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Why is the first value in digital first-order IIR filter impulse response not the largest?
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When I create a digital first-order IIR low-pass filter with scipy (code below), I get the following coefficients:
B: [ 0.1367 0.1367]
A: [ 1. -0.7265]
The impulse response of this filter does not look like most text-book examples in that the first value is not the largest.
Why is this so? Does it have to be so?
If I want to make an FIR filter to mimic such a first-order IIR low-pass filter, should it be a normal exponential decay shape, or does it need the first value "tweaked" in this manner?
import numpy as np
from scipy.signal import butter, lfilter
import matplotlib.pyplot as plt
b, a = butter(1, .1, btype='lowpass')
print('B:', b)
print('A:', a)
x = np.zeros(1000)
x[0] = 1.
xf = lfilter(b, a, x)
plt.figure()
plt.plot(x,'o', label='impulse')
plt.plot(xf,'s-', label='response')
plt.xlim(-1, 10)
plt.ylim(-.1, .3)
plt.legend()
filter-design infinite-impulse-response fir impulse-response
edited Sep 3 at 11:34
Matt L.
45.2k13678
asked Sep 3 at 10:47
adr
404
add a comment |Â
up vote
5
down vote
favorite
When I create a digital first-order IIR low-pass filter with scipy (code below), I get the following coefficients:
B: [ 0.1367 0.1367]
A: [ 1. -0.7265]
The impulse response of this filter does not look like most text-book examples in that the first value is not the largest.
Why is this so? Does it have to be so?
If I want to make an FIR filter to mimic such a first-order IIR low-pass filter, should it be a normal exponential decay shape, or does it need the first value "tweaked" in this manner?
import numpy as np
from scipy.signal import butter, lfilter
import matplotlib.pyplot as plt
b, a = butter(1, .1, btype='lowpass')
print('B:', b)
print('A:', a)
x = np.zeros(1000)
x[0] = 1.
xf = lfilter(b, a, x)
plt.figure()
plt.plot(x,'o', label='impulse')
plt.plot(xf,'s-', label='response')
plt.xlim(-1, 10)
plt.ylim(-.1, .3)
plt.legend()
filter-design infinite-impulse-response fir impulse-response
edited Sep 3 at 11:34
Matt L.
45.2k13678
asked Sep 3 at 10:47
adr
404
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
When I create a digital first-order IIR low-pass filter with scipy (code below), I get the following coefficients:
B: [ 0.1367 0.1367]
A: [ 1. -0.7265]
The impulse response of this filter does not look like most text-book examples in that the first value is not the largest.
Why is this so? Does it have to be so?
If I want to make an FIR filter to mimic such a first-order IIR low-pass filter, should it be a normal exponential decay shape, or does it need the first value "tweaked" in this manner?
import numpy as np
from scipy.signal import butter, lfilter
import matplotlib.pyplot as plt
b, a = butter(1, .1, btype='lowpass')
print('B:', b)
print('A:', a)
x = np.zeros(1000)
x[0] = 1.
xf = lfilter(b, a, x)
plt.figure()
plt.plot(x,'o', label='impulse')
plt.plot(xf,'s-', label='response')
plt.xlim(-1, 10)
plt.ylim(-.1, .3)
plt.legend()
filter-design infinite-impulse-response fir impulse-response
edited Sep 3 at 11:34
Matt L.
45.2k13678
asked Sep 3 at 10:47
adr
404
When I create a digital first-order IIR low-pass filter with scipy (code below), I get the following coefficients:
B: [ 0.1367 0.1367]
A: [ 1. -0.7265]
The impulse response of this filter does not look like most text-book examples in that the first value is not the largest.
Why is this so? Does it have to be so?
If I want to make an FIR filter to mimic such a first-order IIR low-pass filter, should it be a normal exponential decay shape, or does it need the first value "tweaked" in this manner?
import numpy as np
from scipy.signal import butter, lfilter
import matplotlib.pyplot as plt
b, a = butter(1, .1, btype='lowpass')
print('B:', b)
print('A:', a)
x = np.zeros(1000)
x[0] = 1.
xf = lfilter(b, a, x)
plt.figure()
plt.plot(x,'o', label='impulse')
plt.plot(xf,'s-', label='response')
plt.xlim(-1, 10)
plt.ylim(-.1, .3)
plt.legend()
filter-design infinite-impulse-response fir impulse-response
edited Sep 3 at 11:34
Matt L.
45.2k13678
asked Sep 3 at 10:47
adr
404
edited Sep 3 at 11:34
Matt L.
45.2k13678
edited Sep 3 at 11:34
Matt L.
45.2k13678
edited Sep 3 at 11:34
Matt L.
45.2k13678
45.2k13678
asked Sep 3 at 10:47
adr
404
asked Sep 3 at 10:47
adr
404
asked Sep 3 at 10:47
adr
404
404
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
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up vote
6
down vote
accepted
A causal first-order IIR filter is characterized by the following difference equation:
$$y[n]=b_0x[n]+b_1x[n-1]-a_1y[n-1]tag1$$
with $x[n]$ the input signal, and $y[n]$ the output signal. The impulse response of that system can be computed via the $mathcalZ$-transform or otherwise, and it turns out to be
$$h[n]=b_0delta[n]+(-a_1)^n-1(b_1-b_0a_1)u[n-1]tag2$$
where $delta[n]$ is the unit impulse, and $u[n]$ is the unit step sequence. Assuming that the filter is stable, we can conclude that $|a_1|<1$ is satisfied. Consequently, $|h[0]|>|h[n]|$ for $n>0$ is satisfied if and only if
$$|b_0|>|b_1-b_0a_1|tag3$$
holds.
