread signal is (from above) W = 2.34MHz, so how can I be sampling at a rate of 3.84M? This seems to be violating the Nyquist rate? Where have I gone wrong when calculating the relationship between the signal bandwidth and my system sampling rate? Anyone? Col
UMTS sampling freq and bandwidth
Started by ●October 28, 2003
Reply by ●October 28, 20032003-10-28
On 28 Oct 2003 12:27:47 -0800, cb135@hotmail.com (Col Brown) claimed:>All, question for you. I thought I had this sorted, but I think I'm >missing a key point here... > >Okay, UMTS right. We have a chip rate of 3.84Mchip/sec. Now, the >bandwidth of our system is calculated by: > >W = Fc/2 (1+a), where Fc = 3.84M and a = 0.22 (filter roll-off). > >W = 2.34MHz for UMTS at basebandI'm not an expert in WCDMA, but is't the bandwidth of the carrier ~5 MHz (including guard-band)?
Reply by ●October 29, 20032003-10-29
cb135@hotmail.com (Col Brown) wrote in message news:<a254af6b.0310281227.4ab7e88c@posting.google.com>...> All, question for you. I thought I had this sorted, but I think I'm > missing a key point here... > > Okay, UMTS right. We have a chip rate of 3.84Mchip/sec. Now, the > bandwidth of our system is calculated by: > > W = Fc/2 (1+a), where Fc = 3.84M and a = 0.22 (filter roll-off). > > W = 2.34MHz for UMTS at baseband > > Now, question time. To simulate this using complex baseband > representation, I construct the signal as follows > > For a spreading factor (SF) = 4 we have room for 3.84M/4 = 960k > sym/sec. Therefore, I can construct a complex vector (960k elements > long) and then repeat each symbol by the SF and multiply by my > spreading sequence. > > So we would have something like > > Tx Signal = 960k symbols * SF elements > Spread Seq = 3.84M elements long > > Spread Signal = element by element multiplication of the Tx signal and > the Spread seq > > The result of this is then fed into my RRC filter prior to > transmission, so the sequence looks like this for, say, QPSK, where Fc > = chip rate 3.84M, SF = 4 > > QPSK (960K signal at rate Fc/SF) -> Spread (signal at rate Fc) -> > Upsample x 2 -> RRC (signal at rate 2Fc) -> TX^ | | Filtering is done at 2*3.84 Mcps > 2 * 2.34MHz> So, question, the bandwidth of my spread signal is (from above) W = > 2.34MHz, so how can I be sampling at a rate of 3.84M? This seems to > be violating the Nyquist rate? Where have I gone wrong when > calculating the relationship between the signal bandwidth and my > system sampling rate? Anyone?The baseband output is at chip-rate ==> 3.84 Mcps. The output after the RRC filter has a bandwidth of 2.3424 MHz. Somewhere between baseband and RRC filter - you must be doing an upconversion (2X/4X) also - so that the O/P of the RRC filter is sampled at 7.68 Mcps(2X) / 15.36 Mcps(4X). Shouldn't that take care of Nyquist criteria ? Sachin
Reply by ●October 29, 20032003-10-29
I see your point, but I guess my question in return would be: well, even if I have upsampled prior to my RRC filter, then isn't it already too late? By sampling (or rather specifying) my signal at a chip rate of 3.84M, then I'm already aliased, and so my shifting the data upto 2 * 3.84Mchip/sec won't help. I guess my real question is why did they choose a chip rate of 3.84M? How does this relate to the bandwidth of the signal? To answer your question though, I perform the upsampling after the spreading and prior to the RRC filtering. Col scngupta@yahoo.com (Sachin Gupta) wrote in message news:<d2308724.0310290249.27fbc531@posting.google.com>...