How Do You Solve a First Order Laplace Transform with a Ramp Input?

[topcat123]

Homework Statement

A process can be represented by the first order equation

(4δy(t)/δt) + y(t) = 3u(t)

Assume the initial state is steady (y = 0 at t = –0).

(a) Determine the transfer function of this process in the s domain.
(b) If the input is a ramp change in u(t) = 4t, determine the value of y(t)
when t = 10 s.

Homework Equations

Ramp change 1/s²

The Attempt at a Solution

(a)
Transfer function

{sY(s) + y(-0)} + Y(s)/4 = 3U(s)/4

sY(s) + Y(s)/4 = 3U(s)/4

G(s) = Y(s)/U(s) = ¾/(s + ¼) = 4/4(s + ¼)

(b)
This is where i am stuck i realize this is where the ramp change 1/s² comes in.

Y(s) =¾/(s + ¼) × 1/s² = ¾/s(s + ¼)
I think this is correct but I don't know where to go from here.

any help would be appreciated.

 

[Mark44]

Homework Statement

A process can be represented by the first order equation

(4δy(t)/δt) + y(t) = 3u(t)

Assume the initial state is steady (y = 0 at t = –0).

(a) Determine the transfer function of this process in the s domain.
(b) If the input is a ramp change in u(t) = 4t, determine the value of y(t)
when t = 10 s.

Homework Equations

Ramp change 1/s²

The Attempt at a Solution

(a)
Transfer function

{sY(s) + y(-0)} + Y(s)/4 = 3U(s)/4

sY(s) + Y(s)/4 = 3U(s)/4

G(s) = Y(s)/U(s) = ¾/(s + ¼) = 4/4(s + ¼)

(b)
This is where i am stuck i realize this is where the ramp change 1/s² comes in.

Y(s) =¾/(s + ¼) × 1/s² = ¾/s(s + ¼)

Check your algebra. $\frac{3/4}{s + 1/4} \times \frac 1 {s^2} \ne \frac{3/4}{s(s + 1/4)}$

I haven't verified the rest of your work...

I think this is correct but I don't know where to go from here.

any help would be appreciated.

 

[topcat123]

Check your algebra. 3/4s+1/4×1s2≠3/4s(s+1/4)

sorry copied it in wrong
Y(s) =¾/(s + ¼) × 1/s2 = 3/s(s + ¼)

 

[Mark44]

sorry copied it in wrong
Y(s) =¾/(s + ¼) × 1/s2 = 3/s(s + ¼)

That is still incorrect.

Also, when you write 3/s(s + 1/4), the usual interpretation is $\frac 3 s (s + 1/4)$, which is not what you meant.

 

[topcat123]

Y(s) =\frac{\frac{3}{4}}{(s+\frac{1}{4})}\frac{1}{s^2} = \frac{\frac{3}{4}}{s^2(s+\frac{1}{4})}

Is this better?

 

[Mark44]

Y(s) =\frac{\frac{3}{4}}{(s+\frac{1}{4})}\frac{1}{s^2} = \frac{\frac{3}{4}}{s^2(s+\frac{1}{4})}

Is this better?

Yes. To find y(t), split up the fraction using partial fraction decomposition, and then take the inverse Laplace transform of each of the resulting terms.

 

[topcat123]

OK so partial fraction. \frac{a}{s^2 (s + a)} = \frac{(As + B)}{s^2} + \frac{C}{(s + a)}
Multiply through by s^2 (s + a)
Giving a = (As + B) (s + a) + Cs^2
Removing the bracketsAs^2 + Bs + Aas + Ba + C^2
s^0... a = Ba... B = 1 s^1...0 = B + Aa substituting in for B A = \frac{-1}{a} s^2... 0 = A + C substituting in for AC = \frac{1}{a}
gives = \frac{(-\frac{s}{a} + 1)}{s^2} + \frac{\frac{1}{a}}{(s + a)}
simplifying = \frac{(-\frac{s}{a} + 1)}{s^2} to -\frac{\frac{s}{a}}{s^2} + \frac{1}{s^2} = \frac{\frac{1}{a}}{s} + \frac{1}{s^2}
Final result\frac{a}{s^2(s + a)} = -\frac{\frac{1}{a}}{s} + \frac{1}{s^2} + \frac{\frac{1}{a}}{(s + a)}
-\frac{\frac{1}{a}}{s}\text { inverse to } -\frac{1}{a}
\frac{1}{s^2}\text{ inverse to } t
\frac{\frac{1}{a}}{(s + a)}\text{ inverse to }\frac{1}{a}e^{-at}
\left[-\frac{1}{a} + t + \frac{1}{a}e^{-at}\right] as a = ¼ ad t = 10
3\left[-\frac{1}{\frac{1}{4}} + 10 + \frac{1}{\frac{1}{4}}e^{\frac{1}{4}}\right]
So y(t) = 3\left[-4 + 10 +4e^{-\frac{1}{4}}\right] = 27.35

Is that any where near?

