a*e
2 楼
和nature的进来讲讲经验,怎么才能发啊?
g*t
3 楼
稍稍研究了下数字滤波,遇到两个问题,请专家帮忙:
(1)
Sigma Delta A/D的理论,是不是回答不了下面的问题?
对于12位理想的Sigma Delta A/D,如果输入是一个被A,B两边bound的信号,
那么输入输出的误差有界吗? 最大值是多少?
(2)
y(z)=G(z)*u(z)
字长12位,u有界的情况下,是否存在一种digitial filter的实现,
使 |输出的y-无量化误差的理想y| 最小?
Thanks!
(1)
Sigma Delta A/D的理论,是不是回答不了下面的问题?
对于12位理想的Sigma Delta A/D,如果输入是一个被A,B两边bound的信号,
那么输入输出的误差有界吗? 最大值是多少?
(2)
y(z)=G(z)*u(z)
字长12位,u有界的情况下,是否存在一种digitial filter的实现,
使 |输出的y-无量化误差的理想y| 最小?
Thanks!
w*8
5 楼
牛老板就行,如果牛校牛老板更行了
h*5
8 楼
跟对队伍
f*b
10 楼
try science of nature first
s*s
16 楼
journal of neurosurgery
g*e
17 楼
if your work is really good, it does not matter where you work in order to
publish your work in CNS.
publish your work in CNS.
b*g
18 楼
I am not an ADC person. For (2), generally, if one can
characterize statistical properties of quantization error,
which in your case may be dependent on the properties of u(z),
then it is possible to minimize the residual variance.
Alternatively, you can build a recursive filter to accomplish
your goal, although most RLS filters minimize error squared,
not absolute error - but you really want error squared, yes?
【在 g****t 的大作中提到】
: 稍稍研究了下数字滤波,遇到两个问题,请专家帮忙:
: (1)
: Sigma Delta A/D的理论,是不是回答不了下面的问题?
: 对于12位理想的Sigma Delta A/D,如果输入是一个被A,B两边bound的信号,
: 那么输入输出的误差有界吗? 最大值是多少?
: (2)
: y(z)=G(z)*u(z)
: 字长12位,u有界的情况下,是否存在一种digitial filter的实现,
: 使 |输出的y-无量化误差的理想y| 最小?
: Thanks!
g*t
20 楼
谢谢回复.
例如y(t)=a*y(t-1)+b*y(t-2)+c*u(t)+d*u(t-1)
这样一个滤波器.
如果只知道u有界,字长8位,那种实现,能让
|理想的y-滤波器实际产生的y|的
min max或平方积分最小,这个问题我看了下文献,好像是没有答案的.
统计特性的结果有不少.但那个我觉得帮助不大.一来统计特性很难准确.一些
文章上的结论,对不准的统计特性是不是仍然有效,这个目前为止.二来,
data sheet上,我总不能说这个东西99%不会numerical blow up吧.
I am not an ADC person. For (2), generally, if one can
characterize statistical properties of quantization error,
which in your case may be dependent on the properties of u(z),
then it is possible to minimize the residual variance.
Alternatively, you can build a recursive filter to accomplish
your goal, although most RLS filters minimize error squared,
not absolute error - but you really want error squared, yes?
【在 b**********g 的大作中提到】
:
: I am not an ADC person. For (2), generally, if one can
: characterize statistical properties of quantization error,
: which in your case may be dependent on the properties of u(z),
: then it is possible to minimize the residual variance.
: Alternatively, you can build a recursive filter to accomplish
: your goal, although most RLS filters minimize error squared,
: not absolute error - but you really want error squared, yes?
例如y(t)=a*y(t-1)+b*y(t-2)+c*u(t)+d*u(t-1)
这样一个滤波器.
如果只知道u有界,字长8位,那种实现,能让
|理想的y-滤波器实际产生的y|的
min max或平方积分最小,这个问题我看了下文献,好像是没有答案的.
统计特性的结果有不少.但那个我觉得帮助不大.一来统计特性很难准确.一些
文章上的结论,对不准的统计特性是不是仍然有效,这个目前为止.二来,
data sheet上,我总不能说这个东西99%不会numerical blow up吧.
