{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hwsolB05 - Therefore Ts = k*T/3 = k(7/30 sec(k=1,2(iii...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Sheet1 Page 1 HW 5 Solution _____________ (i) Clearly, A = 4. The period T equals 0.7 sec, therefore f = 10/7 Hz and W = 20*pi/7 cos(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 waveform (assuming the cycle begins at a peak). Thus the correct value of phi is -pi/6. (ii) x[n] = x[n+3] for all n if and only if w = 2*pi*(Ts/T) satisfies 3*w = k*(2*pi) for integer k
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Therefore Ts = k*T/3 = k*(7/30) sec (k=1,2,. ..) (iii) Setting w = pi/7 = 2*pi*(Ts/T), we obtain Ts = T/14 = 0.05 sec and more generally, Ts = 0.05 + k*(0.7) sec (k=0,1,2,. ..) (iv) cos((16*pi/21)*n - phi) = cos((-16*pi/21)*n + phi). Using the negative value for w = 2*pi*(Ts/T), we obtain Ts = -(8/21)*T = -8/30 sec Then the positive values of Ts that produce the given x[n] are Ts = (0.7 - 8/30) + k*(0.7) = 13/30 + k*(0.7) sec...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern