MIT 6.002x has officially kicked off... - Page 1
Forum section: electronics / general

author: slateraptor
author_rank: Frequent Contributor
timestamp: March 05, 2012, 02:09:10 am
content: Just thought I'd let everyone know since the supposed e-mail notification wasn't sent out.As an aside, spent the weekend setting up anonline tutoring portalto benefit my little sisters; Skype and Remote Terminal was starting to suck major donkey balls. 6.002x should serve as an excellent opportunity to test drive the platform's capabilities.

author: slateraptor
author_rank: Frequent Contributor
timestamp: March 05, 2012, 05:48:13 am
content: First 2 weeks of content knocked out quicker than I anticipated; content is on the order of high school physics thus far. MIT really set the online interface standard high with their new platform. They also did a good job implementing a simple SPICE GUI and pseudo-random coefficient generator for the posted problems; the latter should serve well to curb uncreative cheaters, but it won't be nearly enough to deter the clever ones.EDIT: Did a bit of browsing around the interface and found that the Profile tab up top (1st attached pic) contains a score breakdown. Then I noticed that a bunch of them showed 0/x 0/y 0/z etc and I was scratching my head a bit trying to figure out what these were pointing to.Turns out that there isn't just a single lame ass 1-minute video with zero content for every topic discussed in a week...there's a bunch of them that I had completely missed (2nd attached pic). =\ Oops.

author: McMonster
author_rank: Frequent Contributor
timestamp: March 05, 2012, 08:47:10 am
content: Thanks for the notice. Introductory videos already mention watching kittens on YouTube and Angry Birds, this is surely going to be an interesting course, but nothing's going to beat some lectures on my university like analyzing a nest of aliens (yeah, those from Alien series).

author: Mint.
author_rank: Frequent Contributor
timestamp: March 05, 2012, 09:46:24 am
content: Thanks for the notice. I had received no email on this, but now I know, so thanks. Looks like I've got a lot of study ahead of me

author: MikeK
author_rank: Super Contributor
timestamp: March 05, 2012, 01:06:26 pm
content: Where are you guys getting this info? I didn't get any email and their 6002x site still shows the "enroll" page.EDIT: Okay, I see the login.

author: McMonster
author_rank: Frequent Contributor
timestamp: March 05, 2012, 03:39:44 pm
content: I finally received course start notification.

author: sacherjj
author_rank: Frequent Contributor
timestamp: March 05, 2012, 04:17:34 pm
content: I received my notification at 10:06AM (15:06 GMT).

author: Zad
author_rank: Super Contributor
timestamp: March 05, 2012, 06:35:40 pm
content: Mine arrived at 15:41 GMT, leaving it until I have some nice quiet time to concentrate on it.

author: 8086
author_rank: Super Contributor
timestamp: March 05, 2012, 08:27:29 pm
content: Would anybody object if I posted a quick question about the basic circuit analysis homework task? Spoilers and all that...

author: Bored@Work
author_rank: Super Contributor
timestamp: March 05, 2012, 08:40:21 pm
content: Quote from: 8086 on March 05, 2012, 08:27:29 pmDoesn't MIT have an own forum for this experiment?

author: 8086
author_rank: Super Contributor
timestamp: March 05, 2012, 08:44:01 pm
content: Quote from: BoredAtWork on March 05, 2012, 08:40:21 pmThere is a "discussion" thing on there but it looks pretty pants. It's only a very basic question. I think there is an error with the system, as I am being told one of my answers is wrong but I really think it's correct. I just need a second opinion to see if I'm right.

author: Bored@Work
author_rank: Super Contributor
timestamp: March 05, 2012, 08:50:17 pm
content: Quote from: 8086 on March 05, 2012, 08:44:01 pmBut isn't that something to discuss with "them"? This is a new system, and didn't they run this pilot to get feedback about all kinds of issues?

author: 8086
author_rank: Super Contributor
timestamp: March 05, 2012, 08:53:51 pm
content: Quote from: BoredAtWork on March 05, 2012, 08:50:17 pmI don't want to waste their time without getting a second opinion. I could very well be overlooking something very basic and not realised it.

author: Bored@Work
author_rank: Super Contributor
timestamp: March 05, 2012, 08:57:43 pm
content: Quote from: 8086 on March 05, 2012, 08:53:51 pmBut that would also be valuable feedback. It might be the lecture was not clear, they missed something in the presentation or didn't present it correctly.And that the forum is "pants" is valuable feedback, too.And even wasting their time is valuable feedback. They need to get information how much effort / time they have to plan to deal with student questions.