I am not an ADC person. For (2), generally, if one can
characterize statistical properties of quantization error,
which in your case may be dependent on the properties of u(z),
then it is possible to minimize the residual variance.
Alternatively, you can build a recursive filter to accomplish
your goal, although most RLS filters minimize error squared,
not absolute error - but you really want error squared, yes?
【在 b**********g 的大作中提到】
:
: I am not an ADC person. For (2), generally, if one can
: characterize statistical properties of quantization error,
: which in your case may be dependent on the properties of u(z),
: then it is possible to minimize the residual variance.
: Alternatively, you can build a recursive filter to accomplish
: your goal, although most RLS filters minimize error squared,
: not absolute error - but you really want error squared, yes?
c*l
22 楼
For the digital filter, there are several components contributing to overall
errors.
quantization error of u
quantization error of filter coefficients
computation error in multiplication and adding
Depending on the bandwidth vs sampling ratio of u, there are different
topologies which can be optimized to reduce filter coefficient quantization
error.
Stability of an IIR filter is a different issue, there are bounds that can
guarantee stability. But these bounds are not tight, which typically results
in over-designing the hardware.
【在 g****t 的大作中提到】
: 谢谢回复.
: 例如y(t)=a*y(t-1)+b*y(t-2)+c*u(t)+d*u(t-1)
: 这样一个滤波器.
: 如果只知道u有界,字长8位,那种实现,能让
: |理想的y-滤波器实际产生的y|的
: min max或平方积分最小,这个问题我看了下文献,好像是没有答案的.
: 统计特性的结果有不少.但那个我觉得帮助不大.一来统计特性很难准确.一些
: 文章上的结论,对不准的统计特性是不是仍然有效,这个目前为止.二来,
: data sheet上,我总不能说这个东西99%不会numerical blow up吧.
:
errors.
quantization error of u
quantization error of filter coefficients
computation error in multiplication and adding
Depending on the bandwidth vs sampling ratio of u, there are different
topologies which can be optimized to reduce filter coefficient quantization
error.
Stability of an IIR filter is a different issue, there are bounds that can
guarantee stability. But these bounds are not tight, which typically results
in over-designing the hardware.
【在 g****t 的大作中提到】
: 谢谢回复.
: 例如y(t)=a*y(t-1)+b*y(t-2)+c*u(t)+d*u(t-1)
: 这样一个滤波器.
: 如果只知道u有界,字长8位,那种实现,能让
: |理想的y-滤波器实际产生的y|的
: min max或平方积分最小,这个问题我看了下文献,好像是没有答案的.
: 统计特性的结果有不少.但那个我觉得帮助不大.一来统计特性很难准确.一些
: 文章上的结论,对不准的统计特性是不是仍然有效,这个目前为止.二来,
: data sheet上,我总不能说这个东西99%不会numerical blow up吧.
:
g*t
23 楼
谢谢回复.
例如如下滤波器,字长只允许8位.
y(t)=a*y(t-1)+b*y(t-2)+c*u(t)+d*u(t-1)
保证稳定性的办法是显然的,最简单的,只要让a=b=c=d都等于零即可.
或者求解方程x^2=ax+b,然后并联实现,如果特征根小于1,
那就是两个moving average的组合.即使有量化误差,稳定性应该也没问题.
所以稳定性似乎没多大实际意义.
如果只知道|u(t)|的上界.有没有一种实现方法,使
|实现的y(t)-理论的y(t)| 的min max或者平方积分最小
另外,如果只知道上界没办法做.那么假定u(t)的频率有界,最优的实现能做吗?
overall
quantization
results
【在 c**l 的大作中提到】
: For the digital filter, there are several components contributing to overall
: errors.
: quantization error of u
: quantization error of filter coefficients
: computation error in multiplication and adding
: Depending on the bandwidth vs sampling ratio of u, there are different
: topologies which can be optimized to reduce filter coefficient quantization
: error.
