nm0919: well good morning everybody er just to start off just like to remind 
you of the misprint we had in the er on the notes on bottom of page eight 
equation four-seven when we're doing the fit to the exponential-Y-A exponential-
K-X what we do is we transform it by taking natural logs of each side which is 
log-Y is log-A and i put l-, Ls so that L-Ns of E to remind ourselves it's logs 
to the base E log-A plus K-X i think there was a log in there by mistake and 
therefore if we plot er of the individual data points Y-I and X-I those pairs 
if we plot the log of Y-I against X-I then if this is a reasonable law we would 
expect it to be something like a straight line the gradient would be K and the 
intercept would give us an idea of er log-A in which case we can take the l-, 
antilog and get A er just realize what might have been the confusion last year 
if we had a power law which could also be a quite a reasonable behaviour for 
some transducer say Y equals A-X-to-the-K now instead of 
E-to-the- X in this case obviously when we take logs we get log-Y is log-A and 
this time it is of course plus K-log-X and in this case one er would indeed 
plot log-Y against log- X-I that should be X-I i suppose for each individual 
data point K would be now the gradient which exponent the the power law here 
and log-A the intercept so hopefully if you could correct your written copy 
there from that misprint er next week is er week four or is it week five 
ss: week five 
nm0919: week five yes it's week four this week isn't it yes next week is week 
five thank you and er this is week four so there's no lecture next Thursday i'm 
actually not in Canada next Thursday i actually arrive back from Canada at 
about seven-thirty A-M so er i m-, thought i might not be tremendously alert so 
we won't have a lecture next Thursday but what will occur next week since the 
lab is going to start in week six will be for the three groups who you know you 
are you will have your lab briefing er next 
week before the week in week five before the labs start proper week six seven 
eight nine ten okay five weeks Tuesday ten A-M Thursday ten A-M and Friday for 
the easy people i'm not quite sure presume you know what time it is yeah right 
and then so next week i will be er recovering from my jet lag in the morning 
and we'll then have some lectures on twenty-second of Feb and the first of 
March okay right let me now we'll go through we've got quite a few of these 
problems to go through so let's go through those er take a c-, couple of 
minutes now to hand these out i'm afraid lo-, namex sorry tho-, those of you 
who are wondering what's going on er if you want some cut on the fee you'll 
have to see my agent okay for this er sorry the last few minutes hasn't been 
terribly interesting television has it namex namex anybody namex no okay let me 
also while we're doing that hand out this attendance sheet okay so let's just 
quickly have a look at these er particular aspects and er if you remember the 
first part of this question is to do with er 
if i can find the er particular sheet i'm looking for is to do with er yes er 
we have er twenty-five measurements we have T-bar er plus-or-minus sigma and 
we're told that that is five-point-five degrees centigrade and the standard 
deviation is equal to one degree centigrade so that means that our distribution 
looks something like this this is five degrees and sixty-eight per cent of our 
values are within one degree and so the standard error which is how accurately 
we can work out this bit at the top there is obviously equal to sigma which is 
the standard deviation divided by root-N and we have N is twenty-five 'cause 
there are actually twenty-five readings in this frequency diagram against the 
temperature that's T-bar and so it's obviously not a smooth curve it's a bit of 
a histogram and therefore that is er root of twenty-five is five so the 
standard deviation standard error i should say is plus-or-minus er fi-, er one 
over five- point-two degrees 
centigrade so we could say that the value of T is is equal to five plus-or-
minus nought-point-two degrees centigrade with N equals twenty-five and we 
would expect sixty-eight per cent with to be within one standard deviation of 
these twenty-five values and what are the assumptions we've made well one or 
two people sort of said things like oh well sort of things are pretty accurate 
and people have been rather careful that's not the the as-, i mean if you were 
not accurate and you were not careful then this sord-, it'd just be a broader 
wouldn't it you'd have a bigger standard deviation so the assumptions that 
you've made are two basically one is that there are no systematic errors it's 
all random in other words it's not as if the the mean is up here and they're 
all shifted there's no systematic errors and th-, that the errors are normally 
distributed in other words they obey this Gaussian curve those are the two er 
normal distribution of errors okay 
those are the two assumptions and all this business about that everyone's been 
a bit accurate and you haven't been too careful that's all in here if you 
weren't very accurate and there was m-, more spread that would just come 
