ECSE-4760 Real-Time Applications in Control & Communications Spring 2019 Exam #2 Thursday 5/2 11:30 a.m. Low-3116 Name ____SOLUTION_________________ Section 1: MR 1:30 (Grades will be posted on SIS when they are available.) This exam is open book, open notes, etc. with calculators, but NO laptops, Internet connections, cell phones, …, and no sharing materials. You need your own copies of everything, including lab reports. If you didn’t bring your own hardcopies, you will be penalized. Answer questions in only 5 out of the 8 sets corresponding to each of the experiments in the lab. You do not need to choose the experiments you performed but you will probably wish to do so. If you answer more than 5 sets, the first 5 will be graded. Each set is worth 20 points (max) with the breakdown shown by the questions. Setting up a solution is more important than finding the actual value if you are running out of time. If you are missing any grades on RPILMS, make sure you bring the reports to me (JEC-6028) before the end of the day today to get your records corrected. I Digital Logic ______ II Voice Processing ______ III Binary Communications ______ IV Digital Filter ______ V Interactive Graphics ______ VI Hybrid Control ______ VII DC Motor ______ VIII Optimal Control ______ TOTAL ______

I Digital Logic 2019 Name _________________________ Create a state machine that follows the state flow diagram given below. The binary values of the states CBA starting with 000 are given.

CBA 000 001 010 011 100 101 110 111

1. (6pt) Two input bits IJ change the flow of the counter as shown in the state diagram. Find the Next State Table for this machine.

2. (14pt) Find the simplified expression for the input to the least significant bit flip-flop A, assuming D-flip-flops are used in the implementation. (Set up 2 Karnaugh Maps: for input J = 0 and for input J = 1 for states ICBA and give 2 expressions for DA, one for J = 0 and one for J = 1 if desired, rather than simplifying a 5-variable Kmap.) A complete expression for DA is worth 1 pt. DA Input: ICBA = states, J = 0 DA Input: ICBA = states, J = 1

BA IC

00

01

11

10

BA IC

00

01

11

10

00 1 0 0 1 00 0 1 1 0 01 1 0 0 1 01 0 1 1 0 11 0 0 0 0 11 d d d d 10 0 0 0 0 10 d d d d

For J = 0: DA = /I/A For J = 1: DA = A DA = /J/I/A + JA

Input IJ P.S.

00 01 11 10

000 001 000 ddd 000 001 010 001 ddd 000 010 011 010 ddd 000 011 100 011 ddd 000 100 101 100 ddd 000 101 110 101 ddd 000 110 111 110 ddd 000 111 000 111 ddd 000

0

1 2 3

7 6 5

4

00

01

00

00

00

00 00 00

01 01

01

01 01 & 10

01 01

10 10 10

10

00

10 10 10

II Voice Processing 2019 Name _________________________ 1. (2pt) TRUE or FALSE: By using more sophisticated encoding such as Linear Predictive Coding, systems may be designed that do better than the 6dB rule for improving the SNR. 2. (10pt) Without an antialiasing filter a sampler can be used to modulate a signal to higher frequencies. List 3 ideal sampling frequencies less than or equal to 25kHz that can modulate a 12kHz sine wave up to 64kHz.

sin(2pf0n×Ts) = sin(2p(f0+k×fs)n×Ts)

2pf0n = 2p(f0+k×fs)n, f0 = 12kHz

f0+k×fs = 64 => 12kHz+k×fs = 64kHz => k×fs = 52kHz

k = 3, fs = 52kHz/3 = 17.33kHz k = 4, fs = 52kHz/4 = 13kHz k = 5, fs = 52kHz/5 = 10.4kHz

3. (4pt) Sketch the approximate input & output signal if the input is a 2Vp-p square wave with of 1.20kHz frequency when the program runs using options D0 – 10 – 48/1. Sketch at least 2 complete cycles and scale both axes.

