- #1
emanuel_hr
- 10
- 0
Hello everyone(my first post here), I hope I have posted in the right section...
Given [tex]x[n][/tex] is a discrete stable(absolutely summable) sequence and its continuous Fourier transform [tex]X(e^{j\omega})[/tex] having the following properties:
[tex]x[n]=0, \ \ \ \forall n<1[/tex] and
[tex]Re\{X(e^{j\omega })\}=\frac{3}{2\cos \omega -\frac{5}{2}}, \ \ \ \forall \omega \in \mathbb{R}[/tex]
find [tex]\inline x[/tex] as good as possible(I don't know how to state this any better, basically one should find x if possible, if not a sequence that resembles x as good as possible)
[tex]X(e^{j\omega})=\sum_{n\in \mathbb{Z}}^{ } x[n]e^{-j\omega n}, \ \ \ \forall \omega \in \mathbb{R };[/tex]
[tex]x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi} X(e^{j\omega})e^{j\omega n}d\omega,\ \ \ \forall n \in \mathbb{Z }.[/tex]
It is straightforward to show that if
[tex]x_{e}[n] = \frac{x[n]+\overline{x[-n]}}{2}, \ \ \forall n \in \mathbb{Z}, \ \textup{then} \ X_{e}(e^{j\omega})= Re\{X(e^{j\omega})\}, \ \forall \omega \in \mathbb{R}[/tex]
Given the fact that [tex]x[n]=0, \forall n<1[/tex] by finding [tex]x_{e}[n] [/tex] we can also find [tex]x[n][/tex]
By applying the inverse Fourier transform of [tex]X_{e}(e^{j\omega})[/tex] we obtain:
[tex]x_{e}[n] = \frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{3}{2\cos \omega -\frac{5}{2}}\cdot e^{j\omega n}d\omega[/tex]
And at this point I'm stuck, i have no idea how to evaluate that integral(I've tried the usual tricks but none seem to work). Maybe the approach is not the best one, I don't know.
Thanks in advance for any advice.
Homework Statement
Given [tex]x[n][/tex] is a discrete stable(absolutely summable) sequence and its continuous Fourier transform [tex]X(e^{j\omega})[/tex] having the following properties:
[tex]x[n]=0, \ \ \ \forall n<1[/tex] and
[tex]Re\{X(e^{j\omega })\}=\frac{3}{2\cos \omega -\frac{5}{2}}, \ \ \ \forall \omega \in \mathbb{R}[/tex]
find [tex]\inline x[/tex] as good as possible(I don't know how to state this any better, basically one should find x if possible, if not a sequence that resembles x as good as possible)
Homework Equations
[tex]X(e^{j\omega})=\sum_{n\in \mathbb{Z}}^{ } x[n]e^{-j\omega n}, \ \ \ \forall \omega \in \mathbb{R };[/tex]
[tex]x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi} X(e^{j\omega})e^{j\omega n}d\omega,\ \ \ \forall n \in \mathbb{Z }.[/tex]
The Attempt at a Solution
It is straightforward to show that if
[tex]x_{e}[n] = \frac{x[n]+\overline{x[-n]}}{2}, \ \ \forall n \in \mathbb{Z}, \ \textup{then} \ X_{e}(e^{j\omega})= Re\{X(e^{j\omega})\}, \ \forall \omega \in \mathbb{R}[/tex]
Given the fact that [tex]x[n]=0, \forall n<1[/tex] by finding [tex]x_{e}[n] [/tex] we can also find [tex]x[n][/tex]
By applying the inverse Fourier transform of [tex]X_{e}(e^{j\omega})[/tex] we obtain:
[tex]x_{e}[n] = \frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{3}{2\cos \omega -\frac{5}{2}}\cdot e^{j\omega n}d\omega[/tex]
And at this point I'm stuck, i have no idea how to evaluate that integral(I've tried the usual tricks but none seem to work). Maybe the approach is not the best one, I don't know.
Thanks in advance for any advice.
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