EE C128 / ME C134 Final Exam Fall 2014

December 19, 2014

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1

EE C128 / ME C134 Fall 2014 Final Exam Name:

1. Laplace transform, controllable canonical form

(a) Derive the following Laplace property for convolution integrals

Lg1(t) = G1(s)

Lg2(t) = G2(s)

L∫ t

0g1(τ)g2(t− τ)dτ

= G1(s)G2(s)

Answer: (4 pts)

There are multiple methods for solving this, the following proof is done by working backwards

from the solution

Lg1(t) =

∫ ∞0

g1(t)e−stdt = G1(s)

Lg2(t) =

∫ ∞0

g2(t)e−stdt = G2(s)

G1G2 =

∫ ∞0

g1(x)e−sxdx

∫ ∞0

g2(y)e−sydy

=

∫ ∞0

g1(x)

∫ ∞0

g2(y)e−sye−sxdydx

=

∫ ∞0

g1(x)

∫ ∞0

g2(y)e−s(y+x)dydx

Let t = x+ y then,

=

∫ ∞0

g1(x)

∫ ∞x

g2(t− x)e−stdtdx

=

∫ ∞0

∫ ∞x

g1(x)g2(t− x)e−stdtdx

Switch order of integration

=

∫ ∞0

∫ t

0g1(x)g2(t− x)e−stdxdt

=

∫ ∞0

∫ t

0g1(τ)g2(t− τ)e−stdτdt

= L∫ t

0g1(τ)g2(t− τ)dτ

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(b) Find the solution to the following ODE with the given initial conditions:

d2x

dt2− dx

dt+ 2x =

∫ t

0δ(τ) sin (t− τ)dτ

x(0) = −2, x(0) = 1

Answer: (5 pts) 12

[sin t+ cos t+ e

t2

(cos

√72 t−

11√7

sin√72 t)]

Taking the Laplace transform of the ODE then solving for X(s),∫ t

0δ(τ) sin (t− τ)dτ = sin(t)

s2X(s)− sx(0)− x′(0)− [sX(s)− x(0)] + 2X(s) =1

s2 + 1

(s2 − s+ 2)X(s) = s− 3 +1

s2 + 1

=s3 − 3s2 + s− 2

(s2 + 1)(s2 − s+ 2)

X(s) =s3 − 3s2 + s− 2

(s2 + 1)(s2 − s+ 2)

Doing partial fraction decomposition we get,

X(s) =As+B

(s2 + 1)+

Cs+D

s2 − s+ 2

=(A+ C)s3 + (−A+B +D)s2 + (−B + 2A+ C)s+ 2B +D

(s2 + 1)(s2 − s+ 2)

Solving for coefficients A,B,C,D

A+ C = 1

(2B +D) = −2

−A+B +D = −3

2A−B + C = 1 = 2

Thus

A = 1/2 B = 1/2 C = 1/2 D = −3

Taking the laplace transform with the coefficients,

X(s) =s+ 1

2(s2 + 1)+

s− 6

2(s2 − s+ 2)

=1

2

[sin t+ cos t+ e

t2

(cos

√7

2t− 11√

7sin

√7

2t

)]

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2. Bode Plotting and Nyquist Stability

Consider the following transfer function:

G(s) =100(s+ 4)

(s2 + 12s+ 20)(s+ 2)

(a) Sketch a Bode plot of the system (magnitude and phase). Label all slopes and points on the

graph. (3 pts)

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(b) Using the Bode plots you created in Part (a), calculate the phase margin and gain margin for

G(s).

Answer: (2 pts) Gain margin =∞ Phase margin = 52.3 For phase margin find where the

gain is 1 and draw a line to where it crosses the phase diagram For gain margin find where the

phase is 180 and draw a line to where it crosses the gain

(c) Draw a Nyquist plot of the system and use the Nyquist stability criterion to determine if the

closed loop system under unity feedback is stable.

Answer: (3 pts) The systems is stable, there are no encirclements of -1

(d) Assume we have a closed loop system below where G(s) is given in Part (a) and C(s) = K.

