New answers tagged finite-impulse-response
1
If by "parallel form" you mean "parallel sections" as compared to "cascaded sections", the answer is "not really".
An FIR is can be considered an IIR with all the poles at $z=0$. You can technically go through the motions of doing partial fractional expansion with multiple roots but you are just going to end up with ...
2
The typical references to "parallel FIR" filters (such as this link: https://download.atlantis-press.com/article/23421.pdf)
are polyphase implementations to reduce the resource (processing rate) requirements on any one filter by being able to run each filter at a fraction of the rate required to realize the same processing result as a single FIR ...
0
but is it possible to get only N point sequence for y[n] ,
Not without losing information. You can truncate the "tail" of the convolution but this does create an error.
In most practical cases of convolution, the signal is much longer than the impulse response and it's often acceptable to discard the extra samples, but that really depends on your ...
1
I think you are using the wrong tool for the job. Semitones are spaced logarithmically but FIR filters have linear frequency resolution. If you want to reliably distinguish between the low E and the low F on the bass guitar you need a frequency resolution of better than 2 Hz which requires 10s of thousands of taps (at 48 kHz sample rate).
That's why most ...
0
Reconstruction is possible so long as NOLA is obeyed - which is an easier criterion (on synthesis information) to meet than what you seek (analysis information).
To discriminate temporal variations finer than $T$, the window's temporal width must be $\leq T$. You can use ssqueezepy's window_resolution with appropriate unit conversion (mult by $f_s$) to ...
2
Of course.
The difference is linear operation so you simply get
$$H(z) = H_1(z)-H_2(z)$$
The Z transform of a moving average filter of length N is simply $H_N(z) = \frac{1-z^{-N}}{1-z^{-1}}$ so in your case you get
$H(z) = \frac{1-z^{-10}}{1-z^{-1}}-\frac{1-z^{-21}}{1-z^{-1}}$
Pop in $z = e^{-j\omega}$ for your frequency of interest and solve for amplitude ...
0
A linear filter just attenuate some frequencies more and others less. If you alter the SNR, the output of the FIR filter should change accordingly.
Many applications use more elaborate noise reduction methods in stead of (or in addition to) linear filtering. If you know that your signal is going to be a sine of known frequency, you may estimate its phase and ...
0
There are routines that will provide the Hilbert coefficients directly, but an approach I like to use given its simplicity and clarity in functionality is to transform a Half Band filter to a Hilbert as follows:
Step 1: Estimate the number of taps needed from the specifications using these commonly used estimators. The one Marcus Mueller provided would be ...
2
You can't implement an ideal Hilbert transformer: its impulse response is non causal and infinite in time.
So you can only implement an "approximate" one and the best way to do this depends on the specific requirement of your application: what's your frequency range of interest, how much magnitude and/or phase deviation can you tolerate, are you ...
Top 50 recent answers are included
