[SI-LIST] Re: frquency limit of a channel

  • From: Kihong Joshua Kim <joshuakh@xxxxxxxxx>
  • To: "Lakshmi N. Sundararajan - PTU" <lakshmi.s@xxxxxxxxxxx>
  • Date: Sun, 7 Mar 2010 22:05:56 -0500

I have reviewed my email fired yesterday.
Oops,...
I realize some info should have been corrected.
A big apolgy to si-liter, I would see it is more than a typo.
Any way 'T' should go as 'tr+dwelling time of any state ('1' or '0')'
So please replace all the 'T' to the above new quantity.

Now it makes sense more.

Regards,

Joshua K. Kim, CEO/CTO
B2KC Technology Inc.
www.b2kc.com




2010/3/7 Kihong Joshua Kim <joshuakh@xxxxxxxxx>

> 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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>

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