| Author |
Message |
Ajay_N
Guest
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 Posted:
Thu Dec 22, 2005 1:17 am Post subject:
Approximating log(x+1/x) |
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Hello,
I am trying to approximate log (x+1/x) where x can be >=1. Is it possibl
to do that?
Thanks,
Ajay |
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robert bristow-johnson
Guest
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| Posted:
Thu Dec 22, 2005 8:15 am Post subject:
Re: Approximating log(x+1/x) |
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Ajay_N wrote:
| Quote: |
I am trying to approximate log (x+1/x) where x can be >=1. Is it possible
to do that?
|
ya know, Google is just a wonderful tool. try typing in "series for
log" and see what you get. if you hit the "I'm feeling lucky" button,
you go to http://hyperphysics.phy-astr.gsu.edu/hbase/math/lnseries.html
.. pick one of the series, i like the third one because it can be used
to derive the Bilinear Transform:
log(v) = 2*( u + (1/3)u^3 + (1/5)u^5 + (1/7)u^7 + (1/9)u^9 + ...)
where u = (v-1)/(v+1)
and plug in your x + 1/x into the "v" above. pick as many terms as you
think you need.
r b-j |
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Horatio Hornblower
Guest
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| Posted:
Thu Dec 22, 2005 9:16 am Post subject:
Re: Approximating log(x+1/x) |
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"robert bristow-johnson" <rbj@audioimagination.com> wrote in message
news:1135217704.502588.229040@g47g2000cwa.googlegroups.com...
| Quote: |
Ajay_N wrote:
I am trying to approximate log (x+1/x) where x can be >=1. Is it
possible
to do that?
ya know, Google is just a wonderful tool. try typing in "series for
log" and see what you get. if you hit the "I'm feeling lucky" button,
you go to http://hyperphysics.phy-astr.gsu.edu/hbase/math/lnseries.html
. pick one of the series, i like the third one because it can be used
to derive the Bilinear Transform:
log(v) = 2*( u + (1/3)u^3 + (1/5)u^5 + (1/7)u^7 + (1/9)u^9 + ...)
where u = (v-1)/(v+1)
and plug in your x + 1/x into the "v" above. pick as many terms as you
think you need.
r b-j
There's an easier way to get the Bilinear TForm. |
z=exp(ST) = exp(-ST/2)/exp(+ST/2)
expanding in a Taylor series num and denom
= 1-ST/2/(1+ST/2) to a first order approx. Though I believe it is a 2nd
order approx since it is the division of two first order approximations.
HH |
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Ajay_N
Guest
|
| Posted:
Thu Dec 22, 2005 9:16 am Post subject:
Re: Approximating log(x+1/x) |
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| Quote: | Thanks a lot. I really appreciate your quick response.
|
Ajay
| Quote: | Ajay_N wrote:
I am trying to approximate log (x+1/x) where x can be >=1. Is i
possible
to do that?
ya know, Google is just a wonderful tool. try typing in "series for
log" and see what you get. if you hit the "I'm feeling lucky" button,
you go to http://hyperphysics.phy-astr.gsu.edu/hbase/math/lnseries.html
. pick one of the series, i like the third one because it can be used
to derive the Bilinear Transform:
log(v) = 2*( u + (1/3)u^3 + (1/5)u^5 + (1/7)u^7 + (1/9)u^9 + ...)
where u = (v-1)/(v+1)
and plug in your x + 1/x into the "v" above. pick as many terms as you
think you need.
r b-j
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dbell
Guest
|
| Posted:
Thu Dec 22, 2005 5:16 pm Post subject:
Re: Approximating log(x+1/x) |
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For what kind of implementation (fixed (how many bits), float (what
precision), limited number of cycles, ...) ? What kind of final
accuracy required? Any realistic upper bound on x?
Dirk
Ajay_N wrote:
| Quote: | Hello,
I am trying to approximate log (x+1/x) where x can be >=1. Is it possible
to do that?
Thanks,
Ajay |
|
|
| Back to top |
|
|
robert bristow-johnson
Guest
|
| Posted:
Thu Dec 22, 2005 5:16 pm Post subject:
Re: Approximating log(x+1/x) |
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Horatio Hornblower wrote:
| Quote: | "robert bristow-johnson" <rbj@audioimagination.com> wrote in message
news:1135217704.502588.229040@g47g2000cwa.googlegroups.com...
....
i like the third one because it can be used
to derive the Bilinear Transform:
log(v) = 2*( u + (1/3)u^3 + (1/5)u^5 + (1/7)u^7 + (1/9)u^9 + ...)
where u = (v-1)/(v+1)
....
There's an easier way to get the Bilinear TForm.
z=exp(ST) = exp(-ST/2)/exp(+ST/2)
|
that should be z^-1 and exp(-sT) (or swap the + and - signs).
| Quote: |
expanding in a Taylor series num and denom
= 1-ST/2/(1+ST/2) to a first order approx.
|
it's equivalent. in fact, you're not done, you gotta solve for s.
| Quote: | Though I believe it is a 2nd
order approx since it is the division of two first order approximations.
|
no, it 1st order. go ahead and solve for s. your way is easier to
remember from scratch, but what if you wanted a *better* approximation
for s in terms of z and you didn't care about preserving order. how do
you do it your way?
r b-j |
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Horatio Hornblower
Guest
|
| Posted:
Fri Dec 23, 2005 12:27 am Post subject:
Re: Approximating log(x+1/x) |
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|
"robert bristow-johnson" <rbj@audioimagination.com> wrote in message
news:1135269023.596533.224340@g49g2000cwa.googlegroups.com...
| Quote: |
Horatio Hornblower wrote:
"robert bristow-johnson" <rbj@audioimagination.com> wrote in message
news:1135217704.502588.229040@g47g2000cwa.googlegroups.com...
...
i like the third one because it can be used
to derive the Bilinear Transform:
log(v) = 2*( u + (1/3)u^3 + (1/5)u^5 + (1/7)u^7 + (1/9)u^9 + ...)
where u = (v-1)/(v+1)
...
There's an easier way to get the Bilinear TForm.
z=exp(ST) = exp(-ST/2)/exp(+ST/2)
that should be z^-1 and exp(-sT) (or swap the + and - signs).
expanding in a Taylor series num and denom
= 1-ST/2/(1+ST/2) to a first order approx.
it's equivalent. in fact, you're not done, you gotta solve for s.
Though I believe it is a 2nd
order approx since it is the division of two first order approximations.
no, it 1st order. go ahead and solve for s. your way is easier to
remember from scratch, but what if you wanted a *better* approximation
for s in terms of z and you didn't care about preserving order. how do
you do it your way?
r b-j
That's true but the BT is a good enough approx if you sample high enough. |
HH |
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