Author Topic: Plotting the upper bound of harmonic content - yes it's assignment time again.  (Read 1606 times)

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Online Simon

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1. A low voltage digital device has a low state of 0 V and a high state of
1.8 V. It generates a signal train of high pulses at a rate of 1 MHz, these
pulses having a 10 ns width and transition times of 100 ps.
(a) Plot the upper bound of its harmonic components up to 10 GHz.
(b) Determine the amplitudes of the third and 101st harmonics.
(c) Determine the signal bandwidth.

So i am stumped on (a). the bandwidth of the signal is deemed to end at 1/(Tr*pi) which is just over 3GHz. what exactly do they want. if it's any help attached is the lesson that is supposed to tell me all. Cryptic at best as usual.
 

Offline dmills

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A is figure 4 in your book scaled appropriately.....

Regards, Dan.
 

Online Simon

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OK so I copy the diagram and fill in the numbers, that simple? So where does the 10GHz come in? or is it that I need an end point to work back from as I have no way of knowing any amplitude coefficients (well I can actually but it feels like the explanation has been less than expansive and i suspect they want to make it easy). The last line should actually be a curve ?
 

Offline MrAl

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Hi,

Looking at your first post only this looks like an exercise in Fourier analysis where you want to know the signal spectrum.  The highest harmonic is due to the rise and fall times because that dictates the highest frequency that would be needed to reconstruct the original signal from the spectrum information.  If the rise and fall times were instantaneous the highest frequency would be infinite, but because they are ramped that limits the upper bound in any practical analysis because a ramp has lower amplitude harmonics than a step wave.

After looking at your pdf briefly, it looks like they may have actually done the analysis for you and found the spectrum function and so you could use that to figure out your answers.  I didnt read the whole thing though so there may be more info in there too you can use.  Study that well.
 

Online Simon

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It's the 10GHz that is confusing me. Looking at figure 4 the graph is not that hard to work out as they give the values. So the first section is due to the fundamental, then you have the ever decreasing harmonics of the fundamental and then the fundamental of the rise time and the harmonics from that that will decrease exponentially. so the Rise time produces a frequency of just over 3GHz. If the rise time dictates the frequency at which the last segment of the graph starts and that line follow a 1/f^2 then how can they decide that I am to plot to 10GHz? the plot will end when it ends......
 

Online Simon

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Oh and it is a Fourier analysis thing but they have tried to simplify it as it's not within the scope of the lesson apparently, that is what is making it harder, we either do the subject or we don't. But this is the level of university education now in the UK.
 

Offline dmills

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The series will continue (probably) to infinity, and certainly there will be SOME energy at above 10GHz, why would there not be? However you seem to have established that the second break point is 3GHz and the gradient from there is 1/f^2 so this is just a geometry problem.

It actually did not strike me as a horrible introduction to the subject at a sort of 'technician level', maybe HNC/D sort of level? Lacking the rigour of "Fields and wave in communications electronics" but not a bad introduction to EMC specifically, for all that it is now a little dated (Who uses DIP in a product these days, and some mention of modern MLCC behaviour would be good).

I have see (and been subjected to) far worse course handouts.

Regards, Dan.
 

Online Simon

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It's a HNC course but I think most of the material is common to the HNC and HND, this module is a level 5 and I am not allowed to do too many of the level 5 ones and have had to do mostly level 4 as level 5's are aimed at the HND course. This section is a bit simpler than the capacitive/inductive cross talk sections that also illustrated filters.

Yes the whole thing seems dated, all of their modules are but I think they are going to shake things up and the EMC module is being removed because this thing where you teach useful stuff really will not do!. I am yet to see some practical advice on real world design having moments ago re-flowed my latest PCB at work that is stuffed full of filtering for EMC with the aim to pass a previously failed test. I designed it before I started this module and i am yet to come across any information that would have helped me do a better job.
 

Offline dmills

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You would be surprised, the "Edge rate is the thing that matters, signal fundamental frequency doesn't" thing in there is something that many, many people fail to grasp.

I saw some stuff that could reasonably be unknown to people doing a course at that level if you were not already an experienced designer, nothing that is not in 'advanced black magic', but things that I would rather my new layout guy had at least seen.

Dated? Possibly a little, but Maxwells eqn's have not been revised in a while.

Regards, Dan.

 

Online Simon

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Yea i like the way they bang on about ground planes like they are the discovery of the century. i have learnt stuff here or at least gotten a feel for the numbers but the principles are not new to me. On the one hand i wish their modules went more in depth but then knowing how they can make a mess of explaining things I'm glad they don't. I just need the piece of paper and the contorted mess they make of things affects my grades when I can't answer that question that never had the theory explained for it or was explained so poorly that i have no chance especially when the question uses previously undiscovered ways of mathematically representing things.
 

Online Simon

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The fact that it is nominally a square wave with given waveform makes it trapezoidal. I's not sure what difference the shape makes on upper frequency. They teach us 1/pi*Tr which seems to make sense as to represent a faster changing signal you need higher frequency harmonics.
 

