Problems and Solutions for Section 3.1

Transcription

1 Chapter 3 Dynamic Response Problems and Solutions for Section 3.. Show that, in a partial-fraction expansion, complex conjugate poles have coefficients that are also complex conjugates. (The result of this relationship is that whenever complex conjugate pairs of poles are present, only oneofthecoefficients needs to be computed.) Consider the second-order system with poles at α ± jβ, H(s) (s + α + jβ)(s + α jβ) P erf orm Partial Fraction Expansion: H(s) C s + α + jβ + C s + α jβ C s + α jβ s α jβ β j C s + α + jβ s α+jβ β j C C. Find the Laplace transform of the following time functions: (a) f(t) +t (b) f(t) 3+7t + t + δ(t) (c) f(t) e t +e t + te 3t (d) f(t) (t +) (e) f(t) sinht 75

2 76 CHAPTER 3. DYNAMIC RESPONSE (a) f(t) +t L{f(t)} L{(t)} + L{t} (b) (c) (d) s + s s + s f(t) 3+7t + t + δ(t) L{f(t)} L{3} + L{7t} + L{t } + L{δ(t)} 3 s + 7 s +! s 3 + s3 +3s +7s + s 3 f(t) e t +e t + te 3t L{f(t)} L{e t } + L{e t } + L{te 3t } s + + s + + (s +3) f(t) (t +) (e) Using the trigonometric identity, t +t + L{f(t)} L{t } + L{t} + L{}! s 3 + s + s s +s + s 3 f(t) sinht et e t L{f(t)} L{ et } L{e t } ( s ) ( s + ) s

3 77 3. Find the Laplace transform of the following time functions: (a) f(t) 3cos6t (b) f(t) sint +cost + e t sin t (c) f(t) t + e t sin 3t (a) (b) (c) f(t) 3cos6t L{f(t)} L{3cos6t} s 3 s +36 f(t) sint +cost + e t sin t L{f(t)} L{sin t} + L{cost} + L{e t sin t} s +4 + s s +4 + (s +) +4 f(t) t + e t sin 3t L{f(t)} L{t } + L{e t sin 3t}! s (s +) +9 s (s +) Find the Laplace transform of the following time functions: (a) f(t) t sin t (b) f(t) t cos 3t (c) f(t) te t +tcos t (d) f(t) t sin 3t t cos t (e) f(t) (t)+tcos t (a) f(t) t sin t L{f(t)} L{t sin t}

4 78 CHAPTER 3. DYNAMIC RESPONSE Use multiplication by time Laplace transform property (Table A., entry #), L{tg(t)} d ds G(s) Let g(t) sint and use L{sin at} L{t sin t} d ds ( s + ) s (s +) s s 4 +s + a s + a (b) f(t) t cos 3t Use multiplication by time Laplace transform property (Table A., entry #), L{tg(t)} d ds G(s) Let g(t) cos3t and use L{cos at} L{t cos 3t} d ds ( s s +9 ) [(s +9) (s)s] (s +9) s 9 s 4 +8s +8 s s + a (c) f(t) te t +t cos t Use the following Laplace transforms and properties (Table A., en-

5 79 tries 4,, and 3), L{te at } (s + a) L{tg(t)} d ds G(s) L{cos at} s s + a L{f(t)} L{te t } +L{t cos t} (s +) +( d s ds s + ) (s +) s s 4 +s + (s +) (s)s (s +) (d) f(t) t sin 3t t cos t Use the following Laplace transforms and properties (Table A., entries, 3), L{tg(t)} d ds G(s) a L{sin at} s + a s L{cos at} s + a L{f(t)} L{t sin 3t} L{t cos t} d 3 ds s +9 ( d s ds s + ) (s 3) (s +9) +) (s)s) ((s (s +) 6s (s +9) + (s ) (s +)

6 80 CHAPTER 3. DYNAMIC RESPONSE (e) f(t) (t)+t cos t L{(t)} s L{tg(t)} d ds G(s) L{cos at} s s + a L{f(t)} L{(t)} + L{t cos t} s +( d s ds s +4 ) (s s +4) (s)s (s +4) s ( s +4) (s +4) 5. Find the Laplace transform of the following time functions (* denotes convolution): (a) f(t) sint sin 3t (b) f(t) sin t +3cos t (c) f(t) (sint)/t (d) f(t) sint sin t (e) f(t) R t cos(t τ)sinτdτ 0 (a) f(t) sint sin 3t Use the trigonometric relation, sin αt sin βt cos( α β t) cos( α + β t) α and β 3 f(t) cos( 3 t) cos( +3 t) cos t sin 4t L{f(t)} L{cos t} L{cos 4t} [ s s +4 s s +6 ] 6s (s +4)(s + 6)

7 8 (b) f(t) sin t +3cos t Use the trigonometric formulas, sin t cos t cos t +cost f(t) cos t +cost L{f(t)} L{} + L{cos t} s + s s +4 3s +8 s(s +4) +3( +cost ) (c) We Þrst show the result that division by time is equivalent to integration in the frequency domain. This can be done as follows, Z s Z s Z s F (s) F (s)ds F (s)ds F (s)ds Z 0 Z s e st f(t)dt Z e st f(t)dt ds 0 Interchanging the order of integration, Z Z e st ds f(t)dt 0 s Z t e st f(t)dt 0 Z f(t) e st dt 0 t U sin g this result then, L{sin t} s +, L{ sin t Z } t s ξ + dξ tan ( ) tan (s) π tan (s) s tan ( s ) where a table of integrals was used and the last simpliþcation follows from the related trigonometric identity.

8 8 CHAPTER 3. DYNAMIC RESPONSE (d) f(t) sint sin t Use the convolution Laplace transform property (Table A., entry 7), L{sin t sin t} ( s + )( s + ) s 4 +s + (e) f(t) Z t L{f(t)} L{ cos(t τ)sinτdτ 0 Z t 0 cos(t τ)sinτdτ} L{cos(t) sin(t)} This is just the deþnition of the convolution theorem, L{f(t)} s s +s + s s 4 +s + 6. Given that the Laplace transform of f(t) is F(s), Þnd the Laplace transform of the following: (a) g(t) f(t)cost (b) g(t) R t 0 R t 0 f(τ)dτdt (a) First write cos t in terms of the related Euler identity (Eq. B.33), g(t) f(t)cost f(t) ejt + e jt f(t)ejt + f(t)e jt Then using entry 4 of Table A. we have, G(s) F (s j)+ F (s + j) [F (s j)+f (s + j)]. (b) Let us deþne e f(t ) R t 0 f(τ)dτ, then g(t) R t f(t e 0 )dt, and from entry 6 of Table A. we have L{ f(t)} e F e (s) s F (s) and using the same result again, we have G(s) e s F (s) s ( s F (s)) s F (s) 7. Find the time function corresponding to each of the following Laplace transforms using partial fraction expansions:

9 83 (a) F (s) s(s+) (b) F (s) 0 s(s+)(s+0) (c) F (s) 3s+ s +4s+0 (d) F (s) (e) F (s) s +4 (f) F (s) (g) F (s) s+ s 3s +9s+ (s+)(s +5s+) (s+) (s+)(s +4) (h) F (s) s 6 (i) F (s) 4 s 4 +4 (j) F (s) e s s (a) Perform partial fraction expansion, F (s) s(s +) C s + C s + C s + s0 C s s F (s) s s + L {F (s)} L { s } L { s + } f(t) (t) e t (t)

