[SI-LIST] Re: TEM Approximation
- From: "Swanson, Dan" <Dan.Swanson@xxxxxxxxxx>
- To: "'norhan@xxxxxxxxxxxx'" <norhan@xxxxxxxxxxxx>,si-list@xxxxxxxxxxxxx
- Date: Thu, 25 Sep 2003 08:41:28 -0400
To put some frequency domain numbers on
this. A 25mil alumina substrate is seldom
used above about 10GHz in the microwave
world because of the dispersion in Zo.
The Zo curve looks like a hocky stick
with the inflection point right around
10GHz.
Dan
Dan Swanson EMAIL: d.swanson@xxxxxxxx
Andrew Corp. PHONE: 978-834-4085
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> -----Original Message-----
> From: norhan@xxxxxxxxxxxx [mailto:norhan@xxxxxxxxxxxx]
> Sent: Thursday, September 25, 2003 1:27 AM
> To: si-list@xxxxxxxxxxxxx
> Cc: Raj Raghuram
> Subject: [SI-LIST] Re: TEM Approximation
>
>
> Hello Steve,
> I should have known that my post will generate questions :-).
>
> The shape of the distorted Gaussian impulse that you saw is very
> typical for microwave substrates. It is all over microwave and
> guided wave literature. There are several reasons for the
> large dispersion and you correctly pointed out one of them:
>
> 1. Large dielectric constant of the microstrip substrate (Alumina).
> 2. Relatively large dimensions (h=0.635 mm, w=0.5 mm, l=123 mm).
> 3. Shape and width of the exciting impulse (Gauss, sigma=19 ps).
>
> These three reasons give the Gaussian impulse the characteristic
> negative tail with some ripples. In addition to these reasons,
> the impulse is also influenced by
>
> 4. Point source excitation (multi mode excitation).
> 5. Material losses.
> 6. Other reasons that I better not mention.
>
> The last three visibly influence the impulse, but the changes
> that they make are not really important here.
>
> The main reason why you do not see the characteristic dispersion
> tails in your typical PCB TDR measurements is that you do not
> have 1-3: your dielectric constant is significantly lower, your
> cross sectional dimensions are smaller and your signal waveform
> has different characteristics. The last reason really includes
> two effects and it requires more elaboration.
> While the signal in this example is not particularly fast, it is
> slightly faster than the signal out of your TDR and it is an
> impulse (not a step). Therefore we have to take the derivative
> of your step waveforms to make a comparison. First, the shape of
> the resulting impulses (derivatives) is important. Different
> impulse shapes are affected by dispersion differently. The TDR
> outputs one impulse and the example uses another (Gaussian).
> Second, the response to a Gaussian signal is an impulse with
> high frequency ripples after it. Similarly the response to the
> TDR's input impulse (derivative of the step) is a step with
> accentuated high frequency components (j*w*step_response(w)).
> But these high frequency components are attenuated by the TDR
> because of the limited bandwidth of the TDR system.
> The end result is that the useful resolution of the TDR is
> reduced when you look at derivatives so impulses do not have
> high frequency tails even when they should.
> The world looks smoother through a TDR than it really is
> (this is not a bad thing :-)).
>
> I think that the losses in the measurements are a secondary
> effect in this type of example.
>
> Best regards,
> Neven
> ---
> Neven Orhanovic
> Applied Simulation Technology
> neven@xxxxxxxxxxxxx
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