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Let me add the following graphic to the great answers already given, with the intention of a specific and clear answer to the question posed. The other answers detail what linear phase is, this details why it is important in one graphic:
When a filter has linear phase, then all the frequencies within that signal will be delayed the same amount in time (as ...
25
A linear phase filter will preserve the waveshape of the signal or component of the input signal (to the extent that's possible, given that some frequencies will be changed in amplitude by the action of the filter).
This could be important in several domains:
coherent signal processing and demodulation, where the waveshape is important because a ...
24
FIR filters contain as many poles as they have zeros. but all of the poles are located at the origin, $z=0$.
because all of the poles are located inside the unit circle, the FIR filter is ostensibly stable.
this is probably not the FIR filter the OP is thinking about, but there is a class of FIR filters called Truncated IIR filters (TIIR) which may have a ...
answered Jan 7 '14 at 15:40
robert bristow-johnson
16.2k3 gold badges29 silver badges68 bronze badges
20
If $N$ is the length of the moving average, then an approximate cut-off frequency $F_{co}$ (valid for $N >= 2$) in normalized frequency $F=f/fs$ is:
$F_{co} = \frac {0.442947} {\sqrt{N^2-1}}$
The inverse of this is
$N = \frac {\sqrt{0.196202 + F_{co}^2}}{F_{co}}$
This formula is asymptotically correct for large N, and has about 2% error for N=2, and ...
19
Just to add to what's already been said, you can see this intuitively by looking at the following sinusoid with monotonically increasing frequency.
Shifting this signal to the right or left will change its phase. But note also that the phase change will be larger for higher frequencies, and smaller for lower frequencies. Or in other words, the phase ...
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My favorite "Rule of thumb" for the order of a low-pass FIR filter is the "fred harris rule of thumb":
$$
N=\frac{f_s}{\Delta f}\cdot\frac{\rm atten_{dB}}{22}
$$
where
$\Delta f$ is the transition band, in same units of $f_s$
$f_s$ is the sample rate of the filter
$\rm atten_{dB}$ is the target rejection in dB
For example if you have a ...
17
I agree that the windowing filter design method is not one of the most important design methods anymore, and it might indeed be the case that it is overrepresented in traditional textbooks, probably due to historical reasons.
However, I think that its use can be justified in certain situations. I do not agree that computational complexity is no issue ...
15
"Can you mathematically show that FIR filters have poles, because I'm not seeing it." – Jim Clay
can we assume this FIR is causal?
filter order is $N$. number of taps is $N+1$
the Finite Impulse Response: $ \quad h[n] = 0 \quad \forall \quad n>N, \ n<0$
transfer function of the FIR:
$$ \begin{align}
H(z) & = \sum_{n=-\infty}^{+\infty} h[n] ...
answered Jan 8 '14 at 5:12
robert bristow-johnson
16.2k3 gold badges29 silver badges68 bronze badges
15
The essence and importance of linear phase property lies in the definition and the effect of group delay $$\tau(\omega) = - \frac {d\phi(\omega)}{d\omega}$$ on the applied signal $x[n]$, where $\phi(\omega)$ is the phase response of the filter; (phase of its frequency response).
Assume that a filter, with a fixed group delay of $n_0$ samples, is applied a ...
13
The answer to this question is already been explained clearly in the previous replies. Yet I wish to give it a try to present a mathematical interpretation of the same
Consider a linear time invariant System whose frequency response is governed by $H(w)$.
i.e if the input to this system is $e^{jw_{0}t}$ the output will be $H(w_{0})e^{jw_{0}t}$
Here $H(w_{...
12
FIR filters contain only zeros and no poles. If a filter contains poles, it is IIR. IIR filters are indeed afflicted with stability issues and must be handled with care.
EDIT:
After some further thought and some scribbling and google-ing, I think that I have an answer to this question of FIR poles that hopefully will be satisfactory to interested parties....
12
Consider a zero-phase moving average of length $N$:
$$\text{y}[n] = \begin{cases} \displaystyle\frac{\text{x}[n] + \displaystyle\sum_{k=1}^{\frac{N-1}{2}}\left(\text{x}[n+k] + \text{x}[n-k]\right)}{N},&n\in\mathbb{Z}&\text{for }N\text{ odd}\\
\displaystyle\frac{\displaystyle\sum_{k=1}^{\frac{N}{2}}\left(\text{x}[n+(k-\frac{1}{2})] + \text{x}[n-(k-\...
12
In reviewing Fred Harris Figures of Merit for various windows (Table 1 in this link) the Hamming is compared to the Hanning (Hann) at various values of $\alpha$ and from that it is clear that the Hann would provide greater stopband rejection (The classic Hann is with $\alpha =2$ and from the table the side-lobe fall-off is -18 dB per octave). I provided the ...
12
A moving average filter can be thought of as a type of low-pass filter that doesn't have any control over its bandwidth for a fixed number of taps.