10 84 CHAPTER 3. DYNAMIC RESPONSE (b) Perform partial fraction expansion, F (s) 0 s(s +)(s + 0) F C s + C s + + C 3 s +0 C 0 (s +)(s + 0) s0 C 0 s(s + 0) s 0 9 C 3 0 s(s +) s 0 9 (s) s 9 s s +0 0 f(t) L {F (s)} (t) 0 9 e t (t)+ 9 e 0t (t) (c) Re-write and carry out partial fraction expansion, F (s) 3s + s +4s +0 (s +) (s +) +4 3(s +) (s +) +4 4 (s +) +4 f(t) L {F (s)} (3e t cos 4t 4e t sin 4t)(t) (d) Perform partial fraction expansion, Equate numerators: F (s) 3s +9s + (s +)(s +5s + ) C s + + C s + C 3 s +5s + C 3(s +3s +4) (s +5s + ) s (s +) + C s + C 3 (s +5s + ) 3s +9s + (s +)(s +5s + ) (C )s +(6+C 3 +C )s +(C )3s +9s +

11 85 Equate like powers of s to Þnd C and C 3 : C C F (s) 3 C 9 5 C (s +) + 5 s 3 5 (s +5s + ) 6 s + 5 (s +) + 5 (s + 5 ) (s ) +( f(t) L {F (s)} ( 6 5 e t + e 5 5 t 9 cos + e t sin (e) Re-writeanduseentry#7ofTableA., F (s) s +4 (s + ) f(t) sin t ) 9 )(t) (f) F (s) C (s +) (s +)(s +4) C (s +) + C s + C 3 (s +4) (s +) (s +4) s 5 Equate numerators and like powers of s terms: ( 5 + C )s +(C + C 3 )s +( C 3) s C 3 4 C 3 5 C + C 3 C C 0

12 86 CHAPTER 3. DYNAMIC RESPONSE F (s) 5 (s +) + 5 s + 5 (s +4) 5 (s +) + 5 s (s + ) (s + ) f(t) 5 e t 5 cos t + 6 sin t 5 (g) Perform partial fraction expansion, F (s) s + s (h) Use entry #6 of Table A., s + s f(t) (+t)(t) (i) Re-write as, F (s) s 6 f(t) L { s 6 } t5 5! t5 60 F (s) 4 s 4 +4 s + s +s + + s + s s + (s +) s (s +) + (s ) s (s ) + Use Table A. entry #9 and Table A. entry #5, f(t) L {F (s)} e t cos(t) d dt {e t sin(t)} e t cos(t) d dt {et sin(t)} e t cos(t) { e t sin(t)+cos(t)e t } e t cos(t)+ {et sin(t)+cos(t)e t } cos(t){ e t + e t } +sin(t){ e t + e t } f(t) cos(t) sinh(t) +sin(t) cosh(t)

13 87 (j) Using entry # of Table A., F (s) e s s f(t) L {F (s)} (t )(t) 8. Find the time function corresponding to each of the following Laplace transforms: (a) F (s) s(s+) (b) F (s) s +s+ s 3 (c) F (s) (s +s+) s(s+) (d) F (s) s3 +s+4 s 4 6 (e) F (s) (s+)(s+5) (s+)(s +4) (f) F (s) (s ) (s +) (g) F (s) tan ( s ) (a) Perform partial fraction expansion, F (s) s(s +) C s + C (s +) + C 3 (s +) C sf (s) s0 (s +) s0 4 C 3 (s +) F (s) s s s C d ds [(s +) F (s)] s d ds [s ] s s s 4 F (s) 4 s + 4 (s +) + (s +) f(t) L {F (s)} ( 4 4 e t te t )(t)

14 88 CHAPTER 3. DYNAMIC RESPONSE (b) Perform partial fraction expansion, F (s) s + s + s 3 s + s + (s )(s + s +) C s + C s + C 3 s + s + C (s )F (s) s s + s + s + s + s 4 3 Equate numerators and like powers of s: 4 3 s + C s + C 3 s + s + s + s + (s )(s + s +) s ( C )+s( 4 3 C + C 3 )+( 4 3 C 3) s + s C C C 3 C 3 3 F (s) 4 3 s s s + 3 s + s + s + (s + ) +( 3 ) f(t) L {F (s)} 4 3 et + 3 e t cos 3 {et + e t cos 3 t}(t) 3 t

15 89 (c) Carry out partial fraction expansion, F (s) (s + s +) s (s +) C s + C (s +) + C 3 (s +) C sf (s) s0 (s + s +) (s +) s0 C 3 (s +) F (s) s (s + s +) s s C d ds [(s +) F (s)] s d + s +) ds [(s ] s s (s +)s (s + s +) 0 s s F (s) s + 0 (s +) + (s +) f(t) L {F (s)} { te t }(t) (d) Carry out partial fraction expansion, F (s) s3 +s +4 s 4 As + B 6 s 4 + Cs + D 3 s +4 4 s + s s s +4 4 sinh(t)+3 d 4 dt { sinh(t)} 4 sin(t) d 4 dt { sin(t)} 4 sinh(t)+3 4 cosh(t) 4 sin(t)+ 4 cos(t) (e) Expand in partial fraction expansion and compute the residues using

16 90 CHAPTER 3. DYNAMIC RESPONSE the results from Appendix A (pages 8-83), F (s) (s +)(s +5) (s +)(s +4) C s + + C s j + C 3 s +j + C 4 (s j) + C 5 (s +j) C (s +)F (s) s C 4 (s j) 83 39j F (s) sj 4.50 j C 5 C j.950 C d (s j) F (s) 8 579j sj ds j.895 C 3 C j.895 These results can also be veriþed with the MATLAB residue command, a[8866]; b [ ]; [r,p,k]residue(b,a) r i i i i p i i i i k [] We then have, f(t).8e t + C cos(t +argc )+ C 4 t cos(t +argc 4 ).8e t cos(t.788) t cos(t.70) where C.96489, C , arg C tan ( ).788, using the atan commandinmatlab,andargc 4 tan ( ).70alsousingtheatan commandinmatlab.

17 9 (f) F (s) (s ) (s +) Using the multiplication by time Laplace transform property (Table A. entry #): d G(s) L{tg(t)} ds We can see that d s ds (s +) s (s +). So the inverse Laplace transform of F (s) is: L {F (s)} t cos t (g) Follows from Problem 5 (c), or expand in series, Then, tan ( s ) s 3s 3 + 5s 5... L {tan ( t )} s 3! + t4 5! sin(t).... t Alternatively, let us assume L {tan ( s )) f(t). We use the identity d ds [tan s] +s which means that L { sin(t) t. s + } tf(t) sin(t). Therefore,f(t) 9. Solve the following ordinary differential equations using Laplace transforms: (a) ÿ(t) + úy(t) + 3y(t) 0; y(0), úy(0) (b) ÿ(t) úy(t)+4y(t) 0;y(0), úy(0) (c) ÿ(t)+ úy(t) sint; y(0), úy(0) (d) ÿ(t)+3y(t) sint; y(0), úy(0) (e) ÿ(t)+úy(t) e t ; y(0), úy(0) (f) ÿ(t)+y(t) t; y(0), úy(0) (a) ÿ(t)+ úy(t)+3y(t) 0; y (0), úy (0)

18 9 CHAPTER 3. DYNAMIC RESPONSE Using Table A. entry #5, the differentiation Laplace transform property, s Y (s) sy (0) úy (0) + sy (s) y (0) + 3Y (s) 0 Y (s) s +3 s + s +3 s s s + s Using Table A. entries #9 and #0, + 5 s + q (b) y(t) e t cos t + 5 e t sin t ÿ(t) úy(t)+4y (t) 0;y (0), úy (0) s Y (s) sy (0) úy (0) sy (s)+y(0) + 4Y (s) 0 Y (s) s +4 s s +4 s +4 Using Table A. entries #9 and #0, (s ) +3 (s ) (s ) (s ) +3 (s ) (s ) (s ) +3 y (t) e t cos 3t et sin 3t (c) ÿ(t)+ úy(t) sint; y (0), úy (0) s Y (s) sy (0) úy (0) + sy (s) y (0) s + Y (s) s3 +3s + s +4 s (s +)(s +) C s + C s + + C 3s + C 4 s +

