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Lecture 30

Lecture 29 | Lecture 31


30.1 Fourier Exponential Series and the Transfer Function

We have seen how we can use the Laplace transform to find the the response of an input signal and we even saw a formula for calculating the Laplace transform of a periodic input signal (forcing function) tex:$$f(t) \leftrightarrow F(s)$$; in which case, given the transfer function tex:$$h(t) \leftrightarrow H(s)$$; we can calculate the response by either finding the inverse Laplace transform of tex:$$H(s)F(s)$$ or by calculating the convolution tex:$$h(t)*f(t)$$.

We also saw that the response of an exponential is simply the transfer function evaluated at the angular frequency (radians per second): tex:$$e^{\omega t} \Rightarrow H(\omega) e^{\omega t}$$.

Now relate this fact to the exponential Fourier series, and you will see that you we can now calculate the response of a periodic function simply by using the Fourier series:

Given a system with a periodic input signal defined by the Fourier series

tex:$$x(t) = \sum_{n = -\infty}^\infty a_n e^{jn\omega t}$$

and given a system which has the transfer function tex:$$H(s)$$, the response of the system to the periodic signal may be easily calculated as

tex:$$x(t) = \sum_{n = -\infty}^\infty a_n H(jn\omega) e^{jn\omega t}$$

and that's it: no Laplace transform and inverse Laplace transform. If you know the exponential Fourier series and you know the transfer function, you know the response without ever having to transform to the frequency space. All you are concerned with are the transients.

30.2 Example 1: Integration and Differentiation

Recall that the transfer function of the derivative is tex:$$H_{\rm diff}(s) = s$$ while the transfer function of an integral is tex:$$H_{\rm int}(s) = \frac{1}{s}$$. Now consider the term-by-term differentiation and integration we saw in a previous lecture. Given the Fourier series

tex:$$x(t) = \sum_{n = -\infty}^\infty a_n e^{jn\omega t}$$

we note that a term-by-term derivative is exactly what you would get if you evaluated the transfer function as described above:

tex:$$\dot{x}(t) \sim \sum_{n = -\infty}^\infty a_n jn\omega e^{jn\omega t} = \sum_{n = -\infty}^\infty a_n H_{\rm diff}(jn\omega) e^{jn\omega t}$$.

Similarly, a term-by-term integration also produces the same result as one would expect evaluating the transfer function as described above:

tex:$$\int x(t) dt = \sum_{n = -\infty}^\infty a_n \frac{1}{jn\omega}e^{jn\omega t} = \sum_{n = -\infty}^\infty a_n H_{\rm int}(jn\omega) e^{jn\omega t}$$.

30.3 Example 2: Three Transfer Functions and Three Input Signals and Responses

We will consider the response of three separate signals due to three systems defined by transfer functions, namely:

  • tex:$$H(s) = \frac{1}{s + 1}$$ (see Figure 1)

  • tex:$$H(s) = \frac{1}{s}$$ (an integrator)

  • tex:$$H(s) = \frac{s}{s^2 + s + 16.25}$$ (an RLC circuit).

  • tex:$$H(s) = \frac{s}{s^2 + 3s + 2}$$ (an RLC circuit).

In the lecture on block diagrams, we saw the circuit in Figure 1 and as part of the exercise you were asked to determine the transfer functions for a number of cases.


Figure 1. A simple RC circuit with the transfer function tex:$$H(s) = \frac{1}{s + 1}$$.

The Impulse Responses

The impulse responses (the inverse Laplace transforms of the transfer functions) are shown in Figures 2a through 2d.


Figure 2a. The impulse response tex:$$h(t) = e^{-t}u(t)$$ to the transfer function tex:$$H(s) = \frac{1}{s + 1}$$.


Figure 2b. The impulse response tex:$$h(t) = u(t)$$ to the transfer function tex:$$H(s) = \frac{1}{s}$$.


Figure 2c. The impulse response tex:$$h(t) = 0.125e^{-0.5t}\left(8\cos(4t) - sin(4t)\right)u(t)$$ to the transfer function tex:$$H(s) = \frac{s}{s^2 + s + 16.25}$$.


Figure 2d. The impulse response tex:$$h(t) = \left(2e^{-2t} - e^{-t}\right)u(t)$$ to the transfer function tex:$$H(s) = \frac{s}{s^2 + 3s + 2}$$.

The signals

The four periodic signals are shown in Figures 3a through 3c.


Figure 3a. The periodic extension of tex:$$f(t) = u(t) - u(-t)$$ extended from the interval tex:$$[-1, 1]$$.


Figure 3b. The periodic extension of tex:$$f(t) = 2t$$ extended from the interval tex:$$[-1, 1]$$.


Figure 3c. The periodic extension of tex:$$f(t) = 2 - 4|t|$$ extended from the interval tex:$$[-1, 1]$$.

Response to tex:$$H(s) = \frac{1}{s + 1}$$

The response of each of these three signals are shown in Figures 4a through 4c.


Figure 4a. The response of the signal in Figure 3a to a system with the transfer function tex:$$H(s) = \frac{1}{s + 1}$$.


Figure 4b. The response of the signal in Figure 3b to a system with the transfer function tex:$$H(s) = \frac{1}{s + 1}$$.


Figure 4c. The response of the signal in Figure 3c to a system with the transfer function tex:$$H(s) = \frac{1}{s + 1}$$.

Response to tex:$$H(s) = \frac{1}{s}$$

The response of each of these three signals are shown in Figures 5a through 5c. You will note that in each case, the response is the integral of the signal.


Figure 5a. The response of the signal in Figure 3a to a system with the transfer function tex:$$H(s) = \frac{1}{s}$$.


Figure 5b. The response of the signal in Figure 3b to a system with the transfer function tex:$$H(s) = \frac{1}{s}$$.


Figure 5c. The response of the signal in Figure 3c to a system with the transfer function tex:$$H(s) = \frac{1}{s}$$.

Response to tex:$$H(s) = \frac{s}{s^2 + s + 16.25}$$

The response of each of these three signals are shown in Figures 6a through 6c.


Figure 6a. The response of the signal in Figure 3a to a system with the transfer function tex:$$H(s) = \frac{s}{s^2 + s + 16.25}$$.


Figure 6b. The response of the signal in Figure 3b to a system with the transfer function tex:$$H(s) = \frac{s}{s^2 + s + 16.25}$$.


Figure 6c. The response of the signal in Figure 3c to a system with the transfer function tex:$$H(s) = \frac{s}{s^2 + s + 16.25}$$.

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

The response of each of these three signals are shown in Figures 7a through 7c.


Figure 7a. The response of the signal in Figure 3a to a system with the transfer function tex:$$H(s) = \frac{s}{s^2 + 3s + 2}$$.


Figure 7b. The response of the signal in Figure 3b to a system with the transfer function tex:$$H(s) = \frac{s}{s^2 + 3s + 2}$$.


Figure 7c. The response of the signal in Figure 3c to a system with the transfer function tex:$$H(s) = \frac{s}{s^2 + 3s + 2}$$.