I have provided a solution manual to some problems I am working on, however; I d

Question

I have provided a solution manual to some problems I am working on, however; I do not understand why in part b and c they chose to multiply the integrals by "2" almost as if they were doing integration over half the bandwidth? but the bounds of integration remain unchanged so when I did the work myself I arrived at mean powers that were half their values because of the extra factor of 2 they have outside the integral. So my question is: who is wrong and if it is me, why the extra 2?

Problem 4.2.1 A certain signal f(t) has the following power spectral density (assume a 1- Ohm resistive load): S (w) 5(w 2) (w 2) 1 wi a.) What is the total mean power in f(t)? b.) What is the mean power in f(t) wi thin the bandwidth 0.9 to 1.1 rad/sec c) What is the mean power in f(t) within the bandwidth 1.9 to 2.1 rad/sec? d. Find a signal f(t), in terms of cosines and exponentials, that will satisfy this power spectral density. Are other solutions possible? Solution: Recall that the power of a signal with Spectral density function S (w) in the frequency interval La, b] is given as 44 S (w) du 0.818W 2T Note that dw arctan (b) arctan(a), & arctan (oo) -arctan(-oo) /2. (170) 2) dw [arctan(1.1) 1 w Note that the delta function is not in the range of the integration. 6(w 2) dw arctan(2.1) arctan(1.9) 1] 0.331W 2T J1.9 1 wi Note that the delta function is in the range of the integration

Explanation / Answer

You are right. They shouldnt be multiplying by 2 because integration is not done over half the bandwidth . the frequency range is between 0 to infinity so problem of half bandwidth should not come into picture as the mean power from - infinity to zero is 0

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