For what values of K is the system stable?

Answer: (2 pts) Stable for all K The closed loop transfer function is:

ClosedLoop =100K(s+ 4)

100K(s+ 4) + (s2 + 12s+ 20)(s+ 2)

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3. Solution in time domain, stability

(a) For the following system, explicitly determine the time-domain solution x(t)

x =

[4 1

0 10

]x+

[1

0

]u, x(0) =

[1

1

]

y =[

1 0]x

where u(t) is a unit step function.

Answer: (5 pts)

x(t) =

[1312e

4t + 16e

10t − 14

e10t

]The solution to a system of this form is

x(t) = eAtx0 +

∫ t

0eA(t−τ)Bu(τ)dτ

We first need to find the eigne vaues of A:

det(λI −A) = (λ− 4)(λ− 10)

λ1 = 4 λ2 = 10

Then we can find the eigen vectors and diagonalize A

v1 =

[1

0

]v2 =

[1

6

]

A = PDP−1

P =

[1 1

0 6

]Taking the inverse of P gives us

P−1 =1

6

[6 −1

0 1

]

The diagonalized D = P−1AP

D =1

6

[6 −1

0 1

][4 1

0 10

][1 1

0 6

]=

[4 0

0 10

]

We can then find eAt:

eAt =1

6

[6e4t −e4t + e10t

0 6e10t

]

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We can then find the solution by taking the integral

x(t) =1

6

[6e4t −e4t + e10t

0 6e10t

][1

1

]+

∫ t

0

1

6

[6e4(t−τ) −e4(t−τ) + e10(t−τ)

0 6e10(t−τ)

][1

0

]dτ

x(t) =

[1312e

4t + 16e

10t − 14

e10t

](b) Determine the transfer function G(s) for the zero initial state response, given the system in

Part (a).

Answer: (3 pts) 1s−4

G(s) = C(sI −A)−1B

(sI −A)−1 =

[s− 4 −1

0 s− 10

]−1=

1

(s− 4)(s− 10)

[s− 10 1

0 s− 4

]

G(s) =[

1 0] 1

(s− 4)(s− 10)

[s− 10 1

0 s− 4

][1

0

](c) BONUS: Does the degree of your state space model and transfer function match? Why or

why not?

Answer: (2 pts) No, the degree of the state space is 1 while the state space is 2nd degree.

This is because systems that aren’t controllable or observable when converted to TF’s will be

of lesser degree.

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4. Electrical Circuit and Root Locus

(a) For the above circuit derive the transfer function C(s) =VoutVin

.

Answer: (3 pts) (R2 + 1C2s

)( 1R1

+ C1s) or s2R1R2C1C2+(R1C1+R2C2)s+1sR1C2

Looking at the first opamp we know

VoutVin

=−R2eq

R1eq

R1eq =1

1R1

+ C1s

R2eq = (R2 +1

C2s)

The second opamp is simply an inverter so the entire transfer function is

=R2 + 1

C2s1

1R1

+C1s

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(b) Assume the system below, G(s), is in unity negative feedback. Determine the value of K such

that the steady state error to a step response is 111 . Also determine the percent overshoot and

settling time of the feedback system at this K.

G(s) =K

(s+ 5)(s+ 15)

Answer: (4 pts) K = 750 OS = 31% Ts = .40

The steady state error of a closed loop systems is given by

ess = lims−>01

1 +G(s)

=1

1 + k75

=1

11

Thus

K = 750

Gcl =750

s2 + 20S + 825

wn =√

825 2ξwn = 20 ξ = .348

OS = e−ξπ√1−ξ2 = 31% Ts =

4

wnξ=

4

(28.7)(.348)= 0.40s

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(c) Now assume that the system is described by the figure below, where C(s) and G(s) are ob-

tained from parts (a) and (b) respectively. Draw the root locus given R1 = 125 MΩ, C1 = 15 µF,

R2 = 625 MΩ and C2 = 0.1 µF. Watch your signs!

