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Note that the inverse of an FIR system is IIR, and the same is true for the inverse of an IIR system, unless it is an all-pole system, the inverse of which would be FIR. So in most cases the ideal equalizer should have an infinitely long impulse response in order to perfectly invert the channel. In practice almost all adaptive equalizers are FIR filters ...
4
Bunch of things:
Your band definition is really odd. Some of the lower bands are really small while some of the mid bands are huge. Ideally you want octaves where the band edges are 1/sqrt(2) and sqrt(2) around the center frequency
You have designed bandpass filters in parallel. These are not "reconstructive" so you don't end up with a flat frequency ...
4
Assuming that you also want to equalize the filter's phase response (not only its magnitude response), you need an equalizer with a transfer function $E(z)$ that is the inverse of the FIR filter's transfer function $H(z)$:
$$E(z)=\frac{1}{H(z)}=\frac{1}{c_0+c_1z^{-1}+\ldots+c_{N-1}z^{-(N-1)}}\tag{1}$$
This is an all-pole filter which can be implemented by ...
4
Ok, there is some misconceptions in your question. I strongly recommend you to read a little more about the topics, but I will try to help you a little. My answers and some comments:
...linear equalizer is a filter that can undo these channel effects. When
the channel coefficients w are unknown, we perform blind equalization.
In this scenario, we ...
3
A filter bank system is generally composed of:
an analysis filter bank, that split an input signal into components with filters and may reduce their rate
a synthesis filter bank (FB), that takes the components, potentially increase their rate, and feed them into filter elements
so that the novel outputs can be combined into a signal. It is perfect ...
3
Thanks for the answer. You are correct - the training sequence is known by the time you establish any communication.
In case if somebody stumbles upon this in the future, here's how to get this done:
The synchronization burst contains an extended training sequence. It's longer than the regular one and there is only one sequence. You are supposed to use ...
2
From the referenced description and also from [1] it seems to me that the training sequence (here: midamble) is assumed to be known already. This knowledge has probably been acquired before in the synchronization burst. The first block in Fig. 2 extracts the training sequence from the received signal. This suggests that the receiver is already synchronized ...
2
result[i] =
band_gain[0] * LPF(t, band_freq[0]) +
band_gain[1] * (LPF(t, band_freq[1]) - LPF(a, band_freq[0])) +
band_gain[2] * (LPF(t, band_freq[2]) - LPF(a, band_freq[1])) +
band_gain[3] * (LPF(t, band_freq[3]) - LPF(a, band_freq[2])) +
band_gain[4] * (IMPULSE(t) - LPF(t, band_freq[3]));
is actually very simple to explain with pictures. First we assume f0 ...
2
A typical communications system is composed of the following parts:
Message (Information Source) --> Encoding --> Modulation --> Channel --> Demodulation --> Decoding --> Recovered Message
A "chaotic modulator" is just another way of doing the Modulation building block. This is depicted in Figure 1 of the paper that is cited in the ...
2
Does an audio equalizer consist of a perfect-reconstruction filterbank?
Short answer: No. It's a bit unsharp what you're asking here, since reconstruction filterbanks are usually things that combine multiple, separate, independent signals back into one signal, but let's assume this makes sense here, and your audio source magically produces one signal stream ...
2
it looks like this eq10q uses cascaded parametric EQs, initially set up to be bell, but switchable to other types (like shelving).
i wouldn't doubt that they're using the audio EQ cookbook. if you're using that, there is a parameter called "shelf slope" or $S$ that is often set to 1, because that gets you the steepest slope without dips or lips or bumps. ...
answered Apr 28 '16 at 13:21
robert bristow-johnson
12.8k3 gold badges20 silver badges53 bronze badges
2
The audible range is about 10 octaves, and usually the center frequencies of a graphic equalizer would be distributed equally spaced on a log scale to cover that range. Common equalizers have either $30$ bands (with $1/3$ octave filters) or $10$ bands (with $1$ octave filters).
If you want $5$ bands, you could choose filters that cover approximately $2$ ...
2
Most likely people use different words to describe the same concept. This happens quite often. The TEQ is the more general term and also used to equalize other methods that OFDM, while CS is related to OFDM and tries to make the channel shorter so that it fits within the CP duration.
You can check online, you will find that almost all articles related to CS ...
2
This depends a bit on what you want to get out of this and how much effort/work you are willing to put in.
