Q & A on ME3242/TM3242
From 3rd November. Last updated 10/11/2004 2200hr
3-1
Dear C,
It's good that you are trying the tutorials. :)
I'll discuss with the class today about setting aside a time for "clinic" sessions.
You can also try to see me after lectures. We should finish with the lectures
soon and it is likely I will end up earlier than the two hours allocated so that
there will be some time for me to answer questions from individual students.
For your question on getting different final answers, it is perfectly okay. In
fact, there will be many different, and completely correct solutions, depending
upon how the state of the flip-flops are assigned, and also, once they are
assigned, how the alternative ways you take care of the remaining races.
Regards,
anpoo
-----Original Message----Sent: Wednesday, November 03, 2004 12:25 AM
To: Poo Aun Neow
Subject: ME3242 Consultation time
Greetings Sir,
I'm currently one of your student from ME 3242 -Industrial Automation. I have
encountered some problems with Huff man method esp the part on mapping
the output and would like to seek consultation to clear my doubts.I would like
to make an appointment with you for clarifications.May i know which day and
time will be convenient for you? Thanks a million.
Sir, I have done the tutorial and have different merged table as your suggested
tutorial question, is it correct to have different S1, R1 and S2 and R2 at the
end ?
Thank you once again.
Have a Nice day.
Regards
C
3-2
Dear P,
My apologies. Here's the attachment, which has also been included in the Q&A
that I have placed on the web for the course. Just a note here: in going from a
stable state to another and passing through two transition states, if the output is
0 in both stable states, then all the intermediate transition cells should also be
zero. The same if both stable states have 1's in which case, all the intermediate
transition cells should have all 1's. If the output changes, the acceptable
situation is when there is only one transition in the change, either from 0 to 1
or from 1 to 0. Thus, for the former, the acceptable transition would be 0001,
0011 or 0111 but not 0101. For the latter going from 1 to 0, acceptable
transitions are 1110, 1100, or 1000 and the unacceptable sequence is 1010.
Therefore, initially when the output changes, the two transition cells should be
placed with conditional don't cares. In the final stage when the K-map is being
covered and the Boolean expression derived, we need to check these
conditional don' care cells to make sure that we don't end up with the
unacceptable sequence. If this should happen, we need then to cover the K-map
in a different way.
For your latest question, I think you are referring to slide 32 of the notes. Since
we decided for the system to stop in cell(4,1) with the unstable1 state, we need
START to be pushed before it moves on to cell(1,1) Stable 1. To move from
cell(4,1) to cell(1,1) requires movement from row 4 to row 1 through a change
in the states of the two flip-flops. Only one flip-flop should change. In slide 32,
from the way the flip-flops have been assigned, Y2 needs to change from 1 to
zero with Y1 remaining unchanged at 0. Thus we need to reset Y2 from 1 to 0.
This means that when we are in cell(4,1), we will be issuing an R2=1 signal to
reset Y2 in order to move from row 4 to row 1 in the merged flow table. To
stop this movement and allow it to occur ONLY when START is actuated, we
AND R2 with START.
To make things a little clearer, suppose that we have assigned the flip-slops
such that row 1 has Y1Y2 as 11 and in row 4 as 10. Then to move from row 4
to row 1, Y2 needs to change from 0 to 1, meaning that we need to issue the S2
signal in cell(4,1). In this case we would then AND START wi th S2.
Hope above is clear enough. :)
regards,
anpoo
-----Original Message----Sent: Wednesday, November 03, 2004 2:56 PM
To: Poo Aun Neow
Subject: RE: ME3242: Qn on Tutorial Set 4 qn 1 and Random Inputs lecture
Dear Prof Poo,
thank you for your prompt reply. However, I did not see any attachment to your email
which you referred to in your reply.
In your lecture notes for sequential systems with random inputs, there is a section on
incorporating START Pushbutton for single cycle fixed sequences. For the given
example in lecture note pages 30, START is ANDed with S1 signal. However, when
the flip-flops assignment were changed, the START button is ANDed with R2 signal. I
do not understand how we determine where we should AND the START button to.
Please advise. Thank you.
