MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems--Fall 2003 PROBLEM SET 9 SOLUTIONS Home Study 1 O&w 9.25(e) We assume that the system we are considering, H(s)takes the following form: b s+a+jwo(s+ s+bs+2as+a2+∞) where b corresponds to the pole and the zero on the real axis and a and wo are characterizing the complex conjugate zeros with wo >a. Thus, we can see that H(s)is a cascade of two LTI systems H1(s) and H2(s each of which is examined more detailed below The pole-zero diagram of Hi(s)is shown below. It has a pole and a zero on the real axis which are spaced equally from the origin m The magnitude response is the ratio of the magnitude of the vectors from the zeros to the vectors from the poles as we traverse the ju axis. From any point along the ju-axis, the pole and zero vectors have equal length, and, consequently, the magnitude of the frequency response, H1Gw)l is 1 and independent of w. This is an all-pass system discussed in section 9. 4.3. Therefore H1(ju)=1 This can be, of course, clearly seen from the equation as well: IHGa)= 2u +bl ,2⊥1 H1(u)=1
� � � � � � � � MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 9 Solutions Home Study 1 O&W 9.25(e) We assume that the system we are considering, H(s) takes the following form: H(s) = s − b (s + a + j�0)(s + a − j�0) s + b s − b 2 + �2 = (s2 + 2as + a 0), s + b where b corresponds to the pole and the zero on the real axis and a and �0 are characterizing the complex conjugate zeros with �0 > a. Thus, we can see that H(s) is a cascade of two LTI systems, H1(s) and H2(s) each of which is examined more detailed below. The pole-zero diagram of H1(s) is shown below. It has a pole and a zero on the real axis which are spaced equally from the origin. ←−p ←−z � →m √e × −b b The magnitude response is the ratio of the magnitude of the vectors from the zeros to the vectors from the poles as we traverse the j� axis. From any point along the j�-axis, the pole and zero vectors have equal length, and, consequently, the magnitude of the frequency response, |H1(j�)| is 1 and independent of �. This is an all-pass system discussed in section 9.4.3. Therefore, |H1(j�)| = 1. This can be, of course, clearly seen from the equation as well: |H1(j�)| = j� − b j� + b = � �2 + b2 = � �2 + b2 � |H1(j�)| = 1. 1 |j� − b| |j� + b|
The pole-zero diagram of H2(s) is shown below. It has 2 zeros in the left half plane, and as we assume at the beginning wo >a Thus, we can see H2(s)as an inverse of a second order system with small damping ratio H2(s) tt While w<< wo, H2(u)remains roughly equal to wo. As w approaches wo, IH2 (ju)I attains minimum since the length of the vector, a' 1 stemming from s= -a+ jwo becomes shortest However, because of the effect of the other zero, IH2(jw)l does not achieve its minimum exactly at w=wo, but at w slightly smaller than wo. In the vicinity of w= wo, the length of the vector from the other zero, z 2 does not change much. Thus, the variation of H2 u)l depends mostly on how the z 1 changes its length. For w>>w0, H2 Gju)l increases quadratically against linear w. Thus, the overall magnitude plot is identical to that of H2 (jw) and is sketched below for w>0 H(元u) Note that the scaling on w and H(w)l axes are not the same
e � The pole-zero diagram of H2(s) is shown below. It has 2 zeros in the left half plane, and as we assume at the beginning �0 > a. →m j�o −a √ −j�o Thus, we can see H2(s) as an inverse of a second order system with small damping ratio: 2 + �2 H2(s) = s2 + 2as + a 0. While � << �0, H2(j�) remains roughly equal to �2 | | 0. As � approaches �0, H2(j�) attains its minimum since the length of the vector, −←z 1 stemming from s = −a + j�0 becomes shortest. However, because of the effect of the other zero, | | H2(j�) does not achieve its minimum exactly at � = �0, but at � slightly smaller than �0. In the vicinity of � = �0, the length of the vector from the other zero, −←z 2 does not change much. Thus, the variation of | | H2(j�) depends mostly on how the −←z 1 changes its length. For � >> �0, | | H2(j�) increases quadratically against linear �. Thus, the overall magnitude plot is identical to that of H2(j�) and is sketched below for � > 0: �0 �2 0 |H( ) j� | Note that the scaling on � and | | H(j�) axes are not the same. 2
Home study 2 O& W 9.40 Because we are dealing with a system which has initial conditions, it is easier to use the unilateral Laplace transform. From the properties of the unilateral Laplace transform, we get the followin y'(t) 3y(s)-y(0-) y"(t) s2y(s)-s(0-)-y(0-) 3yVs)-s2y(0-)-sy/(0-)-y(-) Substituting these into the differential equation and solving for y(s), we get y(s) 3+6:2+1s+6x~(0-)+sb(0-)+60(0-)+"0-)+6y/0-)+110 x(s) ZSR ZIR We have indicated in the previous equation that y(s)can be split into two parts. The first part the ZSR, corresponds to the output when the system is initially at rest and solely responds to the input. The second part, the AIR, corresponds to the output when there is no input and the system ds only to its initial stat (a) To determine the ZSR, we can simply use the equation derived on the previous page, (s3+6s2+1ls+6)(s+4) (s+1)(s+2)(s+3)(s+4) 8+4 Taking the inverse transform gives 1 () 1 u()+2-()-ca( (b) Using the ZiR portion derived previously, we can find the ZIR by substituting in the given i +5s+6 ZIR(S 3+6s2+11s+6 (c) Then, by superposition y(t)= yzs(t) 1 (t)+
