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Laplace Transformations

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I need the time domain response of this system as a unit RAMP input

C(s) = ((2s²) + 20s) / ((s²) + 4s + 20)

I get that the RAMP input is C(s) = A/s² G(s)

And now I think I need to simplify it so I can get it into a form that's on the Laplace Transformation table but this is what I'm having trouble with, I've tried manipulating it, but the 2s² on the top seems to be causing me an issue. The exponentially decaying
cosine wave on the table seems to look the closest after manipulation, but when I have sketched it out, it doesn't cut the y axis at 1.

I think it's my manipulation that's the problem. Any ideas? Cheers.
 

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  • #2
gabbagabbahey
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The 2s2 in the numerator can be dealt with easily enough:

[tex]\frac{2s^2+20s}{s^2+4s+20}=\frac{2(s^2+4s+20)+12s-40}{s^2+4s+20}=2 + 4\frac{3s-10}{s^2+4s+20}[/tex]

The inverse Laplace transform will then be the sum of a delta distribution and an exponentially decaying sinusoid. Why were you expecting a ramp function?
 
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  • #3
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The question said it was a unit RAMP input, sorry, I forgot to put that
 
  • #4
gabbagabbahey
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The question said it was a unit RAMP input, sorry, I forgot to put that
Can you post the entire question verbatim (word for word), because the inverse Laplace transform of the given [itex]C(s)[/itex] is not a unit RAMP input
 
  • #5
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A control system is described by the Laplace equation

C(s) = ((2s²) + 20s) / ((s²) + 4s + 20) R(s)

where R(s) is the input and C(s) is the output. A unit RAMP input is applied to the system.

a) Establish the time domain response of the system c(t).

b) What is the steady state value of the system output.

I've worked out the time domain response to be L^-1 (F)(s) = 1 - (e^-2t) * cos4t

Is this correct and how do you start to find the steady state value?
 
  • #6
rude man
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A control system is described by the Laplace equation

C(s)/R(s) = (2s² + 20s) / (s² + 4s + 20)
where R(s) is the input and C(s) is the output. A unit RAMP input is applied to the system.

a) Establish the time domain response of the system c(t).

b) What is the steady state value of the system output.

I've worked out the time domain response to be L^-1 (F)(s) = 1 - (e^-2t) * cos4t

Is this correct and how do you start to find the steady state value?
a) What was your C(s)? Then factor the denominator so part of it is of the form (s+s1)(s+s2). You'll be dealing with a complex-conjugate pole pair. The inveresion to the time domain is still of the form 1/(s+a) → e-at after you used partial-fraction expansion on the denominator.

b) Remember the final-value theorem? Steady-state value means the value of the output at t = ∞. Or, use your time-domain answer. Get the same either way ...

P.S. looks like your answer is correct but I haven't worked it out in detail. Your pole pair is right and so is the final value in the time domain using your answer.
 
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  • #7
gabbagabbahey
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A control system is described by the Laplace equation

C(s) = ((2s²) + 20s) / ((s²) + 4s + 20) R(s)

where R(s) is the input and C(s) is the output. A unit RAMP input is applied to the system.
That makes sense now. C(s) is the output, so there is no reason to expect it to be the Laplace transform of a unit ramp function.

I've worked out the time domain response to be L^-1 (F)(s) = 1 - (e^-2t) * cos4t
That doesn't look correct (I get [itex]\mathcal{L}^{-1} \left[ C(s) \right] = 2\delta(t) + \left( 12 \cos(4t) - 4 \sin(4t) \right)e^{-2t}u(t)[/itex]), perhaps you should show your steps so we can see where you are going wrong.
 
  • #8
rude man
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That makes sense now. C(s) is the output, so there is no reason to expect it to be the Laplace transform of a unit ramp function.



That doesn't look correct (I get [itex]\mathcal{L}^{-1} \left[ C(s) \right] = 2\delta(t) + \left( 12 \cos(4t) - 4 \sin(4t) \right)e^{-2t}u(t)[/itex]), perhaps you should show your steps so we can see where you are going wrong.
His R(s) is a ramp. You need to include that fact in obtaining L-1{C(s)}. When you do you get his answer, or close to it (I'm still too lazy to perform the entire inversion).
 
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  • #9
gabbagabbahey
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His R(s) is a ramp. You need to include that fact in obtaining L-1{C(s)}. When you do you get his answer, or close to it (I'm still too lazy to perform the entire inversion).
I'm not sure what you mean here. Maybe this is just some weird engineering terminology I'm unfamiliar with (I'm not an engineer), so correct me if I'm wrong here: when the problem says "Establish the time domain response of the system c(t)", do you not just compute the inverse Laplace transform of C(s)? Why and how do I need to include the input R(s) in the calculation?
 
  • #10
rude man
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I'm not sure what you mean here. Maybe this is just some weird engineering terminology I'm unfamiliar with (I'm not an engineer), so correct me if I'm wrong here: when the problem says "Establish the time domain response of the system c(t)", do you not just compute the inverse Laplace transform of C(s)? Why and how do I need to include the input R(s) in the calculation?
Because his system has an input R(s), transfer function H(s), and corresponding output
C(s). The formalism is C(s) = R(s)H(s). He was given H(s) = (2s2 + 20s)/(s2 + 4s + 20) and a unit ramp in the Laplace domain is R(s) = 1/s2. So you need to multiply what you deemed to be C(s), which is actually H(s), by 1/s2 before performing the inversion into the time domain.
 
  • #11
rude man
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Gabba, I might add that what you did was find the impulse response to his system. I.e. R(s) = δ(t), then the output would be as you derived (presumably, I'm too lazy to work the details).
 
  • #12
gabbagabbahey
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Because his system has an input R(s), transfer function H(s), and corresponding output
C(s). The formalism is C(s) = R(s)H(s). He was given H(s) = (2s2 + 20s)/(s2 + 4s + 20)
Okay, looking more closely at post #5, I now see the R(s) in his equation for C(s) (which wasn't in his first post).
 
  • #13
rude man
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Okay, looking more closely at post #5, I now see the R(s) in his equation for C(s) (which wasn't in his first post).
Yes, ands I might add that I should have said L{δ(t)} = transform of the impulse input δ(t) = 1.
 

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