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Dear Zhongshan, Lepidus,The original intent behind the Fs = floor(n * BW / 8000) * 8000 equation was that the result is very close to n * BW, however there is an integer amount of samples in a frame. Since the allowed frame sizes (such a s 2.5, 4, 5 msec etc.) are all a multiple of 0.125 msec, having the Fs a multiple 8 KHz satisfies the requirement of an integer number of samples in a frame. This requirement eases the implementation and relieves the implementer of the temptation to introduce jitter by quantizing the frame start time to the nearest integer sample.Regards,Naftali
From: Chao_Ming Chang [mailto:Lepidus_Chang@MTK.COM.TW]
Sent: Tuesday, October 03, 2006 8:34 AM
To: STDS-802-16@listserv.ieee.org
Subject: Re: [STDS-802-16] two simple questions.
Hello Zhongshan,
1. For practical implementation, as least for IEEE 802.11abgn, we count on analog IIR filter and digital (FIR or IIR) filter
to perform the task of spectrum shaping. As I am not familar with the design of DAC, I am not sure if the output of DAC would
also apply a low-pass filter.
2. Yes, it is correct to say that when the sub-carrier spacing is the inverse of the OFDM symbol duration excluding CP,
the sub-carriers are orthogonal.
It is my pleasure to share some experience with you.
Best regards,
Lepidus.
"Zhongshan Wu" <wuzhongshan@gmail.com>
2006/10/03 12:37 PM
收件人： " Lepidus_Chang@mtk.com.tw" <Lepidus_Chang@mtk.com.tw>
副本抄送： STDS-802-16@listserv.ieee.org
主旨： Re: Re: [STDS-802-16] two simple questions.
Hi Lepidus,
First of all, thank you so much for your clear comments. That is really the difference between text-book theories and real-world implementations. Ideally we should have one "brick-wall" filter to shape the spectrum of transmitted signal. But that is not achieveable. If my understanding is right, the DAC at the transmit side serves as a transmit filiter for spectrum shaping purpose too. Another point is, except those confusing terms for BW and Fs, the subcarrier spacing should be exactly the inverse of the usable signal duration (no CP part). This helps maintain the orthogonality between ajacent or non-ajacent subcarriers. Am I right?
Thanks again,
Zhongshan
On 10/1/06, Lepidus_Chang@mtk.com.tw < Lepidus_Chang@mtk.com.tw> wrote:
Hello Zhongshan,
Your understanding is correct. Please see my further comments between the lines
with red color.
Best regards,
Lepidus.
"Zhongshan Wu" < wuzhongshan@gmail.com> 2006/10/02 01:29 PM
收件人： " Lepidus_Chang@mtk.com.tw" < Lepidus_Chang@mtk.com.tw>
副本抄送： STDS-802-16@listserv.ieee.org
主旨： Re: [STDS-802-16] two simple questions.
Hi Lepidus,
Thank you so much for your helpful input!! Here is my understanding.
1) The BW is (or roughly) the actual bandwidth that the data and pilot sub-carriers occupied. The Fs is kind of virutal bandwidth which includes the guard band at both ends.
[Lepidus]: Correct. But I used to call Fs as the fundamental sampling frequency.
2) I am not very clear about why the Fs is NOT the sampling frequency for ADC after the IFFT. For example, Fs/N is the sub-carrier spacing and so its inverse, i.e., N/Fs is the symbol duration (cyclic prefix is not considered here) and (N/Fs)/N is the sampling period and it is actually 1/Fs.
[Lepidus]: The only reason that I emphasize that Fs is not the sampling frequency of analog-to-digital converter (ADC/DAC) is
for practical implementation consideration only. Theoretically, your understanding is correct. But when we need to ensure the transmit
spectrum after IFFT, we usually need to apply a very sharp digital transmit filter in order to make sure a strong suppression
on signals outside BW. (e.g. for WiBro whose BW = 8.75 MHz and Fs = 10 MHz, the frequency band is only 9 MHz wide. The analog IIR filters usually do not give strong suppression for signal outside [-5, 5] MHz. In this case, we need either a very high order digital filter based on DAC at sampling rate Fs, or a digital filter of moderate order with DAC at sampling rate 2*Fs or 4*Fs.) The similar argument applies to ADC sampling rate at receiver side, by considering adjacent channel interference suppression instead of transmit spectrum mask.
3) why is the bandwidth for data and pilots ROUGHLY equal to BW? some practical considerations?
[Lepidus]: I guess that BW only approximates to the actual occupied bandwidth of data and pilot sub-carriers, because it is not appropriate to name the BW, for example the FUSC of WiBro, as 8.212890625 MHz. This is why I would suggest you to
treat the BW = 8.75 MHz as a tag without true physical meaning for technical development.
Thanks for sharing your knowledge with me. I guess your reply can help some others too.
yours,
Zhongshan
On 9/30/06, Lepidus_Chang@mtk.com.tw <Lepidus_Chang@mtk.com.tw > wrote:
Hello Zhongshan,
For an OFDM (or OFDMA) system whose FFT/IFFT size is N, we usually do not use all of the
N sub-carriers for data and pilots. Instead, we would leave some sub-carriers unused, called
the guard sub-carriers (or guard band), on the left and right side of the data and pilot sub-carriers.
For example, in the WirelessMAN-OFDMA (also known as 16e-2005) with N=1024, there are 86
guard sub-carriers at left side and right side of spectrum.
In this case, in order to utilize the N-point IFFT to generate transmit signal or N-point FFT to demodulate
signal, the sampling rate considered should be Fs. However, the actual null-to-null bandwidth occupied
by data and pilot sub-carriers is usually smaller than Fs and is roughly the value BW in your description.
In order not to be confused by the strange equation, I would suggest you to treat the BW value as the tag
while Fs is the fundamental bandwidth to compose the transmit OFDM signal
s(t) = 1/N \sum_{k=0}^{N-1} d_{k} exp{j 2\pi k Fs t/N}.
e.g . The 5 signal bandwidths considered in the current Mobilie WiMax System Profile (v1.2.0) are
Tag0: BW=3.0 MHz => Fs = 4.0 MHz;
Tag1: BW=5.0 MHz => Fs = 5.6 MHz;
Tag2: BW=7.0 MHz => Fs = 8 MHz;
Tag3: BW=8.75 MHz => Fs = 10 MHz;
Tag4: BW=10.0 MHz => Fs = 11.2 MHz.
In this case, you do not have to remember the value of n and the strange equation between BW and Fs.
Remarks:
1. Please note that Fs is not the sampling frequencies we considered in specifying analog-to-digital converters (ADC/DAC);
2. The actual signal banwidth occupied by data and pilot sub-carriers is also a bit different from the BW.
3. I do not directly answer your two questions, because I think it is easier to interpretate the meaning of BW and Fs this way.
Best regards,
Lepidus.
Zhongshan Wu < wuzhongshan@GMAIL.COM> 2006/09/30 06:45 AM
請回信 給 Zhongshan Wu
收件人： STDS-802-16@LISTSERV.IEEE.ORG
副本抄送：
主旨： [SPAM] [STDS-802-16] two simple questions.
Hi,
I am reading the WiMax standard 802.16-2004 together with 802.16-2005. Here are two simples questions.
In section 8.4.2.4 of 802.16-2004, the sampling frequency Fs is defined as Fs = floor(n * BW / 8000) * 8000. The parameter "n' is defined as sampling factor. So, what is the purpose to use this sampling factor ? and what do we choose the constant 8000 in the previous definition ?
Maybe these are so simple for you. Please give me some hints. Thanks.
yours,
Zhongshan
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