Single transfer lag in an open-loop system

[Keeeen]

Homework Statement

FIGURE 3 shows an open loop system containing a distance velocity lag
and a single transfer lag.

Figure 3 shows -

Xi Xo
---->----[Distance - velocity lag of 4.0s]---->----[single transfer lag if t=5s]---->----

If the system input Xi is subjected to a step disturbance from 2 units to 12 units, plot the response of Xo on a base of time. Determine graphically, and verify mathematically, the time taken for the output to change by 4 units.I know the guidelines but I am at a lost for where to even start with this. I don't know what it means by 'units' and this is not discussed anywhere in the learning materials I have been given. The learning materials also only briefly cover single transfer lags and I can't find any information about 'single' transfer lags on google. Only first-order lag, linear lag and exponential lags. Could it be that it has been worded differently and actually means something else?

If anyone could just give me guidance on how to even start this or any information that might be helpful is appreciated, Thanks!

Homework Equations

output change = input change (1-e^-t/T)

The Attempt at a Solution

 

[FactChecker]

"single transfer lag" is not a term I am familiar with but I see it used in a post that treats it like a pure time delay (see https://www.physicsforums.com/threads/mathematical-relationship-of-a-thermocouple.878367/). I have used other terms like "transport delay" or "transport latency" for that. Does that match the small amount of information that you have in your learning material? If so, the transform is an exponential function, as shown in the link.

 

[Keeeen]

"single transfer lag" is not a term I am familiar with but I see it used in a post that treats it like a pure time delay (see https://www.physicsforums.com/threads/mathematical-relationship-of-a-thermocouple.878367/). I have used other terms like "transport delay" or "transport latency" for that. Does that match the small amount of information that you have in your learning material? If so, the transform is an exponential function, as shown in the link.

That thread is a question that is actually on this assignment. It's still not much help to this question I'm on. I'm beginning to think that 'single' transfer lag may actually be 'first-order' transfer lag.

 

[FactChecker]

Without knowing what your learning material says about "single transfer lag", it is hard to guess. But without that information, a first-order lag is a good guess.

 

[Keeeen]

That thread is a question that is actually on this assignment. It's still not much help to this question I'm on. I'm beginning to think that 'single' transfer lag may actually be 'first-order' transfer lag.

Do you know what the question means when they talk about ‘units’?

 

[FactChecker]

Because Laplace transformations are linear, whatever the units of the input are (feet, inches, etc), the output is naturally in the same units. So they only need to tell you that the input changes from 2 to 12 without specifying what units they are talking about. You can describe the output the same way.

 

[James FC]

I’m after a little guidance regarding this question: Am I right in thinking that the output will not change with regards to distance velocity lag of 4 seconds, it’s merely delayed?However, once the transfer lag starts (time constant 5 seconds) the output will start moving (exponential curve) towards 12 units of output.So…. the step disturbance is 12-2=10 units.Time constant is 5 secondsSo I just want 4 units (40%) of 5 seconds and to add this onto the 4 seconds of dist/velocity lag?4+2= 6 seconds for the output to change by 4 units??

 

[FactChecker]

"distance-velocity lag" and " single transfer lag" are terminology that I am not familiar with. Also, Figure 4 is not shown, so I don't know if there is some hint in the symbols used in the figure. I don't want to guess.

 

[David J]

I think I am doing the same module. Figure 3 is attached. I have a bit of an idea where I am going with this, just trying to understand it at the moment

 

[James FC]

David J - I got the answer correct when I submitted my work. 6 seconds.

 

[David J]

Hello James, I submitted my TMA last week with an answer of 10.2 seconds so I think I may be slightly out

 

[James FC]

you know anyone whose running with this one? As per usual the course notes are nothing like the TMA Q's:

TMA2. Q1)
A 5 to 20 bar reverse acting proportional pressure controller has an output of 4 to 20 mA. The set point is 11 bar. Determine:

A) the measured value pressure which gives an output of 15 mA when the proportional band setting of the controller is 40%

B)the proportional band setting which will give an output of 8 mA when the measured value is 14 bar and the desired value is 11 bar.

 

[FactChecker]

David J - I got the answer correct when I submitted my work. 6 seconds.

Just curious. What did "single transfer lag" mean? Was it a first order lag ($\frac 1 {\tau S + 1}$), a pure delay ($e^{-\tau S}$), or something else?

 

[David J]

I think it was a first order lag. Our notes are not good to say the least. This is an abstract taken from the notes.

