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tohji
Joined: 22 Jul 2004
Posts: 15
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 14 Dec 2004 21:04 SRRC question |
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Hi DSP experts:
I have a question on the square root raised cosine filter (SRRC).
Usually SRRC are used on both sides of the Tx and Rx. Given the channel is 1,
and additive white Gaussian noise, are the noise samples at the output of the
matched filter (SRRC) still white? The output signal are sampled at once per symbol.
Thanks.
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zorro
Joined: 06 Sep 2001
Posts: 324
Helped: 36
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| 15 Dec 2004 21:57 Re: SRRC question |
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Hello tohji,
Suppose that the noise at the input of the receiving SRRC filter is AWGN (additive, white and gaussian noise). The noise Power Spectral Density (PSD) at the output has a shape proportional to |H(f)|^2, this is a raised cosine shape.
The autocorrelation function of the noise is the Fourier transform of the PSD, and this is the shape of the pulse (after SRRC Rx filtering), that has zero-crossings at the times nT (n = any nonzero integer, T = symbol time) because it is designed in order to avoid intersymbol interference (ISI). So, noise samples nT seconds apart are not correlated at the output of this filter. As the noise at the filter output is still gaussian, they are independent.
Conclusion: noise samples corresponding to the times at which the signal is sampled for symbol decisions are not correlated (and independent).
This property is valid for any shape that gives no ISI and that is evenly split between Tx and Rx in order to have a matched filter.
Regards
Z
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tohji
Joined: 22 Jul 2004
Posts: 15
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| 16 Dec 2004 3:14 Re: SRRC question |
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Hi zorro,
Thank you for your response. I agree with you that the auto-correlation function of the noise at the output of SRRC is 0 at t = nT. But it does not mean the noise samples at the t=nT are not correlated. The auto-correlation of a sampled signal is not a the auto-correlation function of the continues signal, i.e.
E(v(t) * v(t+T) ) != interg( v(t) * v(t+T) dt)
Another view of this question: The output noise signal has a spectrum H(w), which is the SRRC spectrum. After sampling at symbol rate, the spectrum H(w) will be shifted by n/T and is then folded into the [-1/2T,1/2T] area. Note only three H(w) will be overlapping becase of the limited tail of the SRRC spectrum over 1/2T. If we add these spectrums together, the sum is not constant over [-1/2T,1/2T]. Instead, it shows a little peak at high freq.
Does this make sense?
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zorro
Joined: 06 Sep 2001
Posts: 324
Helped: 36
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| 16 Dec 2004 13:55 Re: SRRC question |
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Hello tohji,
The noise samples at the symbol-decision times are v(nT), n integer.
The time average integ( v(t) * v(t+T) dt ) does not matter.
The correlation between two different noise samples is E[v(nT) * v(mT)] (with n!=m)
As the noise is stationary (at least in wide sense)........: E[v(nT) * v(mT)] = Rvv[(m-n)*T] = 0,
The noise is zero-mean...........: E[v(nT)] = E[v(mT)] = 0,
so E[v(nT) * v(mT)] = E[v(nT)] * E[v(mT)] : the noise samples are not correlated.
Your other view:
when you combine different noise spectra, the PSD have to be added. The noise spectrum at the output is |H(w)|^2, that is a a raised cosine shape (not SRRC). Adding these spectra, the sum is constant.
Best regards
Z
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tohji
Joined: 22 Jul 2004
Posts: 15
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| 16 Dec 2004 21:35 Re: SRRC question |
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Hi Zorro,
You are absolutely right. I agree with all your points.
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