On Fri, 27 Nov 2015 22:06:24 -0600, "Sharan123" <99077@DSPRelated> wrote:>>The modulation type doesn't matter. The subcarrier width is >>determined solely by the FFT size and sample rate, which determines >>the bin width. >> >>Does a subcarrier width shrink if it's a constant (CW)? If it updates >>at half, or some other fraction, of the OFDM symbol rate? Ans: no. > >Dear Eric, Steve, > >Thanks for the comments. I have one question, especially pertaining to >LTE. > >In case of OFDM, there are multiple sub-carriers and spacing between >consecutive sub-carriers is (say) 15 KHz. Hence isn't bandwidth of each >sub-carrier 15 KHz?Pretty much. Usually people talk about subcarrier spacing rather than bandwidth, but it's not wrong to say an LTE subcarrier occupies 15kHz BW. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
LTE, FFT, OFDM
Started by ●November 15, 2015
Reply by ●November 28, 20152015-11-28
Reply by ●November 28, 20152015-11-28
>On Fri, 27 Nov 2015 22:06:24 -0600, "Sharan123" <99077@DSPRelated> >wrote: > >>>The modulation type doesn't matter. The subcarrier width is >>>determined solely by the FFT size and sample rate, which determines >>>the bin width. >>> >>>Does a subcarrier width shrink if it's a constant (CW)? If it updates >>>at half, or some other fraction, of the OFDM symbol rate? Ans: no. >> >>Dear Eric, Steve, >> >>Thanks for the comments. I have one question, especially pertaining to >>LTE. >> >>In case of OFDM, there are multiple sub-carriers and spacing between >>consecutive sub-carriers is (say) 15 KHz. Hence isn't bandwidth of each >>sub-carrier 15 KHz? > >Pretty much. Usually people talk about subcarrier spacing rather >than bandwidth, but it's not wrong to say an LTE subcarrier occupies >15kHz BW. > > >Eric Jacobsen >Anchor Hill Communications >http://www.anchorhill.comUnofficially I visualise ofdm as multiple fdm orthogonally spaced.The multiple carriers being generated by IFFT. Imagine first you want to transmit 7.68 Msps worth of data using fdm rather than ofdm, then you will need 7.68 MHz (complex QAM assumed). In the case of LTE 5 there are 300 data points + CP. Thus each data runs @ 7.68/(300+CP) = ~ 22.85 KHz if CP = 36. This is the bandwidth of each modulated carrier within ofdm symbol, so some overlap occurs controlled by CP length. Notice how ofdm for lte 5 occupies only ~4.5MHz instead of 7.68MHz if it was FDM. The ofdm symbol itself can then be transmitted together with any other LTE or cdma using usual FDM. Kaz --------------------------------------- Posted through http://www.DSPRelated.com
Reply by ●November 28, 20152015-11-28
Eric Jacobsen <eric.jacobsen@ieee.org> wrote:>On Sat, 28 Nov 2015 00:12:03 +0000 (UTC), spope33@speedymail.org>>Eric Jacobsen <eric.jacobsen@ieee.org> wrote:>>>On Fri, 27 Nov 2015 21:10:11 +0000 (UTC), spope33@speedymail.org>>>>On Fri, 27 Nov 2015 09:23:12 -0600, "Sharan123" <99077@DSPRelated>>>>>> I assume that, specifically referring to LTE/OFDM, when >>>>> we refer to sub-carrier spacing of 15 KHz, we also mean that >>>>> sub-carriers themselves are 15 KHz wide. Is this correct?>>>>Depends what you mean by "width".>>>>If each subcarrier is modulated using QAM (as is typical), then >>>>the details of this modulation define the width of the subcarrier. >>>>[..] You could define width however you like, but the inverse of the >>>>baud length is one possible defintion. The 20 dB bandwidth is >>>>another possible definition. Probably (but not certainly) you >>>>end up with a value slightly less than the subcarrier spacing.>>>It doesn't matter how the subcarriers are modulated because their >>>values are static for the duration of the symbol (and therefore the >>>DFT length).>>I agree it doesn't matter, however, the OP asked what is the "width" >>of the subcarrier. The answer to this is that the "width" (however >>defined) is the same as that of any carrier modulated in the same >>manner.>The modulation type doesn't matter. The subcarrier width is >determined solely by the FFT size and sample rate, which determines >the bin width.>Does a subcarrier width shrink if it's a constant (CW)?Yes, an unmodulated carrier has zero width (if by width you mean bandwidth), and the fact that it's one carrier out of many does not change this. Steve
Reply by ●November 28, 20152015-11-28
