# Power of a speech signal

I have been asked to calculate the power of a speech signal given by the following: $$\ p(x)= \left(\frac a2\right)e^{-a\mid x \mid}$$

I know the formula for the power of a signal to be the following:

$$\frac{1}{T}\int_\frac{-T}{2}^\frac{T}{2}p(x)^2dx$$

Only this is not a periodic signal so I don't know where to set the upper and lower limits. I have tried taking the Fourier transform and evaluating but that gives me $\ P(f) = \frac{a^2}{a^2+2\pi f^2}$. If anyone would be able to guide me through a solution to this I'd really appreciate it. Simon.

• I am assuming 'a' in the equations is constant.The 'a' in your equation should be negative otherwise power is infinite, and for negative 'a' you can limit the integration up-to the 5-6 times of time constant of the dying exponential(as per your equation). this will give you the power approximately equal to actual. – arpit jain Apr 14 '16 at 5:19
• Yes it is negative, I just made that edit. I don't know what you mean by 5-6 times the constant? do you mean 5xa or 6xa. Could you elaborate more please? – burton01 Apr 14 '16 at 5:38
• Please read again your assignment and figure out what $p(x)$ actually is. I'm pretty sure that it is not the signal itself but a function describing its behavior. As soon as you understand the meaning of $p(x)$ you will be able to find the appropriate formula for deriving the average power of the signal. The formula in your question is not applicable in this case. – Matt L. Apr 14 '16 at 7:00

This looks like energy rather than power. The $p(x)$ looks like voltage at a given moment of time. You square it to get the power. You integrate the power to get the energy. You then divide total energy by total time to get the average power. Why periodicy? I would just
$$P_{avg} = {a\over 2T} \int_0^T{e^{-ax}dx}.$$
• Why would you want to do that if you can integrate to $\infty$ without any problem? – Matt L. Apr 14 '16 at 7:01