Optimal Controller Gain for Desired Response in 2nd Order Laplace Transforms

[elijah78]

Homework Statement

Solve the DE for y(t) with the IC's
y(0)=20.8m/s and y'(0)=0

if the input is a step function scaled by the desired velocity V_o.
v_d(t)=V_ou(t).
Assume the desired velocity V_o=27.8m/s

Homework Equations

y''(t) + (D/M)y'(t) + (K/M)y(t) = (K/M)v_d(t)

M = 1,000kg
D = 100kg/s
K = controller gain
y(t) = output velocity
v_d is the input function

The Attempt at a Solution

So I'm Laplace transforming the whole 2nd order equation and I end up with a mess. The next problem is to find the optimal controller gain K for a desired response.

My Laplace transform of the 2nd order equation is:

Y(s) = [ KV_o + Ms²y(0) + Dsy(0) ] / [ s² + (D/M)s + (K/M) ]

if in fact I'm doing it correctly, here is where i am stuck.

 

[elijah78]

plugging in the variables into Y(s) i get:


Y(s) = [ 27.8K + 20800s² + 2080s ] / [ s² + .1s + .001K ]

 

[Chestermiller]

Homework Statement

Solve the DE for y(t) with the IC's
y(0)=20.8m/s and y'(0)=0

if the input is a step function scaled by the desired velocity V_o.
v_d(t)=V_ou(t).
Assume the desired velocity V_o=27.8m/s

Homework Equations

y''(t) + (D/M)y'(t) + (K/M)y(t) = (K/M)v_d(t)

M = 1,000kg
D = 100kg/s
K = controller gain
y(t) = output velocity
v_d is the input function

The Attempt at a Solution

So I'm Laplace transforming the whole 2nd order equation and I end up with a mess. The next problem is to find the optimal controller gain K for a desired response.

My Laplace transform of the 2nd order equation is:

Y(s) = [ KV_o + Ms²y(0) + Dsy(0) ] / [ s² + (D/M)s + (K/M) ]

if in fact I'm doing it correctly, here is where i am stuck.

This Laplace Transform doesn't look correct to me. Please show your work. Please show your Laplace Transform expressions for y'', y', and v_d(t).

Chet

 

[elijah78]

Laplace Transform expressions for y'', y', and vd(t).y''(t):

s²Y(s) - sy(0) - y'(0)

+

(D/M)y'(t):

(D/M)(sY(s) - y(0))

+

(K/M)y(t):

(K/M)Y(s)

=

(K/M)v_d(t):

(K/M)(V_o/s)then I factor out Y(s) and then solve for Y(s). that's when i get that ugly fraction up there that i don't know what to do with. i know how to do simple inverse transforms with tables and partial fractions.

 

[Chestermiller]

Laplace Transform expressions for y'', y', and vd(t).


y''(t):

s²Y(s) - sy(0) - y'(0)

+

(D/M)y'(t):

(D/M)(sY(s) - y(0))

+

(K/M)y(t):

(K/M)Y(s)

=

(K/M)v_d(t):

(K/M)(V_o/s)


then I factor out Y(s) and then solve for Y(s). that's when i get that ugly fraction up there that i don't know what to do with. i know how to do simple inverse transforms with tables and partial fractions.

It looks like the expressions above are correct, but there were algebra errors in solving for Y(s). Please try again. It looks like you are missing a factor of Ms in the denominator.

 

[elijah78]

ok so this time i got:

Y(s) = [ (KV_o/Ms) + y'(0) + sy(0) + (D/M)y(0) ] / [ s² + (D/M)s + (K/M) ]

 

[Chestermiller]

ok so this time i got:

Y(s) = [ (KV_o/Ms) + y'(0) + sy(0) + (D/M)y(0) ] / [ s² + (D/M)s + (K/M) ]

Good. Don't forget that y'(0) = 0.
Now, the next step is to manipulate this into a form that is a linear combination of some of the transforms in your tables. Start out by factoring y(0)/s out of the numerator.

Chet

 

[elijah78]

if it wasn't for that darn K. so:


(y(0)/s)[ (KV_o/My(0)) + s² + (D/M)s ] / [ s² + (D/M)s + (K/M) ]

 

[Chestermiller]

if it wasn't for that darn K. so:


(y(0)/s)[ (KV_o/My(0)) + s² + (D/M)s ] / [ s² + (D/M)s + (K/M) ]

No problem. I'm going to retype what you have:

y(s)=\frac{y(0)}{s}\frac{(s^2+(D/M)s+\frac{KV_0}{My(0)})}{(s^2+(D/M)s+\frac{K}{M})}

Check out the terms in parenthesis in the numerator and denominator. Does this suggest something you can do algebraically to simplify things?

 

[elijah78]

the only option i can see would be to pull 1 s out of the first 2 terms of each polynomial. if the K weren't there i could factor the polynomial using quadratic.

 

[Chestermiller]

the only option i can see would be to pull 1 s out of the first 2 terms of each polynomial. if the K weren't there i could factor the polynomial using quadratic.

Suppose you wrote the term in parenthesis in the numerator as
\left(s^2+(D/M)s+\frac{KV_0}{My(0)}\right)=\left(s^2+(D/M)s+\frac{K}{M}\right)+\left(\frac{KV_0}{My(0)}-\frac{K}{M}\right)

Then the Laplace Transform would become:

y(s)=\frac{y(0)}{s}\left(1+\frac{\left(\frac{KV_0}{My(0)}-\frac{K}{M}\right)}{(s^2+(D/M)s+\frac{K}{M})}\right)

This should be pretty easy to invert.

Chet

 

[elijah78]

A big big big thank you to you guys for your help. i ended up with an A on the project and a B for the semester. Thank you!