Regards,
P
3-3
Dear T S,
PLC is part of the course and could also be examined.
rgds,
anpoo
-----Original Message----Sent: Wednesday, November 03, 2004 8:52 PM
To: Poo Aun Neow
Subject: ME3242 - PLC
Dear Sir,
Would like to ask whether PLC is examinable or not.
I realised it is not tested in the previous years' exam papers.
3-4
Thanks and Best Regards
TS
Dear Eddy,
Very good question and very detailed consideration on your part.
Suppose we do as you did. In Stable State9 when you had 11/1 and a T=1
signal is generated to push the part away, the states of X1X2 will change to 00
once the part is pushed off. There will be some uncertainty here whether X1X2
goes from 11 through 10 to 00, or from 11 through 01 to 00. This depends
upon how the part is being pushed off or even how individual parts falls off.
Still this can be managed.
The point is that the system cannot remain in Stable State 9, which requires
X1X2 to be 11, but needs to move to some other stable state (remember that
from any stable state, only one input can change at any one time and when any
input changes, the system will go to another stable state) with X1X2=00
through either 01 or 10. It is only from input state 00 that it can go to state 01
when a new part arrives, whether standing up or lying down. Thus your
solution with the system going from Stable State 9 back to Stable State 2 will
not work.
Hope the above is clear enough.
Regards,
anpoo
-----Original Message----Sent:
Wednesday, November 03, 2004 8:53 PM
To:
Poo Aun Neow
Subject:
ME3242 Query
Dear Prof Poo,
Refering to your solution of Problem 2 of tutorial set 5 ( Random Systems &
PLC ).
In the flow diagram( Fig 2 ), specifically from stable state 2 to stable state 9.
Your stable state 9 = (11/0) then from stable state 9 to stable state 10.
However I got stable state 9 = (11/1) then from stable state 9 go back to stable
state 2.
Essentially, once it senses a block standing up(X1 = 1), T=1 and the part is push
off. Then when it senses the next block lying down(X2 = 1), it goes back to stable
state 2.
Why is there a need to go through stable state 10 to stable state 0 then back to
stable state 2?
Hope my description is clear. Thanks in advance.
Regards,
Eddy
3-5
Dear K Y,
1) I've placed another example of a sequence with repeating and overlapping
motion on the web to help students. This has solutions and explanations. Please
go through this first and see if that example helps to clear up things. If you still
have question after that, please do not hesitate to get in touch with me again.
2) Interesting point and good question. This was raised by another student right
after the lecture. Much depends upon the physical arrangement. The
arrangement for the solution provided could be as shown below:
In the figures above, the spot with the larger circle containing an "X" would be
the spot where the piston used for pushing off the parts are located. The
first figure shows the point at which an UP part will be pushed off in stable
states 10 and 12 for the UP part when the x1x2 signal changes to 10 from 11.
The second figure is the same for a DOWN part pushed off in stable state 13
when X1X2 changes to 00 from 10.
If we allow T=1 in stable states 9 and 11, then the piston will push the parts off
earlier for an UP part, with the part located approximately in the position
shown in dotted lines in the first figure. Pushing the part off at this point
should also be okay. However, at the start of the pushing action, sensors X1X2
will be in state 11 and after the part is completely pushed off, the state of
X1X2 will be 00. There will thus be a race, meaning that X1X2 will either
change from 11 through 10 to 00, or from 11 through 01 to 00, depending on
how the part falls off. If we were to use 11/1 for stable state 9, we will then
need a flow diagram such as that shown below. The same situation for stable
state 11.
Hope the above is clear enough.
Regards,
anpoo
-----Original Message----Sent: Thursday, November 04, 2004 2:07 AM
To: Poo Aun Neow
Subject: ME3242 Industrial Automation Questions
Dear sir,
I have some questions regarding the tutorial.
1) For additional question exercises, for overlapping motion of B- & C+(Group
2). Since both are air-reset valves, does it means that we need to connect to the
Group 4 manifold one each tru the axuilary valve(B-) and the air reset
valve(C+) ?
2) For the Sequential Systems with random inputs tutorial Q2, I still cannot
undertsnad why at state(9 & 11) 11/0 the T is still 0. And the we must wait till
state(10& 12) 10/1 before the T=1. Can you explain on this?
3-6
Dear Eric,
Yes, the choice of the states of the relays assigned to either the columns
representing the inputs or the rows of the merged flow table will affect all the
K-maps and the final designed circuit. However, all can be equally correct.
Regards,
anpoo
-----Original Message----Sent: Thursday, November 04, 2004 10:52 AM
To: Poo Aun Neow
Subject: ME3242 - Huffman's Method
Dear Sir,
Am I right to say that the combination of the relays chosen (e.g., 00 01 11 10
as compared to 01 00 10 11) will affect the answer to a question?
Regards
3-7
Dear T G,
I would suggest that you read the two Q&A postings I placed on the IVLE site.
1) Please refer to the query from H K S on 27/10/2004.
2) Yes, it is possible. A sample program and its corresponding ladder diagram
is given below:
X1
X2
X4
Y1
X5
Y3
X6
Y4
X3
Y2
Good consideration.
Regards,
LD X1
MPS
AND X2
MPS
AND X4
OUT Y1
MPP
AND X5
OUT Y3
MRD
AND X6
OUT Y4
MPP
AND X3
OUT Y2
END
anpoo
-----Original Message----Sent:
Thursday, November 04, 2004 3:13 PM
To:
Poo Aun Neow
Subject:
ME3242
Hi Prof Poo,
Have a few doubts that I hope you can claify for me for PLC and huffman method.
1. Is there isn't any so called "fixed" solution for the boolean expression dervived
through the huffman method? I mean my will answers differ from yours when I
assign different flip flops identity to the merged paths. Is this ok?
2. For the mps, mrd and mpp functions, is it possible to have this kind of program
sequence?
MPS
AND
MPS
AND
OUT
MPP
AND
OUT
MRD
AND
OUT
MPP
AND
OUT
I know its little hard to understand the sequence, but main question I guess is that
is it possible to call up MPS consecutively twice at the same without removing the
first result?
Thank you for your time sir!!
Regards,
AL
3-8
Dear Bernard,
I had thought that you could be referring to Tut5. However, your statement
"On the output table, extreme right column, the third and 4 row showed 0 and
don't care respectively." confused me. Are you referring to your own solution
or the solution provided? In the solution provided, the output table is reproduced below. Extreme right column 3rd and 4th row both have 0's. ???
00
X1X2
01
11
10
Y1Y2Y3
0
0
0
0
0
-
0
0
010
000
100
0
0
-
1
0
-
1
0
-
0
-
0
0
-
110
011
001
101
111
In the solution provided, the output state in cell(3,4) [cell(row,col)] and
cell(4,4) are determined in this way. Both contains unstable2 but, as you
rightly noted, because of taking care of a race, from cell(3,4) the movement do
not go directly to cell(1,4) [stable2 where it should end up in] but is diverted
first to cell(4,4)
For cell(3,4), movement through this cell comes from stable5 in cell(3,1),
through cell(3,4), then to cell(4,4) and finally to stable2 in cell(1,4). Since
stable5 has T=0 and stable2 has T=0, going from 0 to 0 means that ALL
intermediate transition cells needs to be 0. Thus, the movement from stable5 to
stable2 requires that BOTH cell(3,4) and cell(4,4) have output T=0.
Cell(4,4) is also passed through when stable1 in cell(4,1) moves to stable2 in
cell(1,4). The output T changes from 1 in stable1 to 0 in stable2. Thus, this
movement allows cell(4,4) to have T=don't care.
Cell(4,4) thus has two requirements, one requiring it to be 0 and the other a
don't care (which means either 1 or 0 is okay). Therefore cell(4,4) needs to
have T=0. We do not have the situation of a sequence of 0010 as you indicated
in your first email. The final covering of 1's can only cover cells (4,1), (4,2),
(8,1) and (8,2) giving T as shown above.
Hope the above is clear enough.
Regards,
anpoo
-----Original Message----Sent: Friday, November 05, 2004 1:39 AM
To: Poo Aun Neow
Subject: RE: ME3242: Industrial Automation
Hi Prof Poo,
my first question is regarding the solution to problem 1 for the last set of
tutorial on random systems (Tut 5). Sorry for any confusion caused. Thank you
once again for answering my queries.
-----Original Message----From: Poo Aun Neow
Sent: Fri 11/5/2004 12:22 AM
Subject: RE: ME3242: Industrial Automation
Hi CKS,
Don't know what you are referring to. Are you referring to your own solution
or to the solution provided? Is this for the last set of tutorial on random
systems?
