Solved Problems
Consider a signal [Tex]x(t)[/Tex] whose power spectral density can be given as [Tex]S_x(f) = \frac{1}{4\pi} e^{-2\pi |f|}[/Tex]. Then (a) Find the Autocorrelation function [Tex]R_x(\tau)[/Tex] of signal [Tex]x(t)[/Tex]. (b) Determine the total power of the signal [Tex]x(t)[/Tex].
a) As known that the power spectral density and the autocorrelation function represents Fourier transform pairs. So,
[Tex]\therefore R_x(\tau) = \int_{-\infty}^{\infty} S_x(f) e^{2\pi if\tau} df [/Tex]
[Tex]\therefore R_x(\tau) = \int_{-\infty}^{\infty} \frac{1}{4\pi} e^{-2\pi |f|} e^{2\pi if\tau} df[/Tex]
[Tex]\therefore R_x(\tau) = \frac{1}{4\pi} \int_{-\infty}^{\infty} e^{-2\pi |f|} e^{2\pi if\tau} df[/Tex]
[Tex]\therefore R_x(\tau) = \frac{1}{4\pi} \int_{-\infty}^{0} e^{-2\pi (-f)} e^{2\pi if\tau} df + \frac{1}{4\pi} \int_{0}^{\infty} e^{-2\pi (+f)} e^{2\pi if\tau} df [/Tex]
[Tex]\therefore R_x(\tau) = \frac{1}{4\pi} \int_{-\infty}^{0} e^{2\pi (i\tau f + f)} df + \frac{1}{4\pi} \int_{0}^{\infty} e^{2\pi (i\tau f-f)} df [/Tex]
[Tex]\therefore R_x(\tau) = \frac{1}{4\pi} [ \frac{e^{2\pi (i\tau f + f)}}{2\pi (i\tau +1)} |_{-\infty}^{0}] + \frac{1}{4\pi} [ \frac{e^{2\pi (i\tau f – f)}}{2\pi (i\tau – 1)} |_{0}^{\infty}] [/Tex]
Now, as [Tex]1 >> i\tau[/Tex] so the limit converges to 0 near [Tex]+\infty[/Tex] and also [Tex]lim_{x\to\infty} e^{-x} = 0[/Tex].
[Tex]\therefore R_x(\tau) = \frac{1}{4\pi} [ \frac{1}{2\pi (i\tau +1)}] + \frac{1}{4\pi} [ \frac{-1}{2\pi (i\tau – 1)}] [/Tex]
[Tex]\therefore R_x(\tau) = \frac{1}{8\pi^{2}} [ \frac{1}{ (1+i\tau)} + \frac{1}{ (1-i\tau )}] [/Tex]
[Tex]\therefore R_x(\tau) = \frac{1}{8\pi^{2}} [ \frac{2}{ (1^{2}-(i\tau)^{2})}] [/Tex]
As, [Tex]i^{2} = -1[/Tex]
[Tex]\therefore R_x(\tau) = \frac{1}{8\pi^{2}} [ \frac{2}{ (1-((-1)(\tau)^{2})}] [/Tex]
[Tex]\therefore R_x(\tau) = \frac{1}{4\pi^{2}} [ \frac{1}{ 1+\tau^{2}}] [/Tex]
b) Now, the total power of the signal is given as follows
[Tex]\therefore P_x = \frac{1}{2\pi}\int_{-\infty}^{\infty} S_x(\omega) d\omega = \int_{-\infty}^{\infty} S_x(f) df[/Tex]
[Tex]\therefore P_x = \int_{-\infty}^{\infty} \frac{1}{4\pi} e^{-2\pi |f|} df[/Tex]
[Tex]\therefore P_x = \frac{1}{4\pi} \int_{-\infty}^{\infty} e^{-2\pi |f|} df[/Tex]
As [Tex]e^{-2\pi |f|}[/Tex] is an even function i.e. symmetric about y-axis. So, using the property of even function
[Tex]\therefore \int_{-\infty}^{\infty} e^{-2\pi |f|} df = 2\int_{0}^{\infty}e^{-2\pi f} df[/Tex]
[Tex]\therefore P_x = \frac{1}{4\pi} .2\int_{0}^{\infty} e^{-2\pi f} df[/Tex]
[Tex]\therefore P_x = \frac{1}{2\pi} \int_{0}^{\infty} e^{-2\pi f} df[/Tex]
[Tex]\therefore P_x = \frac{1}{2\pi} [\frac{e^{-2\pi f}}{-2\pi}|_{0}^{\infty}][/Tex]
As, [Tex]lim_{x\to\infty} e^{-x} = 0[/Tex]
[Tex]\therefore P_x = \frac{1}{2\pi} [\frac{-1}{-2\pi}][/Tex]
[Tex]\therefore P_x = \frac{1}{4\pi^{2}}[/Tex] Watt
Power Spectral Density
In terms of electronics, Power is defined as the total amount of energy that is getting transferred or converted per unit measurement of time, or in general terms Power is defined as the strength or the intensity level of the signal. Power is generally measured in watts (W).
In this article, we will be going through Power Spectral Density, First we will start our Article with the Definition of Power Spectral Density with an Example, Then we will go through its derivation Properties and Characteristics, At last, we will conclude our Article with Solved Examples, Applications, and Some FAQs.
Table of Content
- Definition
- Derivation
- Characteristics
- Properties
- Solved Problems
- Applications
Contact Us