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If I have a average normalized power spectrum of noise is like this $W(f)=N_0$ while frequency is $-B$ to $B$.

Why the average normalized power is $1/4 N_0 (2B)$ insteat of $N_0 (2B)$?

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kept staring at it and thought.. that makes no sense! Remark that it is $(1/4)N_0(2B)$. not sure if that is right though. gotta work out the math. –  thang Jan 25 '13 at 9:35
    
Yes, I have completely no idea where is the 1/4 come form. –  Samuel Jan 25 '13 at 10:50
    
Well I think that if you work out the formula, it comes out to 1/4. It will take me awhile to do it, and I am kinda lazy now (it's 3am here... :(). To get an intuition, just look at sin(t). –  thang Jan 25 '13 at 10:52
    
OK thank for the tips –  Samuel Jan 25 '13 at 11:03
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@JasonR, as you can see from the formula, you don't need to know $f$ at every point in time to get the average power. If you do, all the better. In fact, consider for example $g=f+h$ where $h$ is nonzero only on a set of measure 0, then average power of $g$ is the same as that of f. There is a formula that relates the power spectral density and average power. He's trying to get a derivation of this formula. I suggested, for simplicity to save time, just do $sin(t)$ instead of generally. I have been meaning to derive it, but got caught up with a stupid thing called the job :p –  thang Jan 25 '13 at 16:40
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1 Answer

up vote 3 down vote accepted

If a filter with transfer function $H(f)$ has a noise process as input, then the output noise power $P_{\scriptstyle{\text{out}}}$ is given by $$P_{\scriptstyle{\text{out}}} = \int_{-\infty}^\infty |H(f)|^2 S_x(f)\,\mathrm df $$ where $S_x(f)$ is the power spectral density of the input noise process. If the input process is modeled as a white noise process, then $S_x(f)$ is a constant, call it $K$ for now, and so we have that $$P_{\scriptstyle{\text{out}}} = K \int_{-\infty}^\infty |H(f)|^2\,\mathrm df $$ For an ideal low-pass filter with cutoff frequency $B$ Hz, $$ H_{\scriptstyle{\text{lowpass}}}(f) = \begin{cases} 1, &-B \leq f \leq B,\\ 0, & |f| > B,\end{cases}$$ we have $P_{\scriptstyle{\text{out}}} = 2KB$. More generally, for an arbitrary linear time-invariant system with maximum passband gain equal to $1$, the noise-equivalent bandwidth of the system is defined as $$B_{\scriptstyle{\text{ne}}} = \frac{1}{2}\int_{-\infty}^\infty |H(f)|^2\,\mathrm df$$ and so we get $P_{\scriptstyle{\text{out}}} = 2KB_{\scriptstyle{\text{ne}}}$. Put another way, the noise-equivalent bandwidth of an ideal LPF with cutoff $B$ Hz is just $B$ Hz.

It is conventional in the engineering literature to write $N_0/2$ instead of $K$ because then the output noise power expression becomes $N_0B_{\scriptstyle{\text{ne}}}$. The constant of proportionality $N_0$, measured in watts/Hz as Jason R pointed out, is convenient for use in ordinary language when conversing with other engineers. If we have a filter with noise equivalent bandwidth $B_{\scriptstyle{\text{ne}}}$, and the thermal noise power at the input of the filter is $N_0$ watts/Hz, then the output noise power is $N_0B_{\scriptstyle{\text{ne}}}$. That's the way people talk during discussions. Nobody will say that an ideal lowpass filter with cutoff $B$ Hz has "bandwidth" $2B$ even though the transfer function has value $1$ for $f \in [-B,B]$ and thus the support of $H_{\scriptstyle{\text{lowpass}}}(f)$ has length $2B$.

Turning to your question, it is not a good idea to write $W(f) = N_0$ for $f \in [-B,B]$, but if the paper you are reading does insist on this notation, then the output power is $N_0(2B)$. If the paper says that it is $\frac{1}{4}N_0(2B)$, then it is mistaken. Another possibility is that there are other conditions that are mentioned in the paper that make $\frac{1}{4}N_0(2B)$ the correct answer, but you have not included these conditions in your question. For example, if the passband gain of the ideal LPF is $\frac{1}{2}$, then the expression $\frac{1}{4}N_0(2B)$ for the output power is perfectly correct.

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Thanks Dilip for very clearly articulating what I was trying to say above. –  Jason R Jan 26 '13 at 0:41
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