1.2kHz is above the 120Hz DC blocking frequency: 48kHz/1.2kHz = 40 samples/period => 20 sample in ½ period 20 x 40.7mV = ∆V = 0.814V = 4.07/5 x 1V/div in half cycle for scope scaling below => never gets large enough before input changes; just ramps up and down with Vp-p = 814mV.

II Voice Processing 2019 Name _________________________

a) b)

c) d)

4a. (1pt) For the above input sine waves and ideal delta modulator outputs, which plots definitely use a deadband modulator, if any?

a (2 sequential samples with same magnitude) 4b (1pt) Which plots use a 2-bit modulator, if any?

a, b, c, d (2 step sizes: 1V & 2V) 4c (1pt) Which plots have no deadband, if any?

c (forced hunting or granular noise) 4c. (1pt) Other than those in 4a) and 4c), which of the other plots possibly have no deadbands, if any?

b, d (Samples never fall on same voltage consecutively, so can’t determine)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-5

-4

-3

-2

-1

0

1

2

3

4

5In: 20Hz sin, Out: 2-bit delta modulation (fs=160Hz)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-5

-4

-3

-2

-1

0

1

2

3

4

5In: 20Hz sin, Out: 2-bit delta modulation (fs=160Hz)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-6

-4

-2

0

2

4

6In: 20Hz sin, Out: 2-bit delta modulation (fs=160Hz)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-5

-4

-3

-2

-1

0

1

2

3

4

5In: 20Hz sin, Out: 2-bit delta modulation (fs=160Hz)

III Binary Communications 2019 Name _________________________ 1. (6pt) Given a Gaussian noise distribution on a voltage pulse such that A = 3V where A is the amplitude of the pulse, if the standard deviation of the noise s = 1V, find the probability of a bit error from the PDF. (May leave answer in terms of the standard deviation and the Gaussian PDF.)

2a. (1pt) A (9,8) Parity code adds a parity bit to the end of the message bits to guarantee the codeword has odd parity. What is the efficiency of the new code?

8/9 = 88.89% 2b (2pt) By how much would the bit rate of this parity code need to be adjusted (higher or lower) to match the information bit rate of an unencoded 7-bit message system?

message rate (7)/message rate (9,8) => (7)/((7/8) x 9) = 8/9 Rate must be 1.125x or 12.5% faster

Also 100%/88.89% = 9/8 = 1.125 3. (2 pt) What is the SNR for a RTZ 1V pulse train with 25% duty cycle in a channel with 0.01Vrms of white noise?

Signal Vrms = 0.5V 20log &.(

&.&)= 20log(50) = 20(1.7) = 34𝑑𝐵

Pe = F(3) = 0.00135 (area under tail from -¥®-3s)

0 A pulse amplitude Voltage

III Binary Communications 2019 Name _________________________ Below is the Venn diagram of a modified Hamming code using the following configuration: m1 m2 m3 m4 c1 c2 c3 c4 with c1 based on message bits 1, 2, & 3, c2 based on message bits 1, 3, & 4, c3 is based on message bits 2, 3, & 4, and c4 is set to that the parity of all 8 bits is even. 4. (4pt) The received word for this Hamming code was 11101000. What was the transmitted word?

5. (3pt) What Hamming distance is needed between code words to correct double-bit errors and simultaneously detect 3 more bit errors?

O c c d d d c c O Distance = 8 (c = correct, d = detect)

6. (2pt) A channel using only the In-phase component encodes 2 bits to amplitude signals with a minimum separation of 2 volt between any encodings. A second system is designed using 90° Quadrature components that maintains the same minimum amplitude separation between any 2 constellations with the same 2-bit encoding. How does the maximum amplitude voltage in the first system compare to that in the second? Current separation on real-axis = 2V. Amplitude on real or imaginary axis must be √2 for separation of 2V between any 2 points

c1 c4 m1 m2 m3 m4 c2 c3

2

√2

m1⊕m2⊕m3⊕c1 = 1⊕1⊕1⊕1 = 0 (OK) m1⊕m3⊕m4⊕c2 = 1⊕1⊕0⊕0 = 0 (OK) m2⊕m3⊕m4⊕c3 = 1⊕1⊕0⊕0 = 0 (OK) m1⊕m2⊕m3⊕m4⊕c1⊕c2⊕c3⊕c4 = 1⊕1⊕1⊕0⊕1⊕0⊕0⊕0 = 0 (OK) => no errors, transmitted word is 11101000