With $b_0=b_1$ and $a_1<0$ as in your example, condition $(3)$ can never be satisfied, and, consequently, the maximum magnitude of the impulse response is not assumed at index $n=0$.
EDIT:
I add more details referring to your filter design problem. A Butterworth low pass filter's frequency response satisfies $H(1)=1$ (DC) and $H(-1)=0$ (Nyquist). With
$$H(z)=fracb_0+b_1z^-11+a_1z^-1tag4$$
we have the following requirements on the coefficients:
$$H(1)=fracb_0+b_11+a_1=1\
H(-1)=fracb_0-b_11-a_1=0tag5$$
which gives
$$b_0=b_1quadtextrm and quad a_1=2b_0-1tag6$$
From the stability condition $|a_1|<1$ we get
$$|2b_0-1|<1Longrightarrow 0<b_0<1tag7$$
Combining the constraints $(6)$ and $(7)$ with the inequality $(3)$ yields
$$1 >b_0>frac12tag8$$
If $(8)$ is satisfied, $h[0]$ will have the largest magnitude, otherwise it won't. Now it turns out that the exact value of $b_0$ depends on the desired cut-off frequency of the low pass filter. For cut-off frequencies less than $frac12$ (i.e., less than half the Nyquist frequency), you get $b_0<frac12$, and hence $|h[0]|$ is smaller than the maximum magnitude of the impulse response. Only for cut-off frequencies greater than half the Nyquist frequency do you get $|h[0]|>|h[n]|$, $n>0$.
So to directly answer the question in your title: the first impulse response value is not the largest (in magnitude) because the filter coefficients do not satisfy inequality $(3)$, and in your example of a (first-order) Butterworth low pass filter this happens because the cut-off frequency is smaller than half the Nyquist frequency.
answered Sep 3 at 11:33
Matt L.
45.2k13678
I'm afraid this doesn't help me understand the bigger picture. I know how the coefficients produce the observed impulse response. But why are the coefficients the way they are? Does it have something to do with analog vs. digital?
– adr
Sep 3 at 12:37
I believe it is because he has zeros in his filter implementation (it is not completely IIR but FIR/IIR) and a true minimum phase IIR would have trailing echos only. (??) Is that true @MattL?
– Dan Boschen
Sep 3 at 13:31
1
@DanBoschen: The zeros don't really matter, it depends on the value of the cut-off frequency. For cut-off frequencies less than half the Nyquist frequency, the first value of the impulse response has a smaller magnitude than the second value. See my edited answer.
– Matt L.
Sep 3 at 15:18
1
@adr: Sure, any good low pass filter has a zero at Nyquist. Note though that this is just an ordinary IIR filter. IIR filters normally have a recursive and a feed forward path. What you mean is called an "all pole filter", i.e., a filter with only poles away from the origin of the z-plane, with just a constant in the numerator.
– Matt L.
Sep 3 at 15:23
1
@MattL In retrospect, my question should have been: why does the first value not lie on the exponential curve defined by the subsequent values? Anyway, things are clearer to me now, which is what the objective was!
– adr
Sep 3 at 19:00
 |Â
show 6 more comments
up vote
2
down vote
The impulse response of this filter does not look like most text-book
examples in that the first value is not the largest.
That's simply an issue with interpreting the text books the wrong way. There is nothing special about the first sample, and there is no reason at all, why it would have to be the largest. I'm quite sure that none of the text box actually says anything about this.
Some very simple "beginner" filters just happen to have this property, so you looked at a bunch of pictures and formed a conclusion & generalization that's neither supported by the text or the math.
answered Sep 3 at 13:54
Hilmar
9,002916
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Starting from an FIR filter with an absolutely flat frequency response:
[ 1 0 0 0 0 ]has a delay of one sample,[ 0 0 1 0 0 ]has a delay of three samples.
The largest coefficient corresponds to the filter's group delay. For a lowpass filter like [ 1 -2 7 -2 1] it is the group delay for the DC component — if you look at the step [ 0 0 0 1 1 1 1 1 1 1 ] response [ 0 0 0 1 -1 6 4 5 5 5] it is fairly obvious what the filter's DC delay is. If you design an FIR lowpass filter with uniform group delay for all passband frequencies, it will have a single largest coefficient in the middle bin, while the length of the filter corresponds to the filter's selectivity.
The same filter with no delay would be non-causal, as the small coefficients before the large one would have negative indexes.
For IIR filters, it's slightly less obvious, because they have two coefficient sets and are unlikely to have flat group delay, but you can still get better stopband attenuation and sharper falloffs with larger delays.
answered Sep 3 at 19:06
Simon Richter
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The confusion might arise from what people mean by first order. In some places, like:
- SE.DSP: What Is the Best First Order IIR (AR Filter) Approximation to a Moving Average Filter (FIR Filter)?
