· The signal d-h(t).dh(t) dt is of finite duration h(t) has exactly one zero at infinity Determine H(s)and its ROC Solution: To determine H(s) and its ROC, we need to analyze and combine all the informa- n given. The first piece of information we are given is that the system is causal and stable. Because the system is causal, we know that the ROC is right-sided. Because the system is stable, we know that the roc includes the jw-axis The next piece of information, The steady state response to a unit step, i.e., s(oo)=3,gives us information about H(s)ls=0. Note that the step response is s(t)=o h(r)dT. Thus, for t=oo, s(oo)=L_oo h(r)dr. But this is the same Laplace transform equation that can be used to solve for H(SIs=0. Thus, H(O)=3 The next piece of information, When the input is eu(t), the output is absolutely integrable, gives us information about a zero of H(s). We know that if a(t)=etu(t) then X(s=d, es>1 and Y(s=H(sX(s). The ROC for Y(s) will be at least the intersection of the ROC for X(s with the ROC for H(s). If y(t) is absolutely integrable, then we can take a Fourier transform of i. e, the ROC includes the jw-axis. Because of Property 2 in Chapter 9 of O& w, the ROC of any system, Y(s included, does not include any poles. Thus, the pole in Y(s)at s= l is eliminated by having a zero at 8= l in H(s) Jumping to the fourth bullet point, h(t) has exactly one zero at infinity gives us information about the relative orders of the numerator and denominator for a rational tranform. Specifically, the order of the denominator is one greater than the order of the numerator Thus, we know that the denominator has two poles The third bullet point gives us information about the poles of H(s). By Property 3 in Chapter 9 of O&w, if a signal is of finite duration and is absolutely integrable then the roc of the signal is the entire s-plane. We know that d h(t).dh(t) +6h(t)is of finite duration. Is it als integrable? We can show that it is by looking individually at each of the terms in the function We know that h(t) is absolutely integrable because it is stable, i. e, its ROC includes the ju-axis. Multiplying h(t) by 6(a constant)will not change its absolute integrability. From Table 9.1, the ROC of the deriviative of a function includes the roc of the original function. Thus 5- will include the ju-axis and be absolutely integrable. Likewise the second derivative the rOC of the first derivative and thus, it is absolutely integrable also. From Table 9.2, the sum of 3 functions has at least the intersection of the rocs of each of the three functions. Since the ROC of each of these functions includes the jw-axis, the sum will include the jui-axis and thus, the� � | • The signal d2h(t) dh(t) + 5 + 6h(t) dt2 dt is of finite duration. • h(t) has exactly one zero at infinity. Determine H(s) and its ROC. Solution: To determine H(s) and its ROC, we need to analyze and combine all the informa￾tion given. The first piece of information we are given is that the system is causal and stable. Because the system is causal, we know that the ROC is right-sided. Because the system is stable, we know that the ROC includes the j�-axis. The next piece of information, The steady state response to a unit step, i.e., s(∼) = 1 3 , gives us � t information about H(s) s=0. Note that the step response is s(t) = −� h(� )d� . Thus, for t = ∼, s(∼) = h(� )d� . But this is the same Laplace transform equation that can be used to solve for −� H(s)|s=0. Thus, H(0) = 1 3 . The next piece of information, When the input is et u(t), the output is absolutely integrable, gives 1 us information about a zero of H(s). We know that if x(t) = et u(t) then X(s) = s−1 , √e{s} > 1 and Y (s) = H(s)X(s). The ROC for Y (s) will be at least the intersection of the ROC for X(s) with the ROC for H(s). If y(t) is absolutely integrable, then we can take a Fourier transform of it, i.e., the ROC includes the j�-axis. Because of Property 2 in Chapter 9 of O&W, the ROC of any system, Y (s) included, does not include any poles. Thus, the pole in Y (s) at s = 1 is eliminated by having a zero at s = 1 in H(s). Jumping to the fourth bullet point, h(t) has exactly one zero at infinity, gives us information about the relative orders of the numerator and denominator for a rational tranform. Specifically, the order of the denominator is one greater than the order of the numerator. Thus, we know that the denominator has two poles. The third bullet point gives us information about the poles of H(s). By Property 3 in Chapter 9 of O&W, if a signal is of finite duration and is absolutely integrable then the ROC of the signal is d2h(t) dh(t) the entire s-plane. We know that + 5 + 6h(t) is of finite duration. Is it also absolutely dt2 dt integrable? We can show that it is by looking individually at each of the terms in the function. We know that h(t) is absolutely integrable because it is stable, i.e., its ROC includes the j�-axis. Multiplying h(t) by 6 (a constant) will not change its absolute integrability. From Table 9.1, the ROC of the deriviative of a function includes the ROC of the original function. Thus, 5 dh(t) will dt include the j�-axis and be absolutely integrable. Likewise the second derivative, d2h(t) will include dt2 the ROC of the first derivative and thus, it is absolutely integrable also. From Table 9.2, the sum of 3 functions has at least the intersection of the ROC’s of each of the three functions. Since the ROC of each of these functions includes the j�-axis, the sum will include the j�-axis and thus, the 9