Many thanks for the great insight and reply. I will
try to experiment it shortly and let you know what I
get. Sorry I did not see your message sooner.
cheers,
Gordon
--- wrote: >
> Gordon,
> For the low-pass filter below, the alpha can be
> shown to be equal to
> alpha = 1/(1+(Sampling_Interval/Time_Constant)).
> Since you know the time
> constant and sampling interval, you can find out
> this quantity. It is
> approximately inverse of Number Samples in the
> time-constant when the
> time-constant is >> 1.
> Regards
> arun
>
> -----Original Message-----
> From: Jeff Brower [mailto:]
> Sent: Monday, January 24, 2005 9:48 PM
> To: Gordon Ao
> Cc:
> Subject: Re: [matlab] Exponenital Filter design
> Gordon-
>
> > At the moment I am trying to design a one pole
> simple IIR digital
> > filter to simulate thermal heating and cooling
> delay effect with
> > simple equations like 1 - e^(-t/T), where T is a
> known time
> > constant. I think an one pole IIR filter will do
> it: Y[n] = alpha/[1-
> > (1-alpha)Z^(-1)] can simulate the delay effect
> when alpha < 1. But I
> > am quite pullzing how the alpha will be
> determined. i'd appreciate
> > if you shed some light on this.
>
> The time domain implementation (difference equation)
> for your LPF is:
>
> y[n] = (alpha)x[n] + (1-alpha)y[n-1]
>
> If you choose alpha small then you can see the
> filter is "slow" -- it
> responds slowly
> to new input. If you give it an impulse response,
> then the rate of
> decay is
> 1/(1-alpha). From that you can pick alpha to be the
> time constant you
> need -- how
> many samples it should take for the decay to reach
> your desired minimum
> value.
>
> -Jeff
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Reply by Jeff Brower●January 25, 20052005-01-25
Gordon-
Here is an e-mail I got from Jon Lark at L-3 Com (AS Div) that might be
helpful:
> If you choose Alpha = Ts/Tau
> Where
> Ts is the sampling interval
> Tau is the desired Time constant Then the step response will
> closely match the step response of the corresponding analog filter
> with the transfer function
> G(s) = 1/(Tau*s + 1)
This appears similar to what Arun Naik suggested.
-Jeff
-------- Original Message --------
Subject: Re: [matlab] Exponenital Filter design
Date: Mon, 24 Jan 2005 10:17:48 -0600
From: Jeff Brower <>
Organization: Signalogic, Inc
To: Gordon Ao <>
CC:
Gordon-
> At the moment I am trying to design a one pole
simple IIR digital
> filter to simulate thermal heating and cooling delay effect with
> simple equations like 1 - e^(-t/T), where T is a known time
> constant. I think an one pole IIR filter will do it: Y[n] = alpha/[1-
> (1-alpha)Z^(-1)] can simulate the delay effect when alpha < 1. But I
> am quite pullzing how the alpha will be determined. i'd appreciate
> if you shed some light on this.
The time domain implementation (difference equation) for your LPF is:
y[n] = (alpha)x[n] + (1-alpha)y[n-1]
If you choose alpha small then you can see the filter is "slow" -- it
responds slowly
to new input. If you give it an impulse response, then the rate of decay is
1/(1-alpha). From that you can pick alpha to be the time constant you need --
how
many samples it should take for the decay to reach your desired minimum
value.
-Jeff
Reply by ●January 25, 20052005-01-25
A correction:
alpha = 1/(1+(Time_Constant/Sampling_Interval)).
Sorry for the inconvenience.
Regards
arun
-----Original Message-----
From: arun d naik (WT01 - EMBEDDED & PRODUCT ENGINEERING SOLUTIONS)
Sent: Tuesday, January 25, 2005 5:08 PM
To: 'Jeff Brower'; Gordon Ao
Cc:
Subject: RE: [matlab] Exponenital Filter design
Gordon,
For the low-pass filter below, the alpha can be shown to be equal to
alpha = 1/(1+(Sampling_Interval/Time_Constant)). Since you know the time
constant and sampling interval, you can find out this quantity. It is
approximately inverse of Number Samples in the time-constant when the
time-constant is >> 1.