Concerning your other question, try it out yourself. Have an even number of
stable sets that are connected by an even number (say 2, 4, 6 or 8) of arrows in
a closed loop and using three flip-flops try to assign the states such that there is
no race for all the arrows around the loop. You will find that this will be
possible in all cases. Then try to do the same with an odd number (3, 5, or 7) of
stable states connected in a closed loop by an odd number of arrows. You will
find that it will not be possible to assign states so that there is no race.
If the loop is open, then there is no constraint and you should be able to assign
states such that all the arrows connected the stable states are not races. Just
make sure that there is only one flip-flop changing when moving from one
stable state to the next in the open loop.
Rgds,
anpoo
-----Original Message----Sent: Thursday, November 04, 2004 11:22 PM
To: Poo Aun Neow
Subject: ME3242: Industrial Automation
Hi Sir,
wonder if have got this right. Please refer to the PLC tut Prob 1. On the
output table, extreme right column, the third and 4 row showed 0 and don't
care respectively. 0 is being assigned as it is transition from 0 to 0. don;t care is
assigned as it is a transition from 1 to 0. But in the third row, the unstable 2 is
actually moving to the unstable 2 in the fourth row before it goes to the stable 2
in row 1. The Boolean expression will mean the whole row of 1's (row 4) will
be covered. But doesn't that mean for stable 5( in row 3) to move to stable 2,
the output movement will be 0010 instead? Hasn't the output changed twice in
this case?
Can you also please explain to me on the rationale behind why even number of
transitions will lead to no race? I do not quite understand that part. Must it be a
closed loop? What if the loop is open?
Thank you very much for answering my queries.
3-9
Regards,
CKS
Dear Y H,
The situation you refer should not occur at all. When considering the K-map
for the output, you should always consider the complete movement from one
stable state to another. Thus you need to determine the state of the output in the
originating stable state and that in the destination stable state. These states
must be either 0 or 1.
Regards,
anpoo
-----Original Message----Sent: Sunday, November 07, 2004 3:25 PM
To: Poo Aun Neow
Subject: RE: ME3242 Industrial Automation - Tutorial Set 4 Huffman Method
Dear Sir,
I have another question to ask regarding output maps. (question 1 output Z2
map)
From cell
-
Intermediate cell
?
To cell
1
Should '?' be a '1' or a '-'?
My thinking is if it is a '-' it might not be correct since the 'from' cell could be a
1 and the 'intermediate' cell could be a 0 and will become 1->0->1, which we
do not want.
Thank you.
Regards,
YH
3-10
Dear Francis,
Please see Q&A III (3-8) and also others. Many students have
similar problems filling out the output maps.
The cell(2,1) you refer to is passed through when moving from
Stable5 to Stable1 (can be don't care) and also from Stable4 to
Stable1 (needs to be 0). Cell(2,1) needs to be 0 to satisfy
both requirements.
Rgds,
anpoo
-----Original Message----Sent: Sunday, November 07, 2004 3:42 PM
To: Poo Aun Neow
Subject: ME3242 Huffman's Method Tutorial Question 1
Dear Prof Poo,
I have a question regarding the solution to the Z1 output map
for question 1. In row 2 column 1 of the Z1 output map, the
value assigned is 0. Shouldn't it be a don't care instead? The
value assigned there should be for stable 5 moving to stable 1
which is from output 1 to output 0. Since there is a change in
output value, the output map for unstable 1 should be assigned
don't care isn't it? Thank you in advance for answeing my query
best regards,
Francis
3-11
Dear B T,
The circuit as presented ought to work too if cylinder c also has repeating
motions. The main point to note is that there should be a "unique" signal
generated at the end of Group II to shift an auxiliary valve. Air supply is then
channeled through this valve to power up Group III. This valve should be reset
by Group IV. This will properly power up Group III manifold at the end of
Group II motion and keep it on until Group IV is powered.
With some deeper thoughts on this, you could make use of the following
alternative connections instead to power up Group III after the A- and C+
motions in Group II and have this on until Group IV is powered up:
The above circuit will also work if C+ is a repeated motion. You will note that
the signal S shown above is generated only when the system is in Group II and
BOTH A- and C+ have completed their motion. This signal will not occur
anywhere else, even if there is a another situation of an A- and C+ overlapping
motion because S can only be generated when the system is in Group II.
