机械控制理论基础习题答案.doc

机械控制理论基础习题答案习题：2.2，2.62.2. Write the (a) loop, (b) node equations for the circuit shown after the switch S is closed.2.2. 当开关S闭合后，写出电路的(a)回路，(b)节点方程式。321(a) loop(b) nodeNode 1 between R1 and R2, Node 2 between R2 and L, and Node 3 between L and C21. 2. 3. 2.6. (a) Derive the differential equation relating the position y(t) and the force f(t). (b) Draw the mechanical network. (c) Determine the transfer function G (D)=y/f.2.6. (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。K2(b) Draw the mechanical network.BfK1K3(a) node xanode xb(c) where , , 习题：3.83.8 Solve the following differential equations. Assume zero initial conditions. Sketch the solutions.(a) (1)r=1, k=0, w=0 q=k-w=0The steady state output is therefore:xss=b0D2 xss =0.Inserting these values into previous equation(1):16 xss =16 b0=1 xss =b0= (2)The homogeneous equation is formed by letting the right side of the differential equation equal zero: (3)the transient response is the solution of the homogeneous equation, is obtained by assuming a solution of the formxt=Amemt (4)where m is a constant yet to be determinedthe characteristic equation of system: (5)m1=4j, m2=-4jvalues of m are complex, by using the Euler identity and then combining terms, transient solutions are (6)x= xt + xss= (7)Assume zero initial conditions, i.e., t=0, x(0)=0, Dx(0)=0, inserting these values into previous equation(7):, , x= xt + xss=(b) (1)r=9, k=0, w=0 q=k-w=0The steady state output is therefore:xss=b0D xss =0, D2 xss =0.Inserting these values into previous equation(1):3 xss =9, b0=3 xss =b0=3 (2)The homogeneous equation is formed by letting the right side of the differential equation equal zero: (3)the transient response is the solution of the homogeneous equation, is obtained by assuming a solution of the formxt=Amemt (4)where m is a constant yet to be determinedthe characteristic equation of system: (5)m1=-3, m2=-1There are no multiple roots, transient solutions are (6)x= xt + xss= (7)Assume zero initial conditions, i.e., t=0, x=0, Dx(0)=0,inserting these values into previous equation(7) and (1):, x= xt + xss=(c) (1)r=t+1, k=1, w=0 q=k-w=1The steady state output is therefore:xss=b1t+b0D xss = b1, D2 xss =0.Inserting these values into previous equation(1):4.25b1t+b1+4.25b0=t+1b1=0.2353 , b0=0.18 xss =0.2353t+0.18 (2)The homogeneous equation is formed by letting the right side of the differential equation equal zero: (3)the transient response is the solution of the homogeneous equation, is obtained by assuming a solution of the formxt=Amemt (4)where m is a constant yet to be determinedthe characteristic equation of system: (5)m1=-0.5+j2, m2=-0.5-j2values of m are complex, by using the Euler identity and then combining terms, transient solutions are (6)x= xt + xss (7)Assume zero initial conditions, i.e., t=0, x=0, inserting these values into previous equation(7) and (1):x= xt + xss=(d) (1)Assume the steady state output is:xss=Asin(10t+)D xss =10 Acos(10t+), D2 xss =-100 Asin(10t+), D3 xss =-1000 Acos(10t+)Inserting these values into previous equation(1):cos(10t+)=cos10tcos-sin10tsinsin(10t+)=sin10tcos+cos10tsin-960A cos(10t+)-298A sin(10t+)=10 sin10t(960Asin-298Acos)sin10t-(960Acos+298Asin)cos10t=10sin10t960Asin-298Acos=10960Acos+298Asin=0=-arctan0.3104A=xss =sin(10t+) (2)The homogeneous equation is formed by letting the right side of the differential equation equal zero: (3)the transient response is the solution of the homogeneous equation, is obtained by assuming a solution of the formxt=Amemt (4)where m is a constant yet to be determinedthe characteristic equation of system: (5)m1=-2, m2=-1, m2=-1There is a root of multiplicity, transient solutions are (6)x= xt + xss= (7)Assume zero initial conditions, i.e., t=0, x=0, inserting these values into previous equation(7) and (1):4.5 Find the partial-fraction expansions of the following:(a) , , , (b) (c) A=-2, B=-12(d) (e) 4.7 Write the Laplace transforms of the following equations and solve foe x(t); the initial conditions are given to the right.