Regards
arun
-----Original Message-----
From: Jeff Brower [mailto:]
Sent: Monday, January 24, 2005 9:48 PM
To: Gordon Ao
Cc:
Subject: Re: [matlab] Exponenital Filter design
Gordon-
> At the moment I am trying to design a one pole
simple IIR digital
> filter to simulate thermal heating and cooling delay effect with
> simple equations like 1 - e^(-t/T), where T is a known time
> constant. I think an one pole IIR filter will do it: Y[n] = alpha/[1-
> (1-alpha)Z^(-1)] can simulate the delay effect when alpha < 1. But I
> am quite pullzing how the alpha will be determined. i'd appreciate
> if you shed some light on this.
The time domain implementation (difference equation) for your LPF is:
y[n] = (alpha)x[n] + (1-alpha)y[n-1]
If you choose alpha small then you can see the filter is "slow" --
it
responds slowly
to new input. If you give it an impulse response, then the rate of
decay is
1/(1-alpha). From that you can pick alpha to be the time constant you
need -- how
many samples it should take for the decay to reach your desired minimum
value.
-Jeff
Reply by ●January 25, 20052005-01-25
Gordon,
For the low-pass filter below, the alpha can be shown to be equal to
alpha = 1/(1+(Sampling_Interval/Time_Constant)). Since you know the time
constant and sampling interval, you can find out this quantity. It is
approximately inverse of Number Samples in the time-constant when the
time-constant is >> 1.
Regards
arun
-----Original Message-----
From: Jeff Brower [mailto:]
Sent: Monday, January 24, 2005 9:48 PM
To: Gordon Ao
Cc:
Subject: Re: [matlab] Exponenital Filter design
Gordon-
> At the moment I am trying to design a one pole
simple IIR digital
> filter to simulate thermal heating and cooling delay effect with
> simple equations like 1 - e^(-t/T), where T is a known time
> constant. I think an one pole IIR filter will do it: Y[n] = alpha/[1-
> (1-alpha)Z^(-1)] can simulate the delay effect when alpha < 1. But I
> am quite pullzing how the alpha will be determined. i'd appreciate
> if you shed some light on this.
The time domain implementation (difference equation) for your LPF is:
y[n] = (alpha)x[n] + (1-alpha)y[n-1]
If you choose alpha small then you can see the filter is "slow" --
it
responds slowly
to new input. If you give it an impulse response, then the rate of
decay is
1/(1-alpha). From that you can pick alpha to be the time constant you
need -- how
many samples it should take for the decay to reach your desired minimum
value.
-Jeff
Reply by Jeff Brower●January 24, 20052005-01-24
Gordon-
> At the moment I am trying to design a one pole
simple IIR digital
> filter to simulate thermal heating and cooling delay effect with
> simple equations like 1 - e^(-t/T), where T is a known time
> constant. I think an one pole IIR filter will do it: Y[n] = alpha/[1-
> (1-alpha)Z^(-1)] can simulate the delay effect when alpha < 1. But I
> am quite pullzing how the alpha will be determined. i'd appreciate
> if you shed some light on this.
The time domain implementation (difference equation) for your LPF is:
y[n] = (alpha)x[n] + (1-alpha)y[n-1]
If you choose alpha small then you can see the filter is "slow" -- it
responds slowly
to new input. If you give it an impulse response, then the rate of decay is
1/(1-alpha). From that you can pick alpha to be the time constant you need --
how
many samples it should take for the decay to reach your desired minimum
value.
-Jeff
Reply by gordon_ao●January 24, 20052005-01-24
Hello Folks
At the moment I am trying to design a one pole simple IIR digital
filter to simulate thermal heating and cooling delay effect with
simple equations like 1 - e^(-t/T), where T is a known time
constant. I think an one pole IIR filter will do it: Y[n] = alpha/[1-
(1-alpha)Z^(-1)] can simulate the delay effect when alpha < 1. But I
am quite pullzing how the alpha will be determined. i'd appreciate
if you shed some light on this.