Q & A on ME3242/TM3242
Last updated 26/10/2004 1040hr
8/10/2004
Dear Sir,
Refer to the last slide on the Lucas Method lecture. The statements "a 3/2 spring returned limit
valve otherwise" and "3/2 spring returned ausiliary valve". Won't this 2 statements violate Rule
No.2 of the Lucas Method, whereby the last letter in all groups must appear again somewhere
in the next group?
Thank YOU and regards.
eric
*******************
Dear Eric,
I am not sure what you mean but guessed that you may be confused with the rules that are
applicable for non-overlapping (meaning single) motions with overlapping motions that are
last in a Group. For single motions, according to the “Rule 2” applicable for single motions,
the last letter in all Groups must appear again somewhere in the following Group.
However, when you have overlapping motions that are last in a Group, then only one of these
overlapping letters need to appear again in the following Group.
The following rules for dividing the sequence, for Lucas Method, will be applicable for both
single and for overlapping motions:
Rule 1: No letter must appear more than once in any Group.
Rule 2: At least one of the last letters in the last group of motions in all Groups must appear
again somewhere in the following Group. (Note: If there is only one motion that is last in a
Group, then this rule means that that letter must appear again in the following Group. If there
are overlapping motions in the last group, then at least one of these must appear again in the
following Group.)
Rule 3: At lease one letter in the last letters in all Groups, except for the last Group, must be
plus. The last letters in the last Group can be all minus. (Note: If there is only one motion that
is last in a Group, then this rule means that that letter must be plus, except for the last Group
for which the last letter could be plus or minus.)
I am not sure if the above clears your doubts. If not, please let me know. It will help if your
question is a little more specific and clearer.
Regards,
anpoo
8/10/2004
Dear Prof Poo.
your lecture today was very clear. However would it be possible for you to place the
powerpoint slides of Tutorial 2 Solutions on your webpage, also. The step by step method
presented in the powerpoint format made it very clear to understand, and I think that me and
my friends would benefit greatly from it too.
Thank you very much.
Have a good weekend
P
*******************
Dear P,
Will do.
8/10/2004
Rgds,
anpoo
Dear Sir,
With regards to the second question of tutorial 1, can I simply have one valve with a START
button at the top and a STOP button at the bottom instead of a two push buttons with spring
return? My thinking is that when START is pressed and then released, the valve still remains
that position until STOP is pressed from bottom. Pretty much the same as the double pilot
actuation lines of the auxillary valve of your solution. Thank you for your time.
Regards,
Adrian
*******************
Dear Adrain,
Yes, you can have that arrangement with a two-position, doubly manually-operated 3/2 valve.
In this case, however, the physical mounting of the valve may present some problems in some
circumstances because you need to mount the valve in such a way that you can push the
START pushbutton from one side, and also the STOP pushbutton from the other side. On a
table-top, or a panel, for example, this may not be so convenient.
Regards,
anpoo
8/10/2004
Dear Prof Poo,
I have a few enquiries to ask of you.
For Cascade method lecture notes slide 17, is there a mistake made in the labelling of A- and
A+?
Also, for the lab, I understand that there will be a 5% CA component for each lab. Will the
assessment be in the form of lab reports or on-the-spot assessment during the lab session?
Thank you for your attention.
Best Regards,
BT
*******************
Dear B T,
Thank you for pointing out an error in Slide 17 of the Cascade presentation. The labeling of Aand A+ for the 5/2 control valve as connected is correct. However, the red arrows for the two
A+ were pointed to the wrong side of the control vale, and the same for those for the two A-.
I’ve corrected the slide and have the website pdf file updated accordingly. Thanks again.
There will be a CA component for the labs and the lab reports. The “on-the-spot” assessment
will be on how well the sequencing circuit has been designed and patched up so that it works
according to specification.
Regards,
9/10/2004
anpoo
Dear Sir,
Qn2:
According to the solution that you have given us, the SINGLE pushbutton is spring-returned. I
guess this means that the person actuating the pushbutton will have to press and hold down the
button until the group I actions are over? Isn't it better to make it air reset by connecting the air
reset line to the group III manifold? In this case, we won't have to hold down the pushbutton.