√2

IV Digital Filter 2019 Name _________________________ 1. (4pt) A 6th order analog filter with fcutoff = 3kHz was transformed using a scaled bilinear with fs = 18kHz. What will be the actual (digital) cutoff frequency of the filter when it is downloaded to the Speedy-33 DSP board and executed?

𝑠 =2𝑇𝑧 − 1𝑧 + 1 => 𝑧 =

1 + <=>

1 − <=>

𝜔@ =2𝑇 tan

DE 𝑇𝜔F2 = 36000tanDE H E

IJKKK2𝜋3000M =

17365𝑟𝑎𝑑𝑠 = 2.764𝑘𝐻𝑧

2. Design a 5th order Butterworth high pass digital filter with fs = 36kHz, that is 3dB down @ 6kHz and >30dB down @ 3kHz. 2a. (4pt) Sketch its approximate pole-zero diagram 180° => 18kHz, 60° => 6kHz, 5th order => 5 poles, 5 zero at z = 1

2b. (3pt) Sketch |H(ejw)| vs. for 0 ≤ w ≤ 2p with scaled axes in Hz (magnitude plot, not Bode in dB). -30dB = 0.032, Butterworth, monotonic in both pass and stop bands, no ripple -3dB = 0.707

21000.0 26000.0 31000.0 36000.0

IV Digital Filter 2019 Name _________________________ 3. (4pt) For a 6th order Butterworth & Chebyshev I HPF, what is the numerator of H(z)? (Don’t worry about scaling for unity gain at peak output.) Give answer as a 6th order polynomial in z or as a product of second order systems: 𝑎𝑧J + 𝑏𝑧S + 𝑐𝑧U + 𝑑𝑧I + 𝑒𝑧= + 𝑓𝑧E + 𝑔𝑜𝑟(𝑎𝑧= + 𝑏𝑧E + 𝑐) ∙ (𝑑𝑧= + 𝑒𝑧E + 𝑓)⋯

all the zeros at z = 1:

𝐻(𝑧) =(𝑧 − 1)J

(… ) =(𝑧 − 1)=

(… ) ∙(𝑧 − 1)=

(… ) ∙(𝑧 − 1)=

(… )

=(𝑧= − 2𝑧 + 1)

(… ) ∙(𝑧= − 2𝑧 + 1)

(… ) ∙(𝑧= − 2𝑧 + 1)

(… )

=𝑧J − 6𝑧S + 15𝑧U − 20𝑧I + 15𝑧= − 6𝑧 + 1

(… )

4. (3pt) A digital elliptic filter designed with a sampling frequency of 24kHz has its first zero in the magnitude of the output at 7kHz. If the filter is run with the same coefficients, but at a sampling frequency of 54kHz, at what frequency will the first zero (null) in the output signal appear?

f = 7kHz x (54kHz/24kHz) = 15.75kHz 5. (2pt) For 𝐻(𝑠) = E

(>]E)(>]=) where are the zeros in H(z), the impulse invariant transformation of H(s)?

