man111111
Badges: 12
Rep:
?
#1
Report Thread starter 1 year ago
#1
Question what is the inverse laplace of (0.6e^(-0.1s))/ 2s+1

i multiplied the transfer function by 1/s ignoring the e expression for time shift

then
I simplifed it and did partial fractions and I got the following
X(s)= 0.6/s - 0.6/(s+0.5)

Then I did the laplace transform of each term and I got

X(t)=0.6u(t) - (0.6e^(-0.5t))u(t)

then i used this x(t- τ) u(t- τ) which is

to give me

x(t- 0.1) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)

then multiplied the above which is x(t- 0.1) with u(t- 0.1) to get x(t- τ) u(t- τ)

Answer:

[0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)]u(t- 0.1)

But I've got three u(t- 0.1) is this correct.

I don't know what I've done wrong.

Any help will be much appreciated
0
reply
mqb2766
Badges: 19
Rep:
?
#2
Report 1 year ago
#2
(Original post by man111111)
Question what is the inverse laplace of (0.6e^(-0.1s))/ 2s+1

i multiplied the transfer function by 1/s ignoring the e expression for time shift

then
I simplifed it and did partial fractions and I got the following
X(s)= 0.6/s - 0.6/(s+0.5)

Then I did the laplace transform of each term and I got

X(t)=0.6u(t) - (0.6e^(-0.5t))u(t)

then i used this x(t- τ) u(t- τ) which is

to give me

x(t- 0.1) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)

then multiplied the above which is x(t- 0.1) with u(t- 0.1) to get x(t- τ) u(t- τ)

Answer:

[0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)]u(t- 0.1)

But I've got three u(t- 0.1) is this correct.

I don't know what I've done wrong.

Any help will be much appreciated
Multiple u(t-0.1) which multiply are the same as a single one? If you already have a u(t) in the expression, just shift the time. Also note that the shift only applies to the expression on the right, not the left.
Why multiply by 1/s at the start? Can you upload the full question?
Last edited by mqb2766; 1 year ago
0
reply
man111111
Badges: 12
Rep:
?
#3
Report Thread starter 1 year ago
#3
(Original post by mqb2766)
Multiple u(t-0.1) which multiply are the same as a single one? If you already have a u(t) in the expression, just shift the time. Also note that the shift only applies to the expression on the right, not the left.
Why multiply by 1/s at the start? Can you upload the full question?
Thank you for replying


The question is asking me to draw the step response. So i multipied it with 1/s and i tried to change it to get the time domain
Last edited by man111111; 1 year ago
0
reply
mqb2766
Badges: 19
Rep:
?
#4
Report 1 year ago
#4
(Original post by man111111)
Thank you for replying


The question is asking me to draw the step response. So i multipied it with 1/s and i tried to change it to get the time domain
Ok there was no explanation in the original post.
so you have
y(t) = 0.6(1-exp(-t/2))u(t)
Befor e applying the time shift, so just shift the times on the right to get ...

Useful sanity check while learning
https://www.wolframalpha.com/input/?...2s+%2B+1%29%29
Last edited by mqb2766; 1 year ago
0
reply
man111111
Badges: 12
Rep:
?
#5
Report Thread starter 1 year ago
#5
x(t- τ) u(t- τ) so if you use this you don't multiply by u(t- τ) at the end you only shift y(t)

y(t) = 0.6(1-exp(-t/2))u(t)
0
reply
mqb2766
Badges: 19
Rep:
?
#6
Report 1 year ago
#6
(Original post by man111111)
x(t- τ) u(t- τ) so if you use this you don't multiply by u(t- τ) at the end you only shift y(t)

y(t) = 0.6(1-exp(-t/2))u(t)
y(t) = 0.6(1-exp(-(t-0.1)/2))u(t -0.1)
not sure what you're asking.
If you shift y(t), on the left, as well as the signal on the right, it is the same as you started with
0
reply
mqb2766
Badges: 19
Rep:
?
#7
Report 1 year ago
#7
(Original post by man111111)
x(t- τ) u(t- τ) so if you use this you don't multiply by u(t- τ) at the end you only shift y(t)

y(t) = 0.6(1-exp(-t/2))u(t)
Are you good now?
Note
u(t-T)*u(t-T) = u(t-T)
Last edited by mqb2766; 1 year ago
0
reply
man111111
Badges: 12
Rep:
?
#8
Report Thread starter 1 year ago
#8
(Original post by man111111)
x(t- τ) u(t- τ) so if you use this you don't multiply by u(t- τ) at the end you only shift y(t)

y(t) = 0.6(1-exp(-t/2))u(t)
Ok Thank you for your help

Just to clarify when I use x(t- τ) u(t- τ) from the laplace transform table

i ignore the e^(-0.1s) then i do the inverse laplace to get y(t) then I shift it to the right by 0.1