Thank YOU and regards.
eric
*******************
Dear Eric,
No, you are not correct. Air supply to Group I manifold is not through SINGLE at all. The
SINGLE pushbutton needs only to be actuated momentarily and long enough to pass air
pressure through to the + side of the A cylinder control valve to shift this valve. This should
take less then tens of milliseconds. Once this valve is shifted, SINGLE can be released.
10/10/2004
Rgds,
Anpoo
Dear Sir
With regards to the solution given for tutorial set 1 Q2, may I arrange for the input to the
OPERATE and STOPE valves to tap their pressure input from Gp 1 manifold instead?
Regards
CLM
*******************
Dear CLM,
You may, but it will mean that both the OPERATE and STOP pushbuttons will only perform
the functions of switching the system into the different modes when the Group I manifold is
pressurized, and not at any time. By connecting their inputs to air supply, they will function at
any time when they are actuated.
Rgds,
anpoo
Hi Sir,
I would like to further clarify the use of auxiliary valves (AV).
During the lecture break yesterday, I asked you something about the AV. You mentioned that
there could be only one of the AV where the connections are connected to the left position of
the valve while the rest are connected to the right position of the AV. Then, can I assumed that
in such cases, there must always be only one of such AV (connections to left position of valve)
in the system.
Also, in choosing where the connections are connected to the left position of the AV, can we
decide any position of it or must it be one where it is not involved in an overlapping motion or
in the last group or etc.
Thank You
Regards,
Jimmy
*******************************
Dear Jimmy,
Yes, that is correct. Only one, or none, of the AV valves should be connected on the left box of
the valve. How the connections should be done depends upon how you divide the sequence
into groups. At the position "at rest", the sequence should be just before the START
pushbutton. So, which ever Group START belongs to should be powered up at the beginning.
This will usually be either the first Group or the last Group depending upon how the sequence
is divided.
Once the initial Group that is powered up is determined (from the Group in which START
belongs), then the connections must be shown such that this Group is powered up. For
example, if START belongs to Group I, then the connections to the limit valves, or auxiliary
valves, must be shown connected to the appropriate side of the valves to allow air supply to
power up Group I.
In the extra example that I have given, START belongs to Group I. The air supply to power up
Group I results from the last letter in the previous group, i.e. A- in Group IV. Since A- is also a
repeating letter, it has a spring-returned limit valve connected to an AV. This AV is the one
connecting air supply to Group I so the connections to this AV must be on the let box allowing
air supply to power up Group I. All other AVs, which are used to connect air supply to other
Groups, should be shown connected to their right boxes.
I hope the above is clear enough, if not, please raise a question at next week's class and I can
discuss this in class.
Regards,
anpoo
13/10/2004
dear dr. poo
regarding today's slide, how we get those equation for Y1 and Y2 on slide 29?
Y1 = (S1 + y1).R1'
Y1 = (S2 + y2).R2'
thank you
Best regards,
HKS
*******************************
Dear KS,
These are the basic "standard" equations for Reset-Set flip-flops. If I am not wrong, these
should also have been covered in the electromechanical part of your course for all the Cascade
relays used. Each Cascade relay nees an ON, or Set, condition when it will turn on, and an
OFF, or Reset, condition when it turns off. The starting basic equation is
Y = (S + Y).R'
where Y is the output of the flip-flop, S the set input, and R the reset input. You can check that
this equation will give the truth table for the RS fkip-fkop (Pg 15 of Huffman slides). From the
equation we can see that
when S=0, R=0, then Y=(0 + Y).0' = Y
when S=0, R=1, then Y= (0 + Y).1' = (0 + Y).0 = 0
when S=1, R=0, then Y=(1 + Y).0' = (1 + Y).1 = 1
which is what the truth table gives.
Rgds,
anpoo
Dear Sir
With regards to the solution given for tutorial set 1 Q2, may I arrange for the input to the
OPERATE and STOP valves to tap their pressure input from Gp 1 manifold instead?