Regards,
anpoo
-----Original Message----Sent: Monday, November 08, 2004 8:07 AM
To: Poo Aun Neow
Subject: ME3242 enquiry
Dear Prof Poo,
I have an enquiry to ask of you regarding the extra example you gave us on
overlapping and repeated motion.
For Lucas method, in Group II, if C+ is also a repeated motion, then limit
valve c2 will be spring returned, with input channeled from AV2 (instead of
from Group II so that A- and C+ can be in series?) and output to auxiliary
valve AC, whose air reset is connected to Group IV, and output to Group III?
Thank you for your attention.
3-12
Best Regards,
BT
Dear MS,
1. Yes, there could certainly be more than one correct answer. Normally the
problem specification will not be complete and will only specify the "normal"
operational requirements. Your logic flow diagram will need to satisfy all the
specified operational requirements. Beyond this, you need to use your common
sense and judgment to take care of situations that may arise but which are not
specified.
2. Let me try to explain. Each circle in the flow diagram represents a stable
state of the system. For a system with memory, as all sequential systems
normally will have, states with the same inputs and outputs need not be the
same state of the system. So, even though states 4 and 7 happen to have the
same input/output pattern, they need not be the same state. You will note from
the lecture that states 1 through 6 were worked out based on the description of
the system requirements. States 7 and 8 were added to take care of input
changes that could arise but which were not specifically described in the
system requirements. State 4 represents the state of the system following the
normal procedure: after a fault has occurred (State 2), the operator clears the
fault (State 3) and presses on the reset button the first time but before releasing
it. When he/she releases it, the system goes into state 5. State 8 is the state the
system gets into when a fault occurs and the operator presses on the reset
button before he/she clears the fault. When the system is in State 8, two things
can happen: one being the operator releases the reset button and the other
(unlikely but still possible) is the fault being cleared while his finger is still
pressing on the reset button. In the latter case, we could either allow the system
to go into State 4, or into another new state 7. The first alternative with State 4
will mean that the operator's first pressing of the reset button before the fault is
cleared is taken as the first of the two actuation of the button that is required to
restart the machine. This is because when the system is in State 4, when the
operator releases the button the system will go into State 5 and upon pressing
the button one more time, the system will go into State 6 where the machine
comes on. By making the system move to State 7 instead means that the first
press of the button is discounted as the first press to restart the machine. The
operator then needs to release the button and press two more times before the
machine restarts. There is no clear right or wrong in this instance. It is a matter
of the judgment of the designer which way he prefers.
Hope the above clears things up.
Regards,
anpoo
-----Original Message----Sent:
Tuesday, November 09, 2004 10:47 AM
To:
Poo Aun Neow
Subject:
Queeries on random input:
Dear Prof,
1.When drawing the logic flow diagram, could there be more than one answer that
is correct?
2. Referring to example in the notes for the random unput s (slides 35).
Some states were actually repeated,like 4 and 7 which seem a bit
unecessary, why cant u just use 7 and assign more arrows from it to state 5 and
do away with state 4? Actually, I still don’t really understand the logicof the whole
flow.. When it means press twice in a row to be restarted.. Why does the state
goes 00/0 in between (state 4>5>6)?
Sorrry if the questions seem muddled.. But I appreciate your reply.
Thank you so much
MS
3-13
Dear Y H,
You're right in that when the output goes from 0 to 1 or from 1 to 0, the two
intermediate cells can both be conditional don't cares. Please note that cell(1,3)
is also passed through going from Stable5 to Stable3. For Z1, the output goes
from 0 to 0 requiring it to be 0 while for Z2, it goes from o to 1 allowing it to
be an unconditional don't care. The resulting value for cell(1,3) needs to satisfy
both the requirements going from Stable2 to Stable3 as well as going from
Satble5 to Stable3.
Hope the above is clear.
Regards,
anpoo
-----Original Message----Sent: Tuesday, November 09, 2004 12:24 PM
To: Poo Aun Neow
Subject: ME3242 Industrial Automation - Lecture Notes on System with Random
Inputs Page 26
Dear Sir,
I have a question with regards to Lecture Notes on System with Random Inputs
Page 26.