Tutorial on a very simple yet useful filter: the first order IIR filter, J. Arzi, 2016
a first order filter follows the input ($x$)/output ($y$) equation:
$$ y[n] = (1-lambda)y[n-1]+lambda x[n],,$$
or equivalently in tracking form:
$$ y[n] = y[n-1]+lambda (x[n]-y[n-1]),.$$
It has numerous names, including first order exponential averaging low-pass filter or exponentially weighted moving average (EWMA). More details are given in
- What is the name of these simple filter algorithms?
- Is there a technical term for this simple method of smoothing out a signal?
In that case, the impulse response $h$ of the filter is explicit. For $k< 0 $, $h[k] = 0$, and for $kge 0 $, $h[k] = lambda(1-lambda)^k$, where $0<lambda < 1$ is a smoothing factor. In this situation, the "first" value $h[0] $ is indeed the largest, since the exponential sequence $lambda(1-lambda)^k$ is decreasing:
Other filter families are called first order, with additional constraints, like Butterworth's for which Mmatt's answer is complete.
answered Sep 3 at 15:33
Laurent Duval
14.8k31955
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
A causal first-order IIR filter is characterized by the following difference equation:
$$y[n]=b_0x[n]+b_1x[n-1]-a_1y[n-1]tag1$$
with $x[n]$ the input signal, and $y[n]$ the output signal. The impulse response of that system can be computed via the $mathcalZ$-transform or otherwise, and it turns out to be
$$h[n]=b_0delta[n]+(-a_1)^n-1(b_1-b_0a_1)u[n-1]tag2$$
where $delta[n]$ is the unit impulse, and $u[n]$ is the unit step sequence. Assuming that the filter is stable, we can conclude that $|a_1|<1$ is satisfied. Consequently, $|h[0]|>|h[n]|$ for $n>0$ is satisfied if and only if
$$|b_0|>|b_1-b_0a_1|tag3$$
holds.
With $b_0=b_1$ and $a_1<0$ as in your example, condition $(3)$ can never be satisfied, and, consequently, the maximum magnitude of the impulse response is not assumed at index $n=0$.
EDIT:
I add more details referring to your filter design problem. A Butterworth low pass filter's frequency response satisfies $H(1)=1$ (DC) and $H(-1)=0$ (Nyquist). With
$$H(z)=fracb_0+b_1z^-11+a_1z^-1tag4$$
we have the following requirements on the coefficients:
$$H(1)=fracb_0+b_11+a_1=1\
H(-1)=fracb_0-b_11-a_1=0tag5$$
which gives
$$b_0=b_1quadtextrm and quad a_1=2b_0-1tag6$$
From the stability condition $|a_1|<1$ we get
$$|2b_0-1|<1Longrightarrow 0<b_0<1tag7$$
Combining the constraints $(6)$ and $(7)$ with the inequality $(3)$ yields
$$1 >b_0>frac12tag8$$
If $(8)$ is satisfied, $h[0]$ will have the largest magnitude, otherwise it won't. Now it turns out that the exact value of $b_0$ depends on the desired cut-off frequency of the low pass filter. For cut-off frequencies less than $frac12$ (i.e., less than half the Nyquist frequency), you get $b_0<frac12$, and hence $|h[0]|$ is smaller than the maximum magnitude of the impulse response. Only for cut-off frequencies greater than half the Nyquist frequency do you get $|h[0]|>|h[n]|$, $n>0$.
So to directly answer the question in your title: the first impulse response value is not the largest (in magnitude) because the filter coefficients do not satisfy inequality $(3)$, and in your example of a (first-order) Butterworth low pass filter this happens because the cut-off frequency is smaller than half the Nyquist frequency.
answered Sep 3 at 11:33
Matt L.
45.2k13678
I'm afraid this doesn't help me understand the bigger picture. I know how the coefficients produce the observed impulse response. But why are the coefficients the way they are? Does it have something to do with analog vs. digital?
– adr
Sep 3 at 12:37
I believe it is because he has zeros in his filter implementation (it is not completely IIR but FIR/IIR) and a true minimum phase IIR would have trailing echos only. (??) Is that true @MattL?
– Dan Boschen
Sep 3 at 13:31
1
@DanBoschen: The zeros don't really matter, it depends on the value of the cut-off frequency. For cut-off frequencies less than half the Nyquist frequency, the first value of the impulse response has a smaller magnitude than the second value. See my edited answer.
– Matt L.
Sep 3 at 15:18
1
@adr: Sure, any good low pass filter has a zero at Nyquist. Note though that this is just an ordinary IIR filter. IIR filters normally have a recursive and a feed forward path. What you mean is called an "all pole filter", i.e., a filter with only poles away from the origin of the z-plane, with just a constant in the numerator.
– Matt L.
Sep 3 at 15:23
1
@MattL In retrospect, my question should have been: why does the first value not lie on the exponential curve defined by the subsequent values? Anyway, things are clearer to me now, which is what the objective was!
