Transfer Functions


In control theory, a system is some mathematical relation between an input and an output.  Systems are usually represented as a rectangle connecting some input function to an output function. Figure 1  depicts a SISO (single input, single output) system taking in an input function of time $u(t)$ and returning an output function of time $y(t)$.

Figure 1: A SISO system

Generally, ordinary differential equations (ODEs) are the simplest way to represent a system.

  • Note: dependent variables for ODEs (usually, this is time) are generally positive real numbers

Laplace Transform

We use the Laplace transform to switch between a function of real numbers (for our purposes, lets assume this is time $t$) to a function of some complex variable $s$ (Figure 2). Note that our $s$ variable turns out to denote frequency since our original function is a function of time.

Figure 2: Laplace transfrom for our input function $u(t)$ and our output function $y(t)$ from Figure 1.

Specifically, the expression for the Laplace transfrom of a single-variable function of time $f(t)$ is \begin{equation} F(s) = \mathcal{L} \left[ f(t) \right] = \int_{0}^{\infty} f(t)e^{-st} dt \end{equation}

The following equations are simplified expressions of the Laplace transform for $f(t)$ and its corresponding first and second order ODEs:

\begin{equation*} \mathcal{L} \left[ f(t) \right]  = F(s) \end{equation}

\begin{equation*} \mathcal{L} \left[ f'(t) \right]  = sF(s) – f(0) \end{equation}

\begin{equation*} \mathcal{L} \left[ f”(t) \right] = s^2 F(s) – sf(0) – f'(0) \end{equation}

Transfer Function (SISOs)

The transfer function $H(s)$ for a dynamic SISO system relates an input $u(t)$ with an output $y(t)$ as shown in Figure 3:

Figure 3: Transfer function for a SISO system


Example 1: Mechanical System

Find the transfer function for a single translational damped mass-spring system (depicted in Figure 4)

Figure 4: Translational mass-spring system with damping
  • Aside: we can find the equation of motion for such a system using Newton’s and D’Alembert equations
  • The ODE describing our mass-spring system turns out to be…  \begin{equation} F(t)=m \ddot{x}(t) + c\dot{x}(t) + kx(t) \end{equation}
    • Variables:
      • $F(t)=$ an external force $[\mathrm{N}]$ acting on our mass $m \, [\mathrm{kg}]$
        • Note: this is the input for our system
      • $x(t)=$ the displacement $[\mathrm{m}]$ of our object due to $F(t)$
        • Note: this is the output for our system
      • $c=$ the damping coefficient $[\mathrm{Ns/m}]$
      • $k=$ the spring constant (i.e., stiffness) $[\mathrm{N/m}]$

Assume our initial conditions are $x(0)=0$ and $\dot{x}(0)=0$

First, let’s apply the Laplace transform to each individual term in our ODE for the mass-spring system:

\begin{equation*} \mathcal{L}[\ddot{x}]  = s^2 X(s) – sx(0) – \dot{x}(0) = s^2 X(s) \end{equation}

\begin{equation*} \mathcal{L}[\dot{x}]  = sX(s) – x(0) = sX(s) \end{equation}

\begin{equation*} \mathcal{L}[x(t)]  = X(s) \end{equation}

\begin{equation*} \mathcal{L}[F(t)]  = F(s) \end{equation}

Subsituting these expressions into our ODE is equivalent to taking the Laplacian of both sides:

\begin{equation} F(s) = ms^2 X(s) + cs X(s) + k X(s) \  =  X(s) (ms^2 + cs + k)  \end{equation}

Rearragning our $F(s)$ equation as follows gives us our transfer function $H(s)$ for the sytem:

\begin{equation} H(s) = \frac{X(s)}{F(s)} = \frac{1}{ms^2 + cs + k}  \end{equation}

Thus, we have defined our mechanical system as a second order ODE and as a transfer function. Figure 5 depicts the corresponding system diagrams in terms of the ODE and in terms of the transfer function.

Figure 5: Two equivalent system diagrams for our mass-spring system. The top system diagram represents our mass-spring system in terms of a second order ODE. The bottom system diagram represents our mass-spring system in terms of a transfer function.


How to find the transfer function of a system

  • Note: most of the content in the article originally comes from the above link

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