# Sinc function's energy

Given this signal: $$4000\,\mbox{sinc}(4000t)+D1000\,\mbox{sinc}(1000t)$$ with $D=6000$, find its energy.

To find the energy of sinc I use this transform:

$$A\,\mbox{sinc}(Bt) \rightarrow A\,\mbox{rect}(f/B)/B$$

and compute the energy $E$ using: $$E=\frac{A^2}{B}$$

In this case what is the right answer?

• A) $4000^2/4000+D^21000^2/10000$ or
• B) $1/4000+A^2/1000$
• I don't understand the A/B question. So you have an approach, what's the question, then? – Marcus Müller Mar 31 '17 at 19:05
• i have edited the variable. don't know how to apply it. – Santo1991 Mar 31 '17 at 19:14

Neither A nor B.

Hint: the Fourier transforms of the two sinc function can be added in the frequency domain due to the Fourier transform linearity property, but the energies do not since the spectrums overlap.

Note that the energy formula comes from computing the product of the square of the frequency domain amplitude by the frequency range of a rectangular pulse. Applying your formula, the amplitude for a single $\mbox{sinc}$ is given by $A/B$ and covers a range of $B$, yielding an energy of $(A/B)^2 B = A^2/B$.

The amplitude of the first rectangular function (corresponding to the $4000\mbox{sinc}(4000t)$ function in the time-domain) in the frequency-domain is $1$, and covers frequencies in the range $[-2000,+2000]$. Similarly, the amplitude of the second rectangular function (corresponding to the $D 1000\mbox{sinc}(1000t)$ function in the time-domain) in the frequency-domain is $D$ and covers frequencies in the range $[-500,+500]$.

Now, computing the energy of more than one $\mbox{sinc}$ is a little trickier... You should realize that there is an overlap for frequencies in $[-500,+500]$, where the total amplitude is $1+D$. The energy in that range is thus $1000 (1+D)^2$.

For the rest of the ${[-2000,-500], [+500,+2000]}$ frequencies, there is no overlap and the amplitude is that of the first rectangle function, i.e. $1$. The corresponding energy there is thus $3000$.

This then gives you a total energy of $3000 + 1000(1+D)^2$. Substituting $D=6000$ gives $36,012,004,000$.