– adr
Sep 3 at 19:00
 |Â
show 6 more comments
up vote
6
down vote
accepted
A causal first-order IIR filter is characterized by the following difference equation:
$$y[n]=b_0x[n]+b_1x[n-1]-a_1y[n-1]tag1$$
with $x[n]$ the input signal, and $y[n]$ the output signal. The impulse response of that system can be computed via the $mathcalZ$-transform or otherwise, and it turns out to be
$$h[n]=b_0delta[n]+(-a_1)^n-1(b_1-b_0a_1)u[n-1]tag2$$
where $delta[n]$ is the unit impulse, and $u[n]$ is the unit step sequence. Assuming that the filter is stable, we can conclude that $|a_1|<1$ is satisfied. Consequently, $|h[0]|>|h[n]|$ for $n>0$ is satisfied if and only if
$$|b_0|>|b_1-b_0a_1|tag3$$
holds.
With $b_0=b_1$ and $a_1<0$ as in your example, condition $(3)$ can never be satisfied, and, consequently, the maximum magnitude of the impulse response is not assumed at index $n=0$.
EDIT:
I add more details referring to your filter design problem. A Butterworth low pass filter's frequency response satisfies $H(1)=1$ (DC) and $H(-1)=0$ (Nyquist). With
$$H(z)=fracb_0+b_1z^-11+a_1z^-1tag4$$
we have the following requirements on the coefficients:
$$H(1)=fracb_0+b_11+a_1=1\
H(-1)=fracb_0-b_11-a_1=0tag5$$
which gives
$$b_0=b_1quadtextrm and quad a_1=2b_0-1tag6$$
From the stability condition $|a_1|<1$ we get
$$|2b_0-1|<1Longrightarrow 0<b_0<1tag7$$
Combining the constraints $(6)$ and $(7)$ with the inequality $(3)$ yields
$$1 >b_0>frac12tag8$$
If $(8)$ is satisfied, $h[0]$ will have the largest magnitude, otherwise it won't. Now it turns out that the exact value of $b_0$ depends on the desired cut-off frequency of the low pass filter. For cut-off frequencies less than $frac12$ (i.e., less than half the Nyquist frequency), you get $b_0<frac12$, and hence $|h[0]|$ is smaller than the maximum magnitude of the impulse response. Only for cut-off frequencies greater than half the Nyquist frequency do you get $|h[0]|>|h[n]|$, $n>0$.
So to directly answer the question in your title: the first impulse response value is not the largest (in magnitude) because the filter coefficients do not satisfy inequality $(3)$, and in your example of a (first-order) Butterworth low pass filter this happens because the cut-off frequency is smaller than half the Nyquist frequency.
answered Sep 3 at 11:33
Matt L.
45.2k13678
I'm afraid this doesn't help me understand the bigger picture. I know how the coefficients produce the observed impulse response. But why are the coefficients the way they are? Does it have something to do with analog vs. digital?
– adr
Sep 3 at 12:37
I believe it is because he has zeros in his filter implementation (it is not completely IIR but FIR/IIR) and a true minimum phase IIR would have trailing echos only. (??) Is that true @MattL?
– Dan Boschen
Sep 3 at 13:31
1
@DanBoschen: The zeros don't really matter, it depends on the value of the cut-off frequency. For cut-off frequencies less than half the Nyquist frequency, the first value of the impulse response has a smaller magnitude than the second value. See my edited answer.
– Matt L.
Sep 3 at 15:18
1
@adr: Sure, any good low pass filter has a zero at Nyquist. Note though that this is just an ordinary IIR filter. IIR filters normally have a recursive and a feed forward path. What you mean is called an "all pole filter", i.e., a filter with only poles away from the origin of the z-plane, with just a constant in the numerator.
– Matt L.
Sep 3 at 15:23
1
@MattL In retrospect, my question should have been: why does the first value not lie on the exponential curve defined by the subsequent values? Anyway, things are clearer to me now, which is what the objective was!
– adr
Sep 3 at 19:00
 |Â
show 6 more comments
up vote
6
down vote
accepted
up vote
6
down vote
accepted
A causal first-order IIR filter is characterized by the following difference equation:
$$y[n]=b_0x[n]+b_1x[n-1]-a_1y[n-1]tag1$$
with $x[n]$ the input signal, and $y[n]$ the output signal. The impulse response of that system can be computed via the $mathcalZ$-transform or otherwise, and it turns out to be
$$h[n]=b_0delta[n]+(-a_1)^n-1(b_1-b_0a_1)u[n-1]tag2$$
where $delta[n]$ is the unit impulse, and $u[n]$ is the unit step sequence. Assuming that the filter is stable, we can conclude that $|a_1|<1$ is satisfied. Consequently, $|h[0]|>|h[n]|$ for $n>0$ is satisfied if and only if
$$|b_0|>|b_1-b_0a_1|tag3$$
holds.
With $b_0=b_1$ and $a_1<0$ as in your example, condition $(3)$ can never be satisfied, and, consequently, the maximum magnitude of the impulse response is not assumed at index $n=0$.