> cb135@hotmail.com (Col Brown) wrote in message news:<a254af6b.0310281227.4ab7e88c@posting.google.com>... > > All, question for you. I thought I had this sorted, but I think I'm > > missing a key point here... > > > > Okay, UMTS right. We have a chip rate of 3.84Mchip/sec. Now, the > > bandwidth of our system is calculated by: > > > > W = Fc/2 (1+a), where Fc = 3.84M and a = 0.22 (filter roll-off). > > > > W = 2.34MHz for UMTS at baseband > > > > Now, question time. To simulate this using complex baseband > > representation, I construct the signal as follows > > > > For a spreading factor (SF) = 4 we have room for 3.84M/4 = 960k > > sym/sec. Therefore, I can construct a complex vector (960k elements > > long) and then repeat each symbol by the SF and multiply by my > > spreading sequence. > > > > So we would have something like > > > > Tx Signal = 960k symbols * SF elements > > Spread Seq = 3.84M elements long > > > > Spread Signal = element by element multiplication of the Tx signal and > > the Spread seq > > > > The result of this is then fed into my RRC filter prior to > > transmission, so the sequence looks like this for, say, QPSK, where Fc > > = chip rate 3.84M, SF = 4 > > > > QPSK (960K signal at rate Fc/SF) -> Spread (signal at rate Fc) -> > > Upsample x 2 -> RRC (signal at rate 2Fc) -> TX > ^ > | > | > Filtering is done at 2*3.84 Mcps > 2 * 2.34MHz > > > So, question, the bandwidth of my spread signal is (from above) W = > > 2.34MHz, so how can I be sampling at a rate of 3.84M? This seems to > > be violating the Nyquist rate? Where have I gone wrong when > > calculating the relationship between the signal bandwidth and my > > system sampling rate? Anyone? > > The baseband output is at chip-rate ==> 3.84 Mcps. The output after > the RRC filter has a bandwidth of 2.3424 MHz. Somewhere between > baseband and RRC filter - you must be doing an upconversion (2X/4X) > also - so that the O/P of the RRC filter is sampled at 7.68 Mcps(2X) / > 15.36 Mcps(4X). > > Shouldn't that take care of Nyquist criteria ? > > Sachin
Reply by ●October 30, 20032003-10-30
cb135@hotmail.com (Col Brown) wrote in message news:<a254af6b.0310291307.6198e5f8@posting.google.com>...> I see your point, but I guess my question in return would be: well, > even if I have upsampled prior to my RRC filter, then isn't it already > too late? By sampling (or rather specifying) my signal at a chip rate > of 3.84M, then I'm already aliased, and so my shifting the data upto 2 > * 3.84Mchip/sec won't help.Never looked at it in this way. 3.84 Mcps data is modulated as such, forget about the symbol-rates before spreading for time being. So complex symbols @ 3.84 M are being modulated. And I have never had any dilemma about sampling a X bps digital data at more than X bits per second.> I guess my real question is why did they choose a chip rate of 3.84M?Don't know. If you go through the development history and some of the early proposals they do talk about some other chip-rates also, but somehow settled upon 3.84 Mcps finally.> How does this relate to the bandwidth of the signal?The standard specifies a bandwidth of 5 MHz. RF Bandwidth is Chip_rate * (1+a) = 4.68 MHz. And rest is I suppose guard band.> To answer your question though, I perform the upsampling after the > spreading and prior to the RRC filtering.I have a slight confusion as to where do you think aliasing is happening. As I see things there shouldn't be any problem. Sachin
Reply by ●October 30, 20032003-10-30
On 29 Oct 2003 21:06:06 -0800, scngupta@yahoo.com (Sachin Gupta) claimed:>> I guess my real question is why did they choose a chip rate of 3.84M? >Don't know. If you go through the development history and some of the >early proposals they do talk about some other chip-rates also, but >somehow settled upon 3.84 Mcps finally.Originally, they had a chip rate of 4.096 Mcps. However, in order to harmonize with cdma2000, which has a chip rate of 3.6864 Mcps, it was reduced to 3.84 Mcps.
