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Lecture 20

Lecture 19 | Lecture 21

The Transfer Function and the Impulse Response

We will now look at an important means of understanding the behaviour of circuits and other linear systems. Previously, we solved differential equations by finding first the general solution to the homogeneous equation, then finding a particular solution, and then matching the solution to the initial conditions. Unfortunately, the particular solution is not unique and therefore nothing physical can be interpreted from the particular solution or its effect on the coefficients of the general homogeneous solution.

Instead, we will see that the Laplace transform allows us to divide the solution into two components:

  • The response of the system due to the initial conditions, and
  • The response of the system due to the input (or forcing function).

Consider the differential equation for the circuit shown in Figure 1.

Figure 1. An RC circuit.

If the charge on the capacitor (the output or response) is given by tex:$$y(t)$$ then we may relate the forcing function (or input) tex:$$x(t)$$ with the response through tex:$$\dot{y} + \frac{3}{2C}y = x(t)$$. Assume, also, that the charge on the capacitor at time tex:$$t = 0$$ is tex:$$y(0) = y_0$$. Let us define tex:$$a = \frac{3}{2C}$$ and therefore the differential equation simplifies to tex:$$\dot{y} + ay = x(t)$$. Using the Laplace transform, we get that the solution is

tex:$$sY(s) + y_0 + aY(s) = X(s)$$.

We may solve this for tex:$$Y(s)$$ to get

tex:$$Y(s) = \frac{X(s)}{s + a} + \frac{y_0}{s + a}$$.

Notice that the first time depends only on the forcing function (or input) and the second term depends only on the initial conditions. Using the inverse Laplace transform and the convolution theorem, we get that

tex:$$y(t) = \int_0^t e^{-s\tau} x(t - \tau)d\tau + y_0 e^{-st}$$


  • tex:$$\int_0^t e^{-s\tau} x(t - \tau)d\tau$$ is the response to the input, and
  • tex:$$y_0 e^{-st}$$ is the response to the initial conditions.

In a stable system (necessary), the initial conditions are transient, that is, they decay to zero, and therefore the focus is on the response to the input. In this example, the input response in the frequency domain is the Laplace transform of the input (tex:$$U(s)$$) multiplied by the factor tex:$$\frac{1}{s + a}$$. This last term is called the transfer function and is written as

tex:$$H(s) = \frac{1}{s + a}$$.

Notice that the transfer function is defined as tex:$$H(s) = \frac{Y(s)}{X(s)}$$ and may be thought of as the ratio between Laplace transforms of the the output and the input under the assumption that the initial conditions are zero, that is, the system is initially quiescent.

Once we have the transfer function, it becomes straight-forward to find the response to a system:

tex:$$y(t) = \int_0^t x(\tau) h(t - \tau) d\tau = x(t) * h(t) = L^{-1}\{H(s)X(s)}$$.

Example 1

Find the transfer function of a system where the output/response tex:$$y(t)$$ to an input tex:$$x(t)$$ is defined by the differential equation

tex:$$\ddot{y} + 4\dot{y} + 3y = \dot{x} + 2x$$.

The Laplace transform gives us the expression

tex:$$s^2Y(s) + 4sY(s) + 3Y(s) = sX(s) + 2X(s)$$
tex:$$\left(s^2 + 4s + 3\right)Y(s) = \left(s + 2\right)X(s)$$

Therefore, the transfer function, the ratio between the Laplace transforms of the output and the input, is

tex:$$H(s) = \frac{s + 2}{s^2 + 4s + 3}$$.

Notice that for a linear system with constant coefficients, the transfer function is a rational polynomial, that is, it is the ratio of two polynomials.

The degree of the denominator is the order of the system, in this case, two.

The zeros of the numerator are the zeros of the transfer function, while the zeros of the denominator are the roots of the transfer function. In this case:

  • The characteristic equation of the transfer function is tex:$$s^2 + 4s + 3 = 0$$,
  • The poles of the transfer function are -1 and -3, and
  • The zero of the transfer function is -2.

The zeros and the poles of the transfer function determine the stability of the system and this will be a significant focus later in this class and in your signals and systems course.

Suppose now we have a forcing function which has tex:$$X(s) = 1$$, that is, tex:$$x(t) = \delta(t)$$. In this case, the response of the system is the transfer function:

tex:$$Y(s) = H(s) X(s) = H(s)$$

In this case, if we denote the inverse Laplace transform of the transfer function to be tex:$$h(t)$$, that is, tex:$$h(t) \leftrightarrow H(s)$$, it follows that

tex:$$y(t) = \int_0^t h(\tau) \delta(t - \tau) d\tau = h(t)$$.

Thus, the transfer function is the response of the system to the unit impulse forcing function and this impulse response is usually denoted tex:$$h(t)$$.

Put another way, the impulse response and the transfer function both contain all information necessary about the dynamics of a time-invariant system.

Conclusion: to determine the response of any linear time-invariant system, all it is necessary to do is to start in a quiescent state, excite it with an impulse, and measure the response.

Note: the adjective time-invariant simply indicates that it does not matter when you choose time tex:$$t = 0$$; that is, the the system behaves the same way yesterday, today, and tomorrow.

Example 2

Find the impulse response of the system described by the differential equation

tex:$$\ddot{y} + 6\dot{y} + 8y = 3x$$.

The transfer function is

tex:$$H(s) = \frac{3}{s^2 + 6s + 8}$$

and therefore, we can find the inverse Laplace transform by using partial fraction decomposition

tex:$$H(s) = 1.5 \frac{1}{s + 2} - 1.5 \frac{1}{s + 4}$$

and hence

tex:$$h(t) = 1.5 e^{-2t} - 1.5 e^{-4t}$$

Two Important Transfer Functions

Consider the system

tex:$$y = \dot{x}$$.

That is, the response is the derivative of the input function. Taking the Laplace transform tex:$$Y(s) = sX(s)$$ and therefore the transfer function, the ratio of the Laplace transforms of the output over the input, is

tex:$$H(s) = \frac{Y(s)}{X(s)} = s$$.

Consider the system

tex:$$\dot{y}(t) = x$$.

That is, the response is the integral of the input function. Taking the Laplace transform tex:$$sY(s) = X(s)$$ and therefore the transfer function, the ratio of the Laplace transforms of the output over the input, is

tex:$$H(s) = \frac{Y(s)}{X(s)} = \frac{1}{s}$$.

In the next section, we will see how engineers will often refer to an integrator or a differentiator of a signal by their transfer function tex:$$\frac{1}{s}$$ and tex:$$s$$, respectively.