Answer: (3 pts)

In order to draw the root locus we first need to find all the poles and zeros

G(s) has two poles at p = −5,−15 and C(s) has one pole at p = 0

G(s) has no zeros and C(s) has two zeroes at z = − 1R1C1

,− 1R2C2

= − 162.5 ,−

11875

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(d) Label on the root locus a suitable region if the goal is to achieve a settling time (Ts) ≤ 0.4 sec

and percent overshoot (%OS) ≤ 20%.

Answer: (3 pts) The necessary damping to achieve the desired overshoot is:

ξ =−ln(OS)√

(π2 + ln2(OS))= .456

θ = sin−1 ξ = 27

To achieve the desired settling time:

.4 =4

ξwn

ξwn = 10

Re(pole) > 10

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5. Controller design using State-feedback

Consider the mechanical system shown above. Here, V denotes the voltage applied to the motor

(control input) and x(t) is the position of the mass. You may assume the back emf from the motor

is negligible (EMF = 0) and the torque supplied by the motor is equal to T = IKm − Jmθwhere,

Km : Constant relating T and I

Jm : Inertia of the motor

(a) Show that G(s) is the transfer function from V to x. To do this, you MUST derive the governing

equations for the mechanical/electrical system.

G(s) =X(s)

V (s)=N2

N1

Kmr

R+ Ls

[(J1 + J2 + Jm

(N2

N1

)2)s2 + r2(Ms2 + fvs+ k)

]−1

Answer: (5 pts)

V = I(R+ Ls)

T =V

R+ LsKm − Jms2θm

For ease of solving set the angle of the gear to θ2

x = rθ2

θm =N2

N1θ2 =

N2x

N1r

Doing a balance of forces on the rotating masses:

(J1 + J2)θ2s2 + Fr =

N2

N1T

F = (Ms2 + fvs+ k)θ2r

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Substituting

N2

N1

[(KmV

R+ Ls)− Jms2

N2

N1θ2

]= (J1 + J2)θ2s

2 + (Ms2 + fvs+ k)r2θ2

AlgebraN2

N1

KmV

R+ Ls= (J1 + J2 + Jm

(N2

N1

)2

)θ2s2 + (Ms2 + fvs+ k)r2θ2

Since we have θ2 = xr we get the solution:

X(s)

V (s)=N2

N1

Kmr

R+ Ls

[(J1 + J2 + Jm

(N2

N1

)2)s2 + r2(Ms2 + fvs+ k)

]−1

(b) Choosing x =[x x

]Tas the state vector and x as the output, derive a state space model

(matrices A, B, C and D) for the above system. Use the following parameters:

R = 1, Km = 0.1, L = 0, N2/N1 = 10, r = 1, J1 = J2 = 1, Jm = 0, M = 1, k = 1, fv = 1.

Note your input to the system should be V .Answer: (2 pts)

x =

[0 1

−13 −1

3

]x+

[013

]V, y =

[1 0

]x

Plugging the values above into the given transfer function we get

X

V=

1

3s2 + s+ 1

Using this we can formulate the state space model by inspection

x =[

1 0]x

x =[−1

3 −13

]x+

1

3V

Putting the two together

x =

[0 1

−13 −1

3

]x+

[013

]V, y =

[1 0

]x

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(c) Explicitly write the observability and controllability matrices. Is the system controllable? Is it

observable? (Use parameters from Part (b))

Answer: (2 pts) Controllability:

[0 1

313 −1

9

]Observability:

[1 0

0 1

]And it is both control-

lable and observable, full rank

The controllability matrix is given by

[B AB

]=

[0 1

313 −1

9

]

This is full rank so it is controllable The observability matrix is given[C

CA

]=

[1 0

0 1

]

This is full rank so it is observable

(d) Determine the eigenvalues of matrix A you found in Part (b).