: Stability of an IIR filter is a different issue, there are bounds that can
: guarantee stability. But these bounds are not tight, which typically results
例如如下滤波器,字长只允许8位.
y(t)=a*y(t-1)+b*y(t-2)+c*u(t)+d*u(t-1)
保证稳定性的办法是显然的,最简单的,只要让a=b=c=d都等于零即可.
或者求解方程x^2=ax+b,然后并联实现,如果特征根小于1,
那就是两个moving average的组合.即使有量化误差,稳定性应该也没问题.
所以稳定性似乎没多大实际意义.
如果只知道|u(t)|的上界.有没有一种实现方法,使
|实现的y(t)-理论的y(t)| 的min max或者平方积分最小
另外,如果只知道上界没办法做.那么假定u(t)的频率有界,最优的实现能做吗?
overall
quantization
results
【在 c**l 的大作中提到】
: For the digital filter, there are several components contributing to overall
: errors.
: quantization error of u
: quantization error of filter coefficients
: computation error in multiplication and adding
: Depending on the bandwidth vs sampling ratio of u, there are different
: topologies which can be optimized to reduce filter coefficient quantization
: error.
: Stability of an IIR filter is a different issue, there are bounds that can
: guarantee stability. But these bounds are not tight, which typically results
b*g
24 楼
I wasn't clear about your problem. I assumed you've 8-bit ADC, but
processed with much higher precision digital filter. Otherwise, you
would not be able to represent the "ideal y".
I agree with your second comment. In real world, there are very few
situations where the statistical properties are known, and are
stationary, that's why I suggested using RLS filters, i.e. digital
filters with dynamic coefficients.
If u is bounded and band-limited, it should not be hard to design a
stable filter.
BTW, DSP chips do "blow up" (more like "screw up" with fixed-point
arithmetic). You just need to make that known to your potential
customers (put "do not feed a sinusoid of 20kHz into the circuit"
in your application note). I had worked with one.
【在 g****t 的大作中提到】
: 谢谢回复.
: 例如如下滤波器,字长只允许8位.
: y(t)=a*y(t-1)+b*y(t-2)+c*u(t)+d*u(t-1)
: 保证稳定性的办法是显然的,最简单的,只要让a=b=c=d都等于零即可.
: 或者求解方程x^2=ax+b,然后并联实现,如果特征根小于1,
: 那就是两个moving average的组合.即使有量化误差,稳定性应该也没问题.
: 所以稳定性似乎没多大实际意义.
: 如果只知道|u(t)|的上界.有没有一种实现方法,使
: |实现的y(t)-理论的y(t)| 的min max或者平方积分最小
: 另外,如果只知道上界没办法做.那么假定u(t)的频率有界,最优的实现能做吗?
g*t
25 楼
定常差分方程有解析解,就考虑什么实现可以让结果离解析解最近吧。
不考虑固定长度数了。就考虑IEEE浮点数吧。串联,并联,状态方程,
前后滤波等等各种实现里面,哪种实现可以使
|y的解析解-滤波器输出的IEEE浮点数y(t)|^2
10年内的积分最小?
已知u(t)带宽有限,每点有界。x^2=ax+b的解都在单位园内。
这个问题应该是well defined的吧br />
(如果这还不够,可以假设u(t)是三角函数的
线性组合。)
RLS跟我说的(1)好像没有关系。另外说到RLS,就算是浮点数,matlab,任何一种RLS
的实现都无法保证numerical stability。这个更不可靠。必须隔段时间reset或者作别
的经验主义前后处理。
其实我问这问题,是因为我不是数字滤波器的专家.
不了解常用必须的engineer fix,老怕自己弄出来的东西blow up了.
看看有没有fast solution,牺牲点性能也可以.
I wasn't clear about your problem. I assumed you've 8-bit ADC, but
processed with much higher precision digital filter. Otherwise, you
would not be able to represent the "ideal y".
I agree with your second comment. In real world, there are very few
situations where the statistical properties are known, and are
stationary, that's why I suggested using RLS filters, i.e. digital
filters with dynamic coefficients.
If u is bounded and band-limited, it should not be hard to design a
stable filter.