EDIT:
I add more details referring to your filter design problem. A Butterworth low pass filter's frequency response satisfies $H(1)=1$ (DC) and $H(-1)=0$ (Nyquist). With
$$H(z)=fracb_0+b_1z^-11+a_1z^-1tag4$$
we have the following requirements on the coefficients:
$$H(1)=fracb_0+b_11+a_1=1\
H(-1)=fracb_0-b_11-a_1=0tag5$$
which gives
$$b_0=b_1quadtextrm and quad a_1=2b_0-1tag6$$
From the stability condition $|a_1|<1$ we get
$$|2b_0-1|<1Longrightarrow 0<b_0<1tag7$$
Combining the constraints $(6)$ and $(7)$ with the inequality $(3)$ yields
$$1 >b_0>frac12tag8$$
If $(8)$ is satisfied, $h[0]$ will have the largest magnitude, otherwise it won't. Now it turns out that the exact value of $b_0$ depends on the desired cut-off frequency of the low pass filter. For cut-off frequencies less than $frac12$ (i.e., less than half the Nyquist frequency), you get $b_0<frac12$, and hence $|h[0]|$ is smaller than the maximum magnitude of the impulse response. Only for cut-off frequencies greater than half the Nyquist frequency do you get $|h[0]|>|h[n]|$, $n>0$.
So to directly answer the question in your title: the first impulse response value is not the largest (in magnitude) because the filter coefficients do not satisfy inequality $(3)$, and in your example of a (first-order) Butterworth low pass filter this happens because the cut-off frequency is smaller than half the Nyquist frequency.
answered Sep 3 at 11:33
Matt L.
45.2k13678
A causal first-order IIR filter is characterized by the following difference equation:
$$y[n]=b_0x[n]+b_1x[n-1]-a_1y[n-1]tag1$$
with $x[n]$ the input signal, and $y[n]$ the output signal. The impulse response of that system can be computed via the $mathcalZ$-transform or otherwise, and it turns out to be
$$h[n]=b_0delta[n]+(-a_1)^n-1(b_1-b_0a_1)u[n-1]tag2$$
where $delta[n]$ is the unit impulse, and $u[n]$ is the unit step sequence. Assuming that the filter is stable, we can conclude that $|a_1|<1$ is satisfied. Consequently, $|h[0]|>|h[n]|$ for $n>0$ is satisfied if and only if
$$|b_0|>|b_1-b_0a_1|tag3$$
holds.
With $b_0=b_1$ and $a_1<0$ as in your example, condition $(3)$ can never be satisfied, and, consequently, the maximum magnitude of the impulse response is not assumed at index $n=0$.
EDIT:
I add more details referring to your filter design problem. A Butterworth low pass filter's frequency response satisfies $H(1)=1$ (DC) and $H(-1)=0$ (Nyquist). With
$$H(z)=fracb_0+b_1z^-11+a_1z^-1tag4$$
we have the following requirements on the coefficients:
$$H(1)=fracb_0+b_11+a_1=1\
H(-1)=fracb_0-b_11-a_1=0tag5$$
which gives
$$b_0=b_1quadtextrm and quad a_1=2b_0-1tag6$$
From the stability condition $|a_1|<1$ we get
$$|2b_0-1|<1Longrightarrow 0<b_0<1tag7$$
Combining the constraints $(6)$ and $(7)$ with the inequality $(3)$ yields
$$1 >b_0>frac12tag8$$
If $(8)$ is satisfied, $h[0]$ will have the largest magnitude, otherwise it won't. Now it turns out that the exact value of $b_0$ depends on the desired cut-off frequency of the low pass filter. For cut-off frequencies less than $frac12$ (i.e., less than half the Nyquist frequency), you get $b_0<frac12$, and hence $|h[0]|$ is smaller than the maximum magnitude of the impulse response. Only for cut-off frequencies greater than half the Nyquist frequency do you get $|h[0]|>|h[n]|$, $n>0$.
So to directly answer the question in your title: the first impulse response value is not the largest (in magnitude) because the filter coefficients do not satisfy inequality $(3)$, and in your example of a (first-order) Butterworth low pass filter this happens because the cut-off frequency is smaller than half the Nyquist frequency.
answered Sep 3 at 11:33
Matt L.
45.2k13678
edited Sep 3 at 16:28
answered Sep 3 at 11:33
Matt L.
45.2k13678
answered Sep 3 at 11:33
Matt L.
45.2k13678
answered Sep 3 at 11:33
Matt L.
45.2k13678
45.2k13678
I'm afraid this doesn't help me understand the bigger picture. I know how the coefficients produce the observed impulse response. But why are the coefficients the way they are? Does it have something to do with analog vs. digital?
– adr
Sep 3 at 12:37
I believe it is because he has zeros in his filter implementation (it is not completely IIR but FIR/IIR) and a true minimum phase IIR would have trailing echos only. (??) Is that true @MattL?
– Dan Boschen
Sep 3 at 13:31
1
@DanBoschen: The zeros don't really matter, it depends on the value of the cut-off frequency. For cut-off frequencies less than half the Nyquist frequency, the first value of the impulse response has a smaller magnitude than the second value. See my edited answer.
– Matt L.
Sep 3 at 15:18
1
@adr: Sure, any good low pass filter has a zero at Nyquist. Note though that this is just an ordinary IIR filter. IIR filters normally have a recursive and a feed forward path. What you mean is called an "all pole filter", i.e., a filter with only poles away from the origin of the z-plane, with just a constant in the numerator.
– Matt L.
Sep 3 at 15:23
1
@MattL In retrospect, my question should have been: why does the first value not lie on the exponential curve defined by the subsequent values? Anyway, things are clearer to me now, which is what the objective was!
– adr
Sep 3 at 19:00
 |Â
show 6 more comments
I'm afraid this doesn't help me understand the bigger picture. I know how the coefficients produce the observed impulse response. But why are the coefficients the way they are? Does it have something to do with analog vs. digital?