19 93 C s3 +3s + s +4 (s +)(s +) s0 4 C s3 +3s + s +4 s (s s 5 +) s 3 µ 3 + C 3 4 s s (4 + C 3 + C 4 )+s Match coefficients of like powers of s s + + C 3s + C 4 s + µ 3 + C 4 s3 +3s + s +4 s (s +)(s +) +4 s 3 +3s + s +4 C C 4 C C 3 4 s + 5 s + + s s + 4 s + 5 s + s s + Using Table A. entries #, #7, #7, and #8 y (t) 4 5 e t cos t sin t s + (d) ÿ (t)+3y (t) sint; y (0), úy (0) s Y (s) sy (0) úy (0) + 3Y (s) s + Y (s) s3 +s + s +3 (s +3)(s +) C s + C s +3 + C 3s + C 4 s + (C s + C ) s + +(C 3 s + C 4 ) s +3 (s +3)(s +) Match coefficients of like powers of s: s3 +s + s +3 (s +3)(s +) s 3 (C + C 3 )+s (C + C 4 )+s(c +3C 3 )+(C +3C 4 ) s 3 +s + s +3

20 94 CHAPTER 3. DYNAMIC RESPONSE C + C 3 C C 3 C + C 4 C C 4 C +3C 3 C 3 +3C 3 C 3 C C +3C 4 3 ( C 4 )+3C 4 3 C 4 C 3 Y (s) s + 3 s +3 + s + s + s 3 3 s +3 + s +3 + s s + + s + (e) y (t) cos 3t + 3 sin 3t + cos t + sin t ÿ(t)+úy (t) e t ; y (0), úy (0) s Y (s) sy (0) úy (0) + sy (s) y (0) s Y (s) 5s 4 s (s ) (s +) C s + C s + C 3 s + C C C 3 5s 4 (s ) (s +) s0 5s 4 s (s +) s 3 5s 4 s (s ) s 7 3 Y (s) s + 3 s 7 3 s + y (t) + 3 et 7 3 e t (f) Using the results from Appendix A, ÿ (t)+y(t) t; y (0), úy (0)

21 95 s Y (s) sy (0) úy (0) + Y (s) s Y (s) s3 s + s (s +) C s + C s + C 3s + C 4 s + s 3 s + C d ds (s s0 0 +) s 3 s + C (s s0 +) s + C 3s + C 4 s + s + +(C 3 s + C 4 ) s s (s +) s3 s + s (s +) s3 s + s (s +) Match coefficients of like powers of s: C 3 C 4 + C 4 Y (s) s + s s + s + y (t) t +cost sint 0. Write the dynamic equations describing the circuit in Fig Write the equations in both state-variable form and as a second-order differential equation in y(t). Assuming a zero input, solve the differential equation for y(t) using Laplace-transform methods for the parameter values and initial conditions shown in the Þgure. Verify your answer using the initial command in MATLAB. i C dy dt () v L di dt () u(t) L di Ri(t) y(t) 0 dt di dt u L R L i C y (3)

22 96 CHAPTER 3. DYNAMIC RESPONSE Figure 3.50: Circuit for Problem 3.0 Put differential equations () and (3) into state-space form: úi R L L i 0 + u úv C 0 v Substituting the given values for L, R, andc we have for equation (3): di dt u dy dt y Characteristic equation: So: Solving for the coefficients: ÿ +úy + y u s +s + 0 (s +) 0 y(t) A e t + A te t y(t) A e t + A te t y(t 0 ) A e t 0 + A t 0 e t 0 úy(t) A e t + A e t A te t úy(t 0 ) A e t0 + A e t0 A te t0 0 A e t 0 and A ( t 0 )e t 0 y(t) ( t 0 )e t0 t + te t0 t To verify the solution using MATLAB, re-write the differential equation as, ẏ.. 0 ỵ + L u y y 0 y 0 ỵ y

23 y(t) 97 Then the following MATLAB statements, a[0,;-,-]; b[0;]; c[,0]; d[0]; sysss(a,b,c,d); xo[;0]; [y,t,x]initial(sys,xo); plot(t,y); grid; xlabel( Time (sec) ); ylabel( y(t) ); title( Initial condition response ); generate the initial condition response shown below that agrees with the analytical solution above. Initial condition response Time (sec) Problem 3.0: Initial condition response.. Consider the standard second-order system ω n G(s) s +ζω n s + ω. n

24 98 CHAPTER 3. DYNAMIC RESPONSE Figure 3.5: Plot of input for Problem 3. a) Write the Laplace transform of the signal in Fig b). What is the transform of the output if this signal is applied to G(s). c) Find the output of the system for the input shown in Fig (a) The input signal in Figure 3.5 may be written as: u(t) t t[(t )] t[((t )] + t[(t 3)] where (t τ) denotes a unit step. The Laplace transform of the input signal is: U(s) s e s e s e 3s (b) The Laplace transform of the output if this signal is applied is: ω n Y (s) G(s)U(s) s +ζω n s + ω n µ e s s e s e 3s (c) However to make the mathematical manipulation easier, consider only the reponse of the system to a ramp input: ω n Y (s) s +ζω n s + ω n Partial fractions yields the following: Y (s) s ζ ω n s + µ s ζ ω n (s +ζω n ω n ζ ) (s + ω n ζ) +(ω n p ζ ) Use the following Laplace transform pairs for the case 0 ζ < :

25 99 L s + z { (s + a) + ω } where φ tan ³ ω z a s (z a) + ω ω e at sin(ωt + φ) L { s } t ramp L { s } (t) unit step and the following Laplace transform pairs for the case ζ : L { (s + a) } te at L s { } ( at)e at (s + a) L { s } t ramp L { s } (t) unit step the following is derived: t ζ ω n + e ζωnt sin(ω p ζ ω y (t) n ζ n ζ t +tan ζ 0 ζ < ζ t 0 t ω n + ω n e ω nt ω n t + ζ t 0 Since u(t) consists of a ramp and three delayed ramp signals, using superpostion (the system is linear), then: y(t) y (t) y (t ) y (t )+y (t 3) t 0. A rotating load is connected to a Þeld-controlled DC motor with negligible Þeld inductance. A test results in the output load reaching a speed of rad/sec within / sec when a constant input of 00 V is applied to the motor terminals. The output steady-state speed from the same test is found to be rad/sec. Determine the transfer function θ(s)/v f (s) ofthe motor. Equations of motion for a DC motor: J m θm + b ú θ m K m i a,

26 00 CHAPTER 3. DYNAMIC RESPONSE K eθm ú di a + L a dt + R ai a v a, but since there s neglible Þeld inductance L a 0. Combining the above equations yields: R a J m θm + R a b ú θ m K t v a K t K e ú θm Applying Laplace transforms yields the following transfer function: θ(s) V f (s) Kt J m R a s(s + KtKe R a J m + b J m ) where K Kt J m R a and a KtKe R a J m + b J m. K and a are found using the given information: K s(s + a) V f (s) 00 since V f (t) 00V s úθ( ) rad/sec For the given information we need to utilize ú θ m (t) instead of θ m (t): sθ(s) 00K s(s + a) Using the Final Value Theorem and assuming that the system is stable: 00K lim s 0 s + a lim úθ( s 0 )00K a Take the inverse Laplace transform: L { 00K a a s(s + a) } 00K a ( e at )( e at ) 0.5 e a yields a.39 K a yields K θ(s) V f (s) s(s +.39) 3. For the tape drive shown in Fig..48, compute the following, using the numbers given in Problem.0 (a):

