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### hwB20

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 &lt; 0 and n &gt; 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&lt;v1,v2&gt; = 4-8+4 = 0(ii) &lt;v1,a&gt; = 3+12+12 = 27, &lt;v2,a&gt; = 12-12+3 = 3|a| = sqrt(9 + 18 + 9) = sqrt(36) = 6f1 = ( &lt;v1,a&gt; / |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 -&gt; a = -2bIm( v1'*v2 ) = 3b - 18 + 6a = 0 -&gt; 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 &lt;-&gt; R*S = [ 4 -1-2j -5*j 3-j -6 3+j +5*j -1+2j ].'s1 = P*R*s &lt;-&gt; 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 &lt;-&gt; R*S = [ 4 -1-2j -5*j 3-j -6 3+j +5*j -1+2j ].'s1 = P*R*s &lt;-&gt; 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&gt;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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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[(
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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
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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&amp; r = 12e t ft/s,&amp; = 12e t ft/s 2 r&amp; = 2 2 8e 2t rad/s&amp; r
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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)
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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
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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
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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
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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