ENG006 MT1 F09 Key Lagerstrom

ENG006 MT1 F09 Key Lagerstrom - Engineering 6 Midterm...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Engineering 6 Midterm Solutions, Fall 2009 Regrade requests must be submitted in writing to Dr. Lagerstrom no later than the Review/Q&A session on Wednesday, November 18. Regrades will only be considered in cases where it looks like the grader missed something. That is, requests along the lines of "I think I deserve more points" will not get very far, because the the grading scale on each problem was applied consistently for all students. Note also : Your exam may have been photocopied before it was returned, so please don't risk your engineering career here at UCD by changing an answer and submitting it for a regrade. Exam Version A (blue and white copies) Problem 1 (22 points). (a) (6 points) x = linspace(-2,39,873); %Define values of x in row vector a = [7 0 13 -9 16]; %Define polynomial coefficient vector A = polyval(a,x); %Calculate values (b) (6 points) x = linspace(-2,39,873); %Okay to leave this out if have it in part (a) A = 7*x.^4 + 13*x.^2 – 9*x + 16; %Note: only two dot operators needed. (c) (8 points) h = [2 0 0 3 0 0 0 -8]; %Define polynomial coefficient vectors g = [12 5 4 15]; [c r q] = residue(h,g); %Get residues (c), roots (r), and quotient coeffs (q) Note: Okay to write [c,r,q] = etc. The ratio H/G is an improper rational function, so the general form of the result will be (assuming no repeated roots) n n r x c r x c r x c x Q x G x H 2 2 1 1 ) ( ) ( ) ( where Q(x) is the quotient polynomial and the rest of the terms are the partial fraction expansion. In the Matlab code, the variable q will contain the coefficients of Q(x), the variable c will contain the c’s (residues), and the variable r will contain the r’s (roots of the denominator polynomial G(x)). (d) (2 points) There would be 3 terms in the partial fraction expansion, because the denominator polynomial G(x) is third order, and thus has three roots.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Problem 2 (20 points). Consider ) 8 34 cos( 14 ) ( t t x where t is measured in seconds. (a) (3 points) 34 (b) (3 points) radians/sec or degrees/sec (c) (3 points) Period = 2*pi/angularfreq, so 2*pi/34 = pi/17 (d) (3 points) seconds (e) (5 points) 8 34 14 j t j e e , or ) 8 34 ( 14 t j e (f) (3 points) Take the real part of the rotating phasor
Background image of page 2
Problem 3 (22 points). (a) (4 points) 61 = 1x32 + 1x16 + 1x8 + 1x4 + 0x2 + 1x1, so the binary form is 111101 (b) (4 points) 1x32 + 0x16 + 1x8 + 0x4 + 1x2 + 1x1 = 43 (c) (4 points) subplot(2,3,4) (d) (4 points) Grading note: Exact wording not required. An interpreter is software that takes high-level language source code and translates it into machine code
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/22/2010 for the course ENG 6 taught by Professor Largerstrom during the Fall '08 term at UC Davis.

Page1 / 10

ENG006 MT1 F09 Key Lagerstrom - Engineering 6 Midterm...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online