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Course: ENEE 241, Spring 2008
School: Maryland
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241 ENEE 02* HOMEWORK ASSIGNMENT 20 Due Wed 03/26 Exam 2 (Monday 03/31) will cover Assignments 10-20. Let V= v(0) v(1) v(2) v(3) v(4) v(5) be the matrix of Fourier sinusoids of length N = 6. (This is the same matrix as in the last example of Reading Assignment 20.) (i) (7 pts.) If x = 4 4 1 2 2 1 , use projections to represent x in the form x = Vc. Verify that x is a linear combination of three columns of V...

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241 ENEE 02* HOMEWORK ASSIGNMENT 20 Due Wed 03/26 Exam 2 (Monday 03/31) will cover Assignments 10-20. Let V= v(0) v(1) v(2) v(3) v(4) v(5) be the matrix of Fourier sinusoids of length N = 6. (This is the same matrix as in the last example of Reading Assignment 20.) (i) (7 pts.) If x = 4 4 1 2 2 1 , use projections to represent x in the form x = Vc. Verify that x is a linear combination of three columns of V (only). (ii) (7 pts.) Repeat for y = 4 2 2 4 10 10 that y is also a linear combination of three columns of V. T T , expressing it as y = Vd. Verify (iii) (2 pts.) Verify your results in (i) and (ii) using the FFT command in MATLAB (which will generate the vectors 6c and 6d). (iv) (4 pts.) If s = x + y, use your results from (i) and (ii) to obtain the least squares approximation of s in terms of v(1) , v(3) and v(5) . Display the entries of . s s Solved Examples S 20.1 (P 3.4 in textbook). The columns of matrix the 1 1 1 1 1 j 1 j V= 1 1 1 1 1 j 1 j are the complex Fourier sinusoids of length N = 4. Express the vector s= 1 4 2 5 T as a linear combination of the above sinusoids. In other words, nd a vector c = [c0 c1 c2 c3 ]T such that s = Vc. S 20.2 (P 3.1 in textbook). Let 1 = 2 and 3 = 2 (i) Determine the complex number z such that the vector v= equals 1 z z2 z3 z4 z5 1 + j + j 1 j j T T (ii) If s= 3 2 1 0 1 2 T determine the least-squares approximation of s in the form of a linear combination of 1 (i.e., the s . Clearly show the numerical values of the elements of . all-ones vector), v and v s S 20.3 (P 3.2 in textbook). Let v(0) , v(1) and v(7) denote the complex Fourier sinusoids of length N = 8 at frequencies = 0, = /4 and = 7/4, respectively. Determine the least-squares approximation of s s= 43210123 based on v(0) , v(1) and v(7) . Compute the squared approximation error s 2 . s
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Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 21Due Fri 03/28Solve without using MATLAB, except to verify your results. All vectors have length N = 10. (i) (4 pts.) The time-domain vector x(1) has DFT X(1) = 0 1 0 0 0 0 0 0 0 1TWhich two Fourier frequencies does x
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 22Due Wed 04/02The real-valued signal vector s has DFT S= 6 z1 3 + 2j 7 z2 5 + 3jT(i) (2 pts.) What are the values of z1 and z2 ? (ii) (3 pts.) Without using complex algebra (or MATLAB), determine the sum of the entries