27 0 (a) the transfer function from the motor current to the tape position; (b) the poles and zeros for the transfer function in part (a). (a) Tape tension I a (s) T (s) I a (s) From Problem.0: T B( úx úx )+k(x x ) T (s) (Bs + k)(x (s) X (s)) J úω B ω + k t i a + Br ( úx úx )+kr (x x ) J úω B ω + Fr + Br ( úx úx )+kr (x x ) úx r ω úx r ω J s + B 0 (Bs + k)r (Bs + k)r 0 J s + B (Bs + k)r (Bs + k)r r 0 s 0 0 r 0 s ω ω x x k t I a Fr 0 0 X (s) 5(s + 800) I a (s) s(s + 50s ) X (s) I a (s) s(s + 50s ) X (s) X (s) T (s) I a (s) 5(s + 800) s(s + 50s ) I a(s) (0s (s + 800) ) s(s + 50s ) (b) poles at : 0, 65 ± j996.8 zeros at : 800, 000

28 0 CHAPTER 3. DYNAMIC RESPONSE 4. For the system in Fig..50, compute the transfer function from the motor voltage to position θ. From Problem.3: So we have: L di a dt + R ai a + k e ú θ v a k t i a J θ + b( ú θ ú θ )+k(θ θ )+B ú θ J θ + b( ú θ ú θ )+k(θ θ )0 LsI a (s)+r a I a (s)+sk e Θ (s) V a (s) k t I a (s) s J Θ (s)+b[θ (s) Θ (s)]s + k[θ (s) Θ (s)] + BsΘ (s) s J Θ (s)+b[θ (s) Θ (s)]s + k[θ (s) Θ (s)] 0 we have: Θ (s) V a (s) det k t (bs + k) sk e 0 Ls + R a J s + Bs + bs + k bs k k t bs k J s + bs + k 0 k t (bs + k) (Ls + Ra)[J J s 4 +(J b + BJ + bj )s 3 +(J k + Bb + KJ )s + Bks] +k e k t J s 3 + k e k t bs + kk e k t s k t (bs + k) J J s 5 + J [J R a + L(b + B)]s 4 +[J k e k t J L(b + k)+lj k + R a (b + B)J Lb ]s 3 +[L(b + B)(b + k) bkl + J R a (b + k)+r a J k b R a ]s +[k e k t (b + k)+kl(b + k) bk + R a (b + B)(b + k) bkr a ]s + kr a b 5. Compute the transfer function for the two-tank system in Fig..54 with holes at A and C. From Problem.7 but with s a tank area we have: ú h ú h 0 6a h h + ω in a a 0

29 03 h ú h +6ω in 0 6a h ú 6a ( h h ) s h (s) h (s)+6ω in (s) 6a s h (s) 6a [ h (s) h (s)] h (s) h (s) ω in (s) ω in (s) 6a[a( 6a + s)] 6[a( 6a + s)] 6. For a second-order system with transfer function 3 G(s) s +s 3, determine the following: (a) DC gain; (b) the Þnal value to a step input. (a) DC gain G(0) 3 3 (b) lim t y(t)? s +s +30 s, 3 Since the system has an unstable pole, the Final Value Theorem is not applicable. The output is unbounded. 7. Consider the continuous rolling mill depicted in Fig Suppose that the motion of the adjustable roller has a damping coefficient b, and that the force exerted the rolled material on the adjustable roller is proportional to the material s change in thickness: F s c(t x). Suppose further that the DC motor has a torque constant K t and a back-emf constant K e,and that the rack-and-pinion has effective radius of R. (a) What are the inputs to this system? The output? (b) Without neglecting the effects of gravity on the adjustable roller, draw a block diagram of the system that explicitly shows the following quantities: V s (s), I 0 (s),f(s) (the force the motor exerts on the adjustable roller), and X(s).

30 04 CHAPTER 3. DYNAMIC RESPONSE Figure 3.5: Continuous rolling mill (c) Simplify your block diagram as much as possible while still identifying output and each input separately. (a) Inputs: Output: input voltage v s (t) thickness T gravity mg thickness x (b) Dynamic analysis of adjustable roller: mẍ c(t x) mg b úx F m (s mg ct m + sb + c)x(s)+f m (s)+ s 0 () Torque in rack and pinion: T RP RF m NT motor but T motor K t I f i o F m NK ti f R i o ()

31 05 DC motor circuit analysis: di o v s (t) R a i o + L a dt + v a(t) v a (t) u eθ ú θr N x I o (s) V s(s) K en R sx(s) (3) R a + sl a Combining (), (), and (3): 0(s m + sb + c)x(s)+ mg ct s + NK ti f R " Vs (s) K en R sx(s) # sl a + R a

32 06 CHAPTER 3. DYNAMIC RESPONSE Block diagrams for rolling mill Problems and Solutions for Section Compute the transfer function for the block diagram shown in Fig Note that a i and b i are constants. (a) Write the third-order differential equation that relates y and u. (Hint: Consider the transfer function.) (b) Write three simultaneous Þrst-order (state-variable) differential equations using variables x, x,andx 3,asdeÞned on the block diagram in Fig Notice how the same constant parameters enter the transfer function, the differential equations, and the matrices of the state-variable form. (This special structure is called the control

33 07 Figure 3.53: Block diagram for Problem 3.8 canonical form and will be discussed further in Chapter 7.) Repeat for the block diagram of Fig. 3.50(b). This is the observer canonical form for a 3rd order system. Using Mason s rule: Forward paths: Feedback paths: (a) Y U b s + b s + b 3 s 3 a s + a s + a 3 s 3 b s + b s + b 3 s 3 + a s + a s + a 3 s 3 (s 3 + a s + a s + a 3 )Y (b s + b s + b 3 )U d 3 y dt 3 + a ÿ + a úy + a 3 y b ü + b úu + b 3 u (b) DeÞnitions from block diagram: úx 3 x úx x úx u a x a x a 3 x 3 y b 3 x 3 + b x + b x

34 08 CHAPTER 3. DYNAMIC RESPONSE Figure 3.54: Block diagrams for Problem 3.9. x a a a y b b b 3 x x + 0 u 0 9. Find the transfer functions for the block diagrams in Fig (a) Block diagram for Fig (a)

35 09 Forward Path Gain 34 G 54 G Loop Path Gain 3 -G Y R G +G + G (b) Block diagram for Fig (b) Block diagram for Fig (b): reduced Y R G G G 3 G 4 G ( + G G )( + G 4 G 5 ) (c) Block diagrams for Fig (c)

36 0 CHAPTER 3. DYNAMIC RESPONSE Figure 3.55: Block diagrams for Problem 3.0 Forward Path Gain 3456 G G G G 7 +G G4G5 +G 4 G 6 G 4 G 5 +G 4 Y R G 7 + G 6G 4 G 5 +G 4 + G G G 3 +G G 4G 5 +G 4 0. Find the transfer functions for the block diagrams in Fig. 3.55, using the following: (a) the ideas of Figs. 3.6 and 3.7; (b) Mason s rule. Transfer functions found using the ideas of Figs. 3.6 and 3.7: (a) Block diagram for Fig. 3.55(a)

37 G G 3 G G H G 3 G 3 H 3 Y R G ( + G ) +G ( + G )G 3 G ( G H )( G 3 H 3 )+G G ( G 3 H 3 ) ( G H )( G 3 H 3 )+G G 3 ( G H )+G G G 3 (b) Block diagram for Fig. 3.55(b) Mason s rule: forward path gains, b 3 s 3, b s, b s (c) Applying block diagram reduction: Block diagram for Fig. 3.55(c)