Offline MrAl

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The fact that it is nominally a square wave with given waveform makes it trapezoidal. I's not sure what difference the shape makes on upper frequency. They teach us 1/pi*Tr which seems to make sense as to represent a faster changing signal you need higher frequency harmonics.

Hi,

The shape makes a big difference not really on the upper frequency but on the upper frequency chosen as the upper limit.

Recall that for a square wave we have harmonics that have amplitudes that decrease as the inverse of the harmonic number:
An=K*1/n

but for a sawtooth we have harmonic amplitudes that decrease as the inverse sqaure:
An=K*1/n^2

so you can quickly see that the shape makes a big difference because the amplitude decreases much faster for a ramp.
Note that for the 10th harmonic of the square wave we have amplitude:
A10=K/10

while for the sawtooth we have:
A10=K/100

where K is the same for both.
You see the difference now.

In the practical sense if we chose 10 as the highest harmonic number, that would mean for the square wave we would quickly have amplitude that is 10 percent of the fundamental while with the sawtooth we'd only have 1 percent of the fundamental.  1 percent is often acceptable while 10 percent often is not acceptable.
« Last Edit: February 26, 2019, 12:03:58 pm by MrAl »
 

Offline GrimaWormT

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Hello all hope you are all keeping well.

For part (b) of the question I am stumped. It should be fourier analysis but for the life of me I can't seem to get it to click.

I was going to just apply the formula x(t)=An * Sin (n*2*pi*Fo*t) and punch in 3 & 101 for the n figures, but can't tell how I'm meant to have a "t" figure?

Any help from one of you big brained folk would be much appreciated.

 

Online Simon

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Hi, are you doing this module? condolences. I recently got a small 12/5V 1A SMPS to pass military grade EMC with little sheilding and my study of this module did nothing to help in that  :-DD.

So to question section b specifically, my answer was:

Quote
The third harmonic calculated using the formula on page 3 of EC-3-2 is xxxxxxxx. The 101st harmonic is xxxxxx.

I got into the habit of reminding them of their own text that had driven the answer especially when the explanation made no sense and it seemed to be a case of monkey see monkey do like it is here as they have over condensed and over simplified the theory.

Who is the tutor on this now? if it is Andy, he's a lovely chap and is not hard on his students in marking if they demonstrate effort and grasp of the subject. Don't be afraid to answer the way you want to after giving them what they want, showing that you understand beyond the scope of the question may get you some forgiveness for minor errors.

Always remember when answering these questions, forget what you may already know, work out which bit of their theory they are talking about as that is all they are interested in and given the strange way they explain stuff questions can only be correctly answered using their explanations. The stranger the question the more it was targeted at their strange theory. The sad thing about the EMC stuff is that you can't just pick up an alternative book about it as they are several hundred pages long, expensive and very in depth.

Don't bother with the distributed control systems module, it's not worth the electrons used to store it in memory.
 
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Online Simon

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Oh as for "t" ignore it unless you are doing calculations at a specific angle in the cycle. For your fundamental it is "1" when you calculate on harmonics, it's the harmonic number.
 
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Offline GrimaWormT

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Hi Simon, thanks for your reply!!

Andy has been off for the past few months, not that I was told so I have about 10 queries to the tutors waiting... I have found the module learning materials to be not the greatest as well.. Thanks for the advice and heads up on the DCS module!!

I have another tutor James who instead of answering the queries is doing exaclty what you said and just telling me to read 3 300+ page books to answer one little part of a question.

Apologies I just want to clear up which formula you used, the long winded one in the middle of the page or the smaller one for the individual at the bottom of page 3? I was going to use the one at the bottom as per my previous post and sub in t=1 as you said, does that mean x(t) would still be my amplitude I would need? If I were to use the long winded one, sub in 3 for the n figures and 1 for the T figures and 1.8V for the A figure?

Your help is greatly appreciated. Keep Well!
« Last Edit: July 15, 2021, 02:57:29 pm by GrimaWormT »
 

Online Simon

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Well I had Andy suddenly removed from a module as he was doing research which I guess is one other way for the university to try and stay afloat. Yes James is a pain, he just gave me book titles. To be honest they are all useless.

I am currently proof reading a new module for them, my god what a mess, I thought they were cleaning up but this is awful. I mean they can't even be bothered to use images that are not so pixelated that reading them is hard. The sad thing is that these are schematics they could have drawn in KiCad!!!!
 

Offline GrimaWormT

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Having to proof and make material for them my heart goes out to you!! :-DD

Sorry to ask again but on Page 3 did you use the large formula as we have all the figures or should I just use the smaller one as the bottom?

Thanks Simon means alot!
 

Online Simon

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Having to proof and make material for them my heart goes out to you!! :-DD


Yea, I now know who wrote the worse modules......

Quote

Sorry to ask again but on Page 3 did you use the large formula as we have all the figures or should I just use the smaller one as the bottom?

Thanks Simon means alot!

My exact answer was: The third harmonic calculated using the formula on page 3 of EC-3-2 is xxxxxxxx. The 101st harmonic is xxxxxx.

check the formula mentioned, I quoted it as I must have felt there was something contentious about it.
 


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