Stable 2 goes to Stable 3 passing through 2 intermediate cells, and so the
output goes from 1 to 0 for Z1 and 0 to 1 for Z2.
So should the 2 intermediate cells for both outputs be conditional don't cares?
The Lecture Notes shows that the intermediate cells for Z1 is - and 0 and i
don't understand why.
Thank you.
3-14
Regards,
YH
Dear K S,
Please include the course code in the subject title in future. Otherwise this
could be lost, and worse thrown into the recycle bin, among the many spam
mails.
You should be considering the transition from unstable1 to stable1, but as you
say, for considering the flip-clops, only the row matters.
However, it is really in the cell unstable1 that we are making the system stop at
the end of the sequence and to be restarted by actuation of the START button.
This cell originally is allowed to have a don't care because it was supposed to
be a transition cell, meaning that the system is not expected to remain there for
too long, in the order of milliseconds. Because this is so short and the output
changes, it does not matter if the output changes some milliseconds earlier or
later. However, since we have decided to make the system stop in this cell, the
system will remain in it for a long period of time until START is actuated.
Because the system needs to be at rest in this state before START is pushed,
the output for the Solenoid A needs to be 0. Otherwise the piston would have
moved off even if START is not pressed. It is only when START is pressed
and the system moves to Stable1 that we turn Sol Q on.
Regards,
anpoo
-----Original Message----Sent: Tuesday, November 09, 2004 8:48 PM
To: Poo Aun Neow
Subject:
dear dr. poo,
regarding slide 4-28 of huffman method, when we want to incorporate a start
button, we always look at the last stage to the first stage. so we look at the last
row to first row on the y1 and y2. there is a change in y1 ie either we put start
button in s1 or r1. and we can see that the y1 change from 0 to 1 so a set signal
is required, so we put start button in s1. but now i dont understand whether we
look at the last stage to first stage, we look at stable 8 to stable 1 or we look at
unstable 1 to stable 1? because it happen that unstable 1 and stable 8 are both
in same row. and i still dont understand why we need to assign a 0 to the
unstable 1 for the A+ table. thank you.
3-15
hks
Dear K F,
Yes, the result will be the same. The thing to note is that only when both Aand B- motions are complete will the two valves, a- and AV4 be shifted
downwards so that the upper squares come into operation. We need to take air
supply and pass this through BOTH valves to power up Group I manifold. This
is an AND operation and involves passing the air supply through the two
valves in series. It does not matter which valve comes first.
Hope the above is clear enough.
Regards,
anpoo
-----Original Message----Sent: Thursday, November 11, 2004 1:19 AM
To: Poo Aun Neow
Subject: ME3242
Dear Prof Poo,
I have a query regarding your recently uploaded solution for "Two Additional
Problems". For the first question on the Lucas Method, (Figure P1(a)), I would
like to clarify how come limit valve a- has it's own independent input. Would it
be the same if I changed the input to a- limit valve to come from auxillary
valve AV4 and the corresponding output of a- limit valve to power up group
manifold 1? (Assuming AV4 has it's own supply source). Thank you so much
for helping me clarify my doubt.
3-16
Regards
KKF
Dear B T,
Thank you for spotting the errors and informing me about these. Of course
you're right. I'll make the corrections and have the posting updated.
About your question, you can do as you stated. It'll be the same thing. Please
check also query 3-15 in Q&A III.
Regards,
anpoo
-----Original Message----Sent: Thursday, November 11, 2004 1:00 PM
To: Poo Aun Neow
Subject: ME 3242 Enquiry
Dear Prof Poo,
I have several enquiries on the additional problems after comparing the posted
solution with what I have done.
I think there is some typo in the explanation.
Pg 1:
First Para: ‘Solution (1)………..limit valve (LV) a- and AV1.’ It should be
AV4?
2nd Para: For the overlapping………...limit valve b1- with AV3…) It should
be AV2?
Also, referring to Fig P1(a), regarding connection to limit valve a- and AV4,
Instead of connecting output of limit valve a- to AV4, can I connect air supply
to AV4, then output of AV4 to input of limit valve a-, and output of limit valve
a- to Group I. The reset of limit valve a- remains (from Group II).
Thank you for your attention.
Best Regards,
BT