(a) Dx+8x=0 x(0)=-2The Laplace transforms of the equationsX(s)-x(0)+8X(s)=0X(s)(s+8)=-2The inverse Laplace transforms of the equation(p637, appendxA 7)x(t)=-2e-8t (b) D2x+2.8Dx+4x=10 x(0)=2, Dx(0)=3The Laplace transforms of the equations2X(s)-sx(0)- Dx(0)+ 2.8(sX(s)- x(0)+4X(s)=10s-1X(s)(s2+2.8s+4)-(2s+8.6)= 10s-1X(s)(s2+2.8s+4)=The inverse Laplace transforms of the equation(p637, appendxA 36) The inverse Laplace transforms of the equation(p637, appendxA 26)x(t)=2.5-0.5(c) D2x+4Dx+3x=t x(0)=0, Dx(0)=-2The Laplace transforms of the equations2X(s)-sx(0)- Dx(0)+ 4(sX(s)- x(0)+3X(s)= s-2X(s)(s2+4s+3)=The inverse Laplace transforms of the equation(p637, appendxA 7)x(t)=(d) D3x+4 D2x+ 9Dx+10x=sin5t x(0)=-4, Dx(0)=1, D2x(0)=0The Laplace transforms of the equations2X(s) s2x(0)-sD2x(0)- Dx(0)+4s2X(s)-sx(0)- Dx(0)+ 9sX(s)- x(0)+10X(s)=X(s)(s2+4s2+9s+10)=-(4s2+16s+31)= The inverse Laplace transforms of the equation(p637, appendxA 7)Exercises: 5.24，5.25 习题：5.24，5.255.24 Given the following system:(a) Simplify block diagram (or find C (s) /R (s). (b) Draw an equivalent singal flow graph. (c) Apply Masons gain rule to find C(s)/R(s). (a) (1)(2)(3)(4)(5)(b) (c) Applying Masons rule, this system has four loops, whose transmittances are:-1/s, -1/s, -k/s2, and 1/s4. Therefore L1 = -1/s-1/s-k/s21/s4= (1)Only three loops are nontouching; therefore, L2 = (-1/s)( -1/s)+(-1/s)( -k/s2)= (2)Although there are four loops, there is no set of three loops that are nontouching; therefore, L3 = 0 (3)=1-L1+L2-L3+ =1+= (4) (5) (6)5.25 For the following system,(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) L1 =-HG1 (1) L2 = L3 = 0 (2)=1-L1+L2-L3+=1+ HG1 (3)T1=G1 T2=G2 (4) (5) (6) Exercise: 6.2(a), (b), 6.14(b)习题：6.26.6 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) Solution: , The characteristic equation of the system is:The Routhian array:Based Rouths stability criterion, for stable operation of the system, the range of K: the range of values of KFor the roots on the imaginary axis that yied sustained oscilations,，, , (b) Solution: , The characteristic equation of the system is:The Routhian array:Based Rouths stability criterion, for stable operation of the system, the range of K: Sytisfy the range of values of K is not exist.(c) Solution: , The characteristic equation of the system is:The Routhian array:Based Rouths stability criterion, for stable operation of the system, the range of K: the range of values of KFor the roots on the imaginary axis that yied sustained oscilations, , 6.5 A unity-feedback system has the forword tranfer functionIn order to obtain the best possible ramp error coefficient, the highest possible gain K is desirable. Do stability requirements limit this choice of K?Solution: (a) determining e(t)ss , When Kv=K1=, e(t)ss=0(b) finding the value K for which the system is stableOverall transfer function of system:The Routhian array isBased Rouths stability criterion, for stable operation of the system, the range of K: the range of values of K6.14 A unity-feedback control system has (1) (2)where r(t)=2t. (a) If K=1.5, determine e(t)ss; (b) It is desired that for a ramp input e (t)ss1.5, what minimum value K1 have for this conditio