Maryland - ENEE - 241
ENEE 241 02 The signal vector s is given byHOMEWORK PROBLEM 24Due Mon 04/07s=abcdefghT(i) (8 pts.) Display the following vectors: s(1) = P2 Rs s(2) = R2 Ps s(3) = P4 s s(4) = F2 s (ii) (12 pts.) Express the following vectors in terms of P, R, F and
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 26Due Fri 04/11Single-step the MATLAB script below, keeping a gure window open. Answer questions (i)-(viii). Print and submit the last two graphs, i.e., bar(n, real(X4) and bar(n, real(X5), only. (Make sure that your MAT
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 27 y ofTDue Mon 04/14(i) (7 pts.) Determine the circular convolution s = x x= and y= (ii) (7 pts.) If 3 1 1 5 20 1 2TUsing the FFT function in MATLAB, verify that the DFTs X, Y and S satisfy X Y = S. 1 3 2 1 4 x0 x1
Maryland - ENEE - 241
ENEE 241 02 The signalHOMEWORK ASSIGNMENT 28TDue Wed 04/16x= has DFT X given by X=000abc000X0 X1 X2 X3 X4 X5 X6 X7 X8TExpress the following DFTs in terms of the entries of X: (i) (5 pts.) The DFT X(1) of x(1) = (ii) (5 pts.) The DFT X(2) of x(2) =
Maryland - ENEE - 241
ENEE 241 02HOMEWORK PROBLEM 29Due Fri 04/18Generate the 20-point signal vector x using the MATLAB script below. n = (0:19). ; f0 = 0.15 + 0.25*rand(1) ; A = 10*rand(1) ; phi = pi*rand(1) ; x = A*cos(2*pi*f0*n + phi) + 0.3*A*randn(size(n) ; Following th
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 30Due Mon 04/28Each of the innite sequences x = x[ ], y = y [ ] and u = u[ ] (shown below) contains seven nonzero terms.4 3 2 1 1 x 2 1 4 3 y 2 4 3 un=0n=0n=0(i) (3 pts.) Express u[ ] as a linear combination of seve
Maryland - ENEE - 241
ENEE 241HOMEWORK ASSIGNMENT 32Due Wed 04/30Consider the FIR lter given by the input-output relationship y [n] = x[n] x[n 2] + x[n 4] x[n 6] , (Note the missing, i.e., zero-valued, coecients.) (i) (4 pts.) Show that the input sequences dened for all n b
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 33Due Fri 05/02Consider the FIR lter whose input-output relationship is given by 1 3 3 y [n] = x[n] x[n 1] + x[n 3] x[n 4] , 2 4 8 nZ(i) Determine the response y (i) [ ] of the lter to each of the input signals given by
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 34 4 1 0 1 4TDue Mon 05/05 .Consider the FIR lter with coecient vector b =The following MATLAB script computes a segment of the lter output sequence y (1) [ ] for an input sequence x(1) [ ] which has period L = 5. Speci
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 35Due Wed 05/07Use the MATLAB script below to generate the coecient vector b of a bandpass FIR lter of order M = 44. f1 = 0.19 + .03*rand(1) ; f2 = 0.34 + .03*rand(1) ; F = [0 f1 f1+0.03 f2 f2+0.03 0.5]*2 ; G = [1 1 0 0 1
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 36Due Fri 05/09(Part A) (5 pts.) Consider the FIR lter with impulse response given by h1 [n] = [n] [n 1]. Compute the outputs y1 [n] and y2 [n] of this system when the inputs are (A.1) (2 pts.) x1 [n] = cos(n + ), and (A.
