# EEE 184: Introduction to feedback systems

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1 EEE 84: Introduction to feedback systems Summary x Level() time(s) Time (seconds) Fig.. Illustration of BIBO stability: stable system (the input is a unit step) Fig.. step) Illustration of BIBO stability: unstable system (the input is a unit STABILITY Stability is the most important requirement in control systems, we cannot talk about transients or steady state errors if the the system is not stable. Recall that the total response of control systems consists of the natural response and the forced response c(t) = c f (t) + c n (t) () There are two basic definitions for stability. One of them uses the natural response and the second one uses the forced response. Definition System is stable if c n (t) as t System is marginally stable if c n (t) = constant as t System is unstable if c n (t) as t Definition : A system is BIBO (bounded input bounded output) stable if every bounded input yields a bounded output. Drawback of this definition: what happens if c f is bounded but c n? Definition : A system is unstable if any bounded input yields an unbounded output. Figures,,, and 4 illustrate the basic concepts of stability. ROUTH-HURWITZ TEST The Routh Hurwitz test is one of the most traditional methods to determine the necessary and sufficient conditions.5 Natural response Time (seconds) Fig.. Illustration stability: natural response goes to zero with time for stability of linear systems. The Routh Hurwitz stability criterion allows to determine the stability of a closed loop system without solving for the poles. The test consists of two steps: Building the Routh Hurwitz table Interpretation of the table

2 Introduction to Feedback Systems, Fall 7 Summary Time (seconds) Fig. 4. Illustration of a marginally stable system Fig. 5. Block diagrams Routh Hurwitz table Assume the closed loop transfer function can be written as N(s) a 4 s 4 + a s + a s + a s + a () where N(s) is the numerator of the closed loop system. The characteristic polynomial is given by P (s) = a 4 s 4 + a s + a s + a s + a () This is a 4th order polynomial. The table is constructed based on the coefficients of the characteristic polynomial as follows with b = c = d = a 4 a a a a b = a a b b b c = b b c c c d = s 4 a 4 a a s a a s b b b s c c c s d d d a 4 a a a b = a b b b c = b b c c c d = (4) a 4 a (5) a a b (6) b b c (7) c Because it is a fourth order polynomial, the first column contains the coefficients with even subscript and the second column contains the coefficients with odd subscripts. The exact same principle is used for polynomials of higher (and lower) order. Example Assuming that the characteristic polynomial is given by P (s) = s 5 + s 4 + 5s + 6s + s + 4 (8) Construct the Routh Hurwitz table. The Routh Hurwitz table is given by Example s 5 5 s s s 74 4 s s 4 For the block diagram of figure 5-(a) (top), construct the Routh-Hurwitz table. The closed loop transfer function is given by s + 6s 6 + s 5 + s 4 + s + 7s + 4s + 7 The Routh Hurwitz table is given by s s 5 4 s s 6 s 8 7 s 68 8 s 7 The second step in the RH test is to interpret the table. (9) () ()

3 Introduction to Feedback Systems, Fall 7 Summary 6 Theorem The number of the right half plane poles of the characteristic polynomial is equal to the number of sign changes in the first column of the Routh Hurwitz table. Example Is the closed loop system represented by the block diagram of figure 5-(a) (top) stable? From the RH table shown in table (), there are two sign changes, therefore the system has two poles in the right half plane. The system is unstable. Example 4 For the block diagram of figure 5-(b) (bottom), construct the Routh Hurwitz table and determine the interval of the gain that results in a stable system. The closed loop transfer function is given by 4s 4 + s + 7s + s The Routh-Hurwitz table is given by or s s s s + 5 s 4 + Since the first column has to be positive () () + > (4) > (5) > (6) > 4 (7) Therefore, for stability, the gain has to satisfy: Steps for stability test 4 < < (8) Obtain the closed loop transfer function Build the Routh Hurwitz table The number of right half plane poles is the number of sign changes in the first column. Fig. 6. Tank system example Zero only in the first column Having a zero in the first column poses a problem, we cannot divide by zero. To solve this issue we introduce a small number ε to replace the zero and continue the table. Some entries in the first column are function of ε. We calculate the limit ε from either side and we compare with the other numbers in the first column keeping in mind that a sign change corresponds to an unstable system. The second method to address this problem is by reversing the coefficients in the characteristic polynomial. STABILITY IN THE STATE SPACE In state space models, stability is determined using the eigenvalues of the state transition matrix (matrix A) as follows. Positive eigenvalue or complex eigenvalues with positive real part If the state transition matrix has a positive eigenvalue or complex eigenvalues with positive real part, the system is unstable. Zero real part When the real part is zero, the system behaves as an undamped oscillator The plot of response with time would look sinusoidal. This corresponds to a marginally stable system. Eigenvalues with negative real part When the real part is negative, then the system is stable. EXAMPLE 5 Consider the tank system of figure 6 where a is the cross section of the tank, q i is the inflow rate and q o is the outflow rate. Find the closed loop transfer function and determine the range of the gain for which the system is stable. The output is given by H(s) = as + R(s) as + Q o(s) (9) The Routh-Hurwitz table is s a s ()

5 Introduction to Feedback Systems, Fall 7 Summary 6 TEST INPUTS It is possible to study the steady state error for the different test inputs. Step input R(s) = s e step ( ) = lim s s s + G(s) e step ( ) = + lim s G(s) (6) (7) lim s G(s) is called the DC gain for the forward loop. To achieve a zero steady state error for a step input, we need to have lim s G(s) =. For n = : e step ( ) = (8) (s+z + lim )(s+z )... s s (s+p )(s+p )... e step ( ) = + zz... = constant (9) p p... For n = e step ( ) = (4) (s+z + lim )(s+z )... s s (s+p )(s+p )... e step ( ) = + = (4) For n > Using similar logic, it turns out that Ramp input R(s) = s e step ( ) = + = (4) e ramp ( ) = lim s s s + G(s) e ramp ( ) = lim s sg(s) (4) (44) Clearly, the steady state error depends on lim s sg(s), that is lim sg(s) = e ramp( ) = (45) s For n = e ramp ( ) = lim s s (s+z)(s+z)... s (s+p )(s+p )... For n = (46) e ramp ( ) = = (47) e ramp ( ) = lim s s (s+z)(s+z)... s (s+p )(s+p )... (48) e ramp ( ) = z z... p p... = constant (49) For n = e ramp ( ) = lim s s (s+z)(s+z)... s (s+p )(s+p )... (5) e ramp ( ) = = (5) The result for n > is the same as for n =. Parabola input R(s) = s e parabola ( ) = lim s s s + G(s) e parabola ( ) = lim s s G(s) (5) (5) Clearly, the steady state error depends on lim s s G(s), that is lim s s G(s) = e parabola ( ) = (54) For n = e parabola ( ) = lim s s For n = e parabola ( ) = lim s s For n = e parabola ( ) = lim s s (s+z)(s+z)... s (s+p )(s+p )... (55) e parabola ( ) = = (56) (s+z)(s+z)... s (s+p )(s+p )... (57) e parabola ( ) = = (58) e parabola ( ) = z z... p p... For n = e parabola ( ) = lim s s (s+z)(s+z)... s (s+p )(s+p )... (59) = constant (6) (s+z)(s+z)... s (s+p )(s+p )... (6) e parabola ( ) = = (6) The result for n > is the same for as n =. Error constants Three error constants are defined based on the test input: p, v, a. The table below summarizes the relationship between the error constants and the steady state error. error constant steady state error p = lim s G(s) e( ) = + p v = lim s sg(s) e( ) = v a = lim s s G(s) e( ) = a (6) 5

