ECE-320 Equation Sheet Second Order System Properties Percent Overshoot: 21100.%.PO eζπζ−−×=, If 100maxPOβ= then2ln( )ln( )1βπζβπ−=−⎛⎞+⎜⎟⎝⎠ , 1()cosθζ−= Time to Peak: 2,1pdndTπωωζω==− 2% Settling Time: 44snTτζω== Model Matching Assume we have a proper plant() ()/ ()pGs Ns Ds= and we want the closed loop system to have the transfer function 00() ()/ ()oGs Ns Ds=. We can find a controller []000() ()()() () ()cNsDsGsNs D s N s=− under the following conditions: • degree - degree degree - degree 0()Ds0()Ns≥()Ds ()Ns• The right half plane zeros of are retained in ()Ns0()Ns• is stable 0()GsQuadratic Optimal Control Consider a plant with a proper transfer function ()()()pNsGsDs= where • and have no common factors ()Ns ()Ds• The leading coefficient of is 1 (the coefficient of the highest power of s in is 1) ()Ds()DsAn implementable closed loop transfer function that minimizes the performance index 0()Gs{}220[() ()] ()Jqytrtut∞=−+∫dt where (a unit step) is given by () 1rt = 000(0) ( )()(0) ( )qN N ssDDGs= where 00() () ( ) () ( ) () ( )Qs DsD s qNsN s D sD s=−+ −= − Controller Types Proportional (P), ( )cGs k= Integral (I), ()cGkss= Proportional + Integral (PI), ()()ckGszss+= Proportional + Derivative (PD), ( ) ( )cskGsz=+ Proportional + Integral + Derivative (PID), 12()(()cks zGszss)++= Lead, ()() ,()cks zspspG+=>+z Lag, ()() ,()cks zszspG+=>+pDiophantine Equations For plant , controller () ()/ ()pGs Ns Ds=( ) ( ) / ( )cGs Bs As=, and desired characteristic Polynomial we will have to solve the equation0()Ds0() () () () ()AsDs BsN s D s+= This is called the Diophantine equation. We solve this equation by equating powers of s, setting up a system of equations, and then solving. The closed-loop transfer function will be 00() ()()()BsNsGsDs= where()Bs contains the zeros we have added to the system. Theorem (Strictly Proper Plant) Assume we have a strictly proper order plant transfer function, . Since is strictly proper we have the degree of < the degree of . Since is order the degree of . Assume also that and have no common factors. Then for any polynomial of degree n a proper controller thn() ()/ ()pGs Ns Ds=()DsG()Ds( )cGs()pGsthn) / ( )s As()Ns()ps()Ds n=D()Nsm+0()s(B= of degree m exists so that the characteristic polynomial of the resulting closed-loop system is equal to . 0()DsIf , the controller is unique. If , the controller is not unique and some of the coefficients can be used to achieve other design objectives. 1mn=−mn≥ Theorem (Special case: degree = degree ) Assume we have a proper order plant transfer function, ()Ns()pGs()Ds()Dsthn()/Ns=, where the degree of ()Ds= degree = n. Assume also that and have no common factors. Then for any polynomial of degree()Ns ()Nsm()Ds0()Ds n+ a proper controller ( ) ( ) / ( )cGs Bs As= of degree m exists so that the characteristic equation of the resulting closed-loop system is equal to. If , and the controller is chosen to be strictly proper, the controller is unique. If , the controller is not unique and some of the coefficients can be used to achieve other design objectives. 0()Dsn=mn≥m1+Root Locus Construction Once each pole has been paired with a zero, we are done 1. Loci Branches 0kkpzer soles o==∞→ Continuous curves, which comprise the locus, start at each of the n poles of for which . As k approaches , the branches of the locus approach the m zeros of . Locus branches for excess poles extend to infinity. ()Gs0k =∞()Gs The root locus is symmetric about the real axis. 2. Real Axis Segments The root locus includes all points along the real axis to the left of an odd number of poles plus zeros of . ()Gs 3. Asymptotic Angles As $k , the branches of the locus become asymptotic to straight lines with angles k →∞180 360,0,1,2,.ooiinmθ+==±−.± until all (n) angles not differing by multiples of are obtained. n is the number of poles of and m is the number of zeros of . m−()Gs360o()Gs 4. Centroid of the Asymptotes The starting point on the real axis from which the asymptotic lines radiate is given by ijijcpznmσ−=−∑∑ where ip is the pole of , thi ()Gsjz is thethjzero of , n is the number of poles of and m is the number of zeros of . This point is termed the centroid of the asymptotes. ()Gs()Gs (Gs) 5. Leaving/Entering the Real Axis When two branches of the root locus leave or enter the real axis, they usually do so at angles of . 90o±Frequency Domain Phase Lead Consider the following feedback system ()Gs The primary function of the phase lead compensator is to reshape the frequency response curve to add phase to the system, thereby increasing the phase margin. This usually results in an increased bandwidth, and hence a faster response time. Basic Procedure 1) Assume the phase lead compensator has the form 1(1)()1(1)ccsTsTGs K KTssTαα++==++ where cKKα= Determine K to satisfy any steady state error requirements. 2) Using the value of K determined in step 1, draw the Bode plot of() ()KGsHs. 3) Determine the necessary phase to be added to the system to achieve the desired phase margin. Add an additional 5 to 12 to the necessary phase to account for the fact that the phase lead compensator shifts the gain crossover frequency to the right and decreases the phase margin. The total phase we need to add is oomφ. 4) Determine α using the relationship 1sin( )1sin( )mmφαφ−=+. α should be larger than 0.05, or you need two or more compensators in series. +-()Hs()cGs5) Determine the frequency mωat which 10 10 10120log | ( ) ( ) | 20log ( ) 10log ( )mmKG j H jωωαα=− = This is the new gain crossover frequency 1mTωα= . Since we know mωand α, we can compute 1mTωα= 6) Determine the pole and zero of the compensator 1zT=and 11pTα= 7) Determine cKKα= 8) Check the phase margin to see if it is acceptable. We will use sisotool. When you use sisotool to tweak your compensators, you should 1) enter the transfer function for () ()GsHs 2) instruct sisotool to use the Natural Frequency or Time Constant method of displaying the controller. To do this edit → sisotool preferences →options and select Natural Frequency or Time Constant 3) enter the pole and zero of the compensator 4) Set the gain to be K, the required gain for the steady state errors. This gain cannot change! 5) Play with the pole