2
$\begingroup$

I have a system that is modeled by the following differential equation:

$$db/dt = h_aj(t) + k(T_a-b(t))$$

where $db/dt$ is the rate of temperature change, $j(t)$ is an input, $h_a, k, T_a$ are all constants, and $b(t)$ is an output. Note that this is newtonian cooling with a heating input, $h_aj(t)$.

I want to find the transfer function is the Laplace domain, $B(s)/J(s)$. Taking the laplacian of the equation of interest, assuming all IC's are $0$, yields: $$sB(s)+kB(s)-\frac{kT_a}{s} = h_aJ(s).$$

What I can't figure out is the term $kT_a/s$ is not a function of $J(s)$ or $B(s)$ so I can not get a transfer function of purely $B(s)/J(s)$. Does anyone know how to solve this equation or a better way to find the transfer function relating the input to the output?

$\endgroup$
1
  • 1
    $\begingroup$ What about seeing B as a function of J and s: B(J,s). Then, you can look at the partial derivative of B w.r.t. J. $\endgroup$ – Raskolnikov Nov 29 '10 at 19:18
1
$\begingroup$

When you take the Laplace transform and solve for B(s), you should consider both nonhomogeneous terms in the input:

$$ B(s) = \frac{1}{s+k} \left (h_a J(s) + \frac{kT_a}{s} \right ). $$

The transfer function is just first factor, $H(s) = \displaystyle \frac{1}{s+k}$, and the solution $b(t)$ is given by the convolution of $h(t) = e^{-kt}$ with the forcing term $h_a j(t) + k T_a$.

$\endgroup$
0
$\begingroup$

Thanks cch. Another approach a coworker recommended is to create a new time dependent term that includes all the constants:

c(t) = (ha/k)*j(t)+Ta

Note that ha, k, and Ta are all constant so c(t) and j(t) have no dynamics between themselves.

Then the differential equation simplifies to:

db/dt = k*(c(t)-b(t))

And the laplacian becomes:

sB(s) + kB(s) = kC(s)

And the transfer function becomes:

B(s)/C(s) = k/(s+k)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy