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<math> \Rightarrow ~y(t) \le B*(e^5 - e^2) </math>
 
<math> \Rightarrow ~y(t) \le B*(e^5 - e^2) </math>
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Hence <math> y(t) \le B*c \le C~, ~~~ (where~c = e^5 - e^2 ~and ~~C = B*c) </math>
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<math> \therefore y(t) is bounded </math>

Revision as of 12:02, 1 July 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(t)h(t)\, dt $

$ \Rightarrow ~y(t) \le \int_{-\infty}^{\infty} Bh(t)\, dt $

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

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

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

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

$ \therefore y(t) is bounded $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin