Exam 1 - Problem with Solution - Introduction to Computer Engineering | ECE 2030, Exams of Electrical and Electronics Engineering

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ECE PHD PRELIMINARY

EXAMINATION

SOLUTIONS – FALL 2006 EXAM

Problem 1 (Core: CompE-ECE2030)

Prelim Problem based on ECE2030/2031 material Design a synchronous finite state machine that produces the follow (binary-encoded) output sequence when an external control signal (X) is set to zero: Z = 0 (also Reset), 1, 5, 6, 3, 2, 4, (returning to 0 on the next count). Here "Z" represents the decimal equivalent three-bit output code word (Z2,Z1,Z0). For example, Z=6 is represented by (Z2,Z1,Z0)=(1,1,0). The same circuit should count in the following (different) sequence when X=1: 0 (Reset), 4, 2, 3, 6, 5, 1, (returning to 0 on the next count) For this problem, use positive edge-triggered D-type flip flops for the state register and use the state of the register to define the output code. (a) Construct a state transition table. (b) Construct Karnaugh maps for the flip flop input functions. (c) Express the flip-flop input functions in minimum-sum of products (SOP) form.

SOLUTION:

(a) Based on the problem statement, we can construct the state transition table as follows:

_______PS________ _______NS________ DECIMAL X Z2 Z1 Z0 Z2+ Z1+ Z0+ PS NS(+) 0 0 0 0 0 0 1 0 1 0 0 0 1 1 0 1 1 5 0 0 1 1 0 1 0 3 2 0 0 1 0 1 0 0 2 4 0 1 0 0 0 0 0 4 0 0 1 0 1 1 1 0 5 6 0 1 1 1 X X X 7 X(don’t care) 0 1 1 0 0 1 1 6 3 1 1 0 0 0 1 0 4 2 1 1 0 1 0 0 0 5 1 1 1 1 1 X X X 7 X 1 1 1 0 1 0 1 6 5 1 0 0 0 1 0 0 0 4 1 0 0 1 0 0 0 1 0 1 0 1 1 1 1 0 3 6 1 0 1 0 0 1 1 2 3

Problem 2 (Core: CompE-ECE2030)

Problem 3 (Core: CompE-ECE3055)

Problem 5 (Core: E&M-ECE3025)

Problem 6 (Core: E&M-ECE3065)

Problem 8 (Core: EDA-ECE3050)

Problem 9 (Core: Power-ECE3070)

Fall 2006 Prelim Exam – Power Problem #

A delta connected generator provides a balanced 3 phase line-line voltage of 22,000 volts rms at 60 hertz. The load is a balanced 3 phase wye connected load, with each phase having an inductive impedance at 60 hertz of 10/_30o^ ohms.

a. the power delivered to the load and the rms current rating for the generator winding b. The load is sent to England where the line frequency is 50 hertz. Calculate the power delivered to the load for the same line voltage conditions. c. Calculate the value of Y connected capacitors required at 60 hertz to improve the overall power factor of the load to unity.

Answer:

a) Line – neutral voltage across load = 22,000/sqrt(3) = 12,701.7 volts Load impedance Z = 10. cos (30) + j 10. sin (30) = 8.66 + j 5 ohms

Current in load = 12701.7/10 = 1270.1 Amps

Power per phase = (1270.1)^2. 8.66 = 13.97 MW per phase

Power to load = 41.91 MW

Current in generator winding = 1270.1 / sqrt(3) = 733.3 Amperes

b) At 50 hertz, load impedance is Z = 8.66 + j 5. (50/60) ohms = 8.66 + j 4.166 ohms

Load current = 12701.7 volts/ 9.61 ohms = 1321.7 Amps

Power to load = 41.91 * (1321/1270)^2 = 45.38 MW

c) We need capacitive VARs to equal the inductive VARs

(I. I) XL = (V. V)/ Xc

1270 * 1270 * 5 = 12701 * 12701 / Xc

Xc = 20 ohms

C = 132.6 microfarad

Problem 10 (Core: Power-ECE3070)

Fall 2006 Prelim Exam – Power Problem #

SOLUTION

(a) The output frequency of a synchronous generator is proportional to the speed. Therefore since the frequency at 1800 rpm is 60 Hz, the frequency at 2100 rpm is 60 (2100/1800) = 7 0 Hz.

(b) The no load voltage is directly proportional to the frequency and the field current. Therefore,

350 V 8

V (^) no _load= 400 ⋅ ⋅ =

(c) Phasor diagram:

The no load voltage is the internal voltage, Eaf = 350. The terminal voltage, Vt , is the internal voltage(350) minus the drop across Xs. The value of Xs at 70 Hz is 0.8(70/60) = 0.933 Ω.

The triangle ABC is uniquely defined. Various methods may be used to solve it in order to find Vt. Here is one:

Cos -1^ Θ = 0.8 so Θ = 36.9^0 Hence the angle β is 90^0 + 36.9 = 126.

Using the Sine rule: Sin δ = ((100 x 0.933)/350) Sin 126.9, hence δ = 12.3^0

This means α = 180-12.3 -126.9 = 40.8^0

Then Vt = 350 (Sin 40.8)/(Sin 126.9) = 286 Volts (L-N)

Check: Eaf = Vt + jI (^) aXs = 286 + j(100 ∟-36.9)(0.933) = 286 + 93.3∟53.1 0 = 350 Volts which is the correct magnitude value.

jIaXs Vt

I (^) a

Eaf

α

δ^ β

A

B

C

Problem 12 (Core: Microsystems-ECE3080)

Problem 13 (Core: DSP-ECE2025)

Problem 15 (Core: S&C-ECE3085)