Data Science and Computing with Python for Pilots and Flight Test Engineers

Let us have a system with state (and output) function \(x(t)\) and input function \(f(t)\) given by a differential equation such as

\begin{equation} \ddot x(t) + 2 \dot x(t) + 17 x(t) = f(t) \end{equation}

The transfer function \(H(s)\) of a system is defined as the Laplace transform of the system output \(X(s)\) divided by the Laplace transform of the input function, \(F(s)\):

\begin{equation} H(s):=\frac{X(s)}{F(s)} \end{equation}

The solution of the differential equation above in the Laplace domain can therefore be written as

$$ X(s) = H(s) F(s) $$.

Thus, the transfer function is the function we need to multiply the Laplace transform of the input function with to obtain the output function in the Laplace domain.

Let us work through this on an explicit example. Starting with the above differential equation

\begin{equation} \ddot x(t) + 2 \dot x(t) + 17 x(t) = f(t) \end{equation}

we apply the Laplace transform on both sides of the equation and obtain

\begin{equation} s^2X(s) + 2sX(s) + 17X(s) = F(s). \end{equation}

Solving for \(X(s)\) we obtain:

$$ X(s) = \frac{F(s)}{s^2+2s+17} $$

This is the solution of the differential equation in the Laplace domain. Now, if we apply the inverse Fourier transform to \(X(s)\) we obtain the solution \(x(t)\) of the differential equation in the original time domain.

Using the above definition of transfer function, we obtain for the transfer function:

$$ H(s):=\frac{X(s)}{F(s)} = \frac{1}{s^2+2s+17} $$

We also see that we can write the solution of the differential equation in the Laplace domain as

$$ X(s) = H(s) F(s) $$.

Thus, the transfer function is the function we need to multiply the Laplace transform of the input function with to obtain the output function in the Laplace domain.

Equivalent Definition of the Transfer Function

Alternatively, and equivalently, the transfer function can also be defined (or interpreted) as the Laplace transform of the response of the system to a unit impulse input \(f(t)=\delta(t)\).

In order to see this, recall from a table in our lesson on the Laplace transform, that the unit impulse input \(\delta(t)\) is the Dirac delta function, and that its Laplace transform is simply the constant function 1, i.e. \(\delta(s)=1\).

Then, using the above definition of the transfer function, \(H(s):=X(s)/F(s)\) with \(F(s)=\delta(s)=1\) in this particular case, we obtain \(H(s)=X(s)\). Thus, the transfer function is equal to the Laplace transform of the output function of the system – but only in this special case where the input function is the unit impulse input. Therefore, we can say that the transfer function is the Laplace transform of the system’s response to a unit impulse input.

Frequency Response

If we enter the relation \(X(s) = H(s)F(s)\) with the special \(s=i\omega\) corresponding to the Fourier transform, where \(\omega\) is the angular frequency, we obtain

$$ X(i\omega) = H(i\omega) F(i\omega) $$

The transfer function is the multiplicative factor by which an input signal with a certain frequency \(\omega\) gets amplified or attenuated by the system (ratio of output amplitude to input amplitude for a given frequency). Thus, if we want to explore the frequency response of the system, i.e. its response to different angular frequencies \(\omega\) and how much they get amplified or damped by the system, we need to study \(H(i\omega)\), i.e. we need to enter the transfer function \(H(s)\) with \(s=i\omega\).

The two meaningful quantities to extract will be the magnitude \(M(\omega):=|H(i\omega)|\), giving the ratio of amplitudes between the output \(x(t)\) and the input \(f(t)\) of the system, and the argument (angle) \(\phi(\omega:=\arg(H(i\omega))\), giving information about the phase shift between the output and the input of the system. We will revisit magnitude and phase shift (as well as phase delay) later, when we discuss Bode plots.