All at z = -1

V Interactive Graphics 2019 Name _________________________ 1a. (8pt) The transfer function of the DC motor was given as 𝐺E(𝑠) =

SKKK>_

but the actual model is

. If both systems are configured with negative unity

feedback gain, how will the impulse responses differ? Describe the forms of the response if you can’t find the actual expressions. 𝐺E(𝑠) ⟹ a)(>)

E]a)(>)= SKKK

>_]SKKK⟹ 5000sind√5000𝑡g

𝐺hi (𝑠) ⟹ ajk(>)E]ajk (>)

= ==SK>(>]J.=S)]==SK

= ==SK>_]J.=S>]==SK

lm_

>_]=nlm>]lm_⟹ lm

oEDn_𝑒Dnlmp sind𝜔qo1 − 𝜁=𝑡g

𝜔q = 47.43𝑟𝑎𝑑𝜁 = 0.0659𝜁𝜔q = 3.125

1b. (2pt) Classify the stability of the feedback systems with G1(s) and G’p(s) in 1a.

G1 => Marginally Stable (oscillatory), G’p => Stable (underdamped) 2. (2pt) TRUE or FALSE: For f1(x,y) to always be completely covered by f2(x,y) then sx1 must be less than sx2 and sy1 must be less than sy2.

3a. (7pt) Given a Gaussian pdf , C1 = 10, and a second Gaussian pdf

, C2 = 1, with equal a priori probabilities, determine the decision

boundary between the two probability functions, looking down on the x-y plane.

3b. (1pt) Into how many regions is the x-y plane divided by the decision boundary in 3a)?

2, Decision boundary is a straight line

€

ʹ G p (s) = 25Gp (s) = 25 3.986s(0.4839s +1)

= 25 8.2369s(s + 2.0664)

€

f1(x,y)=12π

exp −12 (x2 + y2){ }

€

f2(x,y)=12π

exp −12 [(x−0.2)2 + (y+ 0.1)2]{ }

€

10f1(x,y)=1012π

exp −12

x2 + y2( )⎧ ⎨ ⎩

⎫ ⎬ ⎭

=1f2(x,y)=1 12π

exp −12

(x−0.2)2 + (y+ 0.1)2[ ]⎧ ⎨ ⎩

⎫ ⎬ ⎭

ln(10)− x2

2−y2

2= −

x2

2+0.4x2

−0.042

−y2

2−0.2y2

−0.012

0.2y= 0.4x−2ln(10)−0.05y= 2x−10ln(10)−0.25= 2x−23.28

G1 with feedback is a pure sine wave with w = 70.71rad G’p with feedback is a decaying sine wave with w = 47.22rad and decay of exp(-3.125t)

VI Hybrid Control 2019 Name _________________________ ANSWER ANY COMBINATION OF FULL QUESTIONS THAT ADD UP TO 20 POINTS. 10-POINT QUESTIONS MUST BE FULLY ANSWERED. 1. (5pt) For the plant defined by the rectangle with state feedback, with a1 = 5 and a2 = 20, what are the equivalent A matrices for the plant alone and the controlled system in terms of kc1 and kc2?

𝐴huvqp = w 0 1−𝑎E −𝑎=

x = y 0 1−5 −20z -a1’ = -5 – kc1, -a2’ = -10 – kc2 𝐴{qp|u = y 0 1

−5 − 𝑘𝑐1 −20 − 𝑘𝑐2z

-a1’ has -kc1 and -5 in parallel so -a1’ = (-5) + (-kc1), similarly -a2’

2. (5pt) For the system in 1. find the characteristic equation for the controlled system in terms of the gains kc1 and kc2.

s2 + (a2 + kc2)s + (a1 + kc1) = 0 s2 + (20 + kc2)s + (5 + kc1) = 0

3a. (8pt) SET UP the equations to find the discrete matrices Ad, Bd, Cd, and Dd for a discrete pole-placement controller given:𝐴{ = y 0 1

−3 −5z , 𝐵{ = y04z , 𝐶{ = y1 −10 2 z , 𝐷{ = y00z. If you use an expression eAt where A is a

matrix, include the steps needed to evaluate that expression by hand.