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 37Due Mon 05/12A FIR lter acts on the input sequence given by x[0 : 5] = 1 3 5 2 6 2T;x[n] = 0 for n < 0 and n > 5to produce the output sequence given by y [0 : 11] = 2 3 0 10 23 27 61 51 54 14 2Tand y [n] = 0 for n
Maryland - ENEE - 241
HW 1 SOLUTION_(i) Plot (-1,sqrt(3) and (1,-1) on the (x,y) plane.(ii) |z1| = sqrt(-1)^2+(sqrt(3)^2) = sqrt(1+3) = 2angle(z1) = arctan(-sqrt(3) + pi = 2*pi/3 = 2.0944|z2| = sqrt(1+1) = sqrt(2) = 2.828angle(z2) = arctan(-1/1) = -pi/4(iii) z1^3 =
Maryland - ENEE - 241
HW 2 SOLUTION_(i) The polar coordinates are |u| = 1, angle(u) = pi/7 |v| = 1, angle(v) = 6*pi/7 |w| = 1, angle(u) = 2*pi/5(ii) |z-u| = |z-v| is tha perpendicular bisector of the segmentjoining u and v. This straight line is the y-axis (x = 0) (ii
Maryland - ENEE - 241
HW 3 Solution_(i) We know that the value 2.5 occurs at t=0, 0.160 and 0.180 sec andat no other intermediate times. Since a sinusoid crosses any givenlevel (except its maximum and minimum) twice in each period, it followsthat the period T equals exact
Maryland - ENEE - 241
HW 4 Solution_(i) For x[.], we have w=3*pi/4 = (3/8)*(2*pi). Thusthe fundamental period of x[.] equals N = 8.N = 8;n = 0 : 4*N-1;x = cos(3*pi*n/4 + 2*pi/3);stem(n,x)(ii) The values of w such that cos(w*n) is periodic with fundamental period N
Maryland - ENEE - 241
HW 5 Solution_(i) Clearly, A = 4. The period T equals 0.7 sec, therefore f = 10/7 Hz andW = 20*pi/7cos(phi) = 2*sqrt(3)/A = sqrt(3)/2, therefore phi = pi/6 or -pi/6. By inspection, t=0 corresponds to a point in the last quarter-cycle of the cosine wa
Maryland - ENEE - 241
Maryland - ENEE - 241
HW 7 Solution _ (i) A = [-4 3 0 2 -2 -1 1 2 2 ] A*[1 ; 0 ; 1] = [-4 ; 1 3] (ii) (a) B*x = [2; 3; 1]. B is a transformation matrix that circularly shifts the elements of the vector x. (b) B^2*x = [3; 1; 2]. Multiplying by B circularly shifts t
Maryland - ENEE - 241
HW 8 Solution_(i) The columns of A are obtained by right-multiplying A by thethree unit vectors.We have[1 ; -1 ; 0] + [2 ; 1 ; 0] = [3 ; 0 ; 0]thereforeA*[3 ; 0 ; 0] = A*[1 ; -1 ; 0] + A*[2 ; 1 ; 0]= [-4 ; -5 ; -6] + [ 1 ; -4 ; 0] = [-3 ; -9 ;
Maryland - ENEE - 241
HW 9 Solution_Note that the dimensions of P and Q can be inferred from the result,i.e., if P*A*Q is m*n, then P is m*3 and Q is 4*n.(i) P = 0 0 10 1 01 0 0Q = I(ii) P = I Q =0 0 0 10 1 0 00 0 1 01 0 0 0(iii) P = 0 0 11 0 00
Maryland - ENEE - 241
HW 10 Solution_(i) To solve U*x = b for arbitrary x = [x1 x2 x3 x4].' and b = [b1 b2 b3 b4].':x4 = b4 -2*x4 + x3 = b3 ; therefore x3 = 2*b4 + b3x2 - 2*x3 + x4 = b2 ; therefore x2 = -b4 + 2*(2*b4+b3) + b2 = b2 + 2*b3 + 3*b4x1 - 2*x2 + x3 = b1
Maryland - ENEE - 241
HW 11 Solution_Forward elimination:m x1 x2 x3 x4b_210-1-2-2401-23-1/21-3/263/235-3697-323m x1 x2 x3 x4b_210-1-20-2107-10-262363067029m x1 x2 x3 x4b_210-1-20-2107005229-20010
