J.17 s -> z conversion

Started by January 16, 2004
I am trying to implement J.17 de-emphasis cure in a 32 bit processor.
I have converted J.17 S-domain transfer function using Bilinear of
MATLAB. The maltlab plot of manitude responce matches with the table
biven in the std. But when i verify in Audio precession I get +5dB
error in the range of 2-4 Khz. I have implemented in Directform 2 type
filter. Does nay body has idea why i am getting this difference..?
Regards
On Fri, 16 Jan 2004 02:33:36 +0000, Anand wrote:

> I am trying to implement J.17 de-emphasis cure in a 32 bit processor. > I have converted J.17 S-domain transfer function using Bilinear of > MATLAB. The maltlab plot of manitude responce matches with the table > biven in the std. But when i verify in Audio precession I get +5dB > error in the range of 2-4 Khz. I have implemented in Directform 2 type > filter. Does nay body has idea why i am getting this difference..? > Regards
Give me figures. S and Z transform equations
"Robin Clark" <robinTEETH48gx@hotTEETHmail.com> wrote in message news:<pan.2004.01.17.00.42.04.550136@hotTEETHmail.com>...
> On Fri, 16 Jan 2004 02:33:36 +0000, Anand wrote: > > > I am trying to implement J.17 de-emphasis cure in a 32 bit processor. > > I have converted J.17 S-domain transfer function using Bilinear of > > MATLAB. The maltlab plot of manitude responce matches with the table > > biven in the std. But when i verify in Audio precession I get +5dB > > error in the range of 2-4 Khz. I have implemented in Directform 2 type > > filter. Does nay body has idea why i am getting this difference..? > > Regards > > > Give me figures. S and Z transform equations
Hi, As per J.17 std Insertion Loss between nominal Impedences = 10log10 75+(w/3000)^2/1+(w/3000)^2 That gives H(s) = s-(75)^1/2 * 3000/s-3000; Normalizing with 18.75 dB ( gain at 0Hz as per J.17) and converting into Z gives me H(z) = 0.1423 - 0.0817 z^-1/1-0.9394 z^-1 Both my s and z domain plots match in MATLAB, After implementing in 32 fixed point processor i get 5db differance as i move to frequencies above 2k hz ...!! Anand
"Robin Clark" <robinTEETH48gx@hotTEETHmail.com> wrote in message news:<pan.2004.01.17.00.42.04.550136@hotTEETHmail.com>...
> On Fri, 16 Jan 2004 02:33:36 +0000, Anand wrote: > > > I am trying to implement J.17 de-emphasis cure in a 32 bit processor. > > I have converted J.17 S-domain transfer function using Bilinear of > > MATLAB. The maltlab plot of manitude responce matches with the table > > biven in the std. But when i verify in Audio precession I get +5dB > > error in the range of 2-4 Khz. I have implemented in Directform 2 type > > filter. Does nay body has idea why i am getting this difference..? > > Regards > > > Give me figures. S and Z transform equations
Hi, As per J.17 std Insertion Loss between nominal Impedences = 10log10 75+(w/3000)^2/1+(w/3000)^2 That gives H(s) = s-(75)^1/2 * 3000/s-3000; Normalizing with 18.75 dB ( gain at 0Hz as per J.17) and converting into Z gives me H(z) = 0.1423 - 0.0817 z^-1/1-0.9394 z^-1 Both my s and z domain plots match in MATLAB, After implementing in 32 fixed point processor i get 5db differance as i move to frequencies above 2k hz ...!! Anand
Anand wrote:

