13.7: System Functions and the Laplace Transform

When we introduced the Nyquist criterion for stability we stated without any justification that the system was stable if all the poles of the system function \(G(s)\) were in the left half-plane. We also asserted that the poles corresponded to exponential modes of the system. In this section we’ll use the Laplace transform to more fully develop these ideas for differential equations.

Lightning review of 18.03

Definitions

1. \(D = \dfrac{d}{dt}\) is called a differential operator. Applied to a function \(f(t)\) we have
   \[Df = \dfrac{df}{dt}. \nonumber \]
   We read \(Df\) as '\(D\) applied to \(f\).'

   Example \(\PageIndex{1}\)

   If \(f(t) = t^3 + 2\) then \(Df = 3t^2, D^2f = 6t\).

If \(P(s)\) is a polynomial then \(P(D)\) is called a polynomial differential operator.

Example \(\PageIndex{2}\)

Suppose \(P(s) = s^2 + 8s + 7\). What is \(P(D)\)? Compute \(P(D)\) applied to \(f(t) = t^3 + 2t + 5\). Compute \(P(D)\) applied to \(g(t) = e^{2t}\).

Solution

\(P(D) = D^2 + 8D + 7I\). (The \(I\) in \(7I\) is the identity operator.) To compute \(P(D) f\) we compute all the terms and sum them up:

\[\begin{array} {rcl} {f(t)} & = & {t^3 + 2t + 5} \\ {Df(t)} & = & {3t^2 + 2} \\ {D^2 f(t)} & = & {6t} \end{array} \nonumber \]

Therefore: \((D^2 + 8D + 7I) f = 6t + 8(3t^2 + 2) + 7(t^3 + 2t + 5) = 7t^3 + 24t^2 + 20t + 51.\)

\[\begin{array} {rcl} {g(t)} & = & {e^{2t}} \\ {Dg(t)} & = & {2e^{2t}} \\ {D^2 g(t)} & = & {4e^{2t}} \end{array} \nonumber \]

Therefore: \((D^2 + 8D + 7I) g = 4e^{2t} + 8(2) e^{2t} + 7e^{2t} = (4 + 16 + 7) e^{2t} = P(2) e^{2t}\).

The substitution rule is a straightforward statement about the derivatives of exponentials.

Theorem \(\PageIndex{1}\) Substitution rule

For a polynomial differential operator \(P(D)\) we have

\[P(D) e^{st} = P(s) e^{st}. \nonumber \]

Proof

This is obvious. We ‘prove it’ by example. Let \(P(D) = D^2 + 8D + 7I\). Then

\[P(D) e^{at} = a^2 e^{at} + 8ae^{at} + 7e^{at} = (a^2 + 8a + 7) e^{at} = P(a) e^{at}. \nonumber \]

Let’s continue to work from this specific example. From it we’ll be able to remind you of the general approach to solving constant coefficient differential equations.

Example \(\PageIndex{3}\)

Suppose \(P(s) = s^2 + 8s + 7\). Find the exponential modes of the equation \(P(D) y = 0\).

Solution

The exponential modes are solutions of the form \(y(t) = e^{s_0 t}\). Using the substititution rule

\[P(D) e^{s_0 t} = 0 \Leftrightarrow P(s_0) = 0. \nonumber \]

That is, \(y (t) = e^{s_0 t}\) is a mode exactly when \(s_0\) is a root of \(P(s)\). The roots of \(P(s)\) are -1, -7. So the modal solutions are

\[y_1(t) = e^{-t} \ \ \text{and} \ \ y_2 (t) = e^{-7t}. \nonumber \]

Example \(\PageIndex{4}\)

Redo the previous example using the Laplace transform.

