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# Wcontrols - ECE 301 Detailed Solutions Set 7 Control...

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Unformatted text preview: ECE 301 Detailed Solutions - Set 7: Control Systems 1 7-1. Let a component h´t µ of a larger system have impulse response h´t µ e 5 t u´t µ. Let the component h´t µ be connected to form the larger system shown in the conﬁguration below: p(t) x(t) + − G h(t) y(t) If x´t µ u´t µ and G = 8, then compute and plot y´t µ over 0 t 0 2 second. DETAILED SOLUTION: Work this problem using the s-domain block diagram of the system in the problem statement. This s-domain diagram is shown below: P(s) X(s) + − G H(s) Y(s) From the s-domain diagram, we see that Y ´sµ is given by Y ´sµ G H ´sµ P´sµ X ´sµ Y ´sµ ´aµ where P´sµ is the difference between input and output P´sµ Substituting ´bµ into ´aµ thus gives Y ´sµ Multiply out the terms in ´cµto obtain Y ´sµ G H ´sµ X ´sµ G H ´sµ Y ´sµ ´d µ ´bµ G H ´sµ X ´sµ Y ´sµ ´cµ Using ´d µ, get all the Y ´sµ terms on the left side and all the X ´sµ terms on the right side, producing Y ´sµ 1 · G H ´sµ Solving ´eµ for Y ´sµ then gives Y ´sµ G H ´sµ X ´sµ 1 · G H ´sµ ´fµ G H ´sµ X ´sµ ´eµ ECE 301 Detailed Solutions - Set 7: Control Systems 2 7-1. (cont.) With x´t µ = u´t µ and h´t µ sions: e 5 t u´t µ from the problem statement we obtain the following s-domain expresX ´sµ Using G 1 s H ´sµ 1 s·5 ´g hµ 8 from the problem statement, substitute ´gµ and ´hµ into ´ f µ, obtaining 8 Y ´sµ 1 s·5 8 s 1· s·5 1 8 s·5·8 s 8 s ´s · 13µ ´iµ To ﬁnd y´t µ, ﬁrst expand Y ´sµ from ´iµ in a partial fraction expansion: Y ´sµ 8 s ´s · 13µ A B · s s · 13 ´ jµ Multiply both sides of ´ jµ by s´s · 13µ to obtain 8 Evaluate ´kµ at s 0 to obtain A and evaluate ´kµ at s A ´s · 13µ · B s 8 13 ´kµ 0 615 ´l µ 13 to obtain B 8 13 0 615 0 615 0 615 s · 13 s ´mµ Substituting ´l µ and ´mµ into ´ jµ then gives Y ´sµ ´nµ The time-domain output y´t µ is then given by taking the inverse Laplace transform of ´nµ, giving the desired form for the answer: y´t µ 0 615 1 e 13 t u´t µ ´oµ ECE 301 Detailed Solutions - Set 7: Control Systems 3 7-6. Let an original system H ´sµ have the transfer function given by H ´sµ 1 s2 Let the system H ´sµ receive the bounded amplitude input x´t µ u´t µ. For stability, the amplitude of the output y´t µ should be bounded in response to the bounded amplitude input. Demonstrate that H ´sµ is either stable or unstable. DETAILED SOLUTION: We can use the following property for Laplace transforms: If G´sµ τ τ 0 ° g´t µ then 1 G´sµ s ° t g´τµ d τ ´aµ Suppose a system q´t µ has an impulse response which is a unit step, that is, q´t µ u´t µ ° Q´sµ 1 s ´bµ The transfer function H ´sµ in the problem statement is related to the Q´sµ in ´bµ by H ´sµ 1 Q´sµ s ´cµ Therefore, the right side relation in ´aµ says that h´t µ is related to q´t µ by τ t 0 h´t µ Since ´bµ has given q´t µ u´t µ, we have q´τµ τ τ q´τµ d τ 0 and therefore ´d µ becomes τ τ 0 t ´d µ u´τµ t 0 1τ h´t µ Evaluating ´eµ then gives h´t µ as h´t µ τ u´τµ d τ dτ ´eµ t 0 t0 else t u´t µ ´fµ From ´ f µ we see that the impulse response grows unbounded, implying the system is unstable. Therefore, we do not even have to consider the unit step input to determine stability of the system. However, we will do this on the following page to give us more experience using transform relations. This will show speciﬁcally that the output is amplitude-unbounded, even though the input is the amplitude-bounded unit step. ECE 301 Detailed Solutions - Set 7: Control Systems 4 7-6. (cont.) If the input is the unit step, then the output is given by multiplication in the s-domain : Y ´sµ X ´sµ H ´sµ 1 H ´sµ s ´gµ Taking the inverse transform of ´gµ and using the property expressed in ´aµ thus gives τ t 0 y´t µ From ´ f µ we have h´t µ t u´t µ. Therefore, h´τµ τ t 0 τ h´τµ d τ ´hµ τ u´τµ and ´hµ becomes τ u´τµ d τ τ τ 0 t y´t µ Performing the integral in ´iµ then provides y´t µ τ τ dτ ´iµ 1 2 ¬τ ¬ τ¬ 2 τ t 0 0 5 t2 ´ jµ Equation ´ jµ shows that the output grows unbounded as t increases. Hence we have an unbounded output in response to the bounded unit step input. Therefore, the system is unstable. ECE 301 Detailed Solutions - Set 7: Control Systems 5 7-9. The feedback system below has the closed loop transfer function HC ´sµ X(s) Y ´sµ X ´sµ : + − P(s) 1 _ Y(s) s2 7s+10 The input is x´t µ u´t µ, a unit step. (a) Compute the simplest math form for p´t µ, where p´t µ is the inverse Laplace transform of P´sµ. (b) Plot p´t µ over 0 t 1 . DETAILED SOLUTION : (a) Compute the simplest math form for p´t µ . X ´sµ ´7s · 10µ Y ´sµ Y ´sµ To obtain P´sµ, use s-domain algebra on the ﬁgure in the problem statement. From the ﬁgure we see that P´sµ But the ﬁgure also shows that 1 P´sµ s2 We need an expression for P´sµ in terms of X ´sµ. Therefore, substitute ´bµ for Y ´sµ into ´aµ, giving P´sµ X ´sµ ´7s · 10µ ¢ 1 P´sµ s2 ´bµ ´aµ ´cµ Get all the P´sµ terms on one side of ´cµ and all the X ´sµ terms on the other : P´sµ 1 · Solving ´d µ for P´sµ then gives s2 X ´sµ s2 · 7s · 10 u´t µ from the problem statement, we have X ´sµ 1 s and therefore ´eµ becomes P´sµ P´sµ s2 s ´ s2 · 7s · 10 µ s ´s · 5µ ´s · 2µ ´eµ 7s · 10 s2 X ´sµ ´d µ Since x´t µ 1 67 0 67 s·5 s·2 ´fµ where we have used partial fraction expansion to obain the last result on the right side of ´ f µ. Taking the inverse transform of equation ´ f µ then gives p´t µ L 1 Ò 1 67 s·5 Ó L 1 Ò 0 67 s·2 Ó ´gµ Evaluating each of the inverse transforms in ´gµ then gives the desired solution : p´t µ 1 67 e 5 t 0 67 e 2 t u´t µ ´hµ ECE 301 Detailed Solutions - Set 7: Control Systems 6 7-10. The feedback system below has the closed loop transfer function HC ´sµ Y ´sµ X ´sµ : X(s) + − 1 _ Y(s) s2 F(s) The closed-loop impulse response of the above system is hc ´t µ 0 1429 e 3 t e 10 t u´t µ . Compute the simplest math s-domain form for the feedback processing function F ´sµ. DETAILED SOLUTION: From the notes, we know that the closed-loop transfer function Hc ´sµ is given by HC ´sµ G H ´sµ 1 · G F ´sµ H ´sµ ´aµ From the ﬁgure in the problem statment we observe the following relations : G 1 H ´sµ 1 s2 ´bµ Substituting the relations in ´bµ into the right side of ´aµ then gives G H ´sµ 1 · G F ´sµ H ´sµ 1 s2 F ´sµ 1· 2 s ´cµ Multiply right side of equation ´cµ by s2 to give the following useful form G H ´sµ 1 · G F ´sµ H ´sµ Substituting ´d µ into ´aµ then gives the following : HC ´sµ 1 s2 · F ´sµ ´eµ 1 s2 · F ´sµ ´d µ The closed-loop transfer function Hc ´sµ can be obtained by taking the Laplace transform of the impulse response hc ´t µ given in the problem statement : Hc ´sµ L hc ´t µ 0 1429 s·3 0 1429 s · 10 ´fµ ECE 301 Detailed Solutions - Set 7: Control Systems 7 7-10. (cont.) Therefore, substituting ´ f µ into ´eµ obtains the following : 0 1429 0 1429 s · 10 s·3 1 s2 · F ´sµ ´gµ Adding the two terms together on the left side of ´gµ then gives 1 s2 · 13 s · 30 You can see that equation ´hµ is satisﬁed if F ´sµ 13 s · 30 ´iµ 1 s2 · F ´sµ ´hµ ...
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