With respect to the distance-velocity lag, we said that it took time for the temperature change front to reach point B. Once the temperature front is at
point B, there is usually another delay whilst the temperature of the sensor actually changes to match that of the water temperature. This lag is an
example of a transfer lag and these normally occur whenever energy has to be transferred as part of the measurement process. They are sometimes called ‘first order lags’ because they can generally be described mathematically by first order differential equations

 

[David J]

I have just started looking at TMA2 Q2 now, just reading through the notes and seeing if I can work out from the notes how to approach it but its not easy as the notes are not very informative

 

[JE93]

I have just started looking at TMA2 Q2 now, just reading through the notes and seeing if I can work out from the notes how to approach it but its not easy as the notes are not very informative

you know anyone whose running with this one? As per usual the course notes are nothing like the TMA Q's:

Sorry to drag up an old thread...
I have just joined this forum, stumbled across it whilst trying to find help with assignments.
I think we may be on the same course through open learning?
I have to agree the notes given are a little bare in relation to what the assignment is asking for.
Did you get anywhere with this question?
Thanks

 

[David J]

Hello JE93

Which question are you looking at? Is it Q1 or 2 from the TMA?

 

[JE93]

Hello JE93

Which question are you looking at? Is it Q1 or 2 from the TMA?

Sorry should have been a bit more clear. Looking for help with the question on the original post, for question 4 on TMA 1 .
Sorry again for late replying, juggling this as well as work, only seem to get round to doing anything whilst working nights.
Thanks

 

[David J]

If you look at lesson CSA-1-4 on page 10 there is a bit about step and ramp changes. On that page there is an equation

output change=input change*(1-e to the power of -t/T) ( I don't know how to write that equation in latex form)

Output change is 4 units

input change is 10 units because the disturbance started at 2 and stopped at 12 so 12-2 = 10 units

so 4 = 10(e to the power of -t/T) then re arrange for -t

I know one of the posts in this thread got a correct answer when he submitted 6 seconds, I also got a correct answer for 10.2 seconds.

I tried to use the info given in the lessons although its not very helpful but the equation is the correct one to use.

Give this a try and let me know how you get on.

I will show you how I did the graph later on

 

[JE93]

If you look at lesson CSA-1-4 on page 10 there is a bit about step and ramp changes. On that page there is an equation

output change=input change*(1-e to the power of -t/T) ( I don't know how to write that equation in latex form)

Output change is 4 units

input change is 10 units because the disturbance started at 2 and stopped at 12 so 12-2 = 10 units

so 4 = 10(e to the power of -t/T) then re arrange for -t

I know one of the posts in this thread got a correct answer when he submitted 6 seconds, I also got a correct answer for 10.2 seconds.

I tried to use the info given in the lessons although its not very helpful but the equation is the correct one to use.

Give this a try and let me know how you get on.

I will show you how I did the graph later on

Thanks, I have been looking at this equation and trying to get it to work but seem to be miles off.

Output Change = Input Change (1-e^-t/T)
4 = 10 (1-e^-t/5)
Re-arranged: t = -T ln (Output Change / Input Change)
t = -5 ln(4/10)
t = 2.55
I must be going wrong somewhere...
T (time constant) = 5 as per the assignment question?

Thanks

 

[David J]

Not wrong

t is the elapsed time which is 2.55 seconds for one unit change.

The question is asking for the time taken for the output to change by 4 units so 2.55*4=...

You then have to create a graph.

I created a graph with an x-axis of 5 units and a y-axis of 12 units. I plotted a line from (y=2, x=0) to (y=2, x=2). We are told the step change occurred from 2 units to 12 units so from 0 units to 2 units nothing happened, it flat lined so to speak. It then jumped by 10 units so i plotted a line from (y=2, x=2) to (y=12, x=4).

There is a right angle triangle created here and the answer your trying to verify is the length of the hypotenuse. If you plot the graph then use square root (adjacent^2 + opposite^2) it gives you a very close answer to the one you proved using the math.

I spent ages on this one and I wasn't convinced my answer was right after reading other posts but I submitted it anyway and was told it was correct

 

[JE93]

Not wrong

t is the elapsed time which is 2.55 seconds for one unit change.

The question is asking for the time taken for the output to change by 4 units so 2.55*4=...

You then have to create a graph.

I created a graph with an x-axis of 5 units and a y-axis of 12 units. I plotted a line from (y=2, x=0) to (y=2, x=2). We are told the step change occurred from 2 units to 12 units so from 0 units to 2 units nothing happened, it flat lined so to speak. It then jumped by 10 units so i plotted a line from (y=2, x=2) to (y=12, x=4).

There is a right angle triangle created here and the answer your trying to verify is the length of the hypotenuse. If you plot the graph then use square root (adjacent^2 + opposite^2) it gives you a very close answer to the one you proved using the math.