Solution:

i multiply the transfer function by 1/s ignoring the e expression for time shift

then partial fractions to get the following
X(s)= 0.6/s - 0.6/(s+0.5)

Then do the laplace transform of each term above

X(t)=0.6u(t) - (0.6e^(-0.5t))u(t)

then i use this x(t- τ) u(t- τ) which gives

x(t- 0.1) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)
Last edited by man111111; 1 year ago
0
reply
mqb2766
Badges: 19
Rep:
?
#9
Report 1 year ago
#9
(Original post by man111111)
Ok Thank you for your help

Just to clarify when I use x(t- τ) u(t- τ) from the laplace transform table

i ignore the e^(-0.1s) then i do the inverse laplace to get y(t) then I shift it to the right by 0.1

Solution:

i multiply the transfer function by 1/s ignoring the e expression for time shift

then partial fractions to get the following
X(s)= 0.6/s - 0.6/(s+0.5)

Then do the laplace transform of each term above

X(t)=0.6u(t) - (0.6e^(-0.5t))u(t)

then i use this x(t- τ) u(t- τ) which gives

x(t- 0.1) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)
The last line you have
x(t) = ...
If you shift both sides, there is no delay. Shift only the right hand side.
0
reply
man111111
Badges: 12
Rep:
?
#10
Report Thread starter 1 year ago
#10
(Original post by mqb2766)
The last line you have
x(t) = ...
If you shift both sides, there is no delay. Shift only the right hand side.
So instead of writing

x(t- 0.1) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)

I write
x(t) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)

But how can I distinguish both of the functions in my solution

I though if you had a signal x(t) and you wanted to shift it to the right by 0.1 you do x(t-0.1)
Last edited by man111111; 1 year ago
0
reply
mqb2766
Badges: 19
Rep:
?
#11
Report 1 year ago
#11
(Original post by man111111)
So instead of writing

x(t- 0.1) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)

I write
x(t) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)

But how can I distinguish both of the functions in my solution

I though if you had a signal x(t) and you wanted to shift it to the right by 0.1 you do x(t-0.1)
Not sure what you mean by distinguishing both functions?
The output
y(t) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)
So
y(0) = 0
y(0.1) = 0
y(0.5) ~ 0.1
Which fits with a transfer function / system being time delayed.

If you write
y(t-0.1) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)
We could redefine t as t-0.1 to get the signal
y(t) = 0.6u(t) - (0.6e^(-0.5(t)))u(t),
? Here, you've got to shift time before applying it to the signal. With the former, it's in the definition of the signal.
Last edited by mqb2766; 1 year ago
0
reply
man111111
Badges: 12
Rep:
?
#12
Report Thread starter 1 year ago
#12
(Original post by mqb2766)
Multiple u(t-0.1) which multiply are the same as a single one? If you already have a u(t) in the expression, just shift the time. Also note that the shift only applies to the expression on the right, not the left.
Why multiply by 1/s at the start? Can you upload the full question?
Ok thankful for your help.

When I ignore the e^(-0.1s) and do the laplace inverse and the laplace inverse results with a unit step inside it, do i not multiply it at the end with u(t). But if it doesn't have a u(t) when I do inverse laplace do I multiply it at the end with u(t)?

I was looking at an example on YouTube and they multiplied the inverse laplace at the end with u(t) but the inverse laplace result didnt have a u(t) when they did the inverse laplace.


I'm using x(t- τ) u(t- τ) from the laplace table
0
reply
mqb2766
Badges: 19
Rep:
?
#13
Report 1 year ago
#13
(Original post by man111111)
Ok thankful for your help.

When I ignore the e^(-0.1s) and do the laplace inverse and the laplace inverse results with a unit step inside it, do i not multiply it at the end with u(t). But if it doesn't have a u(t) when I do inverse laplace do I multiply it at the end with u(t)?

I was looking at an example on YouTube and they multiplied the inverse laplace at the end with u(t) but the inverse laplace result didnt have a u(t) when they did the inverse laplace.


I'm using x(t- τ) u(t- τ) from the laplace table
Part of the problem comes from whether you're using normal (one sided) or bilateral Laplace transforms.