Regards
C LM
*******************************
Dear LM,
Good thinking! Looks like you have either tried to work out the solutions by yourself or is
thinking of alternative solutions.
If you connect in the way you suggested. the operation of the circuit will not be the same.
In the solution I have given, both OPERATE and STOP will work at any time they are
depressed, meaning they will put the system into the Operate and Stop modes respectively.
In the solution you suggest, the two pushbuttons will work only during the time when Gp 1
Manifold is pressurised. During other portion of the sequence when other group manifolds are
pressurised, depressing these pushbuttons will not have their intended effect, which is to put
the system into the Operate and Stop modes respectively.
Regards,
anpoo
Dear Prof Poo,
I seek your help in clearing some of the doubts that I have about the design of pneumatic
control circuit.
Usually, we take input of a limit valve from its own Group manifold for both Cascade and
Lucas Method. However, I noticed that some of the circuits (in your lecture notes and the
answers to tutorials) have limit valves which are directly powered by air supply. Why is it so?
I’ve attached 1 pneumatic circuit that I’ve copied from your PowerPoint slides for Tutorial 2
question 1 as an example. The limit valves in question are circled in red.
Please advise. Thank you.
Yours sincerely,
P
*******************************
Dear P,
For BOTH the Cascade Method as well as the Lucas Method, the Group manifolds needs to be
supplied with air pressure one at a time and in sequence. At the end of the last cylinder motion
(or multiple cylinder motions if there are overlapping motions that are last in a Group) in a
(current) Group, air pressure needs to be applied to the following Group manifold. At the same
time, the air pressure that is in the current Group needs to be vented to atmosphere. Note that
the approach towards this is different for the two methods.
In the Cascade Method, an air pressure signal is generated (from taken the current Group
Manifold pressure and passing it through the limit valve for the last cylinder stroke in the
current Group) to switch the Cascade Valves. Once the Cascade Valves switch, air pressure is
re-directed from the current Group Manifold to the one following.
For the Lucas Method, supply air pressure is taken directly as input to the limit valve which
correspond to the cylinder stroke that is last in the current Group. The output is used to supply
the following Group manifold. Once the following Group maifold is pressurised, the air
pressure from this Group manifold is used to reset the valve supply air pressure to the previous
(current) Group manifold. This arrangement is shown in Slide 20 in the Lucas presentation.
Please note that for those limit valves associated with cylinder strokes that are NOT LAST in
any Group, their inputs are taken from their own Group manifold and the outputs used to
initiate the subsequent cylinder motions.
Hope the above is clear enough.
Regards,
ANPoo
Dear Sir,
Sorry to trouble you once again. I have a question on Huffman Method.
These are what I understand:
If there is a race in a column, usually we will divert to a 2- step movements.
- If there is a "don't-care" state in the column with a race, we will divert the route from
the unstable state (with the race) to the "don't-care" state.
- If there is unstable state X with a race and in the same column, there is another
unstable state X ( which is not a race), we divert the unstable X ( with race) to the
other unstable X (without race).
Question: What if there are a "don't-care" state & the other unstable X(without race)
in the same column as the unstable X with the race? Do we divert to the "don't-care"
or to the other unstable X?
I was thinking that it doesn't matter which one we divert to. Both are correct and just
that both will give different Boolean equations for Set & Reset.
Kindly explain to me further. =)
Thank you very much for your time & patience.
Warmest Regards,
K
*********************************
Dear K,
You are correct. Either way is correct although the resulting circuits will not be the
same. Not only are the set and reset not the same, the output K-map may also be
affected. In general, diverting to another unstable state which is not a race is better.
One thing to remember, when a race is broken into two, or sometimes more, steps,
each step must not be a race. This will not happen if there are only two flip-flops
involved, i.e. only four rows. If there are three or more flip-flops, meaning having 8 or
more rows, then care must be taken to ensure that each of these steps are not races
in themselves.
Regards,
anpoo
Dear Sir,
Can you pls advise me on the following circuit? It is suppose to have 2
switches - 1 Auto and the other is Single.
Thank you.