Solution

For this we solve the differential equation with arbitrary initial conditions:

\[P(D) y = y'' + 8y' + 7y = 0; \ \ y(0) = c_1, y'(0) = c_2. \nonumber \]

Let \(Y(s) = \mathcal{L} (y; s)\). Applying the Laplace transform to the equation we get

\[(s^2 Y(s) - sy(0) - y'(0)) + 8(sY(s) - y(0)) + 7Y(s) = 0 \nonumber \]

Algebra:

\[(s^2 + 8s + 7) Y(s) - sc_1 - c_2 - 8c_1 = 0 \Leftrightarrow Y = \dfrac{sc_1 + 8c_1 + c_2}{s^2 + 8s + 7} \nonumber \]

Factoring the denominator and using partial fractions, we get

\[Y(s) = \dfrac{sc_1 + 8c_1 + c_2}{s^2 + 8s + 7} = \dfrac{sc_1 + 8c_1 + c_2}{(s + 1) (s + 7)} = \dfrac{A}{s + 1} + \dfrac{B}{s + 7}. \nonumber \]

We are unconcerned with the exact values of \(A\) and \(B\). Taking the Laplace inverse we get

\[y(t) = Ae^{-t} + Be^{-7t}. \nonumber \]

That is, \(y(t)\) is a linear combination of the exponential modes.

You should notice that the denominator in the expression for \(Y(s)\) is none other than the characteristic polynomial \(P(s)\).

system function

Example \(\PageIndex{5}\)

With the same \(P(s)\) as in Example 13.7.2 solve the inhomogeneous DE with rest initial conditions: \(P(D) y = f(t)\), \(y(0) = 0\), \(y'(0) = 0\).

Solution

Taking the Laplace transform of the equation we get

\[P(s) Y(s) = F(s). \nonumber \]

Therefore

\[Y(s) = \dfrac{1}{P(s)} F(s) \nonumber \]

We can't find \(y(t)\) explicitly because \(f(t)\) isn't specified.

But, we can make the following definitions and observations. Let \(G(s) = 1/P(s)\). If we declare \(f\) to be the input and \(y\) the output of this linear time invariant system, then \(G(s)\) is called the system function. So, we have

\[Y(s) = G(s) \cdot F(s). \nonumber \]

The formula \(Y = G \cdot F\) can be phrased as

output = system function \(\times\) input.

Note well, the roots of \(P(s)\) correspond to the exponential modes of the system, i.e. the poles of \(G(s)\) correspond to the exponential modes.

The system is called stable if the modes all decay to 0 as \(t\) goes to infinity. That is, if all the poles have negative real part.

Example \(\PageIndex{6}\)

This example is to emphasize that not all system functions are of the form \(1/P(s)\). Consider the system modeled by the differential equation

\[P(D)x = Q(D) f, \nonumber \]

where \(P\) and \(Q\) are polynomials. Suppose we consider \(f\) to be the input and \(x\) to be the ouput. Find the system function.

Solution

If we start with rest initial conditions for \(x\) and \(f\) then the Laplace transform gives \(P(s) X(s) = Q(s) F(s)\) or

\[X(s) = \dfrac{Q(s)}{P(s)} \cdot F(s) \nonumber \]

Using the formulation

output = system function \(\times\) input.

we see that the system function is

\[G(s) = \dfrac{Q(s)}{P(s)}. \nonumber \]

Note that when \(f(t) = 0\) the differential equation becomes \(P(D) x = 0\). If we make the assumption that the \(Q(s)/P(s)\) is in reduced form, i.e. \(P\) and \(Q\) have no common zeros, then the modes of the system (which correspond to the roots of \(P(s)\)) are still the poles of the system function.

Comments

All LTI systems have system functions. They are not even all of the form \(Q(s)/P(s)\). But, in the \(s\)-domain, the output is always the system function times the input. If the system function is not rational then it may have an infinite number of poles. Stability is harder to characterize, but under some reasonable assumptions the system will be stable if all the poles are in the left half-plane.

The system function is also called the transfer function. You can think of it as describing how the system transfers the input to the output.