{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapt 8c - 8.7 Power Spectral Density 287 Figure 8.4 Plot...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
8.7 Power Spectral Density 287 Figure 8.4 Plot of S XX ( w ) for Example 8.10 The cross-power spectral density is generally a complex function even when both X ( t ) and Y ( t ) are real. Thus, since R YX ( τ ) = R XY ( τ ) , we have that S YX ( w ) = S XY ( w ) = S XY ( w ) where S XY ( w ) is the complex conjugate of S XY ( w ) . Example 8.10 Determine the autocorrelation function of the random process with the power spectral density given by S XX ( w ) = braceleftBig S 0 | w | < w 0 0 otherwise Solution S XX ( w ) is plotted in Figure 8.4. R XX ( τ ) = 1 2 π integraldisplay −∞ S XX ( w ) e jw τ dw = 1 2 π integraldisplay w 0 w 0 S 0 e jw τ dw = S 0 2 j πτ bracketleftbig e jw τ bracketrightbig w 0 w 0 = S 0 2 j πτ bracketleftbig e jw 0 τ e jw 0 τ bracketrightbig = S 0 πτ parenleftbigg e jw 0 τ e jw 0 τ 2 j parenrightbigg = S 0 πτ sin ( w 0 τ ) trianglesolid Example 8.11 A stationary random process X ( t ) has the power spectral density S XX ( w ) = 24 w 2 + 16 Find the mean-square value of the process.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
288 Chapter 8 Introduction to Random Processes Solution Method 1 (Brute-Force Method) : The mean-square value is given by E bracketleftbig X 2 ( t ) bracketrightbig = 1 2 π integraldisplay −∞ S XX ( w ) dw = 1 2 π integraldisplay −∞ 24 w 2 + 16 dw = 1 2 π integraldisplay −∞ 24 16 [ 1 + ( w / 4 ) 2 ] dw Let w / 4 = tan θ . Then dw = 4sec 2 ( θ ) d θ 1 + ( w / 4 ) 2 = 1 + tan ( θ ) 2 = sec 2 ( θ ) Also, when w = −∞ , θ = − π / 2; and when w = ∞ , θ = π / 2. Thus, we obtain E bracketleftbig X 2 ( t ) bracketrightbig = 24 32 π integraldisplay π / 2 π / 2 4sec 2 ( θ ) d θ sec 2 ( θ ) = 3 π integraldisplay π / 2 π / 2 d θ = 3 π [ θ ] π / 2 π / 2 = 3 π braceleftbigg π 2 parenleftbigg π 2 parenrightbiggbracerightbigg = 3 π braceleftbigg π 2 + π 2 bracerightbigg = 3 Solution Method 2 (Smart-Move Method) : From Table 8.1 we observe that e a | τ | 2 a a 2 + w 2 That is, e a | τ | and 2 a /( a 2 + w 2 ) are Fourier transform pairs. Thus, if we can iden- tify the parameter a in the given problem, we can readily obtain the autocorrela- tion function. Rearranging the power spectral density, we obtain S XX ( w ) = 24 w 2 + 16 = 24 w 2 + 4 2 = 3 braceleftbigg 2 ( 4 ) w 2 + 4 2 bracerightbigg 3 braceleftbigg 2 a w 2 + a 2 bracerightbigg This means that a = 4 and the autocorrelation function is R XX ( τ ) = 3 e 4 | τ | Therefore, the mean-square value of the process is E bracketleftbig X 2 ( t ) bracketrightbig = R XX ( 0 ) = 3 trianglesolid
Background image of page 2
8.7 Power Spectral Density 289 8.7.1 White Noise White noise is the term used to define a random function whose power spectral density is constant for all frequencies. Thus, if N ( t ) denotes white noise, S NN ( w ) = N 0 / 2 where N 0 is a real positive constant. The inverse Fourier transform of S NN ( w ) gives the autocorrelation function of N ( t ) , R NN ( τ ) , as follows: R NN ( τ ) = ( N 0 / 2 ) δ ( τ ) where δ ( τ ) is the impulse function. The two functions are shown in Figure 8.5.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}