– adr
Sep 3 at 12:37
I believe it is because he has zeros in his filter implementation (it is not completely IIR but FIR/IIR) and a true minimum phase IIR would have trailing echos only. (??) Is that true @MattL?
– Dan Boschen
Sep 3 at 13:31
1
@DanBoschen: The zeros don't really matter, it depends on the value of the cut-off frequency. For cut-off frequencies less than half the Nyquist frequency, the first value of the impulse response has a smaller magnitude than the second value. See my edited answer.
– Matt L.
Sep 3 at 15:18
1
@adr: Sure, any good low pass filter has a zero at Nyquist. Note though that this is just an ordinary IIR filter. IIR filters normally have a recursive and a feed forward path. What you mean is called an "all pole filter", i.e., a filter with only poles away from the origin of the z-plane, with just a constant in the numerator.
– Matt L.
Sep 3 at 15:23
1
@MattL In retrospect, my question should have been: why does the first value not lie on the exponential curve defined by the subsequent values? Anyway, things are clearer to me now, which is what the objective was!
– adr
Sep 3 at 19:00
I'm afraid this doesn't help me understand the bigger picture. I know how the coefficients produce the observed impulse response. But why are the coefficients the way they are? Does it have something to do with analog vs. digital?
– adr
Sep 3 at 12:37
I'm afraid this doesn't help me understand the bigger picture. I know how the coefficients produce the observed impulse response. But why are the coefficients the way they are? Does it have something to do with analog vs. digital?
– adr
Sep 3 at 12:37
I believe it is because he has zeros in his filter implementation (it is not completely IIR but FIR/IIR) and a true minimum phase IIR would have trailing echos only. (??) Is that true @MattL?
– Dan Boschen
Sep 3 at 13:31
I believe it is because he has zeros in his filter implementation (it is not completely IIR but FIR/IIR) and a true minimum phase IIR would have trailing echos only. (??) Is that true @MattL?
– Dan Boschen
Sep 3 at 13:31
1
1
@DanBoschen: The zeros don't really matter, it depends on the value of the cut-off frequency. For cut-off frequencies less than half the Nyquist frequency, the first value of the impulse response has a smaller magnitude than the second value. See my edited answer.
– Matt L.
Sep 3 at 15:18
@DanBoschen: The zeros don't really matter, it depends on the value of the cut-off frequency. For cut-off frequencies less than half the Nyquist frequency, the first value of the impulse response has a smaller magnitude than the second value. See my edited answer.
– Matt L.
Sep 3 at 15:18
1
1
@adr: Sure, any good low pass filter has a zero at Nyquist. Note though that this is just an ordinary IIR filter. IIR filters normally have a recursive and a feed forward path. What you mean is called an "all pole filter", i.e., a filter with only poles away from the origin of the z-plane, with just a constant in the numerator.
– Matt L.
Sep 3 at 15:23
@adr: Sure, any good low pass filter has a zero at Nyquist. Note though that this is just an ordinary IIR filter. IIR filters normally have a recursive and a feed forward path. What you mean is called an "all pole filter", i.e., a filter with only poles away from the origin of the z-plane, with just a constant in the numerator.
– Matt L.
Sep 3 at 15:23
1
1
@MattL In retrospect, my question should have been: why does the first value not lie on the exponential curve defined by the subsequent values? Anyway, things are clearer to me now, which is what the objective was!
– adr
Sep 3 at 19:00
@MattL In retrospect, my question should have been: why does the first value not lie on the exponential curve defined by the subsequent values? Anyway, things are clearer to me now, which is what the objective was!
– adr
Sep 3 at 19:00
 |Â
show 6 more comments
up vote
2
down vote
The impulse response of this filter does not look like most text-book
examples in that the first value is not the largest.
That's simply an issue with interpreting the text books the wrong way. There is nothing special about the first sample, and there is no reason at all, why it would have to be the largest. I'm quite sure that none of the text box actually says anything about this.
Some very simple "beginner" filters just happen to have this property, so you looked at a bunch of pictures and formed a conclusion & generalization that's neither supported by the text or the math.
answered Sep 3 at 13:54
Hilmar
9,002916
add a comment |Â
up vote
2
down vote
The impulse response of this filter does not look like most text-book
examples in that the first value is not the largest.
That's simply an issue with interpreting the text books the wrong way. There is nothing special about the first sample, and there is no reason at all, why it would have to be the largest. I'm quite sure that none of the text box actually says anything about this.
Some very simple "beginner" filters just happen to have this property, so you looked at a bunch of pictures and formed a conclusion & generalization that's neither supported by the text or the math.
answered Sep 3 at 13:54
Hilmar
9,002916
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The impulse response of this filter does not look like most text-book
examples in that the first value is not the largest.
That's simply an issue with interpreting the text books the wrong way. There is nothing special about the first sample, and there is no reason at all, why it would have to be the largest. I'm quite sure that none of the text box actually says anything about this.
Some very simple "beginner" filters just happen to have this property, so you looked at a bunch of pictures and formed a conclusion & generalization that's neither supported by the text or the math.
answered Sep 3 at 13:54
Hilmar
9,002916
The impulse response of this filter does not look like most text-book
examples in that the first value is not the largest.
