Here, note that i(t)="wult and i(t)=dult. It is easy to confirm that the abo ve Im tion of the system indeed represents the system P(s by recursively applying Black's formula Now, we would like to use the system above to implement the other system, Z(s). Note that s in Laplace domain corresponds to differentiation in time domain. Thus, y(t) is nothing but a linear combination of different orders of derivatives of w(t); in our case, y(t)=lxw(t)+lxw(t)+oxw(t) Thus, a direct form representation of the overall system H(s)is as shown below r(t) From this representation of the system, the numbers in the gain boxes are picked off simply from the coefficients of the numerator and denominator for a given rational system function. Often, when the gain terms are 1, they are omitted. When the gain terms are 0, the branch is often omitted Problem 3 Consider the cascade of two lti systems as depicted below: (t) System a w(t) System where we have the followin System A is causal with impulse response
Here, note that w(t) = dt ¨ 2 d2w(t) and w˙ (t) = dw(t) . It is easy to confirm that the above implemen dt tation of the system indeed represents the system P(s) by recursively applying Black’s formula. Now, we would like to use the system above to implement the other system, Z(s). Note that s in Laplace domain corresponds to differentiation in time domain. Thus, y(t) is nothing but a linear combination of different orders of derivatives of w(t); in our case, y(t) = 1×w¨(t)+1×w˙ (t)+0×w(t). Thus, a direct form representation of the overall system H(s) is as shown below: x(t) +− 1 s 1 s 7 12 + 1 + 1 + 0 y(t) From this representation of the system, the numbers in the gain boxes are picked off simply from the coefficients of the numerator and denominator for a given rational system function. Often, when the gain terms are 1, they are omitted. When the gain terms are 0, the branch is often omitted. Problem 3 Consider the cascade of two LTI systems as depicted below: x(t) w(t) System A System B y(t) where we have the following: • System A is causal with impulse response h(t) = e−2t u(t) 6
System B is causal and is characterized by the following differential equation relating its input, w(t), and output, y(t) dy(t)+y(t)=dt dw(t) dt aw(t) If the input z(t)=e-t, the output y(t)=0 1. Find the system function H(s)=Y(s/X(s, determine its ROC and sketch its pole-zero pat- tern. Note: Your answer should only have numbers in them (i.e, you have enough information to determine the value of a) 2. Determine the differential equation relating y(t) and 2(t) Solutions 1. The overall system function H(s) is HA(s x HB(s) since the systems'A and B are cascaded together. From O&W Table 9.2 Laplace transforms of elementary functions HA(s) TB(s) is determined by letting a(t)=est in the differential equation given for system B Then y(t)=H(ses and we find 8+a HB(s) 咒e{s}>-1 The ROC was known to be Res>-1 and not Res<-1 because we are told system B is causal. We need to solve for a. Along with the differential equation relating the input to the output for system B, we are told that if c(t)=e-t, an eigenfunction of an LTI system then y(t)=0 Here, the concept we wished to test was the eigenfunction property of Lti sys- tems. However, this problem was ill-posed. Namely, in order to apply the eigen- function property,-3 from e-t, should have been in the ROC of H(s)so that H(3 had existed. However, since the ROC of H(s) is Res)>-1, H(3)does not exist. Therefore, you did not have enough information to determine H(s)for this problem. The solution below is just to illustrate a possible method to obtain H(s) assuming that H(3) existed Thus, HB(s)s=-3=0. This constraint allows us to solve for a Specifically, B(-3) Therefore, the overall cascaded system function is (s+1)(s+2) The ROC must not have any poles in it and so the overall ROC must be to the right of the The pole-zero plot is shown below
• System B is causal and is characterized by the following differential equation relating its input, w(t), and output, y(t): dy(t) dw(t) + y(t) = + �w(t) dt dt If the input x(t) = e−3t • , the output y(t) = 0. 1. Find the system function H(s) = Y (s)/X(s), determine its ROC and sketch its pole-zero pattern. Note: Your answer should only have numbers in them (i.e., you have enough information to determine the value of �). 2. Determine the differential equation relating y(t) and x(t). Solutions: 1. The overall system function H(s) is HA(s) × HB(s) since the systems’ A and B are cascaded together. From O&W Table 9.2 Laplace transforms of elementary functions, 1 HA(s) = , √e{s} > −2. s + 2 HB(s) is determined by letting x(t) = est in the differential equation given for system B. Then y(t) = H(s)est and we find s + � HB(s) = , √e{s} > −1. s + 1 The ROC was known to be √e{s} > −1 and not √e{s} < −1 because we are told system B is causal. We need to solve for �. Along with the differential equation relating the input to the output for system B, we are told that if x(t) = e−3t , an eigenfunction of an LTI system, then y(t) = 0. Here, the concept we wished to test was the eigenfunction property of LTI systems. However, this problem was ill-posed. Namely, in order to apply the eigenfunction property, −3 from e−3t , should have been in the ROC of H(s) so that H(−3) had existed. However, since the ROC of H(s) is √e{s} > −1, H(−3) does not exist. Therefore, you did not have enough information to determine H(s) for this problem. The solution below is just to illustrate a possible method to obtain H(s) assuming that H(−3) existed. Thus, HB(s)|s=−3 = 0. This constraint allows us to solve for �. Specifically, HB(−3) = −3 + � = 0 −← � = 3. −3 + 1 Therefore, the overall cascaded system function is s + 3 H(s) = (s + 1)(s + 2), √e{s} > −1 The ROC must not have any poles in it and so the overall ROC must be to the right of the all poles in the system. The pole-zero plot is shown below. 7