12

The difference is that with your method, if you imagine the histogram, you're simply going to stretch it to span from 0 to 255, but its shape will be preserved. Histogram equalization not only stretches your histogram, but also tries to make it flat, so that you get an approximately even distribution of pixels of every shade of gray. In terms of why one is ...


7

Unless you have tight space constraints, I would stick to lossless formats like good old WAV. Phones do not pass much high-frequency content so you should be fine with a 8KHz Mono WAVE file. You will however want to keep the bit depth high, say to 24 bits, since you want to process the audio. Once you are finished with your processing you can compress with a ...


7

Sticking with linear systems, removing the ringing is nearly the same as adding back some of the spectral content that your really steep transition filters removed. Why use some crazy scheme to add back the stuff in the "softer" transitions that your hard-edged filters cut out? Just use a more reasonable total filter response in the first place. Going to ...


6

The OFDM signal as a whole is affected by frequency selective filtering. It is usually designed such that the subcarrier bandwidth is smaller than the channel coherence bandwidth. This yields you frequency flat fading for each subcarrier which can be described by a single complex multiplication and equalization can equally done with just one tap. ...


5

FDLS requires a causal frequency response. Your prototype frequency response has zero phase everywhere, which is most definitely not causal. An IIR filter order of 50 is humongous. When FDLS has too many poles and zeroes available, it "tries" to cancel excess poles with excess zeroes. Unfortunately, due to numerical limitations, the cancellation is often ...


5

Your code reveals many misconception about what the CMA is supposed to achieve: your step size mu is much too small; note, however, that the optimal step size can only be found through experiment. the variabe noisedB appears to be the desired SNR of the received signal. An SNR of $0\,\text{dB}$ as specified by you is very poor (the noise is as strong as the ...


5

A (linear) equalizer tries to compensate for the (linear) filtering effect of the channel. This filtering introduced by the channel creates intersymbol interference (ISI), so the goal of the equalizer is to reduce (or, ideally, eliminate) ISI, such that the symbol error rate is reduced to an acceptable level. Linear filtering creates ISI by smearing out the ...


5

The reason why almost all linear adaptive equalizers are implemented as FIR filters is that FIR filters are always stable and that there exist relatively simple and effective adaptation algorithms. Note that much work has been done on adaptive IIR filters (e.g., this book by Phillip Regalia), but in practice FIR filters are still the preferred option. Note ...


4

To answer your question histogram equalization is called like this because its function is to produce an equalized histogram (that is an uniform probability density function). There are different algorithms that may approach this function, and obviously there is a problem in the example that is shown: In fact, the algorithm used there will always have ...


4

Well, I took a look at your code, and spent a lot of time on it, and I have discovered some mistakes, some practicals and some theoretics. Here are my answers: (1) You can use the function filter. Just remember that it returns a vector of the same length of the input, while conv returns the $ N_x + N_h - 1 $, where $ N_x $ is the length of your input and $ ...


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 ...


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

In communication systems, transmitted signals are distorted by the physical medium (the channel) charactheristics. The channel estimator tries to identify the transmission channel characteristics, by computing (estimating) various of its model parameters. Then, the channel equalizer tries to restore to distorted (received) signals back into their originals ...


3

Your low-pass filter's pass band is smaller than your signal's bandwidth, so it is destroying a significant portion of your signal. Given that, you cannot reliably reproduce the original signal without using sophisticated techniques like error correction codes. I would try increasing $\alpha$ to 2400 Hz. Then the rolloff point will be 2400 * .5 = 1200. ...


3

If I understood right, you are stuck in matching a given histogram into a desired one and creating a new image from this matched histogram obtained by your filtering method. I would first suggest you to get rid of all the unnecessary stuff (including the python code) above and isolate your problem as "histogram matching" in mathematical terms. Now I will ...


3

The formula you need, which I don't see in the scanned pages, is the N-dimensional Gaussian distribution of the received vector given that the signal $s_m$ was transmitted: $$f(\vec{r}|s_m)=\frac{1}{(\pi N_0)^{N/2}}\exp\left[-\frac{\sum_{j=1}^N(r_j-x_{mj})^2}{N_0}\right]\tag{1}$$ The maximum likelihood receiver seeks to maximize this conditional ...