Reply by ●October 31, 20032003-10-31
1xEV-DO <nospam@incognito.com> wrote in message news:<se73qvodvied8v6lgom26tpu0ubfnhs9su@4ax.com>...> On 29 Oct 2003 21:06:06 -0800, scngupta@yahoo.com (Sachin Gupta) > claimed: > > >> I guess my real question is why did they choose a chip rate of 3.84M? > >Don't know. If you go through the development history and some of the > >early proposals they do talk about some other chip-rates also, but > >somehow settled upon 3.84 Mcps finally. > > Originally, they had a chip rate of 4.096 Mcps.Saw some papers in IEICE which talk about this chip-rate of 4.096 Mcps.> However, in order to > harmonize with cdma2000, which has a chip rate of 3.6864 Mcps, it was > reduced to 3.84 Mcps.Is CDMA2000 chip rate of 3.6864 IS-95 Legacy (3 x 1.2288) ? But how does bringing UMTS chip rate down from 4.096 to 3.84 harmonize the two standards. I don't know much about cdma2000. Sachin
Reply by ●October 31, 20032003-10-31
On 31 Oct 2003 02:06:51 -0800, scngupta@yahoo.com (Sachin Gupta) claimed:>> However, in order to >> harmonize with cdma2000, which has a chip rate of 3.6864 Mcps, it was >> reduced to 3.84 Mcps. > >Is CDMA2000 chip rate of 3.6864 IS-95 Legacy (3 x 1.2288) ? But how >does bringing UMTS chip rate down from 4.096 to 3.84 harmonize the two >standards. I don't know much about cdma2000.That's a question that has always stomped me. I am not too up-to-date with the RF details regarding filter and amplifier designs, but I would guess that a lower chip rate (closer to cdma2000), more components could be reused in a dual-mode (cdma2000/UMTS) handset. Obviously the digital coding is different, but it'd be nice if a lot of the RF parts as similar (cheaper development/manufacturing). The 3.6864 Mcps is derived from the IS-95 chip rate (1.2288 Mcps). IS-2000 (specification for cdma2000) specifies two chip rates, 1.2288 Mcps (cdma2000 1x) and 3.6864 Mcps (cdma2000 3x). This to allow easy overlay when deploying cdma2000 on an IS-95 (cdmaOne) network. Today cdma2000 1x ius widely deployed; if we ever going to see cdma2000 3x or not, that's a different story. The increase in data rates seems to be achieved by using cdma2000 1xEV-DO (as specified by IS-856) and the coming cdma2000 1xEV-DV (as specified by IS-2000 revision C). 1xEV-DO for example supports a forward link data rate of 2.4576 Mbps using the chip rate of 1.2288 Mcps (and a carrier bandwidth of 1.23 MHz w/o guard band). I've been trying to dig deeper into WCDMA but I can't find my way around 3GPP's web-site. What exactly should I look for if I want more details about the physical layer and the MAc layer? For cdma2000, go to http://www.3gpp2.org/Public_html/specs/index.cfm and look for the C.S0001/2/3/4/5/6 specifications...