through on the standard deviation wouldn't it so let's just go through this a 
little bit i've been doing a little bit of re-, research as i indicated i would 
over the weekend so here we have our situation frequency of time w-, of of er 
numbers of er values of X this is our mean and they're spread in this bell-
shape curve sixty-eight per cent of them should be in within sigma what we call 
the standard deviation and of course as we get more and more and this gets 
smoother and smoother we can tell where the peak is better and better that's my 
standard error in the estimate S-M and that's sigma divided by root-N so the 
standard deviation doesn't matter really how many readings you have er but as 
you get more and more you should be s-, define the peak 
better and better so anyway i've done then a little bit of research here 'cause 
as i was indicating last week er there's a slight problem in my opinion anyway 
in the er nomenclature here when you look at these two terms they're not self-
explanatory and i promised you i'd do a bit of research on this so i've asked 
one or two i was at an international meeting er on m-, in London on Monday so i 
asked the Italian people what the Italian terms were to see if they were any 
more self-explanatory and this was absolutely hopeless because the Italian for 
standard deviation is deviazione standard and the Italian for standard error is 
errore standard so those are not very much better however the French term is 
much better the French term for mean is moyenne and for standard deviation is 
écart type so écart actually means a sort of spreading out typical so it's a 
typical spreading out so that's quite a i'm sure you'll agree that's a very 
logical term and nobody as yet i've raised this with several people has been 
able to tell me got a few e-mails on the go what the word for standard error is 
and have you have you managed to find this out 
sm0920: no i haven't 
nm0919: i asked a few people when i was at a dinner in Paris and they gave me 
some funny looks when i raised it and started talking about something else 
right what about the German term now the German for average is quite good i 
think it's durchschnitt durch means er through er cut through right so it's a 
cut through the middle and then the next term is s-, standard abweichnung okay 
which means again deviation or divergence i'm afraid doesn't it and then i 
asked these people that at were big statisticians and all they could come up 
with for standard error was normalisiert abweichnung well i don't think that's 
right actually 'cause i think the normalisiert abweichnung is sometimes you 
express a standard a fractional standard devation in 
other words what's the fraction of sigma- over-X you know if X was ten 
centimetres and this is standard deviation is one you could say the fractional 
standard deviation i-, is is ten per cent which i think would be this 
normalisiert one and nobody could tell me this one either so obviously er it's 
an international problem what to call this thing 
right let's go on to the er next one okay and in the next one number two here 
we have that the area equals pi-R-squa one person actually managed to get that 
wrong pi-R-squared and R D or equals to pi-D-squared over four if you like and 
D apparently is equal to twenty centimetres plus-or-minus point-one centimetre 
er right that's what it says here which is one part one in two-hundred that 
error right so er well we know what the error is it's er a hundred-pi that's 
twenty metres squared sorry we know what the area itself is which is three-
hundred-and-fourteen s-, square centimetres so the question 
now is what's the er error in that and at this stage er we realize here we've 
got a formula of the form Z it's a product isn't it here in that er er my R is 
squared so the problem is if i've got an error in R and i square it what 
happens to the error in R-squared and er basically well what i'm going to do 
now actually i presented these formula before er for combination of errors when 
you had a product or a square or a power as you've got here just er as er out 
of the hat so to speak so let's actually just do these things properly which 
just involves a little bit of calculus and is also what is required for 
question three of the long problems that i'll hand out in a few minutes okay so 
right let's just go through these errors a little bit more formally and those 
formula that are just quoted on the sheet let's just see where they come from 
and let's just have the formula Z i'm going to derive from A a constant time 
some thing i measure called X and something 
i measure called Y which could also be married by B and i'm interested in let's 
suppose there's an so what's i-, the first thing is what's the effect on Z of 
an error D-X okay well i can work that out by differentiating if i just 
differentiate this with respect to X and i get D-Z D-partial D-X which means 
that means differentiate keeping keep Y constant when you do this okay because 
i'm only interested in the effect of the error on X so this is a constant and 
that is of course equal to er A and similarly the effect well i could do the 
