SC · Laplace

The Transfer Function

The transfer function H(s) = output/input describes the system itself, no matter what input you feed it. Because the Laplace transform turns the derivative d/dt into multiplication by s, a differential equation tidies up into a ratio of polynomials. In this lesson you move a single coefficient and watch the differential equation, the transfer function, the pole, and the impulse response move together as one.

One system, four languages

Take the Laplace transform of the differential equation dy/dt + a·y = a·x shown below the slider and you get the transfer function H(s) = a/(s + a); the pole that makes its denominator 0 is s = −a, and the impulse response is h(t) = a·e^(−at). The upper plot marks the current pole (gold ×) and the target pole (gray ×) on the s-plane, and the lower plot shows the current impulse response (gold) and the target response (dashed). The coefficient a starts at 2.4. When you set a to the target with the slider, the pole and the response curve both land on the target at once, because a single coefficient decides the whole system.

s-plane (× = pole −a)

impulse response h(t) = a·e^(−at)

coefficient a2.40

Four representations

H(s) = 2.40 / (s + 2.40)

dy/dt + 2.40 y = 2.40 x   ·   s = −2.40

The gold pole s = −a is away from the gray target pole. The coefficient a of the differential equation is also the constant in the transfer function's denominator and the location of the pole, and it sets how fast the response dies away, so the lower curve misses the dashed one too. A large a moves the pole left and makes the response spike high and die away fast; a small a does the opposite.

Off

Differentiation becomes multiplication

When all initial values are 0, the Laplace transform turns the derivative d/dt into multiplication by s. Transform both sides of the linear differential equation a system obeys, and every derivative becomes a power of s, leaving an algebraic equation in Y(s) and X(s). Rearranging it as the ratio of the output Y(s) to the input X(s) gives a ratio of two polynomials in s, H(s) = N(s)/D(s). For the first screen's dy/dt + a·y = a·x, sY + aY = aX gives H(s) = a/(s + a). A tedious differential equation shrinks to a single fraction.

ObserveH(s) = Y(s)X(s)

The transfer function is the output divided by the input.

ChooseLaplace: d/dt → ?

Under the Laplace transform, d/dt becomes s.

Fill inY(s) = H(s) ?

The output is H(s) times the input.

On your ownpoles: D(s) = ?

A pole is a root that makes the denominator 0.

The transfer function is the system’s ID

H(s) = Y(s)/X(s) is the same function whatever input you apply, and since it does not depend on the input, you can regard it as the system itself. Feed in an impulse and X(s) = 1, so Y(s) = H(s); the transfer function is exactly the Laplace transform of the impulse response, H(s) = L{h(t)}. This carries over to the s-domain what the B1 unit said: knowing the impulse response alone tells you everything about the system. Whatever the input, the output is Y(s) = H(s)X(s), found with a single multiplication in the s-domain. The 'convolution is multiplication' property from the C4 unit holds here with transients included.

Poles are the roots of the denominator

When you build the transfer function from a differential equation, the denominator D(s) is exactly that equation's characteristic polynomial. Its roots, the solutions of D(s) = 0, are the poles, and as the previous unit showed, they set the modes of the natural response and the stability. The roots of the numerator N(s) are the zeros. So once you know the coefficients of the differential equation, you can factor the denominator and numerator of H(s) to get the poles and zeros directly, and read from their locations how the system will move. The first-order system on the first screen has only one pole, −a, so as long as a > 0 its response simply died away with no oscillation.

Back to the first screen

On the first screen, moving the coefficient a moved the differential equation, the transfer function H(s) = a/(s + a), the pole s = −a, and the impulse response a·e^(−at) all together. That is because the four are four ways of writing the same system. Since the Laplace transform turned the derivative into s and the equation into a single fraction, the system became H(s), something you can handle with one multiplication. The roots of its denominator are the poles, and for any input the output is Y(s) = H(s)X(s).

The transfer function H(s) = Y(s)/X(s) is the ratio of the Laplace transforms of output and input, and it is the system itself, independent of the input. Transforming a differential equation turns d/dt into s, giving H(s) = N(s)/D(s), a ratio of polynomials in s. H(s) = L{h(t)} (the transform of the impulse response), and for any input the output is Y(s) = H(s)X(s). The roots of the denominator D(s) = 0 are the poles, and the roots of the numerator N(s) = 0 are the zeros (D2 unit).

On to the next unit

If one system is written as one H(s), then several systems joined together can also be combined into a single H(s). In the next unit, Block Diagrams and Feedback, you draw systems as boxes and learn the rules: in series the transfer functions multiply, in parallel they add, and with feedback, where the output of the forward path H is fed back through a path G, they combine as H/(1 + HG). Feedback is the core tool of control, moving pole locations to make an unstable system stable and a slow system fast. That is where the story of signals and systems turns into control.