Maryland - ENEE - 241
HW 12 Solution_m x1 x2 x3 x4 _ 1* 4 4 4-2 2 2 0 2-1 1 1 3 3-1 1 3 1 2_ 1 4 4 4 0-6* -8-6-1/2 0-3-1-1-1/6 0-1-3-2_ 1 4 4 4 0-6 -8-6 0 0 3* 2 5/9 0 0-5/3 -1_ 1 4 4 4 0-6 -8-6 0 0 3
Maryland - ENEE - 241
HW 13 Solution_Matrix A_p m x1 x2 x3 x4_* 1* 1 1 1 -1 1 -1 1 -1 -1 1 2 4 8 -1 1 -3 6 -9_* 1 1 1 1* 0 -2* 0 -2 1/2 0 1 3 7 -2 0 -4 5 -10_* 1 1 1 1* 0 -2 0 -2* 0 0 3* 6 -5/3 0 0 5 -6_* 1 1 1 1* 0 -2 0 -2* 0 0 3 6 0 0 0
Maryland - ENEE - 241
HW 14 Solution_(i) |v1| = |v2| = sqrt(1 + 8 + 16) = 5<v1,v2> = 4-8+4 = 0(ii) <v1,a> = 3+12+12 = 27, <v2,a> = 12-12+3 = 3|a| = sqrt(9 + 18 + 9) = sqrt(36) = 6f1 = ( <v1,a> / |v1|^2 )*v1 = (27/25)*v1 Angle between a and v1 = arccos(27/(6*5) = 0.4
Maryland - ENEE - 241
HW 15 Solutions _ v1 = [ 5 3 -1 1 ].' v2 = [ 5 -1 7 3 ].' v3 = [ 1 1 -10 0 ].' s = [ 3 -1 4 4 ].' (i) We form the matrix V = [v1 v2]: V = [ 5 5 3 -1 -1 7 1 3 ] Then V'*V = [ 36 18 18 84 ] V'*s = [ 12 56 ] Solve (V'*V)*c
Maryland - ENEE - 241
HW 17 Solution_(i) Using microseconds (and noting that 1 MHz = 1/(1 microsecond) :f1 = 3 ; f2 = 2.5; gamma = 1.45t = (0 : 0.1 : 10).' ;u1 = cos(2*pi*f1*t) ;w1 = sin(2*pi*f1*t) ;u2 = cos(2*pi*f2*t) ;w2 = sin(2*pi*f2*t) ;V = [u1 w1 u2 w2] ;
Maryland - ENEE - 241
HW 18 Solutions_V = [-2.8867 4.0276 2.1607 0.0369 0.3649 -6.1858 0.0999 -6.0291 0.4916 4.0812 0.5427 2.7973 ](i) Rounded to four significant digits:V'*V = [ 16 -32 8-32 73 -7 8 -7 29 ](ii) Forward elimination on V'*V:m x1 x2 x3_ 16*-
Maryland - ENEE - 241
HW 19 Solution_i) V = [ v1 v2 v3 ]v1'*v2 = 3(a+bj) - 18j + 6j(a-jb)v1'*v2 = 0 for orthogonality, so Re( v1'*v2 ) = 3a + 6b = 0 -> a = -2bIm( v1'*v2 ) = 3b - 18 + 6a = 0 -> b = -2, a = 4The same terms are obtained in v2'*v3 and v3'*v1, so these i
Maryland - ENEE - 241
HW 20 Solutions_Let a = sqrt(3) throughout.v0 = [ 1 1 1 1 1 1 ].'v1 = [ 1 (1+j*a)/2 (-1+j*a)/2 -1 (-1-j*a)/2 (1-j*a)/2 ].'v2 = [ 1 (-1+j*a)/2 (-1-j*a)/2 1 (-1+j*a)/2 (-1-j*a)/2 ].'v3 = [ 1 -1 1 -1 1 -1 ].'v4 = [ 1 (-1-j*a)/2 (-1+j*a)/2 1 (-1-j*a)/2 (-1+j
Maryland - ENEE - 241
HW 21 SOLUTION_(i) X1 = [ 0 1 0 0 0 0 0 0 0 1 ].' The Fourier frequencies present in x1 are: w = (2*pi/10) = pi/5 and w = 9*(2*pi/10) = 9*pi/5 The corresponding Fourier sinusoids are v1[n] = exp(j*pi*n/5) and v9[n] = exp(j*9*pi*n/5) = exp(-
Maryland - ENEE - 241
HW 22 SOLUTION_(i) Since s is real-valued, S has circular conjugate symmetry, i.e.,S = [6 -5-3*j 3+2*j -7 3-2*j -5+3*j ].'(ii) s[0] + . + s[6] = S[0] = 6(iii) We compute the modulus and angle of each entry of S to obtain the amplutude and phase spe
Maryland - ENEE - 241
HW 24 Solution_s = [ a b c d e f g h ].'S = [ 4 -1+2j 5*j 3+j -6 3-j -5*j -1-2j ].'S1:R*s <-> R*S = [ 4 -1-2j -5*j 3-j -6 3+j +5*j -1+2j ].'s1 = P*R*s <-> F^(-1)*R*S = S1The entries of R*S are multiplied by those of the k = 8-1 = 7th sinusoi
Maryland - ENEE - 241