Steve Pope <spope33@speedymail.org> wrote:> Eric Jacobsen <eric.jacobsen@ieee.org> wrote:>>On Sat, 28 Nov 2015 00:12:03 +0000 (UTC), spope33@speedymail.org(snip)>>>I agree it doesn't matter, however, the OP asked what is the "width" >>>of the subcarrier. The answer to this is that the "width" (however >>>defined) is the same as that of any carrier modulated in the same >>>manner.>>The modulation type doesn't matter. The subcarrier width is >>determined solely by the FFT size and sample rate, which determines >>the bin width.>>Does a subcarrier width shrink if it's a constant (CW)?> Yes, an unmodulated carrier has zero width (if by width you > mean bandwidth), and the fact that it's one carrier out of many > does not change this.It has zero width if it exists for an infinitely long time. For a finite time, it depends on how fast you turn it on and off, but not less than 1/time it is on. -- glen
Reply by ●November 28, 20152015-11-28
>Steve Pope <spope33@speedymail.org> wrote: >> Eric Jacobsen <eric.jacobsen@ieee.org> wrote: > >>>On Sat, 28 Nov 2015 00:12:03 +0000 (UTC), spope33@speedymail.org > >(snip) > >>>>I agree it doesn't matter, however, the OP asked what is the "width" >>>>of the subcarrier. The answer to this is that the "width" (however >>>>defined) is the same as that of any carrier modulated in the same >>>>manner. > >>>The modulation type doesn't matter. The subcarrier width is >>>determined solely by the FFT size and sample rate, which determines >>>the bin width. > >>>Does a subcarrier width shrink if it's a constant (CW)? > >> Yes, an unmodulated carrier has zero width (if by width you >> mean bandwidth), and the fact that it's one carrier out of many >> does not change this. > >It has zero width if it exists for an infinitely long time. > >For a finite time, it depends on how fast you turn it on and off, >but not less than 1/time it is on. > >-- glenOFDM is made up of a number of packed mini FDM and carrier width is really irrelevant once modulated just like any FDM. Kaz --------------------------------------- Posted through http://www.DSPRelated.com
Reply by ●November 28, 20152015-11-28
On Sat, 28 Nov 2015 11:15:35 -0600, "kaz" <37480@DSPRelated> wrote:>>Steve Pope <spope33@speedymail.org> wrote: >>> Eric Jacobsen <eric.jacobsen@ieee.org> wrote: >> >>>>On Sat, 28 Nov 2015 00:12:03 +0000 (UTC), spope33@speedymail.org >> >>(snip) >> >>>>>I agree it doesn't matter, however, the OP asked what is the "width" >>>>>of the subcarrier. The answer to this is that the "width" (however >>>>>defined) is the same as that of any carrier modulated in the same >>>>>manner. >> >>>>The modulation type doesn't matter. The subcarrier width is >>>>determined solely by the FFT size and sample rate, which determines >>>>the bin width. >> >>>>Does a subcarrier width shrink if it's a constant (CW)? >> >>> Yes, an unmodulated carrier has zero width (if by width you >>> mean bandwidth), and the fact that it's one carrier out of many >>> does not change this. >> >>It has zero width if it exists for an infinitely long time. >> >>For a finite time, it depends on how fast you turn it on and off, >>but not less than 1/time it is on. >> >>-- glen > >OFDM is made up of a number of packed mini FDM and carrier width is really >irrelevant once modulated just like any FDM. > >KazThe difference with usual FDM is that the frequency packing with OFDM can be much tighter and still maintain orthogonality between subcarriers because they are essentially mixed to their relative frequencies by orthogonal oscillators, (i.e., the basis functions of the transform). This is a significant difference with ordinary FDM. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●November 28, 20152015-11-28
glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:>Steve Pope <spope33@speedymail.org> wrote:>> Yes, an unmodulated carrier has zero width (if by width you >> mean bandwidth), and the fact that it's one carrier out of many >> does not change this.>It has zero width if it exists for an infinitely long time.>For a finite time, it depends on how fast you turn it on and off, >but not less than 1/time it is on.True. Of course, turning it on and off couuld be viewed as a form of modulation. Back to OFDM, if for instances one had an unusually very long cyclic prefix, one could look at the signal with a spectrum analyzer and observe that the modulated subcarriers are not filling the entire space between the subcarriers... Steve
Reply by ●November 28, 20152015-11-28
