Ω 1 ω n m f r uu n m t 1 t n m 103 and using the

This preview shows page 238 - 241 out of 338 pages.

) ( ω 1 , . . . , ω n +m ) = F [ R uu ( n +m ) ( t 1 , . . . , t n +m )] (103) and using the system function defined previously gives S ˆ y n y m ( ω 1 , . . . , ω n +m ) = H nsym ( ω 1 , . . . , ω n ) H msym ( ω n + 1 , . . . , ω n +m ) S uu ( n +m ) ( ω 1 , . . . , ω n +m ) (104) Repeating the derivation leading to (79) gives, in the present setting S y n y m ( ω 1 , ω 2 ) = (2 π ) n +m 2 1 _________ −∞ S ˆ y n y m ( γ 1 , . . . , γ n +m ) δ 0 ( ω 1 −γ 1 . . . −γ n ) δ 0 ( ω 2 −γ n + 1 . . . −γ n +m ) d γ 1 . . . d γ n +m (105) Furthermore, for stationary random process inputs, the partial output power spectral density is given by S y n y m ( ω ) = (2 π ) n +m 1 1 _________ −∞ S ˆ y n y m ( γ 1 , . . . , γ n +m ) δ 0 ( ω−γ 1 . . . −γ n ) d γ 1 . . . d γ n +m (106) Of course, this formula checks with (84) for the case m = n . Now assume that the input is real, stationary, zero-mean, Gaussian, and with power spectral density S uu ( ω ). Substituting (87) and (104) into (106) gives S y n y m ( ω ) = 0 , n +m odd (107) and S y n y m ( ω ) = (2 π ) ( n +m 2) / 2 1 ___________ p Σ −∞ H nsym ( γ 1 , . . . , γ n ) H msym ( γ n + 1 , . . . , γ n +m ) δ 0 ( ω−γ 1 . . . −γ n ) j , k Π n +m S uu ( γ j ) δ 0 ( γ j + γ k ) d γ 1 . . . d γ n +m , n +m even (108) The reduction of this expression to more explicit form is a combinatorial problem of some complexity because the integrand lacks symmetry. I will be content to work out the terms that give the output power spectral density for polynomial systems of degree 3. 230
Image of page 238
Example 5.12 To compute S yy ( ω ) for the case of a degree-3 polynomial system, the terms S y n y m ( ω ) must be computed for n , m = 1,2,3. But it is evident that S y 1 y 2 ( ω ) = S y 2 y 1 ( ω ) = S y 2 y 3 ( ω ) = S y 3 y 2 ( ω ) = 0 For n = m = 1,2,3 the partial output power spectral densities have been calculated previously, and are given in (64), (93), and (94). For n = 1 and m = 3, (108) gives S y 1 y 3 ( ω ) = 2 π 1 ___ p Σ −∞ H 1 ( γ 1 ) H 3 sym ( γ 2 , γ 3 , γ 4 ) δ 0 ( ω−γ 1 ) j , k Π 4 S uu ( γ j ) δ 0 ( γ j + γ k ) d γ 1 . . . d γ 4 = 2 π 1 ___ −∞ H 1 ( γ 1 ) H 3 sym ( γ 2 , γ 3 , γ 4 ) δ 0 ( ω−γ 1 ) S uu ( γ 1 ) S uu ( γ 3 ) δ 0 ( γ 1 + γ 2 ) δ 0 ( γ 3 + γ 4 ) d γ 1 . . . d γ 4 + 2 π 1 ___ −∞ H 1 ( γ 1 ) H 3 sym ( γ 2 , γ 3 , γ 4 ) δ 0 ( ω−γ 1 ) S uu ( γ 1 ) S uu ( γ 2 ) δ 0 ( γ 1 + γ 3 ) δ 0 ( γ 2 + γ 4 ) d γ 1 . . . d γ 4 + 2 π 1 ___ −∞ H 1 ( γ 1 ) H 3 sym ( γ 2 , γ 3 , γ 4 ) δ 0 ( ω−γ 1 ) S uu ( γ 1 ) S uu ( γ 2 ) δ 0 ( γ 1 + γ 4 ) δ 0 ( γ 2 + γ 3 ) d γ 1 . . . d γ 4 Performing the integrations gives S y 1 y 3 ( ω ) = 2 π 3 ___ H 1 ( ω ) S uu ( ω ) −∞ H 3 sym ( −ω , γ , −γ ) S uu ( γ ) d γ In a similar fashion S y 3 y 1 ( ω ) can be computed. Alternatively, the easily proved fact that S y 3 y 1 ( ω ) = S y 1 y 3 ( −ω ) can be used to obtain S y 3 y 1 ( ω ) = 2 π 3 ___ H 1 ( −ω ) S uu ( ω ) −∞ H 3 sym ( ω , γ , −γ ) S uu ( γ ) d γ Now, collecting together all the terms gives the expression 231
Image of page 239
S yy ( ω ) = H 1 ( ω ) H 1 ( −ω ) S uu ( ω ) + 2 π 3 ___ H 1 ( ω ) S uu ( ω ) −∞ H 3 sym ( −ω , γ , −γ ) S uu ( γ ) d γ + 2 π 3 ___ H 1 ( −ω ) S uu ( ω ) −∞ H 3 sym ( ω , γ , −γ ) S uu ( γ ) d γ + 2 π 1 ___ δ 0 ( ω ) −∞ −∞ H 2 sym ( γ 1 , −γ 1 ) H 2 sym ( γ 2 , −γ 2 ) S uu ( γ 1 ) S uu ( γ 2 ) d γ 1 d γ 2 + π 1 __ −∞ H 2 sym ( ω−γ , γ ) H 2 sym ( −ω + γ , −γ ) S uu ( γ ) S uu ( ω−γ ) d γ + (2 π ) 2 6 _____ −∞ −∞ H 3 sym ( ω−γ 1 −γ 2 , γ 1 , γ 2 ) H 3 sym ( −ω +
Image of page 240

Want to read all 338 pages?

Image of page 241

Want to read all 338 pages?

You've reached the end of your free preview.

Want to read all 338 pages?

  • Spring '11
  • Jung
  • The Land, LTI system theory, Linear system, Nonlinear system, Homogeneous Systems, kernel, triangular kernel

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern