cross-discipline job descriptions

Dec 11, 2010 112 Replies

a

Yes, as in it won't ring. You would need a negative value in the taps to make the output go downward.

There is an analog in continuous time filtering as well, though it has to do with the impulse response going negative. [In continuous time, you convolve the time signals, very similar concept to the transversal filter.] A common fallacy is a Bessel filter won't ring. In reality, you need to look at the impulse response. Past a certain order, the impulse response goes negative, so it will ring. It will still be linear phase though. Now the Gaussian response is always positive.Not much of a filter!

Okay, sure, it'll have a monotonic step response.

Cheers

Phil Hobbs

Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 email: hobbs (atsign) electrooptical (period) net http://electrooptical.net

...

not sure i agree with that...

... while i *do* agree with that.

i don't think that i would equate a step response of monotonic slope to be "no ringing".

what if the impulse response is

h[n] =3D A + B*cos(w0*n) (for 0 |B| ?

all h[n] are positive, the step response is monotonically increasing. but i might get a sense of ringing going on there.

now i am not yet sure (regarding Phil's observation) what the peak gain would be compared to the gain at DC. i s'pose i could work it out.

but i think there is a sense of ringing in that filter.

r b-j

Umm.. there are almost always some tasks which employees can delay without delaying a project, and accelerating doing them will have no direct influence on the actual completion date. They are therefore not on the critical path for that project, although they may be important. Each project has at least one critical path (there may be more than one, but that increases risk), but seldom except in the very simplest of project planning and management will you have all the tasks for all the people stacked up in a single critical path.

Best regards, Spehro Pefhany

"it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com

and sometimes the path is not clear. i often think that this is the case when you are closer to "R" than "D". sometimes you have to solve one problem before you proceed to know even what the next problem is. it's sequential and in that sense, you're always on the critical path (even though you don't know precisely where it leads).

and i'm not trying to be any touchy-feely "your path lies before you, Grasshopper". i'm just saying that when you're trying to do something new, something that no one in your organization has done before, that sometimes it's Step A that indicates Step B (and you won't know that before finishing A) which, in turn, indicates Step C and so on. you cannot get to C without first B which requires A before that. then it's sequential and it's functionally no different than the critical path.

r b-j

I've had people actually log into my computer via ssh (secure shell). I don't know if Windoze has that capability, i.e., I don't know if Windoze has anything like telnet (which is about as secure as Britney Spears' drawers) or ssh, but if there is a way, I wouldn't be surprised if the techs know it.

Heck, the Black Hats can write trojans - why couldn't they just log in?

Cheers! Rich

t a

For the Bessel filter, I take it you meant _approximately_ linear phase. As for a Gaussian, reading the taps off Pascal's triangle provides decent filtering, far better than boxcar. It's certainly dandy for 2-D filters.

Jerry

...

Define ringing. The derivative rings, and it's possible that the filter's low-passed output would ring. So what?

Jerry

Ok, then most of my project fall under the "seldom" class but they sure are not simple, or they wouldn't have called me in.

Typically, if one guy on the team screws up one of his tasks the whole projects slips into muddy terrain. This is why good project management is key. Many younger engineers consider a project manager a non-essential or a Gantt chart paper pusher but within my first month of employment I learned that this stuff is very important when a team is larger than 2-3 people.

Regards, Joerg http://www.analogconsultants.com/ "gmail" domain blocked because of excessive spam. Use another domain or send PM.

I think you misunderstand. By "simple", I'm NOT talking about the technical difficulties of the project, but the matrix of dependencies between tasks in the work breakdown structure and the interactions of team members, stakeholders and external entities such as regulatory agencies. With a one-man band all that can be practically managed in your head or on a single sheet of paper, and even more so if you only do the same type of project (again, the sequence of tasks such as design- simulate- review- prototype-release), not the technical details of each design) over and over.

To an individual employee or consultant, they need to see clearly the requirements of the tasks they are involved in, although it's good if they have some idea of the big picture too, to the extent that makes business sense to share such proprietary information (more so with employees because they have more presumed loyalty to the company). When you have a whole bunch of employees, consultants, contractors etc. and government departments involved, the complexity increases much faster than linearly, more like exponentially.

That's why I've put a great deal of effort recently into studying and applying professional project management methodology to engineering projects. Lots of people think slapping together a plausible-looking Gantt chart in MS Project and looking at it on meeting days _is_ project management, rather than just the tiniest tip of the iceberg.

We meant the same thing. The projects I am involved in have at least one consultant (me) or more on the team, plus several in-house engineers, plus vendors such as PCB fab and often external layouters. All it takes is one of the participants to drop the ball on a particular task and much of the original plan goes out the window. So, a good project manager is constantly walking around, seeing how things progress. In my case they email or call.

I ran a start-up and later a division, and did that. To me it was important that people understood the big picture and felt to be part of it. The big common goal was usually a fat ECO release. There were incentives tied to that as an additional motivator. And I really feel sorry for countries where stock options are not legal, they are missing out on a lot.

The most impressive accomplishment happened at a company that designed a complete medical ultrasound machine from scratch in under 1.5 years, with half a dozen employees and half a dozen consultants. I have a lot of respect for the guy who ran the show there. Despite the fact that he roots for the wrong team (Manchester United) :-)

It's great that you studied project management, I wish more engineers would show at least a little interest there. I never formally studied it but had my dose of that in the service (army). The millisecond you take a lead role there you better adhere to good planning. One really gets chewed out there if something major gets screwed up during an exercise. Minor oversights can trigger that, even mundane stuff such as forgetting to carry a set of fresh batteries because there won't be a grocery store or anything else where you are going.

[...]
Regards, Joerg http://www.analogconsultants.com/ "gmail" domain blocked because of excessive spam. Use another domain or send PM.

For a transversal filter to ring, you need to make it, shall we say, take a step back. That would equate to a negative tap.

I suppose I can try to explain it in more detail. Think of the filter coefficients as just positive numbers in an array. Assume a unit step. The first output of the filter is just the first tap coefficient. The second output is the sum of the first two coefficients, again both positive numbers, so the second output is larger than the first. As in an inductive proof, you can see that if every tap coefficient is positive, then every succeeding output of the filter is larger than the preceding value. Thus it will not ring.

not a

p

Yes approximate linear phase. Technically, the Bessel is maximally flat linear phase.

...

...

Maximally flat phase, not maximally flat linear phase. In the same sense, a Butterworth has maximally flat amplitude response in the passband. Rune notwithstanding, I think that precise statements are appropriate for technical topics.

Jerry Jerry

where do you get that fact? you are begging the question. you are defining "ringing" as having a step response with monotonic slope. and it doesn't matter whether it's transversal or not.

given your definition, yes it would. i still don't buy that definition of "ringing".

you need not. i understand what FIR filters are. and even though i did not anticipate the answer you were seeking (i really suck, i get filtered out and i have no PhD), given the initial information, i don't feel very ignorant of what FIR filters are or do.

aga>

ot

well, Jerry, i think then the precise statement is that the Bessel is an all-pole filter (at least in the s-plane) that has maximally flat group delay (vs. frequency).

r b-j

he

.Not

I agree with your characterization of Bessel filters (digital ones, anyhow) but I don't see how you can say that a signal that neither overshoots nor oscillates shows evidence of ringing. Can you expand on that please?

Jerry

I think I follow his argument. The ramp-up of the step response, though monotic, is not smooth. The first derivative rings, with an offset, so there is some sense of ringing.

If you add a sinusoid to an increasing ramp that's increasing faster than the decreasing parts of the sinusoid, the result will be monotic. Is it ringing?

Eric Jacobsen Minister of Algorithms Abineau Communications

formatting link

Only if the sinusoid component is declining in amplitude.

Steve

okay, then the impulse response is

h[n] =3D A + B*exp(-alpha*n)*cos(w0*n) for 0

i think you mean "analog" Bessels, no? i don't even know exactly how a digital Bessel is designed since, even though phase is one-to-one matched (as is magnitude) by the frequency warping of the bilinear transform, group delay might get messed up so that it bears no resemblance to the continuous-time group delay when it is mapped to discrete-time via the bilinear transform. impulse invariant would sorta preserve the step response also, but i cannot say that the group delay would have all derivatives (below the Nth) at DC equal to zero. that gets goofy when transformed from s to z.

r b-j

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required