Lowpass filtering for OFDM introduces interference?

>In an OFDM system, the baseband DFT is a matched filter - i.e. it 
>already optimally removes noise.  There is no need to perform low-pass 
>pre-filtering on your oversampled baseband signal - it will not improve
SNR.
>
>To show this, say your received freq-domain values are:
>
>	Y[k] = DFT{y[n]}
>	     = DFT{x[n] + v[n]}
>
>	     = X[k] + V[k]
>
>[Where v[n] is the time-domain noise sequence, and V[k] is the 
>freq-domain noise sequence].
>
>Evaluating this at some arbitrary k = k_0, this relationship is clearly 
>not affected by the size of the DFT, nor by the amount of noise at any 
>other k.
>
>Therefore, there is no performance advantage in low-pass filtering and 
>downsampling at baseband.
>

Assume v[n] to be a wideband white noise. By lowpass filtering the
variance  of v[n] should be reduced leading to a reduced noise variance of
V[k] in the frequency domain. Why is there no SNR gain?

I think you have no doubt that two receive antennas, even if fully
correlated, give you certain antenna gain if you combine them. I mean
there is no essential difference between the equal gain combining of two
RX antennas and low pass filtering oversampled time-domain signals. Both
can denoise.

LBB
  

lanbaba said the following on 18/04/2006 22:21:
>> To show this, say your received freq-domain values are:
>>
>> 	Y[k] = DFT{y[n]}
>> 	     = DFT{x[n] + v[n]}
>>
>> 	     = X[k] + V[k]
>>
>> [Where v[n] is the time-domain noise sequence, and V[k] is the 
>> freq-domain noise sequence].
>>
>> Evaluating this at some arbitrary k = k_0, this relationship is clearly 
>> not affected by the size of the DFT, nor by the amount of noise at any 
>> other k.
>>
>> Therefore, there is no performance advantage in low-pass filtering and 
>> downsampling at baseband.
>>
> 
> Assume v[n] to be a wideband white noise. By lowpass filtering the
> variance  of v[n] should be reduced leading to a reduced noise variance of
> V[k] in the frequency domain.

That doesn't follow.  Consider:

	y[n] = exp(j*2*pi*k1*n/N) + exp(j*2*pi*k2*n/N)

The variance of the total signal y[n] is 2, and the value of the DFT at 
k1 and k2 is:

	Y[k1] = Y[k2] = 1

If we now filter y[n] so that we completely suppress the component at 
k1, we get:

	y'[n] = exp(j*2*pi*k2*n/N)

The variance of this is now 1.  However, the value of the DFT at k2 is 
still:

	Y'[k2] = 1

i.e. a reduction in signal time-domain variance does not imply a 
reduction in variance of all frequency-domain components.


> Why is there no SNR gain?

Because each bin of the DFT is a matched correlation filter.  By 
definition, it is already optimal.

Try it in MATLAB (or whatever).  Do a simple OFDM simulation (IFFT->AWGN 
channel->FFT), and get a BER measure.  Then do it again, but pad your 
frequency-domain vector with zeros, which is equivalent to time-domain 
oversampling.  The results should be identical.


> I think you have no doubt that two receive antennas, even if fully
> correlated, give you certain antenna gain if you combine them. I mean
> there is no essential difference between the equal gain combining of two
> RX antennas and low pass filtering oversampled time-domain signals. Both
> can denoise.

You're right, these two scenarios are mathematically identical.

But equal gain combining isn't, in general, the optimal solution.  For 
any linear combination scenario, the optimal solution is given by the 
Wiener-Hopf equation.  In the case of filters, Wiener-Hopf gives the 
matched correlation filter.  In the case of (non-pulse-shaped) OFDM, the 
matched filter for each sub-carrier is of the form exp{-j.2.pi.k.n/N}, 
which is what the DFT is already doing.  Any further filtering can, at 
best, give zero further improvement in performance.