> "Robin Clark" <robinTEETH48gx@hotTEETHmail.com> wrote in message news:<pan.2004.01.17.00.42.04.550136@hotTEETHmail.com>... > >>On Fri, 16 Jan 2004 02:33:36 +0000, Anand wrote: >> >> >>>I am trying to implement J.17 de-emphasis cure in a 32 bit processor. >>>I have converted J.17 S-domain transfer function using Bilinear of >>>MATLAB. The maltlab plot of manitude responce matches with the table >>>biven in the std. But when i verify in Audio precession I get +5dB >>>error in the range of 2-4 Khz. I have implemented in Directform 2 type >>>filter. Does nay body has idea why i am getting this difference..? >>>Regards >> >> >>Give me figures. S and Z transform equations > > > Hi, > As per J.17 std > Insertion Loss between nominal Impedences = 10log10 > 75+(w/3000)^2/1+(w/3000)^2 > That gives > H(s) = s-(75)^1/2 * 3000/s-3000; > Normalizing with 18.75 dB ( gain at 0Hz as per J.17) and converting > into Z gives me > > H(z) = 0.1423 - 0.0817 z^-1/1-0.9394 z^-1 > > Both my s and z domain plots match in MATLAB, > After implementing in 32 fixed point processor i get 5db differance as > i move to frequencies above 2k hz ...!! > Anand
Whoa! 10log10 is for power. For voltage, you need 20log10. Start there, and you'll get 10 dB at the end. Jerry -- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Jerry Avins <jya@ieee.org> wrote in message news:<400bf6c5\$0\$6083\$61fed72c@news.rcn.com>...
> Anand wrote: > > > "Robin Clark" <robinTEETH48gx@hotTEETHmail.com> wrote in message news:<pan.2004.01.17.00.42.04.550136@hotTEETHmail.com>... > > > >>On Fri, 16 Jan 2004 02:33:36 +0000, Anand wrote: > >> > >> > >>>I am trying to implement J.17 de-emphasis cure in a 32 bit processor. > >>>I have converted J.17 S-domain transfer function using Bilinear of > >>>MATLAB. The maltlab plot of manitude responce matches with the table > >>>biven in the std. But when i verify in Audio precession I get +5dB > >>>error in the range of 2-4 Khz. I have implemented in Directform 2 type > >>>filter. Does nay body has idea why i am getting this difference..? > >>>Regards > >> > >> > >>Give me figures. S and Z transform equations > > > > > > Hi, > > As per J.17 std > > Insertion Loss between nominal Impedences = 10log10 > > 75+(w/3000)^2/1+(w/3000)^2 > > That gives > > H(s) = s-(75)^1/2 * 3000/s-3000; > > Normalizing with 18.75 dB ( gain at 0Hz as per J.17) and converting > > into Z gives me > > > > H(z) = 0.1423 - 0.0817 z^-1/1-0.9394 z^-1 > > > > Both my s and z domain plots match in MATLAB, > > After implementing in 32 fixed point processor i get 5db differance as > > i move to frequencies above 2k hz ...!! > > Anand > > Whoa! 10log10 is for power. For voltage, you need 20log10. Start there, > and you'll get 10 dB at the end. > > Jerry
Hi, I don't use this log part to compute the transfer function. I get it by substituting s= jw , in the equetion given in J.17 std. Can you be more elobarate on this... Regards Anand
Anand wrote:

...

> Hi, > I don't use this log part to compute the transfer function. I get it > by substituting s= jw , in the equetion given in J.17 std. > Can you be more elobarate on this... > Regards > Anand
Steer me to J.17 so I can see what it says. Maybe I didn't understand the problem. A Z transform implies a sampling rate. For a detailed look at your results, I need to know what it is. Remember: you can expect the actual response to depart considerably from the continuous prototype's as you approach fs/2. Jerry -- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Jerry Avins <jya@ieee.org> wrote in message news:<400d5661\$0\$15631\$61fed72c@news.rcn.com>...
> Anand wrote: > > ... > > > Hi, > > I don't use this log part to compute the transfer function. I get it > > by substituting s= jw , in the equetion given in J.17 std. > > Can you be more elobarate on this... > > Regards > > Anand > > Steer me to J.17 so I can see what it says. Maybe I didn't understand > the problem. A Z transform implies a sampling rate. For a detailed look > at your results, I need to know what it is. Remember: you can expect the > actual response to depart considerably from the continuous prototype's > as you approach fs/2. > > Jerry
Hi, In the previous posting i have puts the J.17 std details. The transferfunction etc..
Anand wrote:

...
> > Hi, > In the previous posting i have puts the J.17 std details. The transferfunction etc..
Do you mean "Insertion Loss between nominal Impedences = 10log10 75+(w/3000)^2/1+(w/3000)^2 That gives H(s) = s-(75)^1/2 * 3000/s-3000;" ? What does "Insertion Loss between nominal Impedences" mean? What are those impedances? 10log10 of what? Of 75+(w/3000)^2/1+(w/3000)^2? I wanted to see the document you pulled those from so I could understand your need well enough to help. I don't understand the problem. Jerry -- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;