MASON ECE 220 - Relationship Between Transfer Function Pole Locations and Time-Domain Step Responses

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1Relationship Between Transfer Function PoleLocations and Time-Domain Step ResponsesA. IntroductionThis example looks at the relationship between the locations of the poles of the transfer functionH(s) for alinear, time-invariant (LTI) system and the response of the system in the time domain to a unit step inputu(t).The important characteristics of the step response that will be considered are the general shape of the response,thevalueoftheresponseast →∞, and the time it takes the response to settle near its final value if it does settle.These characteristics will be related to the real and imaginary parts of the closed-loop poles.For a system m odeled by a transfer function with distinct polesp1,p2,...,pnsubjected to a step input, thetransform of the zero-state response isYzs(s)=H(s) ·1s=N(s)s (s − p1)(s − p2) ···(s − pn)=A0s+A1(s − p1)+A2(s − p2)+ ···+An(s − pn)(1)and the output signal in the time domain isyzs(t)=£A0+ A1ep1t+ A2ep2t+ ···+ Anepnt¤u(t)(2)If a polepiis real, then the corresponding term in (2) is a decaying exponential if pi< 0 and a growing exponentialifpi> 0. If a pole is complex, pi= α+jβ, then the corresponding term in (2) is a sinusoid of frequency β rad/secmultiplied by an exponential which is decaying ifα<0 or growing if α>0. If α =0, then the term represents asinusoid of constant amplitude. For either real or complex poles, ifRe [pi] < 0, then the pole lies in the left halfof the complexs-plane (LHP). If Re [pi] > 0, then the pole lies in the right half of the comple x s-plane (RHP),and ifRe [pi]=0, then the pole lies on the jω axis which separates the left and right halves of the s-plane.If the transfer function has a repeated pole, such as(s − p1)2, then in addition to the exponential terms in(2), there will also be a term involvingtep1t. For repeated poles with multiplicity m, (s − p1)m, the time-domainexpression will contain terms in the time variable up totm−1, each multiplied by the e x ponential ep1t.The transfer function in this example has 2 poles (p1,p2) and no zeros. The locations of the poles for the ex ampleare such thatp1× p2=1. The transfer function has the formH(s)=1(s − p1)(s − p2)=1s2− (p1+ p2) s + p1p2=1s2− (p1+ p2) s +1(3)The locations of the poles are shown below.Trial 1 2 3 4 5 6 7 8 9 10p1−10 −4 −1 −0.8+j0.6 −0.5+j0.866 −0.375 + j0.927 j1 0.5+j0.866 1 4p2−0.1 −0.25 −1 −0.8 − j0.6 −0.5 − j0.866 −0.375 − j0.927 −j1 0.5 − j0.866 1 0.25For each of these trials, the pole locations will be shown in the s-plane and the corresponding step response willbe plotted. A discussion of the relationships between the pole locations and the step response will be given.2B. Trial 1Yzs(s)=1s (s +0.1) (s + 10)=1s (s2+10.1s +1)(4)yzs(t)=£1 − 1.0101e−0.1t+0.0101e−10t¤u(t)(5)The system is bounded-input, bounded-output (BIBO) stable since the poles ofH(s) are in the left-half plane(LHP). The system’s step response settles at a final value of1, which is the final value of the unit step input. Theoutput asymptotically approaches the final value, with the length of time to get “close” to the final value beingcontrolled by the pole ofH(s) that is the closest to the jω axis, that is, the pole at s = p1= −0.1. The exponentialdecay ofe−0.1tis much slower than the decay of e−10t. At t =40seconds, the output is within 2% of the finalv alue. The response is well behaved, but slo w. There is no overshoot of the final value, and there are no oscillationsin the response. Fig. 1 shows the pole locations ofH(s) and the step response yzs(t) for Trial 1.C. Trial 2Yzs(s)=1s (s +0.25) (s +4)=1s (s2+4.25s +1)(6)yzs(t)=£1 − 1.0667e−0.25t+0.0667e−4t¤u(t)(7)This system is also BIBO stable since the poles ofH(s) are in the left-half plane. The system’s step responsesettles at a final value of1, which is the final value of the unit step input. The output asymptotically approachesthe final value, with the length of time to get close to the final value being controlled by the pole ofH(s) that isthe closest to thejω axis, that is, the pole at s = p1= −0.25. At t =16seconds, the output is within 2% of thefinal value. As the pole closest to thejω axis moves to the left, the response gets faster. The response is still wellbehaved. There is no ov ershoot of the final value, and there are no oscillations in the response. Fig. 2 shows thepole locations ofH(s) and the step response yzs(t) for Trial 2.D. Trial 3Yzs(s)=1s (s +1)2=1s (s2+2s +1)(8)yzs(t)=£1 − e−t(t +1)¤u(t) (9)This system is also BIBO stable since the poles ofH(s) are in the left-half plane. The system’s step responsesettles at a final v alue of1, which is the final value of the unit step input. The output now has a term inv olvingte−tsince there is a repeated pole. The response is faster than the previous case, with the output getting close toits final v alue in approximately6 seconds. There is no overshoot of the final value, and there are no oscillations inthe response. Fig. 3 shows the pole locations ofH(s) and the step response yzs(t) for Trial 3.E. Trial 4Yzs(s)=1s (s +0.8 − j0.6) (s +0.8+j0.6)=1s (s2+1.6s +1)(10)yzs(t)=£1 − e−0.8tcos(0.6t) − 1.3333e−0.8tsin(0.6t)¤u(t) (11)This system is also BIBO stable since the poles ofH(s) are in the left-half plane. The system’s step responsesettles at a final value of1, which is the final value of the unit step input. The output no longer asymptoticallyapproaches the final value, but now has a slight bit of overshoot of the final value. The maximum value of theoutput is1.0152. The length of time to get close to the final value is controlled by the real part of the complexpoles ofH(s), that is, by s =Re[p1]=−0.8. The output is very close to its final value at approximately t =3.75seconds, and stays close to that value, even with the overshoot. Fig. 4 shows the pole locations of H(s) and thestep responseyzs(t) for Trial 4.3−10 −8 −6 −4 −2 0−1−0.8−0.6−0.4−0.200.20.40.60.81Real AxisImag AxisStep Response: H(s) = 1/[s2 + 10.1s + 1]0 5 10 15 20 25 30 35 40 45 5000.10.20.30.40.50.60.70.80.91Time (s)AmplitudeStep Response: H(s) = 1/[s2 + 10.1s + 1]Fig. 1. H(s) pole locations and step response for Trial 1.4−5 −4 −3 −2 −1 0 1−1−0.8−0.6−0.4−0.200.20.40.60.81Real AxisImag AxisStep Response: H(s) = 1/[s2 + 4.25s + 1]0 2 4 6 8 10 12 14 16 18 2000.10.20.30.40.50.60.70.80.91Time (s)AmplitudeStep Response: H(s) = 1/[s2


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MASON ECE 220 - Relationship Between Transfer Function Pole Locations and Time-Domain Step Responses

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