HW 25 Solution_s = [ a b c d e f g h ].'S = [ 4 -1+2j 5*j 3+j -6 3-j -5*j -1-2j ].'S1:R*s <-> R*S = [ 4 -1-2j -5*j 3-j -6 3+j +5*j -1+2j ].'s1 = P*R*s <-> F^(-2)*R*S = S1The entries of R*S are multiplied by those of the k = 8-1 = 6th sinusoi
Maryland - ENEE - 241
HW 29 Solution_The problem parameters (frequency f0, amplitude A and phase phi) are randomly generated in MATLAB: n = (0:19).' ; f0 = 0.15 + 0.25*rand(1) ; A = 10*rand(1) ; phi = pi*rand(1) ; x = A*cos(2*pi*f0*n + phi) + 0.3*A*randn(size(n) ;% (i
Maryland - ENEE - 241
S 30.1 (P 4.1)_(d[.] = delta[.], the unit impulse function)(i) x[n] = d[n+2] - d[n+1] + 5d[n] - d[n-1] + d[n-2]In other words,x[0] = 5x[1] = x[-1] = -1x[2] = x[-2] = 1x[n] = x[-n] = 0 for n>2The DTFT of d[n-m] equals exp(-j*m*w), soX(exp(j*
Maryland - ENEE - 241
S 32.1 (P 4.5)_i) MATLAB code:b = [1 -3 1 1 -3 1].';H = fft(b,256);A = abs(H);q = angle(H);ii) H(exp(j*w) = 1 - 3*exp(-j*w) + exp(-j*2*w) + exp(-j*3*w) - 3exp(-j*4*w) + exp(-j*5*w)= exp(-j*5*w/2)*(exp(j*5*w/2) - 3*exp(j*3*w/2) + exp(j*w/2)
Maryland - ENEE - 241
S 33.1 (P 4.4)_(i) H(z) = 1 - z^(-1 - *z^(-2) + z^(-3)(ii) H(exp(j*omega) = 1 - exp(-j*omega) - exp(-j*2*omega) + exp(-j*3*omega)= exp(-j*3*omega/2). *(exp(j*3*omega/2) - exp(j*omega/2) - exp(-j*omega/2) + exp(-j*3*omega/2)= exp(-j*3*omega/2)*(
Maryland - ENEE - 241
S 34.1 (P 4.3)_Divide by 2*pi to express the three frequencies in cycles per sample:3/28, 9/35 and 17/48Since these are rational (i.e., integer fractions), thesignal is periodic. The period is the smallest integerL such that each frequency can be
Maryland - ENEE - 241
S 35.1 (P 4.10)_(i)bar(0:36, b1), title('Filter Impulse Response'), xlabel('Time n')The impulse response is symmetric about n=M/2=20.(ii)f = (0:999)/1000;H = fft(b1,1000);Ha = abs(H);plot(f,Ha), title('Amplitude Response'), xlabel('\Omega/
Maryland - ENEE - 241
S 36.1 (P 4.13)_(i) The response to delta[n-m] is h[n-m]. Therefore the response tox = delta[n+1] - delta[n-1]isy = h[n+1] - h[n-1]The first nonzero value of h occurs at n=0 and the last one at n=4.The first nonzero value of y occurs at n=-1 (pro
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Zishan Yousuf Intermediate Accounting 3300 Take Home AssignmentGeneral JournalJuly 1, 2007 Cash Common Stock July 1, 2007 Auto Cash Accounts Payable July 3, 2007 Supplies Accounts Payable Prepaid Insurance Cash Accounts Receivable Fees Earned Accounts P
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 15.Given: Separate variables Integrate using dv = kt2 dt v = 10 m/s when t = 0 and v = 10 m/s when t = 2 s. a= dv = kt 2 dt10 22 10 dv = 0 kt dtv10 10=13 kt 3 1t0[(
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 3.Position:Velocity: x = 5t 4 4t 3 + 3t 2 ft v= a= dx = 20t 3 12t 2 + 3 ft/s dt dv = 60t 2 24t ft/s 2 dtAcceleration:When t = 2 s,x = ( 5 )( 2 ) ( 4 )( 2 ) ( 3)( 2 ) 2
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 195.Differentiate the expressions for r and with respect to time.( ) = 2 ( 2t + 4e ) rad,r = 6 4 2e t ft, 2t& r = 12e t ft/s,& = 12e t ft/s 2 r& = 2 2 8e 2t rad/s& r