� �� � ��� � �� Home study 2 O&W 9.40 Because we are dealing with a system which has initial conditions, it is easier to use the unilateral Laplace transform. From the properties of the unilateral Laplace transform, we get the following relationships: y(t) �← Y(s) y (t) sY(s) − y(0− �← ) y (t) s2Y(s) − sy(0−) − y (0− �← ) y (t) �← s3Y(s) − s2y(0−) − sy (0−) − y (0−) Substituting these into the differential equation and solving for Y(s), we get Y(s) X (s) s2y(0−) + s [y� (0−) + 6y(0−)] + [y��(0−) + 6y� (0−) + 11y(0−)] = + . s3 + 6s2 + 11s + 6 s3 + 6s2 + 11s + 6 � �� � � �� � ZSR ZIR We have indicated in the previous equation that Y(s) can be split into two parts. The first part, the ZSR, corresponds to the output when the system is initially at rest and solely responds to the input. The second part, the ZIR, corresponds to the output when there is no input and the system responds only to its initial state. (a) To determine the ZSR, we can simply use the equation derived on the previous page, 1 YZSR (s) = (s3 + 6s2 + 11s + 6)(s + 4) 1 1 1 1 1 = 6 2 2 6 (s + 1)(s + 2)(s + 3)(s + 4) = s + 1 − s + 2 + s + 3 − s + 4 . Taking the inverse transform gives 1 1 −3t 1 −4t yZSR (t) = 6 e−t u(t) − 1 e−2t u(t) + e u(t) − e u(t). 2 2 6 (b) Using the ZIR portion derived previously, we can find the ZIR by substituting in the given i s2 + 5s + 6 1 Y = = . ZIR (s) s3 + 6s2 + 11s + 6 s + 1 y = e u(t) ZIR (t) −t (c) Then, by superposition, y(t) = yZSR(t) + yZIR (t) 7 −t 1 −2t 1 −3t 1 −4t = e u(t) − e u(t) + e u(t) − e u(t). 6 2 2 6 3
Problem 1 Consider an LTI system for which the system function H (s is rational and has the pole-zero pattern shown below m (a) Indicate all possible ROC's that can be associated with this pole-zero pattern The ROCs are bounded by vertical lines through the location of the poles as in the figure below m Thus, the possible ROCs are Res] 4<咒e{s}<-1 <e{s} <咒e{s} (b) For each ROC identified in Part(a), specify whether the associated system is stable and /or causal Causal systems have ROCs that are to the right of the right-most pole. Stable systems are systems whose ROCs include the jw-axis Res)<-4: Not Causal. Not Stable
Problem 1 Consider an LTI system for which the system function H(s) is rational and has the pole-zero pattern shown below: →m × − 1 2 √e × 4 −2 − × 1 (a) Indicate all possible ROC’s that can be associated with this pole-zero pattern. The ROCs are bounded by vertical lines through the location of the poles as in the figure below: →m −4 √e × −2 −1 × 1 2 × Thus, the possible ROCs are: √e{s} < −4 −4 < √e{s} < −1 −1 < √e{s} < 2 2 < √e{s} (b) For each ROC identified in Part (a), specify whether the associated system is stable and/or causal. Causal systems have ROCs that are to the right of the right-most pole. Stable systems are systems whose ROCs include the j�-axis. √e{s} < −4 : Not Causal. Not Stable 4
4< Res <-1: Not Causal. Not Stable 1< es)<2: Not Causal. Stable 2<e{s} Causal. Not Stable Problem 2 Draw a direct-form representation for the causal lti system with system function H Note that the system function can be represented as follow H(s) (s+3)(s+4) s2+7s+12 w(s)X(s) Thus, we can see the system H(s)as a cascade of two systems, i. e, Z(s)=wo which accounts for the zeros and P(s)=yio which accounts for the poles First, let's draw a block diagram representation of the system P(s). Since the system is of second order, we would like to represent the system using only two integrators in cascade w(t) w(t) w(t) (t)
−4 < √e{s} < −1 : Not Causal. Not Stable −1 < √e{s} < 2 : Not Causal. Stable 2 < √e{s} : Causal. Not Stable Problem 2 Draw a direct-form representation for the causal LTI system with system function s(s + 1) H(s) = . (s + 3)(s + 4) Note that the system function can be represented as follows: H(s) = s(s + 1) (s + 3)(s + 4) s2 + s = s2 + 7s + 12 = Y (s) W(s) W(s) X(s) = � �� � s2 + s 1 s2 + 7s + 12 . Y (s) � �� � W(s) W(s) X(s) Y (s) Thus, we can see the system H(s) as a cascade of two systems, i.e., Z(s) = W(s) which accounts W(s) for the zeros and P(s) = X(s) which accounts for the poles. First, let’s draw a block diagram representation of the system P(s). Since the system is of second order, we would like to represent the system using only two integrators in cascade. x(t) +− w¨(t) 1 s 1 s w˙ (t) 7 w(t) 12 + 5