6 Introduction to Feedback Systems, Fall 7 Summary 6 Clearly, the errors and the error constants are inversely proportional. From the results above, we notice that if a system of a given type has the ability to track a ramp input, then is has the ability to track a step input. In a very similar way, if a system has the ability to track a parabola, the system will be able to track the ramp and step inputs. Example 6 We consider the first order system shown in figure 9-top. ) Find the interval of the gain for which the system is stable. ) What is the order of the closed loop system? ) For =, determine the steady state error when the input is a unit step 4) Find the interval of the gain for the SSE to be less than. ) First we find the closed loop transfer function: G(s) + G(s) = s+ (64) (65) + s+ = (66) + s + For the system to be stable >. ) This is a first order system ) The equation of the steady state error for unity feedback is given by s s R(s) + G(s) Since the input is a unit step, R(s) = /s, we have For =, + s = + s+ 4) We want the SSE to be less than.: Example 7 (67) (68) (69) (7) (7) e( ) =.75 (7) e( ) <. (7) <. + (74) > 7 (75) We consider the first order system shown in figure 9-bottom. ) Find the interval of the gain for which the system is stable. ) What is the order of the closed loop system? ) Show that the steady state error zero and does not depend on the gain. ) First we find the closed loop transfer function: The RH table is G(s) + G(s) s(s+) = + s(s+) = s + s + s s s The RH theorem implies that >. ) This is a second order system ) For a unit step, the steady state error is Example 8 + s s(s+) (76) (77) (78) (79) (8) s(s + ) = lim s s(s + ) + (8) = (8) The error e( ) is zero and does not depend on the gain. Simulink representation is shown in figure for the systems of figure 9 with gains 7 and, respectively. The step response is shown in figure. Compare between the systems in terms of the settling time. The settling time of the first system is about.s. For the second system the settling time is about 9s. IN THE PRESENCE OF DISTURBANCE Disturbance can be defined as an unwanted input. Consider the block diagram of figure where D(s) is the disturbance. It is possible to write and C(s) = G (s)d(s) + G (s)g (s)e(s) (8) E(s) = R(s) C(s) (84) By combining these two equations, we have E(s) = + G (s)g (s) R(s) G (s) D(s) (85) + G G (s) Now s sg (s) R(s) lim D(s) (86) s + G (s)g (s) s + G G (s) e( ) = e R ( ) e D ( ) (87) 6

7 Introduction to Feedback Systems, Fall 7 Summary 6 Fig. 9. Block diagrams for examples 6 and 7 Fig.. In the presence of disturbance where e R ( ) is the steady state error due to the input R(s) e D ( ) is the steady state error due to the disturbance D(s) Unit step disturbance In this case, we have D(s) = s (88) Fig.. Simulink representation for examples 6 and 7 with gains 7 and, respectively and it is possible to write for the steady state error e D ( ) = lim s G + lim (s) s G (s) (89) In order to reduce the steady state error due to disturbance, we need to increase the DC gain of G decrease the DC gain of G (a) Time (seconds) Example 9 For the system of figure, find the steady state error due to step disturance. e D ( ) = lim s G + lim (s) s G (s) = + 5 (9) (9) (b) Time (seconds) Fig.. Step response for the systems shown in figure 9 with gains 7 and, respectively SYSTEMS WITH NON UNITY FEEDBAC The equations we derived previously are valid for feedback systems with unity feedback. However, in general we do not have unity feedback. There are several ways to solve in this case (non unity feedback) Derive new equations for the steady state error Use the equation for the steady state error as a function of the closed loop transfer function 7