Dd = Dc & Cd = Cc Ad = exp(AcT) = L -1{(sI – Ac)-1} Bd = (Ad – I)(Ac)-1Bc

𝑨� = ℒDE �Hy𝑠 00 𝑠z − y

0 1−3 −5zM

DE�𝑩� = H𝑨� − y

1 00 1zM y

0 1−3 −5z

DEy04z

y(t)

r(t)

VI Hybrid Control 2019 Name _________________________ 3b. (2pt) How many outputs does the system in 3a have?

2, C matrix is 2 x 2 => 2 outputs (& 2 states) 4a. (1pt) TRUE or FALSE: Both an ideal ripple free (FST) and ideal minimal prototype controller on a 2nd order system guarantee no error at the sample points (for t > 1⋅T). 4b. (2pt) For any of the PID approximations, if a 3 Volt step response for a set of gains saturates the output of the controller at ±10 Volts, what can be done to evaluate a step response without changing the PID gains?

Reduce the size of the input step to 1 Volt or less if necessary 4c. (2pt) TRUE or FALSE: Increasing the integral gain will slow down the rise time but not change the settling time. 5. (10pt) Set up the polynomial equations to find the PI gains for a controller that minimizes the integrated time absolute error, given 𝐺h(𝑠) =

>]E(>]I)_

. Provide a set of equations with unknowns that can be solved for KP, KI, and 𝜔,butdonotsolve.

Graham-Lathrop=> leads to 3 nonlinear equations in 3 unknowns (may be difficult to solve):

6. (10pt) Given the following system, find the pole placement gains that will place the state-controlled poles at s = -2 ± j1. x˙1(t) = x2 (t) x˙2(t) = −x1(t) − 2x2(t) + 3u(t) y(t) = 3x1(t) − 4x2(t)

€

Gp (s) =s+1

(s+ 3)2 =s+1

s2 + 6s+ 9 Gc (s) = KP +

KI

s

1+Gp (s)Gc (s) =1+s+1

s2 + 6s+ 9KP +

KI

s⎛

⎝ ⎜

⎞

⎠ ⎟ = 0

s3 + (6 + KP )s2 + (9 + KP + KI )s+ (KI ) = 0 = s3 +1.75ω0s2 + 2.15ω0

2s+ω03

6 + KP =1.75ω0 KI =ω03 9 + KP + KI = 2.15ω0

2

€

a1 =1 a2 = 2 b=3m= −(p1+ p2)= −[(−2+ j2)+ (−2− j2)]= +4n = p1p2 = (−2+ j2)⋅ (−2− j2)= 8(s+ 2+ j2)(s+ 2− j2)= s2 + 4s+ 8= 0s2 + (a2 + bkc2)s+ (a1+ bkc1)= 0s2 + (2+ 3kc2)s+ (1+ 3kc1)= 02+ 3kc2 = 4⇒ kc2 = 2

31+ 3kc1 = 8⇒ kc1 = 7

3

VII DC Motor Control 2019 Name _________________________ 1a. (2pt) Find Gc(s) for a lead compensator that has a maximum gain of 5 and a pole at s = 20.

𝛼z=20,𝛼=5,z=4, 𝐺{(𝑠) = 𝛼 H >]�>]��

M = 5 H >]U>]=K

M 1b. (3) Find the maximum and minimum phase angles of Gc(s).

Minimum = 0°, Maximum = sinDE H∝DE∝]E

M = sinDE HSDES]E

M = sinDE HUJM = 0.73𝑟𝑎𝑑 = 41.8°

1c. (4) Sketch the Bode plot for Gc(s).

2. (3pt) MATCHING: Apply the effects a) – f) to the correct compensators below. a) Reduce steady-state error b) Reduce overshoot c) Increase rise time d) Reduce rise time e) Improve damping f) Has integrator-like properties

Phase-lead Compensator: d, e Phase-lag Compensator: a, b, c, f

Mag

nitu

de (d

B)

-2

0

2

4

6

8

10

12

14

Root Locus

Real Axis

Imag

inar

y Ax

is

10-1 100 101 102 103

Phas

e (d

eg)

0

30

60

Bode Diagram

Frequency (rad/s)

VII DC Motor Control 2019 Name _________________________ 3. (7pt) A digital system has 𝐺h(𝑧) =

E(�]E)(�]K.S)

. Find the location of the poles where the system is marginally stable with pure proportional feedback.

