(New page: ==Problem 4 - Mistake in solution posted== I thought that the solution posted in the Bonus 3 for problem 4 is slightly wrong in explaining why System II is Stable. Its given that <math> x...)
 
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==Problem 4 - Mistake in solution posted==
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[[Category: ECE]]
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[[Category: ECE 301]]
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[[Category: Summer]]
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[[Category: 2008]]
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[[Category: asan]]
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[[Category: Exams]]
 
I thought that the solution posted in the Bonus 3 for problem 4 is slightly wrong in explaining why System II is Stable.
 
I thought that the solution posted in the Bonus 3 for problem 4 is slightly wrong in explaining why System II is Stable.
  

Revision as of 10:15, 21 November 2008

I thought that the solution posted in the Bonus 3 for problem 4 is slightly wrong in explaining why System II is Stable.

Its given that $ x(t) \le B $

$ y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau - t)h(\tau)\, d\tau $

$ \Rightarrow ~y(t) \le \int_{-\infty}^{\infty} Bh(\tau)\, d\tau ~~~\leftrightarrow~~~x(\tau - t)~is~stable~as~x(t)~is~stable. $

$ \Rightarrow ~y(t) \le B\int_{-\infty}^{\infty} e^\tau[u(\tau-2) - u(\tau-5)]\, d\tau $

$ \Rightarrow ~y(t) \le B\int_{2}^{5} e^\tau\, d\tau $

$ \Rightarrow ~y(t) \le B(e^5 - e^2) $

Hence $ y(t) \le Bc \le C~, ~~~ (where~c = e^5 - e^2 ~and ~~C = B*c) $

$ \therefore y(t) ~is ~bounded $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010