线性控制系统分析与设计期末考+答案解析试

湖南工业大学期末考试试卷

线性控制系统分析与设计期末考试（七题命中）

一. (a)求位置y(t)与力f(t)有关的微分方程；(b)画出机械网络图；(c)确定传递函数G(D)＝y/f。

(b) Draw the mechanical network.

K2

f

K1 K3

B

(a) node xa

?K1?K2?M1D2?xa?K2xb?f

node xb

?K2?K3?BD?M2D2xb?K2xa?0

K2 4322D?BD?KD?BKa?KaKb?K2?(c) G?D??where Ka?K1?K2, Kb?K2?K3, K?Ka?Kb

二、Solve the following differential equations. Assume zero initial conditions. Sketch the solutions.

D2x?16x?1 (1)

?r=1, k=0, w=0 ? q=k-w=0

The steady state output is therefore:

xss=b0 湖南工业大学期末考试试卷

D2 xss =0.

Inserting these values into previous equation(1): 16 xss =16 b0=1 xss =b0=

1 (2) 16The homogeneous equation is formed by letting the right side of the differential equation equal zero:

D2xt?16xt?0 (3)

the transient response is the solution of the homogeneous equation, is obtained by assuming a solution of the form

xt=Amemt (4) where m is a constant yet to be determined

the characteristic equation of system:

m2?16m0?m2?16?0 (5)

m1=4j, m2=-4j values of m are complex, by using the Euler identity

e?j?dt?cos?dt?jsin?dt and then combining terms, transient solutions are

xt?A1e4jt?A2e?4jt?B1cos4t?B2sin4t?Asin(4t??) (6)

x= xt + xss=Asin(4t??)?1 (7) 16Assume zero initial conditions, i.e., t=0, x(0)=0, Dx(0)=0, inserting these values into previous equation(7):

x(0)?Asin??1?0, Dx(0)?4Acos??0 16???2, A??1 16x= xt + xss=?

1?1sin(4t?)? 16216三、Write the Laplace transforms of the following equations and solve foe x(t); the initial conditions are given to the right.

湖南工业大学期末考试试卷

D2x+2.8Dx+4x=10 x(0)=2, Dx(0)=3 The Laplace transforms of the equation

s2X(s)-sx(0)- Dx(0)+ 2.8(sX(s)- x(0))+4X(s)=10s-1 X(s)(s2+2.8s+4)-(2s+8.6)= 10s-1

2s2?8.6s?10X(s)(s+2.8s+4)=

s2

A12s2?8.6s?102s2?8.6s?10As?B X(s)????2222s(s?2.8s?4)s[(s?1.4)?(2.04))ss?2.8s?4The inverse Laplace transforms of the equation(p637, appendxA 36) x(t)?2.5?1.69e?1.4tsin(2.04t?17.25)

A1?[sX(s)]s?02s2?8.6s?10?2?2.5 s?2.8s?4s?02s2?8.6s?102.5?0.5s?1.6 Xx???22ss(s?2.8s?4)s?2.8s?42s2?8.6s?102.5?0.5s?1.62.51s?3.2 X(s)?????2222s(s?2.8s?4)ss?2.8s?4s2(s?1.4)?2.04??The inverse Laplace transforms of the equation(p637, appendxA 26)

(?3.2?1.4)2?2.04?1.4t2.04x(t)=2.5-0.5esin(2.04t?tan?1)

2.04?3.2?1.4x(t)?2.5?1.69e?1.4tsin(2.04t?17.25)

四、 For the following system,

湖南工业大学期末考试试卷

(a)Draw an equivalent singal flow graph. (b) Derive transfer functions for E(s)/R(s), X(s)/R(s), B(s)/R(s), C(s)/R(s), and Y(s)/R(s). (a)

(b) Σ L1 =-HG1 (1)

Σ L2 =Σ L3 = 0 (2) Δ=1-ΣL1+ΣL2-ΣL3+…=1+ HG1 (3) T1=G1 T2=G2 (4)

?1?1 ?2?1?HG1 (5)

T?

C?s??R?s??T?nn??G1?G2(1?HG1) (6)

1?HG1五、 For each of the following cases, determine the range of values of K for which the response c(t) is stable, where the driving function is a step function. Determine the roots on the imaginary axis that yied sustained oscilations. (a) C?s??K 2s[s(s?2)(s?4s?20)?K]Solution:

r(t)?u?1(t)?1, R?s??L[r(t)]?L[u?1(t)]?1 sKC?s?s[s(s?2)(s2?4s?20)?K]KG(s)??? 21R?s?s(s?2)(s?4s?20)?KsThe characteristic equation of the system is:

湖南工业大学期末考试试卷

Q(s)?s(s?2)(s2?4s?20)?K?s4?6s3?28s2?40s?K

The Routhian array:

s4s3s2s1s016643?6K?25603K28K40K

Based Routh’s stability criterion, for stable operation of the system, the range of K:

K?0? ??6K?2560/3?0?the range of values of K

0?K?1280 9

六、 A unity-feedback control system has (1) G?s??20K

s[(s?1)(s?5)?20]where r(t)=2t. (a) If K=1.5, determine e(t)ss; (b) It is desired that for a ramp input e (t)ss≤1.5, what minimum value K1 have for this condition to be satisfied? (1) Solution:

(a) determining e(t)ss r(t)?2t, R?s??L[r(t)]?E(s)?R(s)2?2?1?H(s)G(s)s2 2s112(s?1)(s?5)?40??

20Kss[(s?1)(s?5)?20]?20K1?s[(s?1)(s?5)?20]湖南工业大学期末考试试卷

12(s?1)(s?5)?40e(t)ss?e(?)?limsE(s)?lims?s?0s?0ss[(s?1)(s?5)?20]?20K

2(s?1)(s?5)?402(0?1)(0?5)?4055?lim???s?0s[(s?1)(s?5)?20]?20K20K2K3 (b) finding the minimum value K1 If e (t)ss≤1.5 is desired, then K?

七、 For each of the transfer functions

G(s)?4(1?0.5) 2s(1?0.2s)(1?0.05s)55?K1?Kmin? 335?1.5, so 2K(a) draw the log magnitude (exact and asymptotic) and phase diarams; (b) draw the polar plot

湖南工业大学期末考试试卷

12(s?1)(s?5)?40e(t)ss?e(?)?limsE(s)?lims?s?0s?0ss[(s?1)(s?5)?20]?20K

2(s?1)(s?5)?402(0?1)(0?5)?4055?lim???s?0s[(s?1)(s?5)?20]?20K20K2K3 (b) finding the minimum value K1 If e (t)ss≤1.5 is desired, then K?

七、 For each of the transfer functions

G(s)?4(1?0.5) 2s(1?0.2s)(1?0.05s)55?K1?Kmin? 335?1.5, so 2K(a) draw the log magnitude (exact and asymptotic) and phase diarams; (b) draw the polar plot

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