Doing a room EQ that actually works and makes it sound consistently better is quite complicated. There are commercial systems available but they tend to be complicated and expensive or tied to a specific product.
A really good freeware option to play ...
2
If the received signal can be written as
$$\mathbf{y} = \mathbf{H}\,\mathbf{x} + \mathbf{n}$$
where $\mathbf{H}$ is the channel matrix, $\mathbf{x}$ is the transmitted vector, and $\mathbf{n}$ is the AWGN of the channel, then a zero forcing equalizer is simply (assuming that the channel matrix is square, and it's estimated perfectly at the receiver)
$$\...
1
You need an estimate of the channel to receive the sequence but the zero-forcing equalizer does not need the channel response as an input. The zero forcing equalizer estimates the channel response. This can be done either with a training sequence, or can be decision directed when signal to noise ratios are high enough.
Given the received signal is the ...
1
In an initial phase, the training sequence is used to determine the optimal equalizer weights. There is usually no reason to leave the weights fixed because you need an adaptation algorithm to compute the initial weights from the training sequence, so you might as well continue running the adaptation algorithm as soon as actual data are transmitted. If the ...
1
The error should be minimized between the equalized signal and what? There is no known sequence in this case.
But there is: if the channel is still good enough so you can decode a packet successfully, you can reconstruct the original signal by simply re-encoding the packet.
1
With a blind equalization technique like the constant modulus algorithm (which is often implemented using a least mean squares (LMS) filter as you indicated), you aren't directly estimating the channel impulse response itself. Instead, the signal model is like this:
The receiver observes the following signal:
$$
x[k] = s[k] * c[k] + n[k]
$$
where:
$s[k]$ ...
1
I will try to answer on your question with this: Least Square equalizer is minimizing the intersymbol interference of your signal (ISI), between the symbols, which does not translate directly to minimization of the bit error rate. When I am saying 'directly' that means that there is no exact linear dependency between the error reduction in the equalizer, and ...
1
The discrete convolution sum operation is not restricted to equal length vectors. You can, and most of the time you do, convolve two different signals of arbitary lengths. Your confusion is probably with something else. The equalizer length can be different than that of the channel model length. That should not pose a problem but it would of course effect ...
1
You're right. LMS equalizer uses a known input to minimize the error. For communication purposes, this is either provided by a training sequence, or in a decision directed mode, the detector decisions are fed back as known data.
The delta function is also correct. Suppose that the channel impulse response is $h(t)$ and frequency response $H(f)$. Then the ...
1
Your time domain waveform after 6 will be zero-padded as in step 3 with the additional "smearing" due to your equalization to remove the channel effects that will go into the zero padded area. The entire waveform in time will be delayed by the group delay of your frequency domain equalization, which may not necessarily match the group delay of the channel ...
1
Let's take a look at the cost function of CMA:
$ J(\mathbf{w}) = E[|y_{k}^{2} - 1|^{2}] $,
onde $ y_{k} = \mathbf{w}^{H} \mathbf{x}_{k} $ and $ \mathbf{x}_{k} $ is the filter input.
It says that the update function will converge to coefficients that approximate the output of the filter to something whose squared value is $ 1 $. So, it is appropriate to ...
1
I had a look at your code, and I know why you get 0 BER.
The thing is that you do not add any noise, you should use Eb/N0 to calculate the noise variance, and add a complex noise on your channel output.
The distortion added by "chanOutput = filter(chCoeffs,1,modSignal);" merely adds linear ISI, by convolving the channel impulse response with the input ...
1
You will want to work with what Wikipedia calls "field quantities."
Your engineer is telling you to work in steps of .1dB, which would be an amplitude ratio of .1=20log10(x). To get x you do 10^(.1/20) = 1.01158
From x dB to ratio (amplitude): 10^(x/20)
From x ratio (amplitude) to dB: 20Log10(x)
If you can calculate the ratio, then you should be able ...
1
Does your system has an Anti-Aliasing filter before the ADC? If not, I would suggest adding one. Depending on the spectra of the signals you are working with, aliasing can create 1/f noise.
1
I guess that your problem may not really be 1/f noise, but rather the effects from using the FFT directly on your time signal (without any windowing). If that is true, then try applying the Hann window to the time signal before the FFT and check the result.
There is plenty of information available about the window functions and how to use them.
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