Two questions about OFDM

Some articles say that an OFDM system will reduce the symbol rate
through parallel data transmission. Is this really true?

Let's say suppose there are 100 subcarriers in an OFDM system and
the data rate before the IDFT is 100 symbols per second.

It seems to me that, after IDFT, each subcarrier only transmits one
of the 100 symbols in 1/100 second like this:

1st subcarrier: 1/100 idle  idle   idle .....
2nd subcarrier: idle  2/100 idle   idle ....
3rd subcarrier: idle  idle  3/100  idle ....
.......


It looks like a frequency hopping system.

Therefore, each subcarrier is not operating for the full one second 
to transmit one symbol.  Do we still say the "symbol rate" is reduced 
(from 100 symbols per second to one symbol per second) 
even in this situation?  

Will this arrangement also reduce the ISI? I doubt it.

At the receiving side, after the demodulation, the signal will first
go through an A to D conversion and then the DFT process.
The (channel) ISI has already affected the analog signal before the 
A to D conversion.

In order to get rid of the ISI, we still have to do an equalization
in the frequency domain after the DFT.  The parallel carrier
arrangement
has nothing to do with the elimination of the ISI.  Am I right?

 
-- Harry Liang

Hi,

> Some articles say that an OFDM system will reduce the symbol rate
> through parallel data transmission. Is this really true?
>
> Let's say suppose there are 100 subcarriers in an OFDM system and
> the data rate before the IDFT is 100 symbols per second.

When you say an OFDM symbol, you refer to the signal generated after
IDFT.(not the amplitudes of individual carriers)

> It seems to me that, after IDFT, each subcarrier only transmits one
> of the 100 symbols in 1/100 second like this:
>
> 1st subcarrier: 1/100 idle  idle   idle .....
> 2nd subcarrier: idle  2/100 idle   idle ....
> 3rd subcarrier: idle  idle  3/100  idle ....
> .......
> It looks like a frequency hopping system.

It is not a frequency hopping system. If we take the N-point IDFT of
X[k]=N*delta(k-k0) we get x[n]=exp(i*2*pi*k*n/N). Since IDFT is a
linear
operation, taking IDFT with all the subcarriers having non-zero
amplitudes
results in a sum of weighted complex exponentials. All the subcarriers
transmit for the whole symbol period.


> Therefore, each subcarrier is not operating for the full one second
> to transmit one symbol.  Do we still say the "symbol rate" is reduced

> (from 100 symbols per second to one symbol per second)
> even in this situation?

Without taking guard period into consideration, you can say that
the symbol rate is reduced by the number of carriers.

> Will this arrangement also reduce the ISI? I doubt it.
> At the receiving side, after the demodulation, the signal will first
> go through an A to D conversion and then the DFT process.
> The (channel) ISI has already affected the analog signal before the
> A to D conversion.
> In order to get rid of the ISI, we still have to do an equalization
> in the frequency domain after the DFT.  The parallel carrier
> arrangement
> has nothing to do with the elimination of the ISI.  Am I right?

We cannot counter ISI without guard period. If guard period is larger
than
the delay spread of the channel, we can get rid of ISI using
equalization.


regards,
venkateswara rao

harry0d88@aol.com (Harry Lin) wrote in message news:<d2563825.0410271451.154d4466@posting.google.com>...
> Some articles say that an OFDM system will reduce the symbol rate
> through parallel data transmission. Is this really true?
> 
> Let's say suppose there are 100 subcarriers in an OFDM system and
> the data rate before the IDFT is 100 symbols per second.
> 
> It seems to me that, after IDFT, each subcarrier only transmits one
> of the 100 symbols in 1/100 second like this:
> 
> 1st subcarrier: 1/100 idle  idle   idle .....
> 2nd subcarrier: idle  2/100 idle   idle ....
> 3rd subcarrier: idle  idle  3/100  idle ....
> .......
> 

Okay, I think you may be looking at things in a slightly odd fashion. 
Your first question regards reducing the symbol rate is really the key
point to your whole thread i think, so let's address that first.

When we talk about OFDM we want to compare it's performance with
another tx format.  So let's take a single carrier (SC) system as an
example.  Now, let's say that the data rate is the important thing
that we want to keep constant, and let's see where that leads us.