8 Introduction to Feedback Systems, Fall 7 Summary Case : a = 6, = 4 Case : a = 8, = Case : a = 8, = Time (seconds) Fig.. Example with disturbance Fig. 5. Step response for the cases of table From which the error constant can be obtained as follows (a) p = (94) a a and e( ) = a a (95) a Using the closed loop transfer function: The closed loop transfer function is given by and s+a (96) Fig. 4. Non unity feedback Reduce the system to unity negative feedback and use the equation that gives the error as a function of the open loop transfer function. Example 4 Consider the block diagram of figure 4. Find the steady state error for a unit step. : Using the equivalent transfer function and error constants: The equivalent transfer function as shows in figure 4- bottom is given by G eq (s) = G eq (s) = s+a+ s+a (s + a) s + a s a (9) (9) and therefore (s + a) s + a e( ) = T () = a a (97) (98) e( ) = a a (99) a Three cases are illustrated for this example. The results are shown in the table below, and the step response as well as error as function of time are shown in figures 5 and 6. error constant steady state error a = 8, = e( ) = 6 a = 8, = e( ) = 5 a = 6, = e( ) = 5 STEADY STATE ERROR IN STATE SPACE () There are two different methods to determine the steady state error in state space Using the final value theorem Using input substitution We assume single input single output systems. 8

9 Introduction to Feedback Systems, Fall 7 Summary 6 4 Case : e( ) =.6 Case : e( ) = 5/ Case : e( ) = Time (seconds) Fig. 6. Error as a function of time for the cases of table s s s By using Matlab commands [ ] ( [ ]) [ ] si 5 () >> syms s () >> C inv(s eye() A) B () we get [ ] ( [ ]) [ ] s + 5 si = 5 s + 7s + 7 therefore (4) e( ) = 5 7 = =.9 (5) 7 Using the final value theorem The closed loop system can be written as We know that Thus ẋ = Ax + Br () y = Cx () X(s)s = AX(s) + BR(s) () Y (s) = CX(s) (4) Y (s) = C(sI A) BR(s) (5) E(s) = R(s) Y (s) = R(s)( C(sI A) B) (6) and finally s sr(s)( C(sI A) B) (7) SENSITIVITY Sensitivity refers to how a small change in the parameters affects the system. For function F and parameter p, the sensitivity is given by and S F :P = That is Example 4 lim = P Change in the function F Change in the parameter P S F :P = lim F = F p P P S F :P = lim p = P F F P S F :P = P F F P Consider the following transfer function (6) (7) (8) (9) Example 4 G(s) = s( ) () Find the steady state error when the input is a unit step for the following case: A = [ ] B = 5 [ ] [ ] C = (8) (9) () Find the sensitivity of the steady state error to the gains and. The velocity error constant can be obtained as follows v = () e( ) = () 9

10 Introduction to Feedback Systems, Fall 7 Summary 6 Therefore, we have S e: = e S e: = e e = e = = () ( ) = (4) A sensitivity of + means that an increase in the parameter results in an increase in the function and a sensitivity of means that the function and its parameter are inversely proportional. This can be easily seen in equation (). ADDITIONAL READING Steady state error by input substitution Input is unit step: Let X ss be the steady state value for the state. We can write v v X ss = = V (5). v n From the first equation, we have V = AW = W = A V = A B (5) AV = B = V = A B (6) By substitution in the equation of Y ss we get Y ss = C(A Bt + A B) (7) and finally we have for the error (t Y ss (t)) (8) t [ = lim ( + C(A B)t + CA B) ] (9) t Note that in this case, in order to have a finite steady state error we need to have CA B =. It is also important to note that the equations are derived for systems with single input and single output. Since X ss = V = constant, we have Ẋss =. The state space equations become We also have = AV + B (6) Y ss = CV (7) Y ss = CA B (8) e( ) = r(t) y ss = y ss = + CA B (9) Example 4 We consider the previous example for a unit step input. We have [ ] [ ] [ ] e( ) = + CA B = + 5 () Finally e( ) =.7 =.9 () Ramp input: In this case, the steady state for the state vector is v t + w v t + w X ss =. = V t + W (). v n t + w n By substitution in the state space model V = A(V t + W ) + Bt () Y ss = C(V t + W ) (4)

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