Your student,
NT
**************************
Dear NT,
I do not know what the circuit is supposed to solve. Is it meant for the exam question
asking for an AUTO and a SINGLE mode?
Whatever it is, the circuit will not do anything useful and will also not work. Firstly,
there is no input for your AUTO pushbutton valve. If there is supposed to be an air
input for the AUTO valve, the circuit will still not work. Pushing SINGLE will cause the
middle 3/2 valve to shift to the right (i.e. left box in operation). This in turn connects air
supply continuously to the A+ input. If an A- input should come later, there will be a
conflict.
Pushing on AUTO will shift the middle valve to the left. And if there is an air supply at
the input to the AUTO valve, this will continuously connected to the A+ input since the
AUTO valve will stay down. Again a conflict situation will arise if a subsequent Asignal comes.
Regards,
Anpoo
Sir,
In your notes for the Lucas method, you state that the last action in the last
group can either be + or -.
However, when i look through the notes, i realise that if the last group is +,
then the last action of the last group will have to be "touched" by the cylinder
in order for it to power up the group 1 manifold. However, the default state of
the cylinder is at the negative end of all limit valves. How do we power up the
group 1 manifold in this case? Do we draw this + limit valve's dafault starting
state as being touched by the cylinder instead?
Also, if the last action of the last group is a positive, that would mean we need
not actuate a negative previous group. Can this happen?
regards,
L
*********************
Dear L,
Good thinking. The following is an example of a sequence where the last letter in all the
Groups are PLUS.
START A+ B+ | B- A- C+ | C- D+ | DI
II
III
I
Note that the last letter D- is part of the first Group so that last letter of Group III, the last
Group, is a PLUS.
In the initial state, the system will be at rest just before the START pushbutton. Group I power
will be on, and this will be supplied by air supply passing through the d+ limit valve which
would still be actuated (even though the piston is in the MINUS position). Why? Because the
d+ limit valve is air-reset and, although the cylinder D has moved away because of the Dmotion at the beginning of Group I, the system has not arrive at Group II yet. The d+ limit
valve is reset only when Group II power comes on after the B+ motion.
In drawing the circuit for this sequence using the Lucas Method, the d+ limit valve should thus
be shown actuated, meaning the connections should be made to the upper box of the 3/2 limti
valve rather than to the bottom box.
Regards,
ANPoo
24/10/2004
dear dr. poo,
regarding tutorial set 1 solution for question 2, I think the limit valve c2 supply
air should be connected to the group 2 manifold instead of group 1. please
verify my answer. thank you.
Best regards,
HKS
*********************
Dear King Siew,
You absolutely correct! Thanks for pointing this out to me. I'm surprised that
no other has found that mistake yet. It could be that your classmates have not
gotten round to working out the problems in detail yet.
Thanks again.
24/10/2004
Regards,
ANPoo
(Note: Solution on the Web has been corrected accordinly.)
Dear Prof Poo,
could you kindly explain how we get the map for the output (Reference Pages
23 and 24 of the notes).
I know you explained it in the lectures, but when I reviewed the notes, I could
no longer remember. I tried also to clarify with my friends, but they all seem to
have different answers too.
Please help
Best regards
Prem
*********************
Dear Prem,
Since you refer specifically to Pages 23 and 24 of the notes, I shall assume that
you know how to fill in the output map for cells containing stable states, i.e.
page 22.
In considering a cell containing an unstable state, you need first to note that a
cell containing an unstable state is just a transition cell which is passed through
by the system in going from a stable state to another stable state. Take for
example the unstable state 3 in cell(1,3), meaning row 1 column 3. The system
gets to this cell from a stable state that is in the same row, meaning either
stable state 1 or stable state 2. From this unstable state 3, the system will
proceed to stable state 3. (Note: from an unstable state n, the system will
always need to proceed to stable state n). From just the merged flow table,
you will not be able to tell whether it is from Stable 1 or Stable 2 state that is
going through the unstable state 3 in cell(1,3). You can only get this
information from the Primitive Flow Table. From the Primitive Flow Table,
you note that from Stable 1, the system does not go to Stable 3 state because
there is no unstable 3 state in the same row as Stable 1. BUt from Stable 2
state, the system does go to Stable 3. Therefore, the cell (1,3) is a transition cell
for the system to go from Stable State 2 in cell(1,2) to Stable State 3 in
cell(2,3). The output for the "from" Stable State 2 in cell(1,2) is 1 and that for
the "to" Stable State 3 in cell(2,3) is 0. Since the output changes from 1 to 0,
the output state for the unstable state 3 in cell(1,3) can be a "don't care".