38 CHAPTER 3. DYNAMIC RESPONSE (d) The method: reduce innermost box, shift b to b 3 node, reduce next innermost box and continue systematically. Y R b 3 + b (s + a )+b (s + a s + a ) s 3 + a s + a s + a 3 Block diagram for Fig. 3.55(d) The system is tightly connected, easy to apply Mason s. Mason s rule: (a) Forward Path 356 p G G 346 p G Loop Path 347 L G G L G G G 3 Y R p + p G ( + G ) +L + L +G G 3 ( + G ) (b) Loop Path: a3 s, a 3 s, a s Y R b3 s + b 3 s + b s + a3 s 3 + a s + a s b 3 + b s + b s s 3 + a s + a s + a 3

39 3 (c) Block diagrams for Fig. 3.55(c) (d) Mason s Rule: Y R D + AB +G(D + AB ) D + DBH + AB +BH + GD + GBDH + GAB. Use block-diagram algebra or Mason s rule to determine the transfer function between R(s) and Y(s) in Fig

40 4 CHAPTER 3. DYNAMIC RESPONSE Figure 3.56: Block diagram for Problem 3. By block diagram algebra: Block diagram for Fig Block diagram for Fig. 3.56: reduced

41 5 Q R PH 3 PH 4 R P (H 3 H 4 ) P G NG 6 Y So: Now, what is N? Y R G NG 6 +(H 3 + H 4 )G NG 6 Block diagrams for Fig Move G 4 and G 3 : Block diagram for Fig. 3.56: reduced Combine symmetric loops as in the Þrst step: Block diagram for Fig. 3.56: reduced

42 6 CHAPTER 3. DYNAMIC RESPONSE Which is: N O I G 4G + G 5 G 3 +H (G 4 + G 5 ) Y (s) G (G 4 G + G 5 G 3 )G 6 R(s) +H (G 4 + G 5 )+(H 3 + H 4 )G (G 4 G + G 5 G 3 )G 6 By Mason s Rule: Signal ßow graph Flow graph for Fig Forward Path Gain 345 G G G 4 G G G 3 G 5 G 6 Loop Path Gain 345 G G G 4 G 6 H G G G 4 G 6 H G G 3 G 5 G 6 H G G 3 G 5 G 6 H G 4 H 343 G 5 H and the determinants are +[(H 3 + H 4 )G (G G 4 + G 3 G 5 )G 6 + H (G 4 + G 5 )] (0) (0) 3 (0) 4 (0) Applying the rule, the transfer function is

43 7 Figure 3.57: Block diagram for Problem 3. Y (s) R(s) X Gi i G (G 4 G + G 5 G 3 )G 6 +H (G 4 + G 5 )+(H 3 + H 4 )G (G 4 G + G 5 G 3 )G 6. Use block-diagram algebra to determine the transfer function between R(s) and Y(s) in Fig Block diagram for Fig Move node A and close the loop:

44 8 CHAPTER 3. DYNAMIC RESPONSE Block diagram for Fig. 3.57: reduced Add signal B, close loop and multiply before signal C. Block diagram for Fig. 3.57: reduced Move middle block N past summer. Block diagram for Fig. 3.57: reduced Now reverse order of summers and close each block separately.

45 9 Figure 3.58: Circuit for Problem 3.3 Block diagram for Fig. 3.57: reduced feedforward Y z } { R (N + G 3 )( ) +NH {z 3 } feedback Y R G G + G 3 ( + G H + G H ) +G H + G H + G G H 3 3. For the electric circuit shown in Fig. 3.58, Þnd the following: (a) the time-domain equation relating i(t) and v (t); (b) the time-domain equation relating i(t) and v (t); (c) assuming all initial conditions are zero, the transfer function V (s)/v (s) and the damping ratio ζ and undamped natural frequency ω n of the system; (d) the values of R that will result in v (t) having an overshoot of no more than 5%, assuming v (t) isaunitstep,l0mh,andc 4µF. (a) v (t) L di dt + Ri + Z C i(t)dt

46 0 CHAPTER 3. DYNAMIC RESPONSE Figure 3.59: Unity feedback system for Problem 3.4 (b) v (t) C Z i(t)dt (c) v (s) v (s) sc sl + R + (d) For 5% overshoot ζ 0.4 sc s LC + src ζ q R R ζ L C r L C ()(0.4) r Ω Problems and Solutions for Section For the unity feedback system shown in Fig. 3.59, specify the gain K of the proportional controller so that the output y(t) has an overshoot of no more than 0% in response to a unit step. Y (s) R(s) K s +s + K ω n K ζ ω n K () ω n s +ζω n s + ω n In order to have an overshoot of no more than 0%: Solving for ζ : M p e πζ/ ζ 0.0

47 Figure 3.60: Unity feedback system for Problem 3.5 Using () and the solution for ζ: s (ln M p ) ζ π +(lnm p ) 0.59 K ζ.86 0 <K For the unity feedback system shown in Fig. 3.60, specify the gain and pole location of the compensator so that the overall closed-loop response to a unit-step input has an overshoot of no more than 5%, and a % settling time of no more than 0. sec. Verify your design using MATLAB. Y (s) R(s) 00K s +(5+a)s +5a + 00K 00K s +ζω n s + ω n Using the given information: R(s) s M p 5% t % 0.sec unit step Solve for ζ : M p e πζ/ ζ s (ln M p ) ζ π +(lnm p )

48 y(t) CHAPTER 3. DYNAMIC RESPONSE Solve for ω n: e ζωnts 0.0 For a % settling time t s ζω n 0. ω n 4.07 Now Þnd a and K : ζω n (5+a) a ζω n ω n (5a + 00K) K ω n 5a ThestepresponseofthesystemusingMATLABisshownbelow..4 Step Response Time (sec) Step Response for Problem 3.5 Problems and Solutions for Section 3.4

49 3 6. Suppose you desire the peak time of a given second-order system to be less than t 0 p. Draw the region in the s-plane that corresponds to values of the poles that meet the speciþcation t p <t 0 p. s-plane region to meet peak time constraint: shaded ω d t p π t p π ω d <t 0 p π t 0 p < ω d 7. Suppose you are to design a unity feedback controller for a Þrst-order plant depicted in Fig (As you will learn in Chapter 4, the conþguration shown is referred to as a proportional-integral controller). You are to design the controller so that the closed-loop poles lie within the shaded regions shown in Fig (a) What values of ω n and ζ correspond to the shaded regions in Fig. 3.6? (A simple estimate from the Þgure is sufficient.) (b) Let K α α. Find values for K and K so that the poles of the closed-loop system lie within the shaded regions. (c) Prove that no matter what the values of K α and α are, the controller provides enough ßexibility to place the poles anywhere in the complex (left-half) plane.

50 4 CHAPTER 3. DYNAMIC RESPONSE Figure 3.6: Unity feedback system for Problem 3.7 Figure 3.6: Desired closed-loop pole locations for Problem 3.7

51 5 (a) The values could be worked out mathematically but working from the diagram: p ω n 4.6 θ sin ζ ζ sinθ From the Þgure: θ 34 ζ θ 70 ζ ζ 0.9 (roughly) (b) Closed-loop pole positions: s(s + α)+(ks + KK I )K α 0 s +(α + KK α )s + KK I K α 0 For this case: s +(+K)s +KK I 0 ( ) Choose roots that lie in the center of the shaded region, (s +(3+j))(s +(3 j)) s +6s +30 s +(+K)s +KK I s +6s +3 +K 6 K 3 4K I K I 3 4 (c) For the closed-loop pole positions found in part (b), in the (*) equation the value of K canbechosentomakethecoefficient of s take on any value. For this value of K a value of K I canbechosenso that the quantity KK I K α takes on any value desired. This implies that the poles can be placed anywhere in the complex plane. 8. The open-loop transfer function of a unity feedback system is G(s) K s(s +). The desired system response to a step input is speciþed as peak time t p secandovershootm p 5%.