Normal Laplace is for t: 0 -> inf. Sometimes you include the initial conditions of the output, but often the system is assumed to be at rest. Also the output is usually expressed as
y(t) = 0.6(1-exp(-t/2))
There is no need to include a step as the one sided laplace is only valid for non-negative time, where the heaviside step signal is 1. You could say the expression is only valid for t>=0.

Adding a time delay would become
y(t) = 0.6(1-exp(-(t-0.1)/2))u(t -0.1)
The step signal is necessary to switch off the response in the interval 0 -> 0.1. It's irrelevant for negative t as the response is not defined. If you'd already included a u(t), then some simple maths or reasoning would give that you threw it away and replaced with u(t-0.1). If you did not include any step multiplier, you could say the time delayed output y(t) was valid for t>=0.1.

If you use the bilateral Laplace (valid for negative time as well), you'd have to include u(t) in the output
y(t) = 0.6(1-exp(-t/2))u(t )
But the time delay would be similar. However, you'd time shift the existing u(t) and not introduce a new one as the signal is valid for all t. As you're probably doing the first case (one sided laplace) you can probably ignore this, even though it's closer to what you're trying to do.
Last edited by mqb2766; 1 year ago
0
reply
man111111
Badges: 12
Rep:
?
#14
Report Thread starter 1 year ago
#14
(Original post by mqb2766)
Part of the problem comes from whether you're using normal (one sided) or bilateral Laplace transforms.

Normal Laplace is for t: 0 -> inf. Sometimes you include the initial conditions of the output, but often the system is assumed to be at rest. Also the output is usually expressed as
y(t) = 0.6(1-exp(-t/2))
There is no need to include a step as the one sided laplace is only valid for non-negative time, where the heaviside step signal is 1. You could say the expression is only valid for t>=0.

Adding a time delay would become
y(t) = 0.6(1-exp(-(t-0.1)/2))u(t -0.1)
The step signal is necessary to switch off the response in the interval 0 -> 0.1. It's irrelevant for negative t as the response is not defined. If you'd already included a u(t), then some simple maths or reasoning would give that you threw it away and replaced with u(t-0.1). If you did not include any step multiplier, you could say the time delayed output y(t) was valid for t>=0.1.

If you use the bilateral Laplace (valid for negative time as well), you'd have to include u(t) in the output
y(t) = 0.6(1-exp(-t/2))u(t )
But the time delay would be similar. However, you'd time shift the existing u(t) and not introduce a new one as the signal is valid for all t. As you're probably doing the first case (one sided laplace) you can probably ignore this, even though it's closer to what you're trying to do.
thank you this has really clarified my problem.

I was looking at a question and it asked me to determine when would the closed system be stable using the following transfer function to help....

T(S) = kp*k(e^(-td*s))/(t*s+1+(kp*k)(e^(-td*s))


I think it is asking me to workout the root locus but i do not know how to workout the poles for the transfer function given.
0
reply
mqb2766
Badges: 19
Rep:
?
#15
Report 1 year ago
#15
(Original post by man111111)
thank you this has really clarified my problem.

I was looking at a question and it asked me to determine when would the closed system be stable using the following transfer function to help....

T(S) = kp*k(e^(-td*s))/(t*s+1+(kp*k)(e^(-td*s))


I think it is asking me to workout the root locus but i do not know how to workout the poles for the transfer function given.
Can you upload a picture of the full question pls?
0
reply
man111111
Badges: 12
Rep:
?
#16
Report Thread starter 1 year ago
#16
(Original post by mqb2766)
Can you upload a picture of the full question pls?
Hi,
I have mentioned everything that the question asked me
0
reply
mqb2766
Badges: 19
Rep:
?
#17
Report 1 year ago
#17
(Original post by man111111)
Hi,
I have mentioned everything that the question asked me
It's the third time I've asked to see the question(s).
Id appreciate seeing the original wording.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Feeling behind at school/college? What is the best thing your teachers could to help you catch up?

Extra compulsory independent learning activities (eg, homework tasks) (18)
7.35%
Run extra compulsory lessons or workshops (37)
15.1%
Focus on making the normal lesson time with them as high quality as possible (44)
17.96%
Focus on making the normal learning resources as high quality/accessible as possible (35)
14.29%
Provide extra optional activities, lessons and/or workshops (64)
26.12%
Assess students, decide who needs extra support and focus on these students (47)
19.18%

Watched Threads

View All