Engineering Write the differential equation that's equivalent to this transfer function

[s3a]

Homework Statement

Write the differential equation that is mathematically equivalent to the transfer function below: G(s) = Y(s) / R(s).

G(s) = C(s) / R(s) = (s^4 + 2s^3 + 5s^2 + s + 1) / (s^5 + 3s^4 + 2s^3 + 4s^2 + 5s + 2)

Relevant Equations

• $G(s) = Y(s) / R(s)$
• $r(t) = t^3 u(t)$

I have the solution to the problem, and I mechanically, but not theoretically (basically, why do the C(s) and R(s) disappear?), understand how we go from

$(s^5 + 3s^4 + 2s^3 + 4s^2 + 5s + 2) C(s) = (s^4 + 2s^3 + 5s^2 + s + 1) R(s)$

to

$c^{(5)}(t) + 3c^{(4)}(t) + 2c^{(3)}(t) + 4c^{(2)}(t) + 5c^{(1)}(t) + 2c^{(0)}(t) = r^{(4)}(t) + 2r^{(3)}(t) + 5r^{(2)}(t) + r^{(1)}(t) + r^{(0)}(t)$.

And, then I understand that the $r(t)$ needs to be replaced by $t^3 u(t)$, but the final answer in the solution is

$c^{(5)}(t) + 3c^{(4)}(t) + 2c^{(3)}(t) + 4c^{(2)}(t) + 5c^{(1)}(t) + 2c^{(0)}(t) = 18δ(t) + (36 + 90t + 9t^2 + 3t^3) u(t)$, and I don't understand the step that leads the previous step to this final answer.

Could someone please help me figure out what it is that the solution did?

Any input would be GREATLY appreciated!

 

[Ssnow]

Hi, are you sure that $r(t)=t^3u(t)$ and not $r(t)=3t^3u(t)$ ?
Ssnow

 

[s3a]

I'm not sure of anything ( ;P ), but that ( $r(t) = t^3 · u(t)$ ) is technically what the solution says.

 

[s3a]

Maybe the solution is wrong?

Since no one else answered, @Ssnow, could you please explain to me what you've done just in case you are in fact correct?

 

[Ssnow]

I simply used $r(t)=3t^3u(t)$ instead yours... , so using the product rules: $r^{(0)}(t)=3t^3u(t)$, $r^{(1)}(t)=9t^2u(t) + 3t^3u^{(1)}(t)$, $r^{(2)}(t)=18tu(t) + ...$ higher derivatives for $u(t)$ ... so putting inside your second member that contains the derivatives of $r(t)$, and collecting the $u(t)$, we arrive to your solution ...
Ssnow

 

[s3a]

Actually, is the going from (s^5 + 3s^4 + 2s^3 + 4s^2 + 5s + 2) C(s) = (s^4 + 2s^3 + 5s^2 + s + 1) R(s) to $c^(5) + 3c^(4) + 2c^(3) + 4c^(2) + 5 c^(1) + 2c^(0) = r^(4) + 2r^(3) + 5r^(2) + r^(1) + r^(0)$ from the L{f^(n) (t)} = $s^n F(s) – s^{n-1} f(0) – s^{n-2} f'(0) - . . . – s f^(n-2)(0) – f^(n-1)(0)$ property, where all derivatives of f(t), including the 0th derivative, f(t) itself are 0 when t = 0 because the problem statement says to assume that "all initial conditions are equal to zero"?

As for going from the before-last step to the last step, thanks for your response; I now see how plugging in 3t^2 into the derivatives of r(t) and then multiplying by the appropriate coefficients yields the (36 + 90t + 9t^2 + 3t^3) part, but I'm not 100% sure why it's just a u(t) and not a bunch of derivatives of u(t) as well. Am I expected to to compute a monstrosity such as this one ( https://www.wolframalpha.com/input/?i=d^3/dt^3+(3t^3+*+u(t)) ) and then plug in t = 0? If it's the case that t = 0, wouldn't u(t) be u(0) instead? Also, what's with the 18 δ(t)? Does the δ(t) have something to do with the initial condition stuff having it be the case that t = 0?

Edit:
P.S.
I recently realized that the problem statement is inconsistent with the solution for the G(s) = C(s) / R(s) part, but it is consistent in stating that r(t) = u(t) t^3 (whether it's correct or not).

 

[s3a]

Basically, take a look at the "expanded form" of this ( https://www.wolframalpha.com/input/...t+(3t^3+*+u(t)))+++1*(d^4/dt^4+(3t^3+*+u(t))) ).

Notice that it doesn't match the solution of the problem.