That's simply an issue with interpreting the text books the wrong way. There is nothing special about the first sample, and there is no reason at all, why it would have to be the largest. I'm quite sure that none of the text box actually says anything about this.
Some very simple "beginner" filters just happen to have this property, so you looked at a bunch of pictures and formed a conclusion & generalization that's neither supported by the text or the math.
answered Sep 3 at 13:54
Hilmar
9,002916
answered Sep 3 at 13:54
Hilmar
9,002916
answered Sep 3 at 13:54
Hilmar
9,002916
answered Sep 3 at 13:54
Hilmar
9,002916
9,002916
add a comment |Â
add a comment |Â
up vote
1
down vote
Starting from an FIR filter with an absolutely flat frequency response:
[ 1 0 0 0 0 ]has a delay of one sample,[ 0 0 1 0 0 ]has a delay of three samples.
The largest coefficient corresponds to the filter's group delay. For a lowpass filter like [ 1 -2 7 -2 1] it is the group delay for the DC component — if you look at the step [ 0 0 0 1 1 1 1 1 1 1 ] response [ 0 0 0 1 -1 6 4 5 5 5] it is fairly obvious what the filter's DC delay is. If you design an FIR lowpass filter with uniform group delay for all passband frequencies, it will have a single largest coefficient in the middle bin, while the length of the filter corresponds to the filter's selectivity.
The same filter with no delay would be non-causal, as the small coefficients before the large one would have negative indexes.
For IIR filters, it's slightly less obvious, because they have two coefficient sets and are unlikely to have flat group delay, but you can still get better stopband attenuation and sharper falloffs with larger delays.
answered Sep 3 at 19:06
Simon Richter
1862
add a comment |Â
up vote
1
down vote
Starting from an FIR filter with an absolutely flat frequency response:
[ 1 0 0 0 0 ]has a delay of one sample,[ 0 0 1 0 0 ]has a delay of three samples.
The largest coefficient corresponds to the filter's group delay. For a lowpass filter like [ 1 -2 7 -2 1] it is the group delay for the DC component — if you look at the step [ 0 0 0 1 1 1 1 1 1 1 ] response [ 0 0 0 1 -1 6 4 5 5 5] it is fairly obvious what the filter's DC delay is. If you design an FIR lowpass filter with uniform group delay for all passband frequencies, it will have a single largest coefficient in the middle bin, while the length of the filter corresponds to the filter's selectivity.
The same filter with no delay would be non-causal, as the small coefficients before the large one would have negative indexes.
For IIR filters, it's slightly less obvious, because they have two coefficient sets and are unlikely to have flat group delay, but you can still get better stopband attenuation and sharper falloffs with larger delays.
answered Sep 3 at 19:06
Simon Richter
1862
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Starting from an FIR filter with an absolutely flat frequency response:
[ 1 0 0 0 0 ]has a delay of one sample,[ 0 0 1 0 0 ]has a delay of three samples.
The largest coefficient corresponds to the filter's group delay. For a lowpass filter like [ 1 -2 7 -2 1] it is the group delay for the DC component — if you look at the step [ 0 0 0 1 1 1 1 1 1 1 ] response [ 0 0 0 1 -1 6 4 5 5 5] it is fairly obvious what the filter's DC delay is. If you design an FIR lowpass filter with uniform group delay for all passband frequencies, it will have a single largest coefficient in the middle bin, while the length of the filter corresponds to the filter's selectivity.
The same filter with no delay would be non-causal, as the small coefficients before the large one would have negative indexes.
For IIR filters, it's slightly less obvious, because they have two coefficient sets and are unlikely to have flat group delay, but you can still get better stopband attenuation and sharper falloffs with larger delays.
answered Sep 3 at 19:06
Simon Richter
1862
Starting from an FIR filter with an absolutely flat frequency response:
[ 1 0 0 0 0 ]has a delay of one sample,[ 0 0 1 0 0 ]has a delay of three samples.
The largest coefficient corresponds to the filter's group delay. For a lowpass filter like [ 1 -2 7 -2 1] it is the group delay for the DC component — if you look at the step [ 0 0 0 1 1 1 1 1 1 1 ] response [ 0 0 0 1 -1 6 4 5 5 5] it is fairly obvious what the filter's DC delay is. If you design an FIR lowpass filter with uniform group delay for all passband frequencies, it will have a single largest coefficient in the middle bin, while the length of the filter corresponds to the filter's selectivity.
The same filter with no delay would be non-causal, as the small coefficients before the large one would have negative indexes.
For IIR filters, it's slightly less obvious, because they have two coefficient sets and are unlikely to have flat group delay, but you can still get better stopband attenuation and sharper falloffs with larger delays.
answered Sep 3 at 19:06
Simon Richter
1862
answered Sep 3 at 19:06
Simon Richter
1862
answered Sep 3 at 19:06
Simon Richter
1862
answered Sep 3 at 19:06
Simon Richter
1862
1862
add a comment |Â
add a comment |Â
up vote
0
down vote
The confusion might arise from what people mean by first order. In some places, like:
- SE.DSP: What Is the Best First Order IIR (AR Filter) Approximation to a Moving Average Filter (FIR Filter)?
