(New page: 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(t) \le B </math> <math> y(t) = x(t) * h(t...) |
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Its given that <math> x(t) \le B </math> | Its given that <math> x(t) \le B </math> | ||
| − | <math> y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(t)h( | + | <math> y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau - t)h(\tau)\, d\tau </math> |
<math> \Rightarrow ~y(t) \le \int_{-\infty}^{\infty} Bh(t)\, dt </math> | <math> \Rightarrow ~y(t) \le \int_{-\infty}^{\infty} Bh(t)\, dt </math> | ||
Revision as of 19:33, 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(\tau - t)h(\tau)\, d\tau $
$ \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 $
