## Transfer Functions

Transfer functions are input to output representations of dynamic systems. One advantage of working in the Laplace domain (versus the time domain) is that differential equations become algebraic equations. These algebraic equations can be rearranged and transformed back into the time domain to obtain a solution or further combined with other transfer functions to create more complicated systems. The first step in creating a transfer function is to convert each term of a differential equation with a Laplace transform as shown in the table of Laplace transforms. A transfer function, *G(s)*, relates an input, *U(s)*, to an output, *Y(s)*.

$$G(s) = \frac{Y(s)}{U(s)}$$

#### First-order Transfer Function

A first-order linear differential equation is shown as a function of time.

$$\tau_p \frac{dy(t)}{dt} = -y(t) + K_p u\left(t-\theta_p\right)$$

The first step is to apply the Laplace transform to each of the terms in the differential equation.

$$\mathcal{L}\left(\tau_p \frac{dy(t)}{dt}\right) = \mathcal{L}\left(-y(t)\right) + \mathcal{L}\left(K_p u\left(t-\theta_p\right)\right)$$

Because the Laplace transform is a linear operator, each term can be transformed separately. With a zero initial condition the value of *y* is zero at the initial time or *y(0)=0*.

$$\mathcal{L}\left(\tau_p \frac{dy(t)}{dt}\right) = \tau_p \left(s \, Y(s) - y(0)\right) = \tau_p s \, Y(s)$$

$$\mathcal{L}\left(-y(t)\right) = -Y(s)$$

$$\mathcal{L}\left(K_p u\left(t-\theta_p\right)\right) = K_p \, U(s) \, e^{-\theta_p s}$$

Putting these terms together gives the first-order differential equation in the Laplace domain.

$$\tau_p s \, Y(s) = -Y(s) + K_p \, U(s) \, e^{-\theta_p s}$$

For the first-order linear system, the transfer function is created by isolating terms with *Y(s)* on the left side of the equation and the term with *U(s)* on the right side of the equation.

$$\tau_p s \, Y(s) + Y(s) = K_p \, U(s) \, e^{-\theta_p s}$$

Factoring out the *Y(s)* and dividing through gives the final transfer function.

$$G(s) = \frac{Y(s)}{U(s)} = \frac{K_p e^{-\theta_p s}}{\tau_p s + 1}$$

#### Combining Transfer Functions

The additive property is used for transfer functions in parallel. The input signal `X_1(s)` becomes `Y_1(s)` when it is transformed by `G_1(s)`. Likewise, `X_2(s)` becomes `Y_2(s)` when it is transformed by `G_2(s)`. The two signals `Y_1(s)` and `Y_2(s)` are added to create the final output signal `Y(s)=Y_1(s)+Y_2(s)`. This gives a final output expression of `Y(s)=G_1(s) X_1(s)+G_2(s) Y_2(s)`.

The multiplicative property is used for transfer functions in series. The input signal `X_1(s)` becomes `X_2(s)` when it is transformed by `G_1(s)`. The intermediate signal `X_2(s)` becomes the input for the second transfer function `G_2(s)` to produce `Y(s)`. The final output signal is `Y(s)=G_2(s) X_2(s) = G_1(s) G_2(s) X_1(s)`.

#### Exercise

Do the following, assuming that *q* and *V* are constant:

- Transform each equation in to the Laplace domain.
- Find the transfer function between the specified variables.
- Define the time constant and the gain for each transfer function in terms of the parameters given.

Find `{C(s)}/{C_i(s)}`:

$$\frac{dc'(t)}{dt} = \frac{q}{V} c_i'(t) - \frac{q}{V} c'(t) $$

Find `{C(s)}/{C_i(s)}`:

$$\frac{dc'(t)}{dt} = \frac{q}{V} c_i'(t) - \left(\frac{q}{V} + 2 k_2 \bar c\right) c'(t) $$

Find `{T(s)}/{T_i(s)}` and `{T(s)}/{Q(s)}`:

$$\frac{dT'(t)}{dt} = \frac{q}{V} \left(T_i'(t)-T'(t)\right) + \frac{Q'(t)}{\rho\,V\,C_p} $$