For a finite impulse response (FIR) filter, the output signal $y[n]$ is given in terms of the input signal $x[n]$ and the filter taps $h[n]$:
$$
y[k] = \sum_{n=0}^{N-1}h[n]x[k-n].
$$
The filter length in this ...
11
Note that for stable IIR filters, the impulse response does approach zero as $n$ goes to infinity. It just never becomes exactly zero. However, the sum of the absolute values is finite. Just as an example, take the exponential impulse response
$$h[n]=a^nu[n],\qquad |a|<1\tag{1}$$
where $u[n]$ is the unit step function. The sum
$$\sum_{n=-\infty}^{\...
11
See How many taps does an FIR filter need?
In your case you'd need more than 1000 taps depending on the allowable ripple, as your cut-off frequency is less than fs/500.
Alternatives :
use an IIR, a simple order-1 DC removal filter could work great
Average your signal and subtract the average in order to remove the DC
Rick Lyons proposes a clever ...
10
Note the difference between the zeros at $0.3 \pi$ and at $0.8 \pi$.
The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.
At $\theta = 0.8 \pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.
So your zeros are:
2 zeros at $z = e^{\pm j 0.3 \pi}$
2 double zeros at $z = e^{\...
9
This depends a lot on how you implement it.
A single biquad takes about 10 arithmetic operations. (To be precise
a Transposed Form II takes 4-5 multiplies and 3 adds, depending on how the gain management is done).
Arithmetic operation translates into clock cycles of your processor. That depends a lot on the efficiency of your instruction set and how good yo ...
9
Computing the polynomial coefficients from the roots of the polynomial is a potentially ill-conditioned problem. However, it turns out that the order of the roots supplied to the poly function can change things considerably. I learned this from these lecture notes by Ivan Selesnick. The so-called Leja ordering [1] helps to reduce numerical errors even for ...
8
Actually it seems MATLAB implementation of the filter() function is pretty straight forward and not fast.
For a fast implementation, have a look at FilterM by Jan Simon.
Update
In the latest releases of MATLAB (From R2016b and above) the performance of the filter() function has improved.
The methods to accelerate those operations are usually based on:
...
8
In fact you have two signals, and it depends on what you want to achieve, but usually you would just filter both signals (the real and the imaginary part) with the same (real-valued) low pass filter. So you either need two (identical) low pass filters, or you filter both signals sequentially with the same filter.
In the general case, filtering a complex ...
8
What you do in step 1 is simply truncate the infinite impulse response to approximate it by an FIR filter. If you use sufficiently many filter taps, the approximation becomes arbitrarily accurate. This means that the resulting FIR filter approximates the magnitude and the phase characteristic of the original IIR filter. So with this approach the phase will ...
8
I'm convinced that depending on the problem we're trying to solve, we can and should use both approaches: the transformation of classic analog filter designs, and the direct design in the digital domain using optimization methods. Note that the properties of the classic designs are very restricted: piecewise constant desired magnitude responses, minimum-...
7
To be precise the group delay of a linear phase FIR filter is $(N-1)/2$ samples, where $N$ is the filter length (i.e. the number of taps). The group delay is constant for all frequencies, because the filter has a linear phase, i.e. its impulse response is symmetrical (or asymmetric). A linear phase means that all frequency components of the input signal ...
7
I would say that the answer to your question - if taken literally - is 'no', there is no general way to simply convert an FIR filter to an IIR filter.
I agree with RBJ that one way to approach the problem is to look at the FIR filter's impulse response and use a time domain method (such as Prony's method) to approximate that impulse response by an IIR ...
7
Windowed Sinc filters can be adaptively generated on the fly on processors barely powerful enough to run the associated FIR filter. Windowed Sinc filters can be generated in finite bounded time.
The generation of some simple windowed Sinc filters can be completely described (and inspected for malware, etc.) in a few lines of code, versus blind use of some ...
7
I'll show here one benefit of a windowed design and a trick to get the same benefit from Parks–McClellan.
For half-band, quarter-band etc. filters windowing retains the time-domain zeros of the scaled sinc function, which is the prototypical ideal low-pass filter. The zeros end up in the coefficients, reducing the computational cost of the filters. For a ...
7
The unit circle on the z-plane represents the frequency axis, similar to the imaginary axis $j\Omega$ on the s-plane for the Laplace Transform in the continuous time case. So the frequency response of the system is given by $H(z)$ when $z= e^{j\omega}$ with $\omega$ going from $0$ to $2\pi$ representing the normalized fractional radian frequency (which is ...
7
Kalman filters really aren't that special, and you seem to be missing the point of a Kalman filter. A Kalman filter is really just a generally time-varying, generally IIR, generally multi-input multi-output filter that's been designed using a specific procedure.
Can we deem that traditional filters such as FIR and low-pass filter are designed to be used ...
7
The logical implications are the following:
"non-recursive" $\Longrightarrow$ FIR
IIR $\Longrightarrow$ "recursive"
But the opposites are not necessarily true because a FIR system can be implemented recursively (transfer function poles can be cancelled by zeros).
Of course, when referring to "recursive" or "non-recursive" we always talk ...
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