Input-output system

MIT OpenCourseWarehttp://ocw.mit.edu 18.034 Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. � � LECTURE 21. STEP FUNCTIONS Step and impulse signals: motivation. In the analysis of P(D)y:=y(n) +a1y(n−1) +···+any=f we often view the differential equation as an input-output system, as shown below. f input P (D) y output Figure 21.1. The input-output system. Coefficients of P(D)are parameters of the system, e.g. the spring constant in a spring-mass system or the resistance in the circuit analysis. In real life, we do not know the parameters in advance. Instead, we learn about the system by watching how it responds to various input signals. The simpler the input signal is, the clearer we should expect the signature of the system parameeter to be, and the easier we predict how the system will respond to other more complicated signals. The simplest is the null signal, which corresponds to the homogeneous equation. We study two other standard and very simple signals: the unit step function and the unit impulse. Here, we study step signals. Impulsive signals will be discussed in later lectures. Step Signals. The unit step function or the Heaviside∗ unit function is defined as 0, t<0(21.1) h(t)= 1,t� 0. It has a jump discontinuity at t=0. The graph of h(t)is given below. Figure 21.2. Graph of h(t). By definition, 0, t 0. That is, f(t + p)= f(t) for all t. The function (21.3) f0(t)= f(t)h(t) − f(t)h(t − p) agrees with f(t) on the interval t ∈ [0,p) and is zero elsewhere. In this sense, f0 represents f on a single period, and f is the periodic extension of f0. Since f is periodic, we can replace f(t) in the second term of (21.3) by f(t − p). Taking the transform, then gives Lf0 = Lf − e−psLf. Solving for Lf, we obtain (21.4) Lf = Lf0 , where Lf0 = p e −stf(t)dt.1 − e−ps 0 Example 21.4. The square-wave function is defined by the equation 1,t ∈ [2n, 2n +1) f(t)= −1,t ∈ [2n − 1, 2n), where n � 0 is an integer. 1 2 3 4 1 -1 Figure 21. 5. Graph of the square-wave function. Let f0(t)= h(t) − 2h(t − 1) + h(t − 2) = � 1, −1, t ∈ [0, 1) t ∈ [1, 2). Lecture 21 3 18.034 Spring 2009 1 − 2e−s + e−2s (1 − e−s)2 Then, L[f0(t)](s)= = . Therefore, by (21.4) ss 1 (1 − e−s)2 1 1 − e−s Lf = = . s 1 − e−2s s 1+ e−s Lecture 21 4 18.034 Spring 2009