 

[Mark44]

OK so partial fraction. \frac{a}{s^2 (s + a)} = \frac{(As + B)}{s^2} + \frac{C}{(s + a)}
Multiply through by s^2 (s + a)
Giving a = (As + B) (s + a) + Cs^2
Removing the bracketsAs^2 + Bs + Aas + Ba + C^2
s^0... a = Ba... B = 1 s^1...0 = B + Aa substituting in for B A = \frac{-1}{a} s^2... 0 = A + C substituting in for AC = \frac{1}{a}
gives = \frac{(-\frac{s}{a} + 1)}{s^2} + \frac{\frac{1}{a}}{(s + a)}
simplifying = \frac{(-\frac{s}{a} + 1)}{s^2} to -\frac{\frac{s}{a}}{s^2} + \frac{1}{s^2} = \frac{\frac{1}{a}}{s} + \frac{1}{s^2}
Final result\frac{a}{s^2(s + a)} = -\frac{\frac{1}{a}}{s} + \frac{1}{s^2} + \frac{\frac{1}{a}}{(s + a)}
-\frac{\frac{1}{a}}{s}\text { inverse to } -\frac{1}{a}
\frac{1}{s^2}\text{ inverse to } t
\frac{\frac{1}{a}}{(s + a)}\text{ inverse to }\frac{1}{a}e^{-at}
\left[-\frac{1}{a} + t + \frac{1}{a}e^{-at}\right] as a = ¼ ad t = 10
3\left[-\frac{1}{\frac{1}{4}} + 10 + \frac{1}{\frac{1}{4}}e^{\frac{1}{4}}\right]
So y(t) = 3\left[-4 + 10 +4e^{-\frac{1}{4}}\right] = 27.35

Is that any where near?

By my calculation, you're low by about a factor of 3.

What you show as y(t) above is actually y(10). What do you have for y(t)? You need to check that it is a solution of the diff. equation y' + (1/4)y = 3t, and that y(0) = 0. The solution I found satisifies this diff. equation and initial condition, and I get y(10) ≈ 75.94.

 

[Mark44]

Transfer function
{sY(s) + y(-0)} + Y(s)/4 = 3U(s)/4
sY(s) + Y(s)/4 = 3U(s)/4
G(s) = Y(s)/U(s) = ¾/(s + ¼) = 4/4(s + ¼)

I think you have a mistake in the last line above.
3u(t) = 12t, so after dividing by 4 you should get 3t. The Laplace transform of this is $\frac 3 {s^2}$, not $\frac {3/4}{s^2}$

 

[topcat123]

Yes I see the mistake there is also another.
Y(s) = \frac{\frac{12}{4}}{s^2(s + \frac{1}{4})}
also the second error.
\frac{1}{a}e^{at} = \frac{1}{\frac{1}{4}}e^{-\frac{10}{4}}
giving y(10) =12 \left[ -4 + 10 + 4e^{-\frac{10}{4}}\right ] = 75.94

Is this better?

 

[Mark44]

Yes I see the mistake there is also another.
Y(s) = \frac{\frac{12}{4}}{s^2(s + \frac{1}{4})}
also the second error.
\frac{1}{a}e^{at} = \frac{1}{\frac{1}{4}}e^{-\frac{10}{4}}
giving y(10) =12 \left[ -4 + 10 + 4e^{-\frac{10}{4}}\right ] = 75.94

Is this better?

Your answer agrees with mine.
One thing though that you're a little sloppy on: You wrote $\frac{1}{a}e^{at} = \frac{1}{\frac{1}{4}}e^{-\frac{10}{4}}$. The left side is a function of t, but the right side is a number. You should have $y(t) = \frac{1}{a}e^{-at} = 4e^{-\frac{t}{4}}$. From this you get y(10) = 75.94 (approx.).

I also don't see much advantage of bringing a into things as you did in your work.