Q4) Bandlimited process does change with time: In lecture we have upper bounded the change in value of a bandlimited process over a small time T. Here we obtain a lower bound (and a new upper bound). First, two intermediate results are obtained as below. 4.1) Assume that the time is bounded by |T| < (π/σ) and that the frequency is bounded by |w| ≤ σ. Then, using the fact that if 0 < p < (π/2), then (24/π) < sin 4 < 4, find a lower bound and an upper bound on sin ² (WT/2): < sin² (wt/2) ≤

Q4) Bandlimited process does change with time: In lecture we have upper bounded the change in value of a bandlimited process over a small time T. Here we obtain a lower bound (and a new upper bound). First, two intermediate results are obtained as below. 4.1) Assume that the time is bounded by |T| < (π/σ) and that the frequency is bounded by |w| ≤ σ. Then, using the fact that if 0 < p < (π/2), then (24/π) < sin 4 < 4, find a lower bound and an upper bound on sin ² (WT/2): < sin² (wt/2) ≤

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

Q4) Bandlimited process does change with time: In lecture we have upper bounded the
change in value of a bandlimited process over a small time t. Here we obtain a lower bound
(and a new upper bound). First, two intermediate results are obtained as below.
4.1) Assume that the time is bounded by |t| < (T/0) and that the frequency is bounded by
|w| <o. Then, using the fact that if 0 < p < (T/2), then (2º/T) < sin p < p, find a
lower bound and an upper bound on sin 2 (wt/2):
< sin?(wt/2) s
4.2) Let the power spectrum of x(t) be Sxx(w). If x(t) is passed through a differentiator
(frequency response H(w) = jw), then you have already found Sx'x@), the power
spectrum of the output x'(t), in terms of Sxx (w), in Q(2.b). Copy this Sx (w) below to
obtain the autocorrelation of the output x'(t) as
Ryi (t) =
Sxx (w)ejwrdw/2n
Now, putting T = 0, the average power of the output x'(t) is:
E{\x'(t)l?} =
Sxx (w)dw/2n
- 00

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