3

1) Can I just casacade the filters : input signal -> processed through f1 -> ... -> processed through fn -> output signal ? Yep, that's how you do it. 2) Given that filters are not perfect, does the order matters ? Probably not. You're probably working with ≥32-bit floats on a computer processor? So you don't really need to worry about clipping ...


3

In general equalizing a channel with deep fades is a problem for all equalizers. CMA equalizer is no exception. Absence of a training signal makes things worse. This is obviously an example of the CMA stuck in a local minimum. The inital condition of the CMA equalizer was probably in the vicinity of the local minimum. The equalizer filter to which the CMA ...


3

Simply put, you need a bank of passband filters. You feed your signal through each of the filters, and sum up the outputs from the filters. Designing the filters is where the fun comes in. First off, assuming this is just audio (music or the like) then there's no need of special filters. You can use the simplest and fastest and not worry too much about ...


3

From the first to the sixth OFDM-Symbol you have spread virtual carriers equally across [0,..,51] with a distance of 3. With this way you have spread your 'energy' loss over a wide spectrum. In difference to having a single carrier serving as virtual carrier all the time you can transmit information over every 'data' carrier and maybe have a constant mean ...


3

The confusion comes from the fact that what is tagged as "transmitted" isn't the real transmitted waveform but its baseband representation, which are $\left\{+1,-1\right\}$ symbols in 1 dimension. For 2-FSK, that representation would be $\left\{\left(+1,0\right), \left(0,+1\right)\right\}$ symbols in 2 dimensions. A common convenient way to handle such 2 ...


3

1. This is a question investigated by many researchers decades ago. They discovered that the bandwidth limitation of R/2 Hz is not a fundamental limit set by nature. It is just a criterion if we don't want to do any further processing at the Rx (that also doesn't mean that we can infinitely break this limit). As long as we are willing to pay the cost in ...


3

The generalized eigenvalue problem is given by $$Bw=\lambda Cw\tag{1}$$ where $\lambda$ is the generalized eigenvalue of the matrices $B$ and $C$. Multiplying $(1)$ from the left with $w^H$ (with $^H$ denoting the Hermitian conjugate) and dividing both sides by $w^HCw$ (assuming that this term is non-zero), we obtain $$\frac{w^HBw}{w^HCw}=J(w)=\lambda\tag{...


2

A typical ("non-fractional") equalizer operates at the symbol rate. That is, the inputs that you apply to the equalizer are generally soft decisions on what the unequalized received symbols are. A linear equalizer then takes this stream of soft symbol metrics and applies a linear filter that (hopefully) corrects for any non-ideal frequency response found in ...


2

The answer depends a bit on how fancy you want to get. Let's say you want to do a quick and dirty octave band graphic EQ: The easiest way to do this is to cascade individual filters. Using the "peakingEQ" type described in http://www.musicdsp.org/files/Audio-EQ-Cookbook.txt you can actually get a fairly decent octave band filter with a Q of about 2.8 (2*...


2

The choice of algorithm depends on your application scenario - i.e. there is no best solution per se. If your focus is on minimum computational requirements, a comb filter will probably be optimum. If you want optimum sound quality, a notch filter based approach could be a better choice. To gain an overview about advantages and disadvantages of the ...


2

Yes it is true. A simple histogram equalization may fail for certain image intensity distributions. In your case there are some bins with excessive number of pixels. For other methods, try for example gamma and contrast corrections.


2

I haven't used the technique, but an IIR filter order of 50 sounds remarkably high, and possibly prone to numerical problems. Try starting out with much smaller values for M and filter order, and slowly ramping them up. The paper you quote uses $M=8$ and $N=9$.


2

If I hand you a uniform distribution spanning say, 0 to 1, then I can say that the probability of a variable with this uniform distribution taking on value of say, 0.3 is, equal to the probability that it takes on a value of 0.7, or 0.01, or 0.93, etc. In other words, since the uniform distribution is flat, all probability values are equal. Now forget ...


2

To make a signal with that distribution have a truly uniform distribution would probably require that the transformation used be time-varying: a value $10$ at time $t_0$ would be moved to, say, a value of $15$, but a value $10$ at time $t_1$ (later) would be moved to, for example, $20$. So, what the equalizer tries to do is to make the cumulative ...


Only top voted, non community-wiki answers of a minimum length are eligible