52 6 CHAPTER 3. DYNAMIC RESPONSE (a) Determine whether both speciþcations can be met simultaneously by selecting the right value of K. (b) Sketch the associated region in the s-plane where both speciþcations are met, and indicate what root locations are possible for some likely values of K. (c) Pick a suitable value for K, and use MATLAB to verify that the speciþcations are satisþed. (a) T (s) Y (s) R(s) Equate the coefficients: G(s) +G(s) K s +s + K ζω n K ω (*) n ω n K ζ K ω n s +ζω n s + ω n We would need: πζ M p % e ζ ζ 0.69 t p sec π π p ω ω d ω n ζ n 4.34 But the combination (ζ 0.69, ω n 4.34) that we need is not possible by varying K alone. Observe that from equations (*) ζω n (b) Now we wish to have: Mp πζ r 0.05 e ζ (**) t p r sec π ω d where r relaxation factor. Recall the conditions of our system: ω n K ζ K replace ω n and ζ in the system (**): π K r 0.05 sec π K

53 7 r 0.05 e r r. K + π r K 3.0 then with K 3.0 we will have: Mp rm p t p rt p. sec. sec Note: * denotes actual location of closed-loop roots. s-plane regions % Problem 3.8 K3.0; num[k]; den[,, K]; systf(num,den); t0:.0:3; ystep(sys,t); plot(t,y); yss dcgain(sys); Mp (max(y) - yss)*00; % Finding maximum overshoot msg overshoot sprintf( Max overshoot %3.f%%, Mp); % Finding peak time idx max(find(y(max(y)))); tp t(idx);

54 y(t) 8 CHAPTER 3. DYNAMIC RESPONSE msg peaktime sprintf( Peak time %3.f, tp); xlabel( Time (sec) ); ylabel( y(t) ); msg title sprintf( Step Response with K%3.f,K); title(msg title); text(., 0.3, msg overshoot); text(., 0., msg peaktime); grid on;.4 Step Response with K Max overshoot 0.97% 0. Peak time Time (sec) Problem 3.8: Closed-loop step response 9. The equations of motion for the DC motor shown in Fig..6 were given in Eqs. (.63-64) as µ J m θm + b + K tk e úθ m K t v a. R a R a Assume that J m 0.0 kg m, b 0.00 N m sec, K e 0.0 V sec, K t 0.0 N m/a, R a 0Ω.

55 9 (a) Find the transfer function between the applied voltage v a and the motor speed θ ú m. (b) What is the steady-state speed of the motor after a voltage v a 0V has been applied? (c) Find the transfer function between the applied voltage v a and the shaft angle θ m. (d) Suppose feedback is added to the system in part (c) so that it becomes a position servo device such that the applied voltage is given by v a K(θ r θ m ), where K is the feedback gain. Find the transfer function between θ r and θ m. (e) What is the maximum value of K thatcanbeusedifanovershoot M p < 0% is desired? (f) What values of K will provide a rise time of less than 4 sec? (Ignore the M p constraint.) (g) Use MATLAB to plot the step response of the position servo system for values of the gain K 0.5,, and Þnd the overshoot and rise time of the three step responses by examining your plots. Are the plots consistent with your calculations in parts (e) and (f)? (a) µ J m θm + b + K tk e úθ m K t v a R a R a µ J m Θ m s + b + K tk e Θ m s K t V a (s) R a R a sθ m (s) V a (s) s + b J m K t R a J m + K tk e R aj m J m 0.0kg m, b 0.00N m sec, K e 0.0V sec, K t 0.0N m/a, R a 0Ω. sθ m (s) V a (s) 0. s +0.04

56 30 CHAPTER 3. DYNAMIC RESPONSE (b) Final Value Theorem (c) (d) úθ( ) s(0)(0.) s(s +0.04) s Θ m (s) V a (s) 0. s(s +0.04) Θ m (s) 0.K(Θ r Θ m ) s(s +0.04) Θ m (s) 0.K Θ r (s) s +0.04s +0.K (e) M p ζ e πζ/ 0. (0%) ζ ω n Y (s) s +ζω n s + ω n ζω n 0.04 ω n rad/ sec (0.4559) ω n 0.K K < (f) ω n.8 t r ω n 0.K K.0 (g) MATLAB clear all close all K[ e-]; t0:0.0:50; for i::length(k) K K(i); titletext sprintf( K %.4f, K); wn sqrt(0.*k);

57 3 numwnˆ; den[ 0.04 wnˆ]; zeta0.04//wn; sys tf(num, den); y step(sys, t); % Finding maximum overshoot if zeta < Mp (max(y) - )*00; overshoottext sprintf( Max overshoot %3.f %, Mp); else overshoottext sprintf( No overshoot ); end % Finding rise time idx 0 max(find(y<0.)); idx 09 min(find(y>0.9)); t rt(idx09) - t(idx 0); risetimetext sprintf( Rise time %3.f sec, t r); % Plotting subplot(3,,i); plot(t,y); grid on; title(titletext); text( 0.5, 0.3, overshoottext); text( 0.5, 0., risetimetext); end

58 3 CHAPTER 3. DYNAMIC RESPONSE K K Max overshoot Rise time 4.0 sec K Max overshoot 69.3 Rise time.5 sec K Max overshoot 59.3 Rise time 3.70 sec K Max overshoot 77.7 Rise time.73 sec Max overshoot 9.99 Rise time 3.66 sec Problem 3.9: Closed-loop step responses For part (e) we concluded that K< in order for M p < 0%. This is consistent with the above plots. For part (f) we found that K.0 in order to have a rise time of less than 4 seconds. We actually see that our calculations is slightly off and that K can be K 0.5, but since K.0 is included in K 0.5, our answer in part f is consistent with the above plots. 30. You wish to control the elevation of the satellite-tracking antenna shown in Figs and The antenna and drive parts have a moment of inertia J and a damping B; these arise to some extent from bearing and aerodynamic friction but mostly from the back emf of the DC drive motor. The equations of motion are J θ + B ú θ T c, where T c isthetorquefromthedrivemotor.assumethat J 600, 000kg m B 0, 000 N m sec (a) Find the transfer function between the applied torque T c and the antenna angle θ. (b) Suppose the applied torque is computed so that θ tracks a reference command θ r according to the feedback law T c K(θ r θ),

59 33 Figure 3.63: Satellite Antenna (Courtesy Space Systems/Loral) Figure 3.64: Schematic of antenna for Problem 3.30

60 34 CHAPTER 3. DYNAMIC RESPONSE where K is the feedback gain. Find the transfer function between θ r and θ. (c) What is the maximum value of K thatcanbeusedifyouwishto have an overshoot M p < 0%? (d) What values of K will provide a rise time of less than 80 sec? (Ignore the M p constraint.) (e) Use MATLAB to plot the step response of the antenna system for K 00, 400, 000, and 000. Find the overshoot and rise time of the four step responses by examining your plots. Do the plots conþrm your calculations in parts (c) and (d)? J θ + B ú θ T c (a) JΘs + BΘs T c (s) Θ(s) T c (s) Js + B J 600, 000kg m Θ(s) T c (s) B 0, 000N m sec s(s + 30 ) (b) Θ(s) K(Θ r Θ) s(s + 30 ) Θ(s) Θ r (s).667k 0 6 s + 30s +.667K 0 6 (c) M p ζ e πζ/ 0. (0%) ζ 0.59 ω n Y (s) s +ζω n s + ω n ζω n ω n 0.08 rads/ sec (0.59) ω n.667k 0 6 K < 477

