Consider a periodic function x(t). with fundamental period To, where a single pe

Question

Consider a periodic function x(t). with fundamental period To, where a single period is specified by x(t)=cos(pi/To t) for-T alpha/2 inequality t inequality To/2 (This problem is similar to the one solved in the book in Appendix C, section C-2.3 with important differences.) Draw a labelled sketch of x(t) for at least 3 periods. The function should never go negative. x{t) represents a "full-wave rectified" cosine, where the underlying function is: | cos (pi/To t) |. Use the fact that the "D.C." coefficient, ao. is simply given by the integral of x(t) over a single period, divided by its period, to compute alpha o. Also, this DC value does not depend on To so feel free to pick something convenient. Use the Fourier series analysis integral to compute the F.S. coefficients a* for the periodic function x(t) for all k. Simplify your answer as much as possible. (There is no easy way to do this problem given your current knowledge, but look for symmetries where you can. However, the coefficients do not depend on To, so pick a convenient value.) If you get something with an in determine form (i.e., 0/0), evaluate it using L'Hopital's rule or go back to the integral with the offending value of k plugged in directly. Make sure that the values you get for a0 in parts (b) and (c) match. [some constant] time sign (-1)k/pi(1-[some integer]k^2) We're not telling what "some constant" and "some integer" are. you'll need to find that out. If your answer isn't as simple as this, keep working at it Suppose y(t)=alpha_0-x(t).Sketch y(t)

Explanation / Answer

Note that for geometric and binomial distirbutions, the probability of success is constant.

a)

Neither. The probability of success for each ball is different as we continue to draw balls without replacement.

*****************

b)

Neither. The probability of getting it right gets updated when you answer incorrectly because the computer gives you instructions.

***********************

c)

This is geometric. Here,

p = 1/5 = 0.2
q = 1 - p = 0.8

Hence,

P(5) = q^(5-1) p = 0.8^4 * 0.2 = 0.08192 [ANSWER]

*********************

d)

Neither, as the number of successes is 3 (instead of one, which would make it geometric). It is not binomial because the number of trials is not limited.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.