New answers tagged digital-filters
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The RRC output only takes the exact constellation point values at exactly the right timing instant. That's what I meant with "coherent detector feeding a synchronously sampled matched filter" in an answer to a previous question of yours.
You'll have to deal with the inter-symbol interference with previous symbols. Which is no big deal – all symbols ...
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Derivation seems to be wrong. I am guessing from derivation that it is power spectral density that you are talking about. Your derivation seems to be wrong, cross density would be $$S_{xy}(x)S_in=G1(x)G2\times(G2^*)(x)S_in$$ or $$S_{yx}(x)S_in=G1\times(G1^*)(x)G2(x)S_in.$$ If you take magnitude spectrum both will be same, phase spectrum will be of 180 shift.
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HINTS:
What value does $H(z_0)$ have if there's a phase shift of $180$ degrees (for $|z_0|=1$)?
Equate $H(z_0)$ to that value and write down the resulting equation for $a_1$ and $a_2$ and $z_0$.
You want a stable all-pass filter, so choose $a_2$ such that a given desired pole radius $r$, $0<r<1$, is achieved if you assume two complex conjugate poles.
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I have worked out H(z) to be (0.4 + 0.6z^-1) / (1-0.8z^-2) is this correct?
Close but one detail is wrong
Is this filter an FIR or IIR type?
Look at the pole locations
How do you determine the poles and zero?
Poles are the roots of denominator polynomial. Zeros are the roots of the numerator polynomial.
How do you determine if the filter is stable or ...
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One problem with your design is that the quantized numerator coefficients don't add up to zero anymore, so you lose the desired notch at DC. That can be done differently, because even with quantized coefficients you can have a (double) zero at DC. Just make sure that $b[0]=b[2]$ and $b[1]=-2b[0]$.
The real problem is the denominator polynomial. It is well ...
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This filter amounts to a cutoff frequency of 24 Hz @ 44.1 kHz. That means your poles are extremely close to the unit circle so you require way more numerical precision than you can get with 16 bits.
You need more resolution: even 24 bits are iffy for that low a cutoff frequency, but 32-bit should do nicely. This also depends on your requirements: how much ...
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Any filter with such a low cutoff (as compared to the sample rate) will lots of time domaing ringing.
The best way of adressing this is to properly initialize the state of the filter using known properties of your system & signal, but that's fairly tricky.
In your case, there may be a shit cut though: I'm guessing that you have a large bias or DC offset ...
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