61 35 (d) ω n.8 t r ω n.667k 0 6 K K K Max overshoot Rise time 6.0 sec Max overshoot Rise time sec K K Max overshoot Rise time 36.7 sec Max overshoot Rise time.64 sec (e) Problem 3.30: Step responses (e) The results compare favorably with the predictions made in parts c and d. For K<477 the overshoot was less than 0, the rise-time was less than 80 seconds. 3. (a) Show that the second-order system ÿ +ζω n úy + ω ny 0, y(0) y o, úy(0) 0, has the response e σt y(t) y o p sin(ω ζ dt +cos ζ). (b) Prove that, for the underdamped case (ζ < ), the response oscillations decay at a predictable rate (see Fig. 3.65) called the logarithmic decrement δ, where δ ln y o στ d y ln y y ln y i y i,

62 36 CHAPTER 3. DYNAMIC RESPONSE Figure 3.65: DeÞnition of logarithmic decrement and τ d is the damped natural period of vibration τ d π. ω d (a) The system is second order Q(s) s +ζω n s + ω n. The initial condition response can be obtained by plugging a dirac delta at the input at the time 0 (this charges the system immediately to its initial condition and after that the system evolves by itself). Input effective y 0 δ(t) L[Input effective ] y 0 We do not know whether the transfer function has Þnite zeros or not, but further thought will reveal the presence of at least one Þnite zero in the H(s). lim s sh(s)y 0 y(t) 0+ where H(s) P (s) Q(s) P (s) s +ζω n s + ω. n If P (s) were a constant (no zeros in the H(s)), then the limit in the initial value theorem would give always zero (which is wrong because

63 37 we know that the initial value must be y 0.) So we need a zero. We suggest using the following H(s): where H(s) s s +ζω n s + ω n Y (s) sy 0 H(s)y 0 s +ζω n s + ω n R + + R s P + s P q P + ζω n + jω n ζ q P ζω n jω n ζ R + ω ne j(πcos ζ) ω n p ζ e jπ/s R R + Note: The residues can be calculated graphically. R + lim [(s P + )Y (s)] s P + y(t) R + e P +t + R e P t y(t) e ζωnt p ζ t+π/ cos ζ) ζ [e+j(ωn + e j(ω n ζ t+π/ cos ζ) ] y(t) y 0 e σt p ζ sin(ω dt cos ζ) (b) dy(t) dt 0 t nπ ω d t Max π n ω d e σnτ d y(t) tmax y n y 0 p ζ sin(cos ζ) (n is any integer) Note: sin( cos ζ) q ζ

64 38 CHAPTER 3. DYNAMIC RESPONSE y n y p 0 ζ p ζ e σnτ d ( ) (Proof of the Þrst line) From (*) (Proof of the second line) σ ln y 0 y n στ d y y 0 e στ d ln y 0 y n στ d y n y n y n y n y 0 e σnτ d y 0 e (n )στ d y 0 e σnτ d ( e στ d ) y n y n y n y n y 0e σnτ d y 0 e σnτ d ( eστ d ) y i yi for all i, n Problems and Solutions for Section In aircraft control systems, an ideal pitch response (q o )versusapitch command (q c ) is described by the transfer function Q o (s) Q c (s) τω n(s +/τ) s +ζω n s + ω. n The actual aircraft response is more complicated than this ideal transfer function; nevertheless, the ideal model is used as a guide for autopilot design. Assume that t r is the desired rise time, and that ω n.789 t r τ.6 t r ζ 0.89 Show that this ideal response possesses a fast settling time and minimal overshoot by plotting the step response for t r 0.8,.0,., and.5 sec. The following program statements in MATLAB produce the following plots: % Problem 3.3

65 39 tr [ ]; t[:40]/30; tbackfliplr(t); clf; for I:4, wn(.789)/tr(i); %Rads/second tautr(i)/(.6); %tau zeta0.89; % btau*(wnˆ)*[ /tau]; a[ *zeta*wn (wnˆ)]; ystep(b,a,t); subplot(,,i); plot(t,y); titletextsprintf( tr%3.f seconds,tr(i)); title(titletext); xlabel( t (seconds) ); ylabel( Qo/Qc ); ymax(max(y)-)*00; msgsprintf( Max overshoot%3.f%%,ymax); text(.50,.30,msg); ybackflipud(y); yindfind(abs(yback-)>0.0); tstback(min(yind)); msgsprintf( Settling time %3.f sec,ts); text(.50,.0,msg); grid; end

66 Qo/Qc Qo/Qc Qo/Qc Qo/Qc 40 CHAPTER 3. DYNAMIC RESPONSE Figure 3.66: Fig.3.66: Unity feedback system for Problem tr0.8 seconds.4 tr.0 seconds Max overshoot.3% 0. Settling time.3 sec t (seconds) 0.4 Max overshoot.3% 0. Settling time.9 sec t (seconds).4 tr. seconds.4 tr.5 seconds Max overshoot.3% 0. Settling time 3.5 sec t (seconds) Problem 3.3: Ideal pitch response 0.4 Max overshoot.3% 0. Settling time 4.4 sec t (seconds) 33. Consider the system shown in Fig. 3.66, where K(s + z) G(s) and D(s) s(s +3) s + p. () Find K, z, andp so that the closed-loop system has a 0% overshoot to a step input and a settling time of.5 sec (% criterion). For the 0% overshoot: ζ M p e πζ/ 0% s (ln M p ) ζ π +(lnm p ) 0.

67 4 For the.5sec (% criterion): ω n ζt s (0.6)(.5) 5. The closed loop transfer function is: Y (s) R(s) K s+z s+p s(s+3) +K s+z s+p s(s+3) K(s + z) s(s +3)(s + p)+k(s + z) Method I. From inspection, if z 3, (s + 3) will cancel out and we will have a standard form transfer function. As perfect cancellation is impossible, assign z a value that is very close to 3, say 3.. But in determining the K and p, assumethat(s + 3) and(s +3.) cancelled out each other. Then: Y (s) R(s) K s + ps + K As the additional pole and zero will degrade the system, pick some larger damping ration. Let ζ 0.7 ω n ζt s (0.7)(.5) 4.38, so let ω n 4.5 p ζω n K ω n 0.5 Method II. There are 3 unknowns (z,p,k) and only speciþed conditions. We can arbitrarily choose p large such that complex poles will dominate in the system response. Try p 0z Choose a damping ratio corresponding to an overshoot of 5% (instead of 0%, to be safe). ζ From the formula for settling time (with a % criterion) ω n ζt s adding some margin, let ω n 4.88 The characteristic equation is Q(s) s 3 +(s + p) s +(3p + K)s + Kz (s + a)(s +ζω n s + ω n)

68 4 CHAPTER 3. DYNAMIC RESPONSE We want the characteristic equation to be the product of two factors, a couple of conjugated poles (dominant) and a non-dominant real pole far form the dominant poles. Equate both expressions of the characteristic equation. ω na Kz ζω n a + ω n 30z + K ζω n + a 3+0z Solving three equations we get z 5.77 p 57.7 K.45 a Sketchthestepresponseofasystemwiththetransferfunction G(s) s/+ (s/40 + )[(s/4) + s/4+]. Justify your answer based on the locations of the poles and zeros (do not Þnd inverse Laplace transform). Then compare your answer with the step response computed using MATLAB. From the location of the poles, we notice that the real pole is a factor of 0 away from the complex pair of poles. Therefore, the reponse of the system is dominated by the complex pair of poles. G(s) (s/+) [(s/4) + s/4+] This is now in the same form as equation (3.58) where α,ζ 0.5 and ω n 4. Therefore, Fig. 3.3 suggests an overshoot of over 70%. The step response is the same as shown in Fig. 3.3, for α,withmorethan 70% overshoot and settling time of 3 seconds. The MATLAB plots below conþrm this. % Problem 3.34 num[/, ]; den[/6, /4, ];

69 y(t) 43 systf(num,den); t0:.0:3; ystep(sys,t); denconv([/40, ],den); systf(num,den); ystep(sys,t); plot(t,y,t,y); xlabel( Time (sec) ); ylabel( y(t) ); title( Step Response ); grid on;.8 Step Response Time (sec) Problem 3.34: Step responses 35. Consider the two nonminimum phase systems, (s ) G (s) (s +)(s +) ; () G (s) 3(s )(s ) (s +)(s +)(s +3). (3) (a) Sketch the unit step responses for G (s) andg (s), paying close attention to the transient part of the response.

