Dynamical Signficance of Poles and Zeros: In the second lecture, we discussed how the poles of the transfer function determine the modes of the system and the form of its natural response. We also saw how the zeros impact the amplitudes of both the forced response and the natural response of the system. We also discussed the significance of the exponential inputs est as the eigenfunctions of LTI systems. The objective of this problem is to summarize the key points on the dynamical significance of poles and zeros. Consider the following proper rational transfer function: H() = where b(s) and a(s) are polynomials in s and the order of b(s) is equal or smaller than the order of a(s) (hence a proper transfer function) and (3) and a(s) have no common factors (i. e., no shared roots). Recall that the output of the system to an input r(t) 5X() can be written as: y(t) SY(8) = 68 X (s) +18) where I(S) is another polynmial in s of order n-1 where n is the order of a(s), and whose coefficients are all determined by the initial conditions of input and output, i. e., y(0),y(0), 7(0),, 2(0), i(0), (0),... (a) Assume an exponential input is applied to the sytem at time t = 0, i. e., (t) = exou(t), where u(t) is the unit step function, and so is not a pole of the tranfer function. Show that the Laplace transform of the output can be written as follows: Y(a)- K B(8) 1(3) Y(S) = 3-Sea(s) a(s) where B(s) is a polynomial in s of order n - 1. What is the value for Ko? Discuss how you can conclude that the output initial conditions y(0), (0-),... can indeed be set such that system response is obtained as: y(t) = (30)esot u(t) (b) What will now happen if so happens to be a zero of the transfer function, i. e., b(30) = 0? This is the main reason why transfer function zeros are sometimes called transmission zeros. (c) Now, assume the input to the system is identically 0, i. e., 2(t) = 0, Vt. Therefore, the Laplace transform of the zero-input system response can be written as: Y(8) - I(8) Y() = als) Assume s = p is a pole of the transfer function, i. e, a(s) sp = 0 (for simplicity, lets assume it is not a repeated pole). Show that the output initial conditions can be set such that: y(t) = Keptu(t) where K is a constant, and, as before, u(t) is the unit step function.