[SI-LIST] Re: frquency limit of a channel
- From: "Clewell, Craig" <Cclewell@xxxxxxxxxxxxx>
- To: 'Kihong Joshua Kim' <joshuakh@xxxxxxxxx>
- Date: Mon, 8 Mar 2010 08:50:55 -0500
Here is a link to a file I have in my library from Clayton Paul that should
clear things up for you.
http://www.mediafire.com/?k0wvdamwz1n
Craig
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of Kihong Joshua Kim
Sent: Sunday, March 07, 2010 11:22 AM
To: Lakshmi N. Sundararajan - PTU
Cc: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: frquency limit of a channel
Hi Lakshimi,
The question is on transmission media so called 'channel'.
Hard to see the reason why the explanation come from RC circuit unless the
author elaborate its intention.
There can be many ways to model the channel capacity.
Anyway, the channel flows starting from dc current upto certain frequency point
which we call cut-off frequency.
So the channel you have submitted is LPF (low pass filter) not BPF.
I think someone in the loop used 'fc'. Let's use fc to keep our discussion live.
We know that the channel (or your serial link copper trace) would not be cut
off as if ideal LPF does.
So then what is the reasonable frequency you want it to be?
Remember we are focusing on the channel associated with a specific signal type
rather than random noise in which case we may need to focus on Shannon-Hartley.
If you want to stick to k=0.35, then you unintentionally assumed that you are
interested in trapezoidal waveform.
Trapezoidal waveform has two time parameters; rise/fall time and period.
(let's simplify this way).
Some text book mentioned about k=0.35 to 0.45 depending on the overshoot amount
of your signal under consideration. ('Transmission Line...' Richard E. Matick)
Now when you focus on a bit period 'T' and rise/fall 'tr', then your spectral
density will have two significant frequencies under consideration.
One of my old collection papers (copied from the magazine 'Electronics"
Sep.2 1968) explains this topic quiet intuitively.
I summarized it for your here.
Upto the frequency of 1/(pi*(T+tr)) the PSD upper envelope (UE) goes flat from
that point to 1/(pi*tr) the PSD UE goes -20dB/decade and further from the
1/(pi*tr) goes -40dB/decade
So the minimum meaningful number that reflect rise/fall time in this PSD
approach is 1/(pi*tr) in which case k=0.32.
Then why delta = 0.35-0.32 comes? I don know. But I do not care either. The
above mentioned text book says considering 5% of overshoot for high speed
amplifier (those days probably MHz) which is reasonable back then to compensate
that high frequency uncertainty. Well anyhow I think we could tolerate that
delta portion at least today.
Two cents...
Joshua K. Kim, CEO/CTO
B2KC Technology Inc.
www.b2kc.com
2010/3/6 Lakshmi N. Sundararajan - PTU <lakshmi.s@xxxxxxxxxxx>
> Hi Joshua,
> Here is one derivation from searching the internet...
>
> ====================================================================
> This is how it works: the circuit is assumed or approximated to behave
> like a simple RC integrator.
>
> The time constant will then be: ¦Ó=RC
> The bandwidth is then BW=1/(2*¦Ð*RC)=1/(2*¦Ð*¦Ó)
>
> On the other hand, the rise time for this simple RC circuit is
> calculated starting from the equation of the voltage across the cap
> when a step impulse is applied at the input:
> u(t)=U*(1-exp(-t/¦Ó))
>
> For u(t)=0.1U you get 0.1U=U*(1-exp(-t10/¦Ó)), where t10 is the time
> when the voltage reaches 10% of the final value.
>
> After simplification
> 0.9=exp(-t10/¦Ó)
>
> Applying the natural log to the equation you get:
> ln0.9=-t10/¦Ó or t10=-¦Ó*ln0.9
>
> Similarly, for u(t)=0.9U, we have t90=-¦Ó*ln0.1
>
> The rise time is then:
> tr=t90-t10=¦Ó*(ln0.9-ln0.1)¡Ö2.2*¦Ó or ¦Ó=tr/2.2
>
> Substitute this value of ¦Ó into the expression for bandwidth:
> BW=1/(2*¦Ð*tr/(2.2))=2.2/(2*¦Ð*tr)¡Ö0.35/tr
>
> or
>
> BW*tr¡Ö0.35
>
> ====================================================================
>
> Is the above not true?
>
> Thanks,
> -LN
>
>
> -----Original Message-----
> From: Joshua Kim [mailto:joshuakh@xxxxxxxxx]
> Sent: Saturday, March 06, 2010 1:06 PM
> To: Lakshmi N. Sundararajan - PTU; si-list@xxxxxxxxxxxxx
> Subject: RE: [SI-LIST] frquency limit of a channel
>
> The equation is not from RC circuit. I know there will be many to
> explain this. But if not try to contact me.
>
> -----Original Message-----
> From: Lakshmi N. Sundararajan - PTU <lakshmi.s@xxxxxxxxxxx>
> Sent: March 6, 2010 3:59 PM
> To: si-list@xxxxxxxxxxxxx
> Subject: [SI-LIST] frquency limit of a channel
>
> Hi Gurus,
> Suppose assume I have a high speed serial link at 6Gbps. The nominal
> rise time of the signals on this channel is 150ps.
>
> Given this rise time, the bandwidth required to transmit this signal
> is 0.35/tr = 2.33Ghz.
>
>
>
> So, to study this channel behavior, is it correct to only look at
> s-param frequency output till say 3Ghz.
> Can any higher frequency data points on this s-param be ignored and
> still correctly model the channel behavior?
>
>
>
> I also looked up BW * tr = 0.35. This equation is derived from a
> simple RC integrator circuit.
>
> How true can this model any channel, since we seem to be using this
> equation for all our studies.
>
> Please clarify.
>
>
> Thanks,
>
> -LN
>
>
>
>
>
>
>
>
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