Answer: (1 pts)λ = −16 ±

√116 i

det(Is−A) = det

([s −113 s+ 1

3

])= s2 − 1

3s+

1

3

The roots of the polynomial are the eigen values

s = −1

6±√

11

6i

(e) We will now control the system using a state feedback controller as shown in the diagram below

where K =[k1, k2

]and r =

[r1

r2

].

Write the dynamics of the closed-loop system as x = Ax+ Br. That is, find both A and B in

terms of the system parameters given in Part (b) and the elements of the controller gain matrix

K.

Answer: (2 pts) A =

[0 1

−13(1 + k1) −1

3(1 + k2)

]B =

[0 0

−13k1 −1

3k2

]x = Ax+Bu

= Ax+BK(r − x)

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EE C128 / ME C134 Fall 2014 Final Exam Name:

= (A−BK)x+BKr = Ax+ B

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6. State-feedback and observer design

Consider the following system:

x = Ax+Bu, y = Cx with A =

[7 1

−2 4

], B =

[0

1

], C =

[0 1

](a) Compute the eigenvalues and eigenvectors for A.

Answer: (3 pts)λ1 = 5 λ2 = 6 v1 =

[1

−2

]v2 =

[1

−1

]To determine the eigenvalues we do the following

det(Is−A) = det

([s− 7 −1

2 s− 4

])= s2 − 11s+ 30

s2 − 11s+ 30 = (s− 5)(s− 6)

To find the eigenvectors we do the following

det(λ1I −A)v1 = 0[−2 −1

2 1

][v11

v12

]=

[0

0

]v11 = 1 v12 = −2

det(λ2I −A)v2 = 0[−1 −1

2 2

][v21

v22

]=

[0

0

]v21 = 1 v22 = −1

(b) Use state-feedback of the form of u = −Kx. Determine the gain K = [k1 k2] such that the

poles of the closed loop system are located at s1,2 = −2± 5j.

Answer: (2 pts)K =[

104 15]

Give the state-feedback we find the equivalent system is

equal to

A = A−BK =

[7 1

−(2 + k1) 4− k2

]The eigenvalues of A are the desired closed loop poles, thus k1 and k2 can be determined

det(Is− A) = det

([s− 7 −1

2 + k1 s+ k2 − 4

])= s2 + (k2 − 11)s+ (30 + k1 − 7k2)

(s+ 2 + 5i)(s+ 2− 5i) = s2 + 4s+ 29

k2 − 11 = 4 k2 = 15

30 + k1 − 7k2 = 29 k1 = 29− 30 + 7 ∗ 15 = 104

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(c) Unfortunately for this system we are unable to measure all the states. In order to do state

feedback we must use a Luenberger observer of the form:

˙x = Ax+Bu+ L(y − y)

y = Cx

and the system is controlled using state feedback, given by

u = −Kx

Determine the error dynamics of the system, e, where e = x− x. The result must be in terms

of e only.

Answer: (2 pts)e = (A− LC)e

e = ˙x− x

= A(x− x) + L(y − y)

y − y = C(x− x)

e = A(x− x)− LC(x− x)

= (A− LC)e

(d) Determine the observer matrix L = [l1 l2]T such that the error dynamics have poles at

s1,2 = −2± 5j.

Answer: (2 pts) L =[−52 15

]The desired closed loop characteristic equation is

s2 + 4s+ 29

Using the results from the previous problem

A = A− LC

det(A− LC) = s2 + (l2 − 11)s+ (30− 2l1 − 7l2)

l2 − 11 = 4 l2 = 15

30− 2l1 − 7l2 = 29 l1 = −52

(e) Comment on the performance of the state observer given the previously placed poles. What

are we interested in when designing an observer and how could we improve the observer?