You can consider the DFT as mixing the signal down so that the k-th 
sub-carrier is at DC, and then time-domain averaging across the symbol.


-- 
Oli

lanbaba wrote:
>>Averaging two samples gives you an extra bit of significance. If you 
>>discard that bit, you haven't gained. Even if you save it, averaging 
>>reduces noise only if the noise in the two samples is substantially 
>>uncorrelated. After a bandpass filter that removes out-of-band noise, 
>>only the low-frequency components of in-band noise will be uncorrelated.
>>
>>Jerry
>>-- 
>>Engineering is the art of making what you want from things you can get.
>>�����������������������������������������������������������������������
>>
> 
> 
> Correct. I don't have a bandpass filter to remove the out-of-band noise.
> All  I want to do is to reduce the out-of-band noise by a lowpass filter
> in the baseband.

If you have filtered properly considering the band of interest and the 
sample rate*, then you can do that. Averaging won't be the best low-pass 
filter, but it may be the cheapest.

Jerry
____________________________________
Aliasing when oversampling doesn't necessarily hurt provided it doesn't 
extend into the band of interest.
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

>That doesn't follow.  Consider:
>
>	y[n] = exp(j*2*pi*k1*n/N) + exp(j*2*pi*k2*n/N)
>
>The variance of the total signal y[n] is 2, and the value of the DFT at 
>k1 and k2 is:
>
>	Y[k1] = Y[k2] = 1
>
>If we now filter y[n] so that we completely suppress the component at 
>k1, we get:
>
>	y'[n] = exp(j*2*pi*k2*n/N)
>
>The variance of this is now 1.  However, the value of the DFT at k2 is 
>still:
>
>	Y'[k2] = 1
>
>i.e. a reduction in signal time-domain variance does not imply a 
>reduction in variance of all frequency-domain components.
>
>
>> Why is there no SNR gain?
>
>Because each bin of the DFT is a matched correlation filter.  By 
>definition, it is already optimal.
>
>Try it in MATLAB (or whatever).  Do a simple OFDM simulation (IFFT->AWGN

>channel->FFT), and get a BER measure.  Then do it again, but pad your 
>frequency-domain vector with zeros, which is equivalent to time-domain 
>oversampling.  The results should be identical.
>
>
>> I think you have no doubt that two receive antennas, even if fully
>> correlated, give you certain antenna gain if you combine them. I mean
>> there is no essential difference between the equal gain combining of
two
>> RX antennas and low pass filtering oversampled time-domain signals.
Both
>> can denoise.
>
>You're right, these two scenarios are mathematically identical.
>
>But equal gain combining isn't, in general, the optimal solution.  For 
>any linear combination scenario, the optimal solution is given by the 
>Wiener-Hopf equation.  In the case of filters, Wiener-Hopf gives the 
>matched correlation filter.  In the case of (non-pulse-shaped) OFDM, the

>matched filter for each sub-carrier is of the form exp{-j.2.pi.k.n/N}, 
>which is what the DFT is already doing.  Any further filtering can, at 
>best, give zero further improvement in performance.
>
>You can consider the DFT as mixing the signal down so that the k-th 
>sub-carrier is at DC, and then time-domain averaging across the symbol.
>
>
>-- 
>Oli
>

I think I understand what you want to say now. You're right, if you
enlarge the DFT basis to cover all frequency band that oversampling
provides then it's not necessary to do a prefiltering since the decision
variables on usful subcarriers are not affected by out-of-band noise,
which are separated by DFT. In my scheme, prefiltering is followed by
downsampling to keep the DFT basis as small as possible. As we know from
multirate processing, downsampling converts out-of-band spectrum to
in-band. If there is no prefiltering the in-band noise variance will
increase after downsampling. Maybe I was not precise enough to say
prefiltering can denoise. Actually it avoids the out-of-band noise from
comming into in-band after downsampling.

BTW, both y[n] and y'[n] have variance 0 since they are deterministic :)