same thing here on Z of an error D-Y if X is constant right and that of course 
is D-Z D-Y is lo and behold equal to B right in other words i've got my error 
one shall we call it D-Z here is equal to A-D-X and the error two D-Z is equal 
to B-D-Y okay now we make the er assumption that er supposing er if it's true 
that the errors in X and Y are independent then if one is high the other can be 
low et cetera et cetera then 
these two error terms don't add up algebraically they add up as the sums of the 
squares don't they okay so er whoops that's not too good the er so we're now 
going to have these two error terms that's not much better either error one and 
error two ah we'll go back to this one i think for the moment error one and 
error two are independent therefore total-error-squared equals error-one-
squared plus error-two-squa you can see what's coming here can't you two-
squared and therefore D-Z- squared equals A- D-X-squared plus B- D-Y-squared 
okay that's the first formula right now we can now go on to the second one 
which i'll do under here and this is a slightly more er complicated one er 
let's write up here Z equals er A which A is a constant X- to-the-alpha Y- to-
the-beta so this is the general term now where we have a product of two terms 
there could be an X-variable and a Y-variable raised perhaps to some term alpha 
and beta or there could be just one of 
those which is the situation we've got there isn't it right where alpha is two 
and now we're going to do exactly the same thing right we're going to do D-Z D-
X we're going to find the error due to the error error in Z due to X and error 
in X and the error due to an error in Y okay so we do the partials again here D-
Z how does e-, Z change if there's a change in X keeping Y constant well that's 
A and then X-to-the-alpha goes as al-, X- alpha-minus-one Y- B and let's write 
that as A- alpha-X over X i haven't done very much there Y- to-the-B except i 
now realize i'm back where i started with X-A X-S this thing here X whoops X-to-
the-alpha Y-to-the-beta so what i've got there therefore is alpha- Z over X 
okay in other words D-Z-one error one is equal to er alpha-Z- D-X over X 
there's my fractional error coming in can you see that's why it's a fractional 
error in this term when these two are multiplied okay and i've actually er so 
that's my 
error there straight away and in fact this is the one i want over here isn't it 
i could do this one now straight away if there's only an error in X term here 
then i've got that the error D er error in A the fractional error in A is equal 
to twice the fractional error in D shall we say or er B-B over D okay so D-D 
the error in D is one part in two-hundred so the error in A is worse it's twice 
as bad 'cause we've squared it okay so it's one in one-hundred so i can say my 
area is equal to three-hundred-and-fourteen square centimetres plus-or-minus 
three-point- one square centimetres okay er the point about it as you'll see 
when you get to question three on the other sheet is that i gave you some 
special formulae these two here we haven't quite finished this one yet but in 
question three on the second sheet we have a slightly different arrangement 
don't we for the er platinum resistance thermometer so if you understand what 
we're doing here no let's so that's 
this is the now let's next one oh good this one works D-Z D-Y we'll do now 
what's the error change in Z for a change in Y okay and that is er A-X-to-the-
alpha 'cause that's constant for the moment beta- Y-to-the-beta-minus-one okay 
er if i can i can now write this as A-X-to-the-alpha-Y er we'll have a beta out 
here Y-to-the-beta over Y and i realize this thing here is what i started with 
Z so the D-Z error two if you like the second error term due to a change in Y 
'cause X and Y can have errors on is now beta er and i'll i'll bring the Y over 
here Z D-Y over Y 'cause there's a Y under there 'cause i've had to put a Y 
back in because i lost a Y when i differentiated the beta-minus-one you see 
that's where the Y underneath comes from okay so now if these are independent 
in other words if usual thing if Y goes high with an error or high there's no 
reason for X to have a high error they're totally independent then i can say 
that the total error in Z-squared is those squared is alpha- Z D-X over X all 
squared plus beta- Z D-Y over Y all squared and therefore i could write this 
bring i'll take the Zs out here and therefore D-Z over Z all squared is equal 
to alpha- D-X over X all squared that's the term we were left with there when 
there was only X and of course if two things squared are equal to each other we 
can then the things themselves are equal if A-squared equals B-squared A equals 
B plus we've got the other one beta if that's the power D-Y over Y all squared 
and the we notice here as we said before when we get these product terms 
involved or power law it's the fractional error that's important can you see it 
here D-Z over Z the 
fraction D-X over X D-Y over Y okay so er rather than producing them out of the 
hat as before it's little bit of calculus a nice little exercise in calculus 
okay just differentiating X-to-the-N is N-X minus X-to-the N-X-to-the-N-minus-
