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>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�) 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
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, 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, 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
2. We can determine the differential equation relating y(t) and a(t)from the system function found in(a). Since H(s)= xo, we multiply the denominator of H(s) by y(s) and we multiply the numerator of H (s by X(s) Y(s)(s2+3s+2)=X(s)(s+3) Distributing on both sides s Y(s+3sY(s+2Y(s=sX(s)+3X(s) Because of linearity, we can take the inverse Laplace transform of each term above and get the differential equation y(t) (t) 3r(t) Problem 4 Suppose we are given the following information about a causal and stable LTI system ith impulse response h(t)and a rational function H (s) The steady state response to a unit step, i.e., s(oo)=3 When the input is eu(t), the output is absolutely integrable
−1 × √e →m 2. We can determine the differential equation relating y(t) and x(t) from the system function found in (a). Since H(s) = Y (s) X(s) , we multiply the denominator of H(s) by Y (s) and we multiply the numerator of H(s) by X(s): Y (s)(s2 + 3s + 2) = X(s)(s + 3). Distributing on both sides: s2Y (s) + 3sY (s) + 2Y (s) = sX(s) + 3X(s) Because of linearity, we can take the inverse Laplace transform of each term above and get the differential equation: d2y(t) dt2 + 3dy(t) dt + 2y(t) = dx(t) dt + 3x(t). Problem 4 Suppose we are given the following information about a causal and stable LTI system with impulse response h(t) and a rational function H(s): • The steady state response to a unit step, i.e., s(∼) = 1 3 . • When the input is et u(t), the output is absolutely integrable. 8
· 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 information 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
sum is absolutely integrable. Because the ROC of this signal, d ht+5-n(t+ 6h(t)is the entire s-plane, there can be no poles except at oo. However, we know that H(s)has at least two poles. In order for the signal to have no poles, we must make sure that the poles of H(s)are cancelled by eros of the signal. Taking the Laplace transform of the signal gives us s2H(s)+5sH(s)+6(s)=(s+2)(s+3)H(s) Thus, H(s) can only have two poles, at s=-2 and s=-3 Combining all the information about the poles and zeros gives us H(s=K (s+2)(s+3) e{s}>-2. Finally, we choose K such that H(O)=3. That is, K=-2 and s-1 s) (s+2)(s+3) 界e{s}>-2. Problem 5 Consider the basic feedback system of Figure 11.3 (a)on p 819 of O&w. Determine the closed-loop system impulse response when 2 H(s)= G(s) We can use Black's Formula to compute the system function for the entire system, Q(s. This (s+5)(+2) (s+2) s2+7s+12 Next, we can use partial fraction expansion to write Q(s)as a sum of lst order terms Q(s) +78+12 (s+4)(s+3) Since this is a feedback system, we know it is causal. Thus, we find the inverse Laplace transform for the system function using the causal part to obtain the inpulse response as follows q(t)=2e u(t)-e-sut
d2h(t) dh(t) sum is absolutely integrable. Because the ROC of this signal, + 5 + 6h(t) is the entire dt2 dt s-plane, there can be no poles except at ∼. However, we know that H(s) has at least two poles. In order for the signal to have no poles, we must make sure that the poles of H(s) are cancelled by zeros of the signal. Taking the Laplace transform of the signal gives us s2H(s) + 5sH(s) + 6H(s) = (s + 2)(s + 3)H(s). Thus, H(s) can only have two poles, at s = −2 and s = −3. Combining all the information about the poles and zeros gives us, H(s) = K s − 1 (s + 2)(s + 3), √e{s} > −2. Finally, we choose K such that H(0) = 1 3 . That is, K = −2 and H(s) = −2 s − 1 (s + 2)(s + 3), √e{s} > −2. Problem 5 Consider the basic feedback system of Figure 11.3 (a) on p.819 of O&W. Determine the closed-loop system impulse response when 1 2 H(s) = , G(s) = . s + 5 s + 2 We can use Black’s Formula to compute the system function for the entire system, Q(s). This is given by: 1 Q(s) = s+5 1 + 2 (s+5)(s+2) = (s + 2) s2 + 7s + 12 Next, we can use partial fraction expansion to write Q(s) as a sum of 1st order terms. s + 2 Q(s) = s2 + 7s + 12 s + 2 = (s + 4)(s + 3) 2 1 = s + 4 − s + 3 Since this is a feedback system, we know it is causal. Thus, we find the inverse Laplace transform for the system function using the causal part to obtain the inpulse response as follows: −3t q(t) = 2e−4t u(t) − e u(t) 10