You do the same for each of the cells containing the unstable states 5, 7 and 1.
As the state of the output changes in all these cases in going from one stable
state to the next, all these cells get a "don't care".
I hope that above helps you to understand the process. If things are still not
clear, then I think you will need to arrange for us to meet so that I can try to
explain in person on this.
25/10/2004
Regards,
ANPoo
Dear Sir,
I have some queries regarding the last slide of your "Sequential Systems with
Random Inputs" (identical to Qn 2 of Tutorial set 3) which you have gone
through with us on last Friday.
Refering to the last slide on your lecture notes, you have drawn an arrow from
state 13 back to state 6. Why is this arrow not drawn from state 13 back to state
5 since the next input should be on its way passing through X1X2. i.e 00/0 and
not alreadi activated sensor X2 i.e 01/0 ?
The arrow drawn back to state 6 means that the next "part" immediately
touches X2 , i.e 01/0 but from earlier sequence DOWN DOWN, you have
drawn arrows from state 10 back to state 1 & state 12 back to state 3 which
means X1X2 sensor has not yet been activated.
Thanks for answering my query.
regards,
wl
*********************
Dear WL,
Good thoughts and questions. Seems to indicate that you have been pondering
over the matter and that is good. :)
The first thing to note is that the system is completely driven by the inputs X1
and X2, meaning that these are the only two bits of information that the system
has in performing its operations. The other thing to note is that from one stable
state, you change to another stable state whenever one, and only one, of the
inputs changes.
You will note that in State 13, the inputs X1X2 are 00. In State 5, the states of
these two inputs are the same at 00. The system first gets to State 13 from State
6 when the inputs changed from 01 to 00. This tells the system that a DOWN
part has passed through and it thus make T=1 to activate a pistion to push the
part away. Once this is done, there is no way by which the system can get out
of State 13 if the inputs do not change (unless of course we use a timer which
cause T to go back to 0 after a certain time) In a combinatorial logic circuit, as
long as the inputs do not change, there is no way the outputs can change. The
system can thus change, and needs to change, to another stable when the inputs
change, and that will only be from 00 to 01. Note that in State 5, the inputs
must be 00 and not 01. State 6 has the inputs as 01.
I hope the above is clear enough. If it is not, please let me know and I will try
to explain this further in class so that others, who will likely have the same
questions, will also benefit.
25/10/2004
Regards,
ANPoo
Dear Prof Poo,
1) In deriving the system equations say, for the flip-flops where
for e.g
F1 = (S1 + F1).R1'
= (YF1' + F2).(YF1 + XY'F1')'
= (YF1'+ F2).(Y' + F1').(X' + Y + F1) OR is it = (YF1 + F1).(Y' + F1'.X' +
Y + F1)
Is the 1st expression correct or is it the 2nd one ?
2) In your solution for Huffman Tutorial Qn 2, your assigned states for F1F2 in
merged flow table is 01, 00, 10 , 11(reading downwards). And you go on to
derive the boolean expressions for S1 R1 S2 R2. However, the boolean
expressions i got was different(& hence my ladder diagram) from yours if my
assigned states for F1F2 in merged flow table is 00, 01, 11 , 10(reading
downwards). I would like to clarify if this is normal.
Thank you for answering my query.
regards,
wl
*********************
Dear W L,
1) The first expression is correct, although F2 in (YF1' + F2) should be F1
instead. DeMorgan's Theorem states
(a.b.c)' = (a' + b' + c')
(a + b + c)' = a'.b'.c'
2) In assigning the states of the flip-flops, there are many choices and all are
correct even though the final K-maps and ladder diagram may not be the same.
Regards,
anpoo