one isn't it right everybody happy everybody read that yeah okay no yes no 
sorry shall we do number three 
sf0921: excuse me 
nm0919: yes 
sf0921: how did you get error in A as one in one-thousand 
nm0919: sorry 
sf0921: 
sf0922: i got the answer as twenty-two 
sf0921: i got the error in B as one in two-hundred 
nm0919: okay i-, what 
sf0921: and error in A is one in one-thousand 
nm0919: oh sorry it was one in a hundred then sorry that's okay A equals pi-R-
squared twice over four D-squared so D-A over A equals twice D-D over D okay 
this one here was one in two-hundred so this is one in one-hundred so A equals 
three-hundred-and-fourteen plus-or-minus three did i write one in a thousand 
there 
sf0923: yeah 
nm0919: okay so once i've done this right this is half a per cent and twice it 
is one per cent so the other this is the inverse isn't it of the one we did 
last week where we had the t-, the the period 
of a pendulum was proportional to L to the h-, the length to the half in which 
case the period to the fractional error in the period was half the error in the 
length here since it's length's squared it's twice just goes as the power law 
doesn't it well i think most people who got there got number three correct got 
that far it's V equals K-U-minus-C and K here this is a velocity in metres per 
second er K is er three-point-two volts per metre per second so no volts sorry 
that's a voltage in volts and C was equal to six-point-four volts and if you 
think about that you ou-, you've got a that means there's a a characteristic 
looking something like this if this is V then er it's a straight line er if we 
put er er so we can substitute our value in here we're also given so that er 
yeah this er if if U is equal this is U U equals to nought that's minus-six-
point-four down there comes up here and we're interested in a point we're given 
the point U equals what is it 
U e-, U-I equals er ten metres per second and we measure a voltage I equal to 
twenty-six volts and if you substitute that point in there what value do you 
get in the formula and then whereabouts is my actual data point does it lie on 
that line or not and that's the residual i'm interested in so if i substitute U 
equals ten metres per second i get V that's the one on the line is equal to 
thirty-two minus six-point-four is twenty-five-point-six volts so that's the er 
formula which is this point here twenty-five-point-six and the actual data 
point is up here at twenty-six volts so the residual which is how far my data 
point is off the line is obviously equal to equals data point minus er er fit 
and that's obviously equal to twenty-six minus twenty-five-point-six equals 
nought-point-four volts i think everybody actually got that one right apart 
from the odd person who got an arithmetic er substitution er in error right so 
next time let's go now we've got quite a few of these 
to go through this time some more problem sheets here these are the longer ones 
er well i think er everybody who did question one which is really er er i don't 
know if i really need to go through this in all its gory detail because i think 
most people did get this one correct i think everybody who tackled this clearly 
could er work out we had these ten val-, what did i do with my er good one we 
had our ten values of er T 
and people most people got that T-bar was equal to ten-point-two-four degrees 
centigrade and that they got the standard deviation was equal to what did i get 
er plus- nought-point-one-five degrees centigrade okay one or two people made 
an error but i think i've identified that when they got the er arithmetic wrong 
and then er the stan-, the er standard then we also found that seven out of ten 
were within sigma of T-bar and that seems it's fairly reasonable and were 
supposed to mention there that when we had this distribution we would 
normally expect if you had ten readings here you'd expect sixty-eight per cent 
you'd expect seven within that so six you know seven you'd expect sixty-eight 
per cent so seventy is pretty good and then the next thing you had was that er 
the standard error was equal to nought-point-one-five over root- ten which is 
something like nought- point-o-five degrees centigrade so you could then say 
that your final error if you like is ten-point-two-four plus-or-minus nought-
point-nought-five that's the standard error that's how accurately you think 
you've got that mean even though quite a lot most of the points are further 
away from that what is the assumption you've been making here and again the 
assumption er is not that people are being careful with the thermometer et 
cetera et cetera hopefully that's all in these standard errors and standard 
deviations the sys-, the er assumetion is that it's a normal distribution and 
that there's no 
systematic error if i was using the thermometer that was one degree out because 
there was a break in the er mercury then of course this would all be invalid so 
these are the two that they're distributed according to this bell-shape curve 
they're random like the property of noise and then the last question was if if 
you have you have an eleventh reading T is eleven-point- two degrees what do 
you do with it and some people sort of wrote and said whoa looks a bit far out 
you know this looks a bit dodgy that one it's a bit far from there i'd probably 
throw it away that's not correct the correct version here is to say what is the 
difference of that from that and the answer it's one degree centigrade 
different the standard deviation is point-one-five so this is six standard 
deviations outside there so this is one standard deviation six is right out 
here we know there's a ninety-nine-point-nine-and-a-half per cent chance of 
everything being within 
three standard deviations so you must say if this is six standard deviations 
away from the mean therefore i will reject it that's correct just to see ooh it 
looks a bit dodgy it's a bit different from all the others is not correct the 
whole point is we can do this analytically okay that's what we're learning 
hopefully here right now we get to the rather amusing little bit number two 
number one most people didn't have too much bother with that one okay and i'm 
afraid no it wouldn't i suppose the advantage of number two is if you struggled 
over it and got it wrong and i and i explain or i've explained on the sheet 
where you got it wrong y-, you'll certainly get the message as to what the 
difficulty is so basically we have R equals A- exponential- B- T and what we 
have in this particular situation is the resistance of a thermistor yes 
sm0924: would it be over T too 
nm0919: it is indeed yes thank you right okay so we're we're trying to measure 
temperature and we've got 
something in resistance and it's not linear okay in fact er if we plot R 
against T it looks something like er this if it was B-T it would look something 
like that so thanks very much for pointing out i've got it i've written i 
copied it down wrong right so the question is what is the sensitivity of the 
device what do we mean by sensitivity we mean what's the change in resistance 
for a change in temperature in other words it's going to be in ohms per K shall 
we say it's in K is absolute temperature there T big-T is the absolute 
temperature right now everybody realizes that's what's we mean but the problem 
is that most people have er remember a little example we did before where we 
had V against T and V was equal to B-T and that was a straight line through the 
origin so they then said the sensitivity is actually er the change in voltage 
here over the change in temperature and they then said that's V-over-T well 
in fact what you're interested in here clearly for a given temperature is the 
change in temperature resulting in a change in voltage which is the gradient 
okay and so you're interested in that gradient a little change in V over a 
little change in T and it so happens that this is a straight line through the 
origin so er if you actually also work out er V- -over-T you will also get the 
right answer for the gradient so something like er half the class here said the 
sensitivity is gr-, equal equal to R- over-T and as we take a point here what 
we're actually interested in is a change delta-T here what does that give as a 
change delta-R that's what we really want isn't it is D-R D-T okay now if it's 
linear that's constant and it's the straight line through the origin but if you 
then suddenly decide oh well let's say that's R-over-T what you're actually 
calculating the sensitivity i think everybody can see is this local gradient 
what most if you calculate R-over-T 
you're actually calculating that gradient which is not the same thing so if you 
look back at the notes when we talked about that i think we did actually say 
that if it's a non-linear relationship then it's n-, then this gradient you 
have to work out the gradient and the gradient as you can see here is changing 
that's because it's non-linear isn't it so as we we got different gradients 
here and different gradients there right so basically hopefully those who 
ploughed through this using the wrong values when you see that diagram you can 
see what you've done wrong okay right so what we actually have to do now is to 
differentiate this thing here in other words we have to work out D-R D-T okay 
and we realize that this is er we have to use the chain rule here if it was D-D-
X of exponential-X we get exponential-X okay but it's exponential-B-over-T so 
we first of all differentiate the exponential and get what we started with 
which is A- exponential- B-over-T 
and then we have to differentiate what's in the exponential okay and so the D-D-
T of B-over-T is equal to minus-B over T-squared so this is minus-B over T-
squared and i could if i wanted write that back as minus-B over T-squared and 
this is back what i started with times R probably makes some people left it 
like that but this is a bit easier to see this looks correct doesn't it because 
the gradient is in fact negative okay when D er when D-T goes up okay and we 
increase er er T when we decrease R that is negative right hopefully people if 
you can't follow the maths there you'd better have a word with your maths 
people integrating differentiating a function of a function right so at this 
stage now er we're asked what is the sensitivity so we have to work out D-R D-T 
at two temperatures so D-R D-T the first one we need to do is at er two-hundred-
and-seventy-three and then we also need D-R D-T at er three-hundred-and-seventy-
three okay so we're 
interested in the gradient here at two-seven-three and the one down here which 
we think is probably a bit less at three-seven-three what i need to do also is 
to work out what the value of R is and if i substitute in there er er the 
values i get two-thousand twenty-three-thousand- and-eighty-two ohms is the 
value here right at two-s er whereas at R th-, just putting A and B in with the 
values on the sheet leave you to do that R three-seven-three is quite a lot 
less it's only four-hundred-and-fifty-four ohms okay so we could put those in 
in green i suppose here this one down here is actually four-hundred-and- fifty-
four this one up here is twenty-three-thousand so it's a big change isn't it 
and the question is now so by changing a hundred degrees i seem to have changed 
the resistance by about twenty-three-thousand ohms so it looks as if i'm going 
to get some rather large changes in er for a degree temperature change i'm 
going to get a lot of ohms change so it looks potentially as if i've got rather 
a good er sensitivity 
so this is equal to er B is equal to er what is B equal to four-thousand and T 
is equal to in this one two-seven-three so the one the value here is equal to 
er let's do this er this one here R which is two-three this is two-seventy-
three so that's the value of the resistance twenty-three-thousand-and-eighty-
two then it's B four-thousand over two-seven-three two-seven-three minus and 
that comes out to equal minus er one-two-three-eight ohms per K okay this is 
four-thou that's about twen-, yeah that's about a twentieth of that number 
right this one here what i'm doing is i'm substituting here obviously an R 
minus R- B over T-squared so in this case the resistance is at lot less four-
five-four so i think my gradient's also going to be a bo-, lot less it's not 
quite in the ratio because my T is now three-seven-three- squared instead of 
two-seven-three and if i put those numbers you can see that one there cancels 
with that that goes into that it's going to be 
about ten isn't it in fact it's minus- thirteen ohms per K right now the next 
thing i get here is what happens if A changes by one degree okay so let's have 
a look at this first one here A changed by one degree then this means A changes 
by one per cent right so what's going to happen here if A changes by one per 
cent is that my value of R is going to change by one per cent so R changes by 
one per cent of twenty-three-thousand ohms and that is going to equal therefore 
two-hundred-and-thirty ohms and what is that equivalent to in temperature well 
twelve-thirty-eight ohms is equivalent to one K so two-hundred-and-thirty ohms 
is equivalent to er two-hundred-and-thirty over twelve-thirty-eight which i got 
to be nought-point-one-eight degrees centigrade okay now you c-, you can you 
see that it's the A changing the R that's the big effect we've also changed D-R 
D-T by one part in a hundred in other words this number here er twelve-thirty-
eight has changed 
by one per cent but that's rather a small change changing twelve-thirty-eight 
by one per cent the main thing is the change in the value of R er giving a 
drift which looks like a change of temperature of point-one-eight degrees 
centigrade and then the last one here let's have a look at this one now R 
changes by one per cent of a four-hundred-and-fifty-four and that to my 
reckoning is four-point-five ohms and we've got thirteen ohms is e-, equivalent 
to one K so four-point-five ohms here is equal to er er four-point-five over 
thirteen and that's actually almost it's a it's a little bit worse isn't it 
it's about a quar-, a third that okay so basically this region here at the 
lower temperature with the higher resistance and the steeper slope it's more 
sensitive and less sensitive at the higher temperatures i think you can see 
here that allegedly if you er you know if you had an ohmeter that read to an 
ohm here then you might think you'd be able to measure this 
temperature that's a resolution problem isn't it suppose i could reme-, measure 
a change of one ohm i might say to myself oh i can measure the temperature to a 
thousandth of a degree K 'cause that's the resolution of the instrument but in 
fact suppose A due to ageing and this is what happens it changes by one per 
cent and the fact i can resolve these ohms to one ohm one part in a thousand 
doesn't really tell me it's just like having a voltmeter and putting a few more 
extra decimal places at the end that doesn't mean to say i might have got a 
more accurate instrument might have more resolution but supposing that A is 
changing as the thing ages from month to month then that's my limit there 
rather than just saying i can measure ohms incredibly accurately it's a 
difference between resolution and accuracy right let's take a little bit of a 
break then now hopefully er having tried these and struggled with them going 
through them is helping you to follow a little bit what's going on