On unit circle for x = -0.75, 𝑦 = ±√1 − 𝑥= = ±√1 − 0.75= = ±𝑗0.661 4. (1pt) An analog 2nd order DC motor system with a pole at the origin and another pole just to the left of the imaginary axis, and no finite zeros has a root locus that runs parallel to the imaginary axis in the left half plane and never crosses it. This should be (marginally – decaying oscillations) stable for all proportional feedback gains, but in reality, this is not the case. Why is this motor not marginally stable?

With a rotational system, rather than linear systems, when the error exceeds 180° the correction force continues rotation in the same direction rather than attempting to reverse direction. This results in a rotation that speeds up instead of slowing down and reversing, and ultimately sets up a constant spin. Linear systems would exhibit decaying oscillations.

Root Locus

Real Axis

Imag

inar

y Ax

is

-1.5 -1 -0.5 0 0.5 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Root Locus

Real Axis

Imag

inar

y Ax

is

VIII Optimal Control 2019 Name _________________________ 1. (6pt) Given 𝐺h(𝑠) =

E(>]E)

+ E(>]=)

find the A, B, C, and D matrices and set up the equation for the Performance

Index with no terminal state penalty and 𝑸 = y2 00 5z , 𝑹 = 0.5.

𝐺h(𝑠) =

E(>]E)

+ E(>]=)

= =>]I(>]E)(>]=)

= =>]I>_]I>]=

𝐴 = y 0 1−2 −3z 𝐵 = y01z 𝐶 =

[3 2]𝐷 = 0 𝐽 = ∫ �𝑥i(𝑡) y2 0

0 5z 𝑥(𝑡) + 0.5𝑢=(𝑡)� 

K 𝑑𝑡 2. (5pt) A system has been optimized with Q = 1.5 and R = 0.5 without a terminal state penalty. The controlled input and output are measured as:

k 0 1 2 3 4 5 6 7 … u(k) 7 3.7 2.3 1.3 0.8 0.4 0.2 0 … x(k) 3 1.8 1.1 0.7 0.4 0.2 0.1 0 …

Find the value of the performance index for this Riccati solution.

Given x(k) & u(k), use the discrete form of the equations without H final state penalty:

𝐽 = ¡{𝑥i(𝑘)𝑄𝑥(𝑘) + 𝑢i(𝑘)𝑅𝑢(𝑘)} 

¦§K

= ¡{𝑥=(𝑘)1.5 + 𝑢=(𝑘)0.5} 

¦§K

𝐽 = (1.5)(3= + 1.8= + 1.1= + 0.7= + 0.4= + 0.2= + 0.1=) + (0.5)(7= + 3.7= + 2.3= + 1.3= + 0.8= + 0.4= + 0.2=) 𝐽 = 56.48 3. (2pt) TRUE or FALSE: small changes in the weights of the Q matrix will produce large changes in the response plots for the system states. 4. (7pt) For the following values, find the feedback gains for the system:

𝑨{ = ¨0 0 10 1 0−3 −2 −5

© , 𝑩{ = ¨002© , 𝑪{ = [−0.3 1.5 «)¬], 𝑫{ = √2,𝑸 = ¨

5 0 00 6 00 0 7

© , 𝑹 = 8

𝑃 = ¨12 0 30 9 −23 −2 2

© , ¨12 0 30 1 23 2 4

©

|P| = 87, -9 => use positive definite P: 𝐺 = −𝑅DE𝐵i𝑃 = D)¯[0 0 2] ¨12 0 30 9 −23 −2 2

© = [D)°)_ «)_]