Answer: (1 pts)The observer is slow and converges at the same rate as the system. We

want the observer to converge faster than the system so we have an accurate reperesnetation

of the current states. We can improve the observer by making the real part of the poles more

negative

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(f) Complete the following block diagram of the system described in part(c). (2 pts)

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7. Linear Quadratic Regulator

Consider the LTI system

x = Ax+Bu x = [x1, x2]T

where

A =

[1 1

0 −1

], B =

[1

0

]

We would like to solve an LQR problem for the system. That is, we want to find the optimal

control u∗(t) that minimizes the cost functional

J =

∫ ∞t=0

(x21(t) + u2(t)) dt

(a) Solve the Algebraic Riccati Equation for the infinite horizon LQR. Hint: the solution of the

Algebraic Riccati equation is a positive semi-definite matrix. A (2 × 2) matrix P is positive

semi-definite everywhere when:

p11 ≥ 0 p22 ≥ 0 p212 ≤ p11p22 for P =

[p11 p12

p12 p22

]

Answer: (4 pts) P =

[1 +√

2 1

1 12

]The algebraic riccati equation is

P +ATP + PA− PBR−1BTP +Q = 0

For the infinit case P = 0

Looking at the cost function we can see that

Q =

[1 0

0 0

]R = 1

Solving for P we get a system of 4 equations

P =

[p1 p2

p3 p4

]

2p1 − p21 + 1 = 0

p1 − p1p2 = 0

p1 − p3p1 = 0

p2 − 2p4 + p3 − p3p2 = 0

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Solving we get P =

[1 +√

2 1

1 12

](b) Determine the optimal feedback matrix K1 such that the optimal control is u∗(t) = −K1x(t).

Answer: (2 pts)K =[

1 +√

2 1]

K = R−1BTP = 1[

1 0] [ 1 +

√2 1

1 12

]=[

1 +√

2 1]

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8. Linear Quadratic Regulator

Consider the system where the dynamics are scalar:

x = ax+ bu x is a scalar

We want to create a finite horizon optimal controller given the cost function:

J =

∫ tf

t=0(qx2(t) + ru2(t))dt

(a) Write the Ricatti Equation for this system as well as the terminal condition.

Answer: (2 pts) p+ 2ap− p2b2

r + q = 0

(b) Find P for the static case where tf =∞. Your answer should be in terms of q, r, a, and b. Note

the P should be positive semi-definite everywhere.

Answer: (2 pts)p = arb2

+√

a2r2

b4+ ra

b2

During the infinite case p = 0 thus

p+ 2ap− p2b2

r+ q = 0

Becomes

2ap− p2b2

r+ q = 0

This is a quadratic equation with the roots

ar

b2±√a2r2

b4+rq

b2

Since P is positive semi-definite we know the answer is

ar

b2+

√a2r2

b4+rq

b2

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(c) For this scalar system, given any tf , the Riccati equation can be analytically solved:

P (τ) =(aP (tf ) + q) sinh(βτ) + βP (tf ) cosh(βτ)(

b2P (tf )r − a

)sinh(βτ) + β cosh(βτ)

where τ = tf − t, β =

√a2 +

b2q

rand sinh(.) and cosh(.) are the hyperbolic trigonometric

functions. Taking the limit as tf →∞, the solution becomes:

P (τ) =q

−a+ β

Show that this is equivalent to your solution from Part (b).

Answer: (2 pts)

P (τ) =q

−a+ β=

q

−a+

√a2 +

b2q

r

P (τ) =q

−a+ β=

q

−a+

√a2 +

b2q

r

√a2 +

b2q

r+ a√

a2 +b2q

r+ a

=q

√a2 +

b2q

r+ aq

−a2 +b2q

r+ a2

=ar

b2+r

b2

√a2 +

b2q

r

This result is the same as the solution found in part b

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f(t) F (s)

δ(t) 1

u(t) 1s

tu(t) 1s2

tnu(t) n!sn+1

sin(ωt)u(t) ωs2+ω2

cos(ωt)u(t) ss2+ω2

e−αt sin(ωt)u(t) ω(s+α)2+ω2

e−αt cos(ωt)u(t) s+α(s+α)2+ω2

Table 1: Laplace transforms of common functions

sinh(θ)eθ − e−θ

2

cosh(θ)eθ + e−θ

2

tanh(θ)sinh(θ)

cosh θ=

1− e−2θ

1 + e−2θ

Table 2: Trigonometric functions

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