Reply by ●November 1, 20032003-11-01
1xEV-DO <nospam@incognito.com> wrote in message news:<g6l5qv8rk9m1juvsv5h8trp2j43irse4fj@4ax.com>...> On 31 Oct 2003 02:06:51 -0800, scngupta@yahoo.com (Sachin Gupta) > claimed: > > >Is CDMA2000 chip rate of 3.6864 IS-95 Legacy (3 x 1.2288) ? But how > >does bringing UMTS chip rate down from 4.096 to 3.84 harmonize the two > >standards. I don't know much about cdma2000. > > That's a question that has always stomped me. I am not too up-to-date > with the RF details regarding filter and amplifier designs, but I > would guess that a lower chip rate (closer to cdma2000), more > components could be reused in a dual-mode (cdma2000/UMTS) handset. > Obviously the digital coding is different, but it'd be nice if a lot > of the RF parts as similar (cheaper development/manufacturing).Makes more sense now. Yes if RF parts are similar it would make the handset a lot cheaper.> Today cdma2000 1x ius widely deployed; if we ever going to see > cdma2000 3x or not, that's a different story. The increase in data > rates seems to be achieved by using cdma2000 1xEV-DO (as specified by > IS-856) and the coming cdma2000 1xEV-DV (as specified by IS-2000 > revision C). 1xEV-DO for example supports a forward link data rate of > 2.4576 Mbps using the chip rate of 1.2288 Mcps (and a carrier > bandwidth of 1.23 MHz w/o guard band).What do 1xEV-DO and 1xEV-DV stand for ? Is CDMA2000 still being driven by Qualcomm ?> I've been trying to dig deeper into WCDMA but I can't find my way > around 3GPP's web-site. What exactly should I look for if I want more > details about the physical layer and the MAc layer?The best way to start W-CDMA is to get hold of a book. There is one by Harri Holma and Antti Toskala which is very popular and then there is one by Juha Korhonen. Then you can read some papers, look into IEEE Comm Magazine, Sept 1998 Issue. Also there are lots of web-sites which provide a good overview of W-CDMA like <http://www.umtsworld.com> <http://www.fortunecity.com/business/tisch/1301/id64.htm> <http://www.palowireless.com/3g/cdma.asp> W-CDMA Specs are divided into a lot of releases. To the best of my knowledge Rel-99 was the first important release (there are Rel-4 and Rel-5 also). You can get Rel-99 specs from: <http://www.3gpp.org/ftp/Specs/latest/R1999/25_series/> PHY Specs are a part of 25_series. For more info on what spec is what go to <http://www.3gpp.org/ftp/Specs/html-info/25-series.htm> A couple of PHY specs that I would suggest for starting would be 25.201, 25.211, 25.212, 25.213, 25.214, 25.215. All these talk about the FDD Mode. If you are interested in TDD Mode - there are different specs corresponding to the ones above. There are some email-lists on 3GPP web-site that you can subscribe to <http://www.3gpp.org/email/lists.htm> and as a last resort comp.dsp is always there for specific queries.> For cdma2000, go to http://www.3gpp2.org/Public_html/specs/index.cfm > and look for the C.S0001/2/3/4/5/6 specifications...Thanks ! Sachin
Reply by ●November 1, 20032003-11-01
On 31 Oct 2003 23:55:14 -0800, scngupta@yahoo.com (Sachin Gupta) claimed:>What do 1xEV-DO and 1xEV-DV stand for ? Is CDMA2000 still being driven >by Qualcomm ?(cdma2000) 1x EVolution Data Only, or (cdma2000) 1x EVolution Data Optimized (new Qualcomm term apparently), and (cdma2000) 1xEVolution Data and Voice 1x because of the 1.25 Mhz (1.23 MHz) carrier. Everyone keeps arguing which technology is the most spectrum efficient, cdma2000 or WCDMA, and everone presents valid arguments for their claim. However, it seems as if it's widely accepted that 1xEV-DO can carry about 3 times as many bits per Hertz as the other technologies. The drawback being of course that it's a data only technology... Qualcomm is still the driver. You may follow the news and notice that Verizon Wireless will launch a 1xEV-DO network in Washington and Seattle (equipment provided by Lucent and Nortel I believe). There is also a network in Korea (equipment may be provided by Samsung, but I'm not sure). www.cdg.org may have more info.>> I've been trying to dig deeper into WCDMA but I can't find my way >> around 3GPP's web-site. What exactly should I look for if I want more >> details about the physical layer and the MAc layer? > >The best way to start W-CDMA is to get hold of a book. There is one by >Harri Holma and Antti Toskala which is very popular and then there is >one by Juha Korhonen.<snip, snip> Thanks mate. I will visit a bookstore shortly :) BTW: The current revisions of cdma2000 commercially implemented are IS-2000-A and IS-856-1 (1xEV-DO). Also, Agilent has some good application notes where certain aspects of the technologies are explained. From the top of my head, I can recommend note #1335 regarding the HPSK spreading and the impact on RF amplifiers.