BTW, DSP chips do "blow up" (more like "screw up" with fixed-point
arithmetic). You just need to make that known to your potential
customers (put "do not feed a sinusoid of 20kHz into the circuit"
in your application note). I had worked with one.
【在 b**********g 的大作中提到】
:
: I wasn't clear about your problem. I assumed you've 8-bit ADC, but
: processed with much higher precision digital filter. Otherwise, you
: would not be able to represent the "ideal y".
: I agree with your second comment. In real world, there are very few
: situations where the statistical properties are known, and are
: stationary, that's why I suggested using RLS filters, i.e. digital
: filters with dynamic coefficients.
: If u is bounded and band-limited, it should not be hard to design a
: stable filter.
不考虑固定长度数了。就考虑IEEE浮点数吧。串联,并联,状态方程,
前后滤波等等各种实现里面,哪种实现可以使
|y的解析解-滤波器输出的IEEE浮点数y(t)|^2
10年内的积分最小?
已知u(t)带宽有限,每点有界。x^2=ax+b的解都在单位园内。
这个问题应该是well defined的吧br />
(如果这还不够,可以假设u(t)是三角函数的
线性组合。)
RLS跟我说的(1)好像没有关系。另外说到RLS,就算是浮点数,matlab,任何一种RLS
的实现都无法保证numerical stability。这个更不可靠。必须隔段时间reset或者作别
的经验主义前后处理。
其实我问这问题,是因为我不是数字滤波器的专家.
不了解常用必须的engineer fix,老怕自己弄出来的东西blow up了.
看看有没有fast solution,牺牲点性能也可以.
I wasn't clear about your problem. I assumed you've 8-bit ADC, but
processed with much higher precision digital filter. Otherwise, you
would not be able to represent the "ideal y".
I agree with your second comment. In real world, there are very few
situations where the statistical properties are known, and are
stationary, that's why I suggested using RLS filters, i.e. digital
filters with dynamic coefficients.
If u is bounded and band-limited, it should not be hard to design a
stable filter.
BTW, DSP chips do "blow up" (more like "screw up" with fixed-point
arithmetic). You just need to make that known to your potential
customers (put "do not feed a sinusoid of 20kHz into the circuit"
in your application note). I had worked with one.
【在 b**********g 的大作中提到】
:
: I wasn't clear about your problem. I assumed you've 8-bit ADC, but
: processed with much higher precision digital filter. Otherwise, you
: would not be able to represent the "ideal y".
: I agree with your second comment. In real world, there are very few
: situations where the statistical properties are known, and are
: stationary, that's why I suggested using RLS filters, i.e. digital
: filters with dynamic coefficients.
: If u is bounded and band-limited, it should not be hard to design a
: stable filter.
L*r
26 楼
听着好像你把一简单问题搞复杂了。要不直接说说啥应用吧。听意思好像是一个sigma-
delta加低通的应用?
【在 g****t 的大作中提到】
: 稍稍研究了下数字滤波,遇到两个问题,请专家帮忙:
: (1)
: Sigma Delta A/D的理论,是不是回答不了下面的问题?
: 对于12位理想的Sigma Delta A/D,如果输入是一个被A,B两边bound的信号,
: 那么输入输出的误差有界吗? 最大值是多少?
: (2)
: y(z)=G(z)*u(z)
: 字长12位,u有界的情况下,是否存在一种digitial filter的实现,
: 使 |输出的y-无量化误差的理想y| 最小?
: Thanks!
delta加低通的应用?
【在 g****t 的大作中提到】
: 稍稍研究了下数字滤波,遇到两个问题,请专家帮忙:
: (1)
: Sigma Delta A/D的理论,是不是回答不了下面的问题?
: 对于12位理想的Sigma Delta A/D,如果输入是一个被A,B两边bound的信号,
: 那么输入输出的误差有界吗? 最大值是多少?
: (2)
: y(z)=G(z)*u(z)
: 字长12位,u有界的情况下,是否存在一种digitial filter的实现,
: 使 |输出的y-无量化误差的理想y| 最小?
: Thanks!
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