New answers tagged finite-impulse-response
2
What you want is theoretically impossible: if $h(t) \neq 0$ only on a finite interval, then it is not bandlimited. Fortunately, all real-world signals and systems have the property that $H(f) \rightarrow 0$ as $|f| \rightarrow \infty$. This allows you to approximate $h(t)$ very closely with only a finite number of samples. For engineering purposes, this is ...
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What you want is basically a delay
$$x(t-\tau) \Rightarrow e^{-j\omega \tau}$$
If you can live with $k$ being quantized you can simply use a single tap delay FIR filter. If you need more granularity you need to implement a fractional dealy. I would cascade a shot fractional FIR filter with a long single tab bulk delay filter.
If $k$ is positive the filter ...
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You can have all the taps be zero except for the last tap and the filter will delay the signal by that number of taps (as an all-pass). This doesn't do any filtering, and it is the actual filtering requirement that would drive the number of taps needed in the filter. For that requirement alone a dual port memory is a good solution for very long delays.
The ...
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This depends a lot on whether your constraints on the number of taps is computation or memory.
If you have a enough memory, you can simply use an FIR with all taps zero except for the last one (Which is $1$). That's equivalent to using a ring buffer and no filter at all.
If you don't have enough memeory, things are more tricky. YOu can get delays that are ...
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Imagine an incoming signal with 60dB SNR over the whole spectrum.
The 60 dB isn't what matters here – what matters is that you've got the same power spectral density all over your spectrum, including the 39/40 that you'll alias onto your remaining band.
So, if you don't want the SNR of that remaining band to be affected, you'd need infinite attenuation; can'...
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In essence filtfilt does the the following
$$y[n] = x[n] \ast h[n] \ast h[-n]$$
It's convolution with the impulse response and then the time reversed impulse response (in any order since the convolution is commutative).
So in the frequency domain it's simply
$$Y(z) = X(z) \cdot H(z) \cdot H'(z) = X(z) \cdot |H(z)|^2 $$
which obviously is zero-phase since $|H(...
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