I spent ages on this one and I wasn't convinced my answer was right after reading other posts but I submitted it anyway and was told it was correct

I understand it now!
2.55 is just 1 unit change so 4 unit changes would be 10.2
drawing the step change in graph form and getting the triangle works out to 10.19

Thanks for the help in understanding this, appreciate it.
Presuming you are taking HND Electrical electronic engineering? how are you getting on with it?
I have just started my HND with control systems being my first topic.

 

[David J]

Thats the answers I got and submitted and the feedback was that it was correct. Yes I am doing the HND now, completed the HNC at the beginning of the year. This is my 3rd topic and its by far the hardest I have had so far but I am discovering that the methods required to answer most of these TMA questions are well hidden in the lessons. The first 2 I did were electrical protection and electrical transmission and distribution. I was hoping to finish control and automation by xmas but I could be pushing my luck a bit there i think

 

[David Deal]

If you look at lesson CSA-1-4 on page 10 there is a bit about step and ramp changes. On that page there is an equation

output change=input change*(1-e to the power of -t/T) ( I don't know how to write that equation in latex form)

Output change is 4 units

input change is 10 units because the disturbance started at 2 and stopped at 12 so 12-2 = 10 units

so 4 = 10(e to the power of -t/T) then re arrange for -t

I know one of the posts in this thread got a correct answer when he submitted 6 seconds, I also got a correct answer for 10.2 seconds.

I tried to use the info given in the lessons although its not very helpful but the equation is the correct one to use.

Give this a try and let me know how you get on.

I will show you how I did the graph later on

Hello everyone.

I am new to Physics Forums. I seem to be doing the same course studying as you all too. I am having difficulty on how you managed to get

t = -5 ln(4/10)
t = 2.55

I am coming out with

t = 4.58

Any help would be greatly appreciated.

Thanks

 

[Jason-Li]

Sorry to resurrect this post.. @JE93 @David J

I also got 2.554 by rearranging 4=10*(1-e^-(t/5))

However I don't see how this is can 10.2s as the time constant (which is time to reach 63.2% which is greater than a 4 unit change) is 5s?

I was going to assume that 2.554 added to the delay of 4 seconds would give an overall of 6.554 seconds to change 4 units?

Sorry I am just confused how this can be correct, also the graph should be an exponential curve so drawing a right-angle triangle should also not work?

If you could shed some light that would be greatly appreciated, if you don't remember its no issue.

 

[Tromso80]

Sorry to resurrect this post.. @JE93 @David J

I also got 2.554 by rearranging 4=10*(1-e^-(t/5))

However I don't see how this is can 10.2s as the time constant (which is time to reach 63.2% which is greater than a 4 unit change) is 5s?

I was going to assume that 2.554 added to the delay of 4 seconds would give an overall of 6.554 seconds to change 4 units?

Sorry I am just confused how this can be correct, also the graph should be an exponential curve so drawing a right-angle triangle should also not work?

If you could shed some light that would be greatly appreciated, if you don't remember its no issue.

Hi there,

I'm also stuck on this one, can't seem to find out how we get 2.55? so I'm clearly transposing the formula wrong? any help?

 

[Jason-Li]

Hi there,

I'm also stuck on this one, can't seem to find out how we get 2.55? so I'm clearly transposing the formula wrong? any help?

Hi looking at this again, if you look up exp. / Log rules like in the video below will help

 

[Tromso80]

Ah ok, thanks.

I think I have the formula correct, but it was the number that has been used that has confused me.

If you use:

t = -5 ln(4/10)

you get:
t = 4.58

to get 2.55 its actually:

t = -5 ln(6/10)

Which makes sense when putting into a graph.

Unless I missed something?

 

[Eng studies]

Ah ok, thanks.

I think I have the formula correct, but it was the number that has been used that has confused me.

If you use:

t = -5 ln(4/10)

you get:
t = 4.58

to get 2.55 its actually:

t = -5 ln(6/10)

Which makes sense when putting into a graph.

Unless I missed something?

Hi I have been getting the same issue as yourself with the 4.58 figure, How did you get on with above ? Many Thanks

 

[kuruman]

Please note that this thread is bit old and that @Tromso80 was last seen here in August 2021. The thread itself is more than 4 years old. It is unlikely that you will get a response to your query. This question is outside my expertise, otherwise I would help you. Your best bet would be to start a new thread in one of the engineering forums.

 

[Eng studies]

Please note that this thread is bit old and that @Tromso80 was last seen here in August 2021. The thread itself is more than 4 years old. It is unlikely that you will get a response to your query. This question is outside my expertise, otherwise I would help you. Your best bet would be to start a new thread in one of the engineering forums.

Ok Thanks for the heads up, I was not sure if duplicate threads were ok, cheers

 

[kuruman]

Ok Thanks for the heads up, I was not sure if duplicate threads were ok, cheers

We often see the same problems with different numbers in the homework forums. That's OK as long as the same problem is not posted by the same user in two separate threads.