On Sat, 28 Nov 2015 21:18:45 +0000 (UTC), spope33@speedymail.org (Steve Pope) wrote:>glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote: > >>Steve Pope <spope33@speedymail.org> wrote: > >>> Yes, an unmodulated carrier has zero width (if by width you >>> mean bandwidth), and the fact that it's one carrier out of many >>> does not change this. > >>It has zero width if it exists for an infinitely long time. > >>For a finite time, it depends on how fast you turn it on and off, >>but not less than 1/time it is on. > >True. Of course, turning it on and off couuld be viewed as >a form of modulation. > >Back to OFDM, if for instances one had an unusually very long cyclic >prefix, one could look at the signal with a spectrum analyzer >and observe that the modulated subcarriers are not filling the >entire space between the subcarriers...How do such fractionally-narrowed subcarriers maintain orthogonality in the receiver? The receiver depends on each subcarrier's sidelobes being spaced exactly such that they will be orthogonal wrt to the other subcarriers via the basis functions in the receiver FFT (i.e., each subcarrier sits in the nulls of the other subcarrier's sidelobes). If they've narrowed, how can the orthogonality be maintained? Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●November 28, 20152015-11-28
On Sat, 28 Nov 2015 10:27:08 +0000 (UTC), spope33@speedymail.org (Steve Pope) wrote:>Eric Jacobsen <eric.jacobsen@ieee.org> wrote: > >>On Sat, 28 Nov 2015 00:12:03 +0000 (UTC), spope33@speedymail.org > >>>Eric Jacobsen <eric.jacobsen@ieee.org> wrote: > >>>>On Fri, 27 Nov 2015 21:10:11 +0000 (UTC), spope33@speedymail.org > >>>>>On Fri, 27 Nov 2015 09:23:12 -0600, "Sharan123" <99077@DSPRelated> > >>>>>> I assume that, specifically referring to LTE/OFDM, when >>>>>> we refer to sub-carrier spacing of 15 KHz, we also mean that >>>>>> sub-carriers themselves are 15 KHz wide. Is this correct? > >>>>>Depends what you mean by "width". > >>>>>If each subcarrier is modulated using QAM (as is typical), then >>>>>the details of this modulation define the width of the subcarrier. >>>>>[..] You could define width however you like, but the inverse of the >>>>>baud length is one possible defintion. The 20 dB bandwidth is >>>>>another possible definition. Probably (but not certainly) you >>>>>end up with a value slightly less than the subcarrier spacing. > >>>>It doesn't matter how the subcarriers are modulated because their >>>>values are static for the duration of the symbol (and therefore the >>>>DFT length). > >>>I agree it doesn't matter, however, the OP asked what is the "width" >>>of the subcarrier. The answer to this is that the "width" (however >>>defined) is the same as that of any carrier modulated in the same >>>manner. > >>The modulation type doesn't matter. The subcarrier width is >>determined solely by the FFT size and sample rate, which determines >>the bin width. > >>Does a subcarrier width shrink if it's a constant (CW)? > >Yes, an unmodulated carrier has zero width (if by width you >mean bandwidth), and the fact that it's one carrier out of many >does not change this. > >SteveBut aren't all subcarriers unmodulated for the duration of the FFT symbol? They're all constant for the entire length of the DFT, so in any given symbol, ALL subcarriers are CW, just with, perhaps differing phase and amplitude. In other words, how could one tell from looking at a single OFDM symbol, which subcarriers are modulated and which aren't? (Ans: you can't tell for certain without outside information.) Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●November 28, 20152015-11-28
On Sat, 28 Nov 2015 17:06:48 +0000 (UTC), glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:>Steve Pope <spope33@speedymail.org> wrote: >> Eric Jacobsen <eric.jacobsen@ieee.org> wrote: > >>>On Sat, 28 Nov 2015 00:12:03 +0000 (UTC), spope33@speedymail.org > >(snip) > >>>>I agree it doesn't matter, however, the OP asked what is the "width" >>>>of the subcarrier. The answer to this is that the "width" (however >>>>defined) is the same as that of any carrier modulated in the same >>>>manner. > >>>The modulation type doesn't matter. The subcarrier width is >>>determined solely by the FFT size and sample rate, which determines >>>the bin width. > >>>Does a subcarrier width shrink if it's a constant (CW)? > >> Yes, an unmodulated carrier has zero width (if by width you >> mean bandwidth), and the fact that it's one carrier out of many >> does not change this. > >It has zero width if it exists for an infinitely long time. > >For a finite time, it depends on how fast you turn it on and off, >but not less than 1/time it is on. > >-- glenGlen's observation is to the point. If the FFT length in the receiver matches the length in the transmitter, then the observation times are equal and there is no difference between a modulated and unmodulated carrier. They're all unmodulated for the duration of the OFDM symbol, which is the length of the observation in the receiver, so they're all effectively the same BW. Synchronization errors in the receiver that cause ICI bear this out; a constant subcarrier will distort the same and cause the same distortion as a modulated subcarrier. They're all the same since they're constant for the duration of the symbol. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