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 18.Note that a is a given function of xUseUsing the limits andv dv = a dx = 600 x 1 + kx 2 dx = 600 x + 600kx3 dx v = 7.5 ft/s v = 15 ft/s when x = 0, when x = 0.45 ft,
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COSMOS: Complete Online Solutions Manual Organization SystemSolve for x and v. x = 0.25 v= Evaluate at t = 0.2 s. 1 cos10t 201 sin10t 2 1 cos ( (10 )( 0.2 ) ) 20x = 0.271 m v = 0.455 m/sx = 0.25 v=1 sin ( (10 )( 0.2 ) ) 2Vector Mechanics for Engine
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 17.a is a function of x:a = 100 ( 0.25 x ) m/s 2Use v dv = a dx = 100 ( 0.25 x ) dx with limits v = 0 when x = 0.2 m 0 v dv = 0.2100 ( 0.25 x ) dxv x12 1 2 v 0 = (100
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 16.Note that a is a given function of x. a = 40 160 x = 160 ( 0.25 x )(a) Note that v is maximum when a = 0, or x = 0.25 m Use v dv = a dx = 160 ( 0.25 x ) dx with the lim
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 14.Given: Separate variables and integrate.v t 2 0 dv = a dt = 0 ( 9 3t ) dt = 9a = 9 3t 2v 0 = 9 t t3 (a) When v is zero. t (9 t 2 ) = 0 t = 0 and t = 3 s (2 roots) (b)
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 13.Determine velocity.v t t 0.15 dv = 2 a dt = 2 0.15 dtv ( 0.15 ) = 0.15t ( 0.15 )( 2 ) v = 0.15t 0.45 m/sAt t = 5 s,When v = 0, For 0 t 3.00 s, For 3.00 t 5 s, Determ
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 12.Given: At t = 0,v = 400 mm/s; a = kt mm/s 2 at t = 1 s, where k is a constant. x = 500 mmv = 370 mm/s, 1v t t 2 400 dv = 0 a dt = 0 kt dt = 2 ktv 400 =12 kt 2orv
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 11.Given:a = 3.24sin kt 4.32 cos kt ft/s 2 ,x0 = 0.48 ft,tk = 3 rad/sv0 = 1.08 ft/st tv v0 = 0 a dt = 3.24 0 sin kt dt 4.32 0 cos kt dt v 1.08 = = 3.24 cos kt kt 0
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 10.Given:a = 5.4sin kt ft/s 2 ,t tv0 = 1.8 ft/s, x0 = 0, 5.4 cos kt ktk = 3 rad/sv v0 = 0 a dt = 5.4 0 sin kt dt = v 1.8 = Velocity:05.4 ( cos kt 1) = 1.8cos kt 1.8
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 9.a = 3e 0.2t 0 dv = 0 a dtv 0=t 3e 0.2t dt 0vt3 = e 0.2t 0.2t0v = 15 e 0.2t 1 = 15 1 e 0.2t At t = 0.5 s,()()v = 1.427 ft/sv = 15 1 e 0.1()t 0 dx = 0 v
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 8.x = t 2 ( t 2 ) ft3v= (a) Positions at v = 0.dx 2 = 2t 3 ( t 2 ) ft/s dt2t 3 ( t 2 ) = 3t 2 + 14t 12 = 02t=14 (14) 2 (4)( 3)(12) (2)( 3)t1 = 1.1315 s and t2 = 3.5
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 7.Given:Differentiate twice. x = t 3 6t 2 + 9t + 5 v= a= (a) When velocity is zero. dx = 3t 2 12t + 9 dt dv = 6t 12 dtv=03t 2 12t + 9 = 3 ( t 1)( t 3) = 0 t = 1 s and t
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 6.Position: Where Letx = 50sin k1t k2t 2 mm k1 = 1 rad/s d = (1 t ) rad/s dt x = 50sin mm v= and k2 = 0.5 rad/s 2() = k1t k2t 2 = t 0.5t 2 radand d 2 = 1 rad/s 2 dt 2