70 44 CHAPTER 3. DYNAMIC RESPONSE (b) Explain the difference in the behavior of the two responses as it relates to the zero locations. (c) Consider a stable, strictly proper system (that is, m zeros and n poles, where m<n). Let y(t) denote the step response of the system. The step response is said to have an undershoot if it initially starts off in the wrong direction. Prove that a stable, strictly proper system has an undershoot if and only if its transfer function has an odd number of real RHP zeros. (a) For G (s) : Y (s) s G (s ) (s) s(s +)(s +) Q j (s z j ) H(s) k Q l (s p l ) R pi lim s pi [(s p i )H(s)] lim s pi k Q j Q j (s z j ) Q l l6i (s p l) k (p i z j ) Q l l6i (p i p l ) Each factor (p i z j )or(p i p l ) can be thought of as a complex number (a magnitude and a phase) whose pictorial representation is avectorpointingtop i and coming from z j or p l respectively. The method for calculating the residue at a pole p i is: () Draw vectors from the rest of the poles and from all the zeros to the pole p i. () Measure magnitude and phase of these vectors. (3) The residue will be equal to the gain, multiplied by the product of the vectors coming from the zeros and divided by the product of the vectors coming from the poles. In our problem: Y (s) (s ) s(s +)(s +) R 0 s + R (s +) + R (s +) s 4 s s + y (t) 4e t +3e t

71 y(t) 45 Step Response for G Time (sec) Problem 3.35: Step response for a non-minimum phase system For G (s) : Y (s) 3(s )(s ) s(s +)(s +)(s +3) s + 9 (s +) + 8 (s +) + 0 (s +3) y (t) 9e t +8e t 0e 3t

72 y(t) 46 CHAPTER 3. DYNAMIC RESPONSE Step Response for G Time (sec) Problem 3.35: Step response of non-minimum phase system (b) The Þrst system presents an undershoot. The second system, on the other hand, starts off in the right direction. The reasons for this initial behavior of the step response will be analyzed in part c. In y (t): dominant at t 0theterm 4e t In y (t): dominant at t 0theterm8e t (c) The following concise proof is from [] (see also []-[3]). Without loss of generality assume the system has unity DC gain (G(0) ) Since the system is stable, y( ) G(0), and it is reasonable to assume y( ) 6 0. Let us denote the pole-zero excess as r n m. Then, y(t) anditsr derivatives are zero at t 0,and y r (0) is the Þrst non-zero derivative. The system has an undershoot if y r (0)y( ) < 0. The transfer function may be re-written as Q m i G(s) ( s z i ) Q m+r i ( s p i ) The numerator terms can be classiþed into three types of terms: (). The Þrst group of terms are of the form ( α i s) with α i > 0. (). The second group of terms are of the form (+α i s) with α i > 0. (3). Finally, the third group of terms are of the form, (+β i s+α i s ) with α i > 0, and β i could be negative. However, β i < 4α i, so that the corresponding zeros are complex.

73 47 All the denominator terms are of the form (), (3), above. Since, y r (0) lim s sr G(s) it is seen that the sign of y r (0) is determined entirely by the number of terms of group 3 above. In particular, if the number is odd, then y r (0) is negative and if it is even, then y r (0) is positive. Since y( ) G(0),then we have the desired result. [] Vidyasagar, M., On Undershoot and Nonminimum Phase Zeros, IEEE Trans. Automat. Contr., Vol. AC-3, p. 440, May 986. [] Clark, R., N., Introduction to Automatic Control Systems, John Wiley, 96. [3] Mita, T. and H. Yoshida, Undershooting phenomenon and its control in linear multivariable servomechanisms, IEEE Trans. Automat. Contr., Vol. AC-6, pp , Consider the following second-order system with an extra pole: H(s) Show that the unit step response is where ω np (s + p)(s +ζω n s + ω n). y(t) +Ae pt + Be σt sin(ω d t θ), ω n A ω n ζω np + p p B q (p ζω n p + ω n)( ζ ) p p ζ ζ θ tan +tan. ζ p ζω n (a) Which term dominates y(t) asp gets large? (b) Give approximate values for A and B for small values of p. (c) Which term dominates as p gets small? (Small with respect to what?) (d) Using the explicit expression for y(t) above or the step command in MATLAB, and assuming ω n andζ 0.7, plot the step response of the system above for several values of p ranging from very small to very large. At what point does the extra pole cease to have much effect on the system response?

74 48 CHAPTER 3. DYNAMIC RESPONSE Second-order system: Unit step response: H(s) ω np (s + p)(s +ζω n s + ω n) Y (s) s H(s), y(t) L {Y (s)} s +ζω n s + ω n (s + σ + jω d )(s + σ jω d ) p where σ ζω n, ω d ω n ζ. Thus from partial fraction expansion: Y (s) k s + k s + p + k 3 s + σ + jω d + solving for k,k,k 3, and k 4 : k H(0) k k k 4 s + σ jω d ω np s(s + σ + jω d )(s + σ jω d ) ω n s p k ω n pζω n + p k 3 (s + σ + jω d )Y (s) s σ jωd p k 3 q e iθ k 3 e iθ ( ζ )(p pζω n + ω n) k 4 k 3 where Thus Ãp! Ã p! ζ θ tan +tan ω n ζ ζ p ζω n Y (s) s + k µ s + p + k e iθ e +iθ 3 + s + σ + jω d s + σ jω d Inverse Laplace: y(t) +k e pt + k 3 ³e iθ e (σ+jωd)t + e +iθ e d)t (σ jω or ω n y(t) + ω n pζω n + p {z e pt p + q e σt cos(ω d t+θ) } ( ζ )(p pζω n + ω n) A {z } B

75 49 (a) As p gets large the B term dominates. (b) For small p: A,B 0. (c) As p gets small A dominates. (d) The effect of a change in p is not noticeable above p 0. Step Response for p Step Response for p Step Response for p Step Response for p Problem 3.36: Step responses The block diagram of an autopilot designed to maintain the pitch attitude θ of an aircraft is shown in Fig The transfer function relating the elevator angle δ e and the pitch attitude θ is θ(s) δ e (s) G(s) 50(s +)(s +) (s +5s + 40)(s +0.03s +0.06), where θ is the pitch attitude in degrees and δ e is the elevator angle in degrees. The autopilot controller uses the pitch attitude error ε to adjust the elevator according to the following transfer function: δ e (s) ε(s) D(s) K(s +3) s +0. Using MATLAB, Þnd a value of K that will provide an overshoot of less than 0% and a rise time faster than 0.5 sec for a unit step change in θ r. After examining the step response of the system for various values of K, comment on the difficulty associated with making rise-time and overshoot

76 50 CHAPTER 3. DYNAMIC RESPONSE Figure 3.67: Block diagram of autopilot measurements for complicated systems. G(s) Θ(s) δ e (s) 50 (s +)(s +) (s +5s + 40) (s +0.03s +0.06) D(s) δ e(s) e(s) K(s +3) (s + 0) where e(s) Θ r Θ Θ G(s)D(s) Θ r +G(s)D(s) 50K (s +)(s +)(s +3) (s +5s + 40) (s +0.03s +0.06) (s + 0) + K (s +3) 50K s 3 +6s +s +6 s s s s +(7.4+K) s +(4+3K) Θ(s) Θ r(s) Output must be normalized to the Þnal value of for easy computation of the overshoot and rise-time. In this case the design criterion for overshoot cannot be met which is indicated in the sample plots.