author: MikeK
author_rank: Super Contributor
timestamp: March 05, 2012, 09:00:03 pm
content: I would waste their time if I were you. I've already sent them several messages about errors in the problem descriptions.

author: 8086
author_rank: Super Contributor
timestamp: March 05, 2012, 09:03:46 pm
content: I didn't read any of their material, as this is stuff I  already know. It's just that everything is correct except one question, which leads from another question, using the same values, that it says I have got correct. To give you an idea of how trivial it is, it's a DC voltage source in series with two resistors. The question is to calculate "The sum of the powers (in Watts) entering all of the elements".I just don't want to go and tell them their system is broken without being sure.

author: Mint.
author_rank: Frequent Contributor
timestamp: March 05, 2012, 09:52:10 pm
content: Instead of worrying, just go tell them you waste more time on this forum when you could of already done it.

author: slateraptor
author_rank: Frequent Contributor
timestamp: March 05, 2012, 10:15:32 pm
content: Quote from: 8086 on March 05, 2012, 09:03:46 pm...but you didn't consider the independent source, which is likely why you think the sol'n is incorrect. The question is intended to place emphasis on conservation of energy.

author: 8086
author_rank: Super Contributor
timestamp: March 05, 2012, 10:31:49 pm
content: Sorry, I left out "voltage" when I meant to type "DC voltage source" before. It's just a DC voltage source with two resistors in series. I have the source voltage, the total combined resistance, the voltages across each resistor, and the current. This should be more than enough to work out the power, and I have tried both P=IV (current*source voltage) and P=V^2/R (resistor voltage drop squared/resistor value) with the values I know are correct (the site tells me they're right), they of course agree with each other, but the website still says the answer is wrong...?

author: IanB
author_rank: Super Contributor
timestamp: March 05, 2012, 10:50:58 pm
content: If it's a closed loop, it sounds like the answer should be "zero"?

author: Zad
author_rank: Super Contributor
timestamp: March 05, 2012, 10:53:22 pm
content: I am not even out of week "zero" and there are problems aplenty.The introductory videos have background music that threatens to swamp the speech. They have used student "actors" to make a point instead of just using a sheet with half a dozen bullet points, thus making a 30 second item last 10 minutes and obfuscating the simple points.To copy the simple circuit in Lab 0, I have to scroll my screen up and down all the time.The simulator instructions state that voltage sources will show current flowing out of the (+) pin. Which it sort of does. The arrow indicates current flowing into the pin, but with a negative value. Great way to confuse newbies.It requires knowledge of Kirchhoff's Voltage Law (KVL) more than a week before it is dealt with in lectures - or should I say a week before it was dealt with in the 2007 videos.It tells us to use the sign of the first terminal of the first component we come to, to provide the basis of voltage differences. The resistors we have just added have no sign. This would be added arbitrarily in a hand sketch, the polarity being insignificant so long as it is consistent. I am using a negative voltage for a drop and a positive for the supply, but if they chain resistor polarity in the same order as they have denoted the source, my signs will be wrong.Measuring voltages on the SPICE transient plot, the measured (offset sinusoidal) voltage never reaches 2V, or 0V at 90 and 270 degrees. This is obviously due to analysis step sizes, but as a total greenhorn student I would think something was wrong.They really are over-doing "hey engineering is fun and exciting" thing. And saying the same things 3 times.The question and answer area shows a massive 5 questions per screen. I thought MIT were leaders in researching usability?Not to mention saying the same things 3 times.Oh and saying the same things 3 times.

author: 8086
author_rank: Super Contributor
timestamp: March 05, 2012, 10:56:22 pm
content: You are correct, IanB.I interpreted the question as "Whats the power dissipated by the resistors". I don't know, it seems badly worded. But there you go.Cheers IanB and slateraptorHow silly I would have looked telling them their website is broken

author: harnon
author_rank: Regular Contributor
timestamp: March 06, 2012, 12:07:01 am
content: Currently working my way through week 1 lectures at 1.5x videos... fairly basic stuff and some bugs on the lecture questions (i.e. type in the answer, click check, it says its wrong... hit "show answer" and the answer is identical... copy and paste their answer... still wrong).But hey, its week 1 of a pilot program, and its free, what do you expect?

author: slateraptor
author_rank: Frequent Contributor
timestamp: March 06, 2012, 12:10:08 am
content: Quote from: 8086 on March 05, 2012, 10:56:22 pmI wouldn't say badly worded because I think the question was intended to get the student to focus on the lesser noticed abstraction of conservation of energy of an LTI system with independent sources rather than the ability to mechanical apply P=IV.Quote from: Zad on March 05, 2012, 10:53:22 pmKeep in mind that this is an upper-division college level course where a student is expected to have a strong rudimentary background and solid independent study habits, viz. KVL is a concept taught in high school physics and if a student is unfamiliar with the abstraction, then it's a good indication that he/she is lagging behind the prerequisite knowledge curve and needs to consult the (conveniently) free textbook to catch up. Despite any perceived shortcomings, believe me when I say the interface is really quite advanced compared to what students at other major US universities would have at their disposal (it blows my uni's platform out of the water and we're not some small fish in the pond either). Nevertheless, this program is without a doubt a "beta test" for their new platform cleverly marketed in such a way that brings massive "philanthropic" credit to the institute and will gain them exponentially more user feedback than the closed-doors iterative approach that every other university employs today.

author: Mint.
author_rank: Frequent Contributor
timestamp: March 06, 2012, 01:51:45 am
content: Quote from: Zad on March 05, 2012, 10:53:22 pmTell them that!

author: metalphreak
author_rank: Frequent Contributor
timestamp: March 06, 2012, 02:02:06 am
content: I've had no issues finishing all the labs and homeworks up to (and including) week 2. The method of answering questions like that is similar to what I did in the equivalent course at my university. The SPICE circuit stuff is quite cool, although it is a bit... limited in its scope and usability. There's only so much you can do in a browser and only so much they can do in terms of developing a program like that from scratch.

author: MikeK
author_rank: Super Contributor
timestamp: March 06, 2012, 02:25:00 am
content: Quote from: Zad on March 05, 2012, 10:53:22 pmUniversity isn't for babysitting.Quote from: Zad on March 05, 2012, 10:53:22 pmIt bothered me so little that I didn't even notice it.Quote from: Zad on March 05, 2012, 10:53:22 pmIt is a computer based class afterall. I've got a 17" monitor at 1280x1024 and had little trouble.Quote from: Zad on March 05, 2012, 10:53:22 pmGreat way to get newbies to pay attention to signs and directions. Current flowing into the positive terminal is negative by definition. This was explained in the text and at least one place in the introductory descriptions.Quote from: Zad on March 05, 2012, 10:53:22 pmKVL is dealt with in the assigned readings. Again, no babysitting here. Read the text!Quote from: Zad on March 05, 2012, 10:53:22 pmI have zero schooling in engineering and had little trouble with it. I've seen a couple confusing descriptions in the exercises, but nothing I didn't learn from.Try harder, and good luck, mate!

author: 8086
author_rank: Super Contributor
timestamp: March 06, 2012, 02:31:46 am
content: Quote from: slateraptor on March 06, 2012, 12:10:08 amOkay, but my issue was really with the use of the word "entering" in the question. I mean, power doesnt "enter" a battery while it's discharging. It doesn't really "enter" the resistors either, but that made more sense to me at the time. It just seems like quite a non-intuitive use of language.But anyway, other than that I think the whole thing works pretty nicely. I'm sure the discussion area will liven up as the course goes on, too.

author: IanB
author_rank: Super Contributor
timestamp: March 06, 2012, 03:38:51 am
content: Quote from: 8086 on March 06, 2012, 02:31:46 amYou really have to think of it this way in engineering, otherwise energy balances will mess up your day.Power enters the resistor as electrical energy (and leaves the resistor as heat). If you draw a control envelope around the resistor then an energy balance is satisfied: power in minus power out equals rate of accumulation of energy in the resistor.Electrical power leaves the battery while it is discharging, which means, according to the chosen sign convention (power entering has a positive sign), that power leaving the battery has a negative sign. If you draw a control envelope around the battery, then again an energy balance is satisfied: power in minus power out equals rate of accumulation of energy in the battery. In this case power in is zero, power out is positive, meaning the rate of accumulation of energy in the battery is negative--the battery is losing energy to the circuit. Specifically, the battery has a store of chemical energy which is being depleted. In engineering you always have to remember the sign of each term, as a wrong sign is another thing that will mess up your day.If you now draw a control envelope around the whole circuit and consider only electrical energy, then the power leaving the battery (negative sign) is exactly balanced by the power entering the resistors (positive signs). The sum of these is zero. Which means in effect that the chemical energy in the battery is being converted to heat energy in the resistors with the electricity acting as the transfer medium. This "electrical energy balance" only works because there are no other ways the circuit can generate electrical energy or dissipate electrical energy or accumulate electrical energy. If, for example, the circuit could radiate energy as electromagnetic waves then the balance would need to be extended to be complete. That's the last thing in engineering that will mess up your day: making the wrong assumptions about which terms to include and which terms can be neglected.

