Derive ODE for $\phi (u)$

Question

Given a system which is described by $$10 \phi'(t)+\phi(t)=\phi_H(t)$$ $$\phi_H'(t)+0.5\phi_H(t)=0.5i(t)$$ $$i(t) = 0.5u^2(t).$$

Derive a ODE for $\phi(u(t)).$

My approch is du just put $u(t)$ into the system, but this leads to nothing. Hints would be very nice.

Greetings.

Answer 1

You can solve it by taking the derivative of the first equation and substitute all the other equation into it until you are only left with derivatives of $\phi$ and $u$.

And if your last equation would have $i'$ instead of $i$ on the left hand side then you would have to take the derivative one extra time.

Answer 2

You have the following equations:

$$10 \phi'(t)+\phi(t)=\phi_H(t)$$ $$\phi_H'(t)+0.5\phi_H(t)=0.5i(t)$$ $$i(t) = 0.5u^2(t)$$

First, determine the derivative of the first equation. $$10 \phi''(t)+\phi'(t)=\phi_H'(t)$$

Add this equation and $0.5$ times your first equation: $$[10 \phi''(t)+\phi'(t)]+0.5[10 \phi'(t)+\phi(t)]=\phi_H'(t)+0.5\phi_H(t).$$

Note, that the right-hand side is the second of your equations, hence the right-hand side is $0.5i(t)=0.5(0.5u^2(t))=0.25u^2(t)$. So we obtain:

$$[10 \phi''(t)+\phi'(t)]+0.5[10 \phi'(t)+\phi(t)]=0.25u^2(t).$$

$$\implies 10 \phi''(t)+6\phi'(t)+0.5\phi(t)=0.25u^2(t)$$ $$\implies \phi''(t)+0.600\phi'(t)+0.050\phi(t)=0.025u^2(t)$$

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An alternative approach is using transfer functions. The Laplace transform (zero initial conditions) of your equations is given by:

$$(10s+1)\phi(s)=\phi_H(s) \implies F_1(s)=\phi(s)/\phi_H(s)=1/(10s+1)$$ $$(s+0.5)\phi_H(s)=0.5i(s) \implies F_2(s)=\phi_H(s)/i(s)=0.5/(s+0.5)$$ $$i(s) = 0.5u^2(s) \implies F_3(s)=i(s)/u^2(s)=0.5$$

Note, that $F_1F_2F_3=\frac{\phi(s)}{u^2(s)}=\frac{0.5^2}{(10s+1)(s+0.5)}$. From this it is easy to get the differential equation for $\phi(t)$, with $u^2(t)$ as input.

Remark: I am using $u^2(s)$ as a symbolic representation for the Laplace transform of $u^2(t)$.