Tutorial on a very simple yet useful filter: the first order IIR filter, J. Arzi, 2016
a first order filter follows the input ($x$)/output ($y$) equation:
$$ y[n] = (1-lambda)y[n-1]+lambda x[n],,$$
or equivalently in tracking form:
$$ y[n] = y[n-1]+lambda (x[n]-y[n-1]),.$$
It has numerous names, including first order exponential averaging low-pass filter or exponentially weighted moving average (EWMA). More details are given in
- What is the name of these simple filter algorithms?
- Is there a technical term for this simple method of smoothing out a signal?
In that case, the impulse response $h$ of the filter is explicit. For $k< 0 $, $h[k] = 0$, and for $kge 0 $, $h[k] = lambda(1-lambda)^k$, where $0<lambda < 1$ is a smoothing factor. In this situation, the "first" value $h[0] $ is indeed the largest, since the exponential sequence $lambda(1-lambda)^k$ is decreasing:
Other filter families are called first order, with additional constraints, like Butterworth's for which Mmatt's answer is complete.
answered Sep 3 at 15:33
Laurent Duval
14.8k31955
add a comment |Â
up vote
0
down vote
The confusion might arise from what people mean by first order. In some places, like:
- SE.DSP: What Is the Best First Order IIR (AR Filter) Approximation to a Moving Average Filter (FIR Filter)?
Tutorial on a very simple yet useful filter: the first order IIR filter, J. Arzi, 2016
a first order filter follows the input ($x$)/output ($y$) equation:
$$ y[n] = (1-lambda)y[n-1]+lambda x[n],,$$
or equivalently in tracking form:
$$ y[n] = y[n-1]+lambda (x[n]-y[n-1]),.$$
It has numerous names, including first order exponential averaging low-pass filter or exponentially weighted moving average (EWMA). More details are given in
- What is the name of these simple filter algorithms?
- Is there a technical term for this simple method of smoothing out a signal?
In that case, the impulse response $h$ of the filter is explicit. For $k< 0 $, $h[k] = 0$, and for $kge 0 $, $h[k] = lambda(1-lambda)^k$, where $0<lambda < 1$ is a smoothing factor. In this situation, the "first" value $h[0] $ is indeed the largest, since the exponential sequence $lambda(1-lambda)^k$ is decreasing:
Other filter families are called first order, with additional constraints, like Butterworth's for which Mmatt's answer is complete.
answered Sep 3 at 15:33
Laurent Duval
14.8k31955
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The confusion might arise from what people mean by first order. In some places, like:
- SE.DSP: What Is the Best First Order IIR (AR Filter) Approximation to a Moving Average Filter (FIR Filter)?
Tutorial on a very simple yet useful filter: the first order IIR filter, J. Arzi, 2016
a first order filter follows the input ($x$)/output ($y$) equation:
$$ y[n] = (1-lambda)y[n-1]+lambda x[n],,$$
or equivalently in tracking form:
$$ y[n] = y[n-1]+lambda (x[n]-y[n-1]),.$$
It has numerous names, including first order exponential averaging low-pass filter or exponentially weighted moving average (EWMA). More details are given in
- What is the name of these simple filter algorithms?
- Is there a technical term for this simple method of smoothing out a signal?
In that case, the impulse response $h$ of the filter is explicit. For $k< 0 $, $h[k] = 0$, and for $kge 0 $, $h[k] = lambda(1-lambda)^k$, where $0<lambda < 1$ is a smoothing factor. In this situation, the "first" value $h[0] $ is indeed the largest, since the exponential sequence $lambda(1-lambda)^k$ is decreasing:
Other filter families are called first order, with additional constraints, like Butterworth's for which Mmatt's answer is complete.
answered Sep 3 at 15:33
Laurent Duval
14.8k31955
The confusion might arise from what people mean by first order. In some places, like:
- SE.DSP: What Is the Best First Order IIR (AR Filter) Approximation to a Moving Average Filter (FIR Filter)?
Tutorial on a very simple yet useful filter: the first order IIR filter, J. Arzi, 2016
a first order filter follows the input ($x$)/output ($y$) equation:
$$ y[n] = (1-lambda)y[n-1]+lambda x[n],,$$
or equivalently in tracking form:
$$ y[n] = y[n-1]+lambda (x[n]-y[n-1]),.$$
It has numerous names, including first order exponential averaging low-pass filter or exponentially weighted moving average (EWMA). More details are given in
- What is the name of these simple filter algorithms?
- Is there a technical term for this simple method of smoothing out a signal?
In that case, the impulse response $h$ of the filter is explicit. For $k< 0 $, $h[k] = 0$, and for $kge 0 $, $h[k] = lambda(1-lambda)^k$, where $0<lambda < 1$ is a smoothing factor. In this situation, the "first" value $h[0] $ is indeed the largest, since the exponential sequence $lambda(1-lambda)^k$ is decreasing:
Other filter families are called first order, with additional constraints, like Butterworth's for which Mmatt's answer is complete.
answered Sep 3 at 15:33
Laurent Duval
14.8k31955
edited Sep 3 at 15:44
answered Sep 3 at 15:33
Laurent Duval
14.8k31955
answered Sep 3 at 15:33
Laurent Duval
14.8k31955
answered Sep 3 at 15:33
Laurent Duval
14.8k31955
14.8k31955
add a comment |Â
add a comment |Â
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