author: metalphreak
author_rank: Frequent Contributor
timestamp: March 06, 2012, 04:13:27 am
content: All of those questions were also "practise" questions. They do not contribute to your final mark, and you are given the correct answer if you get it wrong. This is so you can LEARNThe homework and lab questions, will tell you if you have a wrong answer, but will not give you the solution. You can continue trying until you understand the answer and get it correct. You can attempt the questions as many times as you like, and still achieve a 100% mark. So if you misread a question and get it wrong because of that, don't worry

author: slateraptor
author_rank: Frequent Contributor
timestamp: March 06, 2012, 04:19:00 am
content: Quote from: metalphreak on March 06, 2012, 04:13:27 amActually, "homework" and "lab" questions count for a combined 30% of your final grade. This can be verified in the Profile section.

author: metalphreak
author_rank: Frequent Contributor
timestamp: March 06, 2012, 04:21:55 am
content: If you look in the Profile section you will perhaps spot the difference between "Practice Scores" and "Problem Scores"Many of the questions people are complaining about are practice questions.

author: 8086
author_rank: Super Contributor
timestamp: March 06, 2012, 04:23:39 am
content: Well I was definitely talking about a "problem score" question. It seems that any lab or homework question counts, except Lab0.

author: slateraptor
author_rank: Frequent Contributor
timestamp: March 06, 2012, 04:25:58 am
content: Quote from: metalphreak on March 06, 2012, 04:21:55 amThe only question I've seen a complaint on was a "homework" problem.

author: metalphreak
author_rank: Frequent Contributor
timestamp: March 06, 2012, 04:36:48 am
content: I thought the sum of all powers question was one in between the lecture videos, but its not. In any case, don't feel apprehensive about trying a solution as you have nothing to loseThe lab in week 2 is a bit tricky, but the solution is quite trivial. It should help people learn about using resistor dividers for signal attenuation as well as resistor balancers for high impedance inputs (like a SPICE probe).

author: MrPlacid
author_rank: Frequent Contributor
timestamp: March 06, 2012, 04:55:59 am
content: 8086, was your question similar to this? Is it 27mWor zero?

author: 8086
author_rank: Super Contributor
timestamp: March 06, 2012, 05:08:10 am
content: Quote from: MrPlacid on March 06, 2012, 04:55:59 amThat's pretty similar to the question, yeah.

author: MrPlacid
author_rank: Frequent Contributor
timestamp: March 06, 2012, 05:12:25 am
content: Quote from: 8086 on March 06, 2012, 05:08:10 amCool, since I didn't sign up I was just curious about what it looked like. Someone should have recorded the whole thing and upload it for the rest of us.edit: Imagine the youtube hits.

author: aluck
author_rank: Regular Contributor
timestamp: March 12, 2012, 11:15:39 pm
content: Went onto Week 3. Kewl!

author: Zad
author_rank: Super Contributor
timestamp: March 13, 2012, 05:39:21 am
content: I'm about 75% of the way through this week's stuff.Videos S6V6 and S6V7 both seem to be the same, not sure if one is missing. I'm certainly struggling to solve S6E2 using the techniques described. Not least because the graph in S6E2 has different scales for the load than for the source. The intercept implies that a (i=v^3) component has a load point of somewhere around 30V, impressive for a 6A source.

author: MikeK
author_rank: Super Contributor
timestamp: March 16, 2012, 11:24:24 pm
content: If you ask in their discussion forum you'll get plenty of help. I noticed that the graph was not to scale, but it really doesn't matter since you don't use need it to solve the problem. All you need to do is construct the equations for the two lines, set them equal to each other, and solve. I'll point out that the technique shown in the lecture video requires that the solution value converges, which it does not in this example...You just have to try a bunch of values and see which direction it's heading and then narrow it down.