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-----Original Message-----
From: Dimitry Taich
Sent: Thursday, February 13, 2003 1:09 AM
To: 'larry rennie'
Cc: Ze'ev Roth; Dimitry Taich
Subject: RE: Correction to your 12/02 CX4 PresentationLarry,Finally I got you... Stupid mistake - I used V_low/V_high ratio itself instead of subtract it from 1... My son would say - "And that guy dare to teach me math?!"BTW, methods' 2 results still seem to be OK - b=a/(1+a), 1-2B=(1-a)/(1+a). But - Pre-emphasis def should be as you state:Pre-emphasis=2b=2a/(1+a)Unless Ze'ev will find something new tonight, I'll update CX4 group about that correction tomorrow morning.Thanks a lot for your review and contribution.Regards,DimitryP.S. That's really amazing that currently selected value (1/3) yields same pre-emphasis value for both - correct one and wrong one - definitions!-----Original Message-----Dimitry,
From: larry rennie [mailto:Larry.Rennie@nsc.com]
Sent: יום רביעי 12 פברואר 2003 14:15
To: Dimitry Taich
Cc: Ze'ev Roth
Subject: Re: Correction to your 12/02 CX4 PresentationYour reply has the following equations under Method 1 ( I have italicized them):
Step 3: Using pre-emphasis def with a parameter (Slide 5) we get next values for V_low & V_high:
V_low = V_peak*(1-a)/(1+a)V_peak -->
1-V_low/V_high = (1-a)/(1+a)
V_high = V_peak -->but the first 2 lines above are not correct,
V_low = V_peak*(1-a)/(1+a)V_peak should be V_low = V_peak*(1-a)/(1+a). You have an extra V_peak term.
and
1-V_low/V_high = (1-a)/(1+a) should be [1-V_low/V_high ]= [1- (1-a)/(1+a)] = 2a/(1+a)More serious (unless I am missing something) is that from slide 8 and method 2 below you say,
pre-emphasis = [1-V_low/V_high] = 1-2beta], I don't think this is correct. I calculate,
pre-emphasis = [1-V_low/V_high] and from your slide 4,
V_low = (1-2beta)V_peak, and
V_high = V_peak, therefore, substituting we get,pre-emphasis = [1-V_low/V_high] = [1- (1-2beta)V_peak)/Vpeak]= 2beta.
I am confusing b and beta?
Regards,
Larry
Dimitry Taich wrote:
Larry,Probably I miss something, but our version seems me to be the correct one. You pay our attention on contradiction between "1-2Beta = (1-alpha)/(1+alpha)" and some "earlier equations" - could you specify which equations exactly do you mean?
Anyway, here is how we got "1-2b = (1-a)/(1+a)" equation:
Method 1:
Step 1: Pre-emphasis is finally defined as "1-V_low/V_high" (Slide 6)
Step 2: Using pre-emphasis def with b parameter (Slide 4) we get next values for V_low & V_high:
V_low = (1-2b)V_peak -->
1-V_low/V_high = 1-2b
V_high = V_peak -->Step 3: Using pre-emphasis def with a parameter (Slide 5) we get next values for V_low & V_high:
V_low = V_peak*(1-a)/(1+a)V_peak -->
1-V_low/V_high = (1-a)/(1+a)
V_high = V_peak -->Step 4: combining Step 2 & Step 3 results for "1-V_low/V_high " equation, we get above ratio - 1-2b = (1-a)/(1+a)
Method 2:
Let's analyze time-domain equations for pre-emphasis' filter definition - using b parameter (slide 4) and a parameter (slide 5). We know that coefficients correspondent to different samples - x_n and x_n-1 - are independent (non-fraction equalizer!), so we can compare them one by one. For example, let's treat x_n-1 coefficients - should be equal, right?
b = a/(1+a) --> 2b = 2a/(1+a) --> 1-2b = 1 - 2a/(1+a) = (1+a-2a)/(1+a)=(1-a)/(1+a)
Same result might be achieved for X_n coefficients comparison.
Let me know if you still have problem - and thanks for your patient to read so far :-)
Regards,
Dimitry-----Original Message-----
From: larry rennie [mailto:Larry.Rennie@nsc.com]
Sent: יום רביעי 12 פברואר 2003 12:15
To: 'dimitryt@mysticom.com'; Zev Roth
Subject: Correction to your 12/02 CX4 PresentationDimitri and Zev,
I was looking at your pre-emphasis definition presentation from the Dec
02 CX4 meeting. I notice that you have a slight error on slide 8,
"Correspondence of pre-empahisis definitions"you have 1-2Beta = (1-alpha)/(1+alpha)
Based on your earlier equations it should be:
1-2Beta = 2*alpha/(1+alpha)
By coincidence, for alpha=1/3, (1-alpha)/(1+alpha) = 2*alpha/(1+alpha)
Is this a correct correction?
Thanks,
Larry
pre_emphasis_to_alpha_ conversion.xls