For SC we tx 100sym/sec.  Therefore the time for each symbol is Ts =
10msec.

For OFDM we still want to tx 100sym/sec (constant data rate) but we
have 100 subcarriers.  Therefore we can put 1 symbol on each carrier
and tx in parallel.  Therefore the time for each symbol is extended
such that the symbol duration is Ts = 1sec.  Remember we wanted to
keep the data rate the same to see what happens and indeed we have.

Now, suppose our channel has multipath delay spread of 30msec. 
Straight away, we can see that the channel delay components will
corrupt 3 symbols in the SC case and will fall within the duration of
the first symbol of the OFDM scheme.  In this simplistic case
therefore, we have eliminated the channel ISI by simply lengthening
the duration of the symbols.  This is one advantage of tx in parallel
with OFDM.  Now, I'm not certain that we want to get into the cyclic
prefix just yet, because it seems to me that you haven't quite got to
grips with the technology yet.  I could be wrong, so keep asking!

col

> 
> It looks like a frequency hopping system.
> 
> Therefore, each subcarrier is not operating for the full one second 
> to transmit one symbol.  Do we still say the "symbol rate" is reduced 
> (from 100 symbols per second to one symbol per second) 
> even in this situation?  
> 
> Will this arrangement also reduce the ISI? I doubt it.
> 
> At the receiving side, after the demodulation, the signal will first
> go through an A to D conversion and then the DFT process.
> The (channel) ISI has already affected the analog signal before the 
> A to D conversion.
> 
> In order to get rid of the ISI, we still have to do an equalization
> in the frequency domain after the DFT.  The parallel carrier
> arrangement
> has nothing to do with the elimination of the ISI.  Am I right?
> 
>  
> -- Harry Liang

"venkateswara rao" <venkat.mandela@gmail.com> wrote in message news:<1098948200.107067.160460@f14g2000cwb.googlegroups.com>...
> 
> It is not a frequency hopping system. If we take the N-point IDFT of
> X[k]=N*delta(k-k0) we get x[n]=exp(i*2*pi*k*n/N). Since IDFT is a
> linear
> operation, taking IDFT with all the subcarriers having non-zero
> amplitudes
> results in a sum of weighted complex exponentials. All the subcarriers
> transmit for the whole symbol period.
> 
> 

I get it.... Actually, the IDFT has combined all the 100 signals multiplied
by exp(j*2*pi*n/N), n = 0, ....,99 and all the subcarriers are tranmitting
for the whole symbol period.

Thank you very very much!

-- Harry Liang.


> regards,
> venkateswara rao

Hi Col,

Thank you very much for your explanation.  

See also the following article: 

http://www.ert.rwth-aachen.de/Projekte/Theo/OFDM/node2.html

-- Harry


cb135@hotmail.com (Col Brown) wrote in message news:<a254af6b.0410281131.696f53eb@posting.google.com>...
> 
> Okay, I think you may be looking at things in a slightly odd fashion. 
> Your first question regards reducing the symbol rate is really the key
> point to your whole thread i think, so let's address that first.
> 
> When we talk about OFDM we want to compare it's performance with
> another tx format.  So let's take a single carrier (SC) system as an
> example.  Now, let's say that the data rate is the important thing
> that we want to keep constant, and let's see where that leads us.
> 
> For SC we tx 100sym/sec.  Therefore the time for each symbol is Ts =
> 10msec.
> 
> For OFDM we still want to tx 100sym/sec (constant data rate) but we
> have 100 subcarriers.  Therefore we can put 1 symbol on each carrier
> and tx in parallel.  Therefore the time for each symbol is extended
> such that the symbol duration is Ts = 1sec.  Remember we wanted to
> keep the data rate the same to see what happens and indeed we have.
> 
> Now, suppose our channel has multipath delay spread of 30msec. 
> Straight away, we can see that the channel delay components will
> corrupt 3 symbols in the SC case and will fall within the duration of
> the first symbol of the OFDM scheme.  In this simplistic case
> therefore, we have eliminated the channel ISI by simply lengthening
> the duration of the symbols.  This is one advantage of tx in parallel
> with OFDM.  Now, I'm not certain that we want to get into the cyclic
> prefix just yet, because it seems to me that you haven't quite got to
> grips with the technology yet.  I could be wrong, so keep asking!
> 
> col
>