New answers tagged lowpass-filter
1
There are several problems with your approach:
The expression $\displaystyle\frac{1+z^{-1}}{2}$ is the transfer function $H(z)$, NOT its magnitude $|H(z)|$.
Consequently, the expression $\displaystyle \frac{1+e^{-j\Omega}}{2}$ is the complex frequency response $H(e^{j\Omega})$, not its magnitude.
So now you have to compute the squared magnitude of $H(e^{j\...
1
If you talk about a moving average filter then you only have a single degree of freedom, namely the filter length (i.e., the number of taps). Consequently, given the number of taps, the magnitude of the frequency response is fixed and the cut-off frequency is given.
Note that you could also define the delay of the moving average filter as another degree of ...
1
You are right in your concerns here; as is obvious from the magnitude of the frequency response, a moving-average system does not block any portion of the spectrum except at those $M-1$ null frequencies, where $M$ is the length of the moving-average impulse response.
However, it's also quite evident from the shape of the spectrum that high frequencies are ...
1
In discrete time, a frequency of zero is just a constant sequence, in complete analogy with continuous time. The main difference between discrete time and continuous time is that in discrete time you have a maximum frequency, whereas in continuous time you (theoretically) don't have one. That maximum frequency in discrete time is achieved for a sequence with ...
0
I feel that someone ought to advise the questioner against thinking that this process is automagically going to give an extra 4 bits of resolution. IF the noise on the DC signal is uncorrelated, gaussian-distributed noise, then yes it will. In the vast majority of real cases, the noise is not like that. In particular, for DC type signals with low-cost ADCs, ...
3
As mentioned in the comments, an impulse response with the shape shown in your question can only be obtained by a system with two real-valued poles. In continuous time, with two distinct poles, the total impulse response is
$$h(t)=\frac{e^{-\alpha_1t}-e^{-\alpha_2t}}{\alpha_2-\alpha_1}u(t)\tag{1}$$
It's a basic exercise to determine the location of the ...
2
If both filters have the same time constant, i.e., if their individual impulse responses are
$$h(t)=e^{-\alpha t}u(t),\qquad \alpha>0\tag{1}$$
then the impulse response of the cascade is
$$h_{tot}(t)=te^{-\alpha t}u(t)\tag{2}$$
If both filter have different time constants with individual impulse responses given by
$$h_i(t)=e^{-\alpha_it}u(t),\qquad \...
1
I would build a tensor of those matrices and use low rank or some thresholding methods on it.
You may have a look for Tensor SVD.
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One simple way to ensure that a filtered signal is always larger (or smaller) than the original is to use a moving maximum (or minimum respectively) window, followed by a non-negative filter with the same support and unity gain at DC.
In other words, you can first perform gray-scale dilation with a flat structuring element, then filter the result using a ...
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