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<math> y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(t)h(t)\, dt </math> | <math> y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(t)h(t)\, dt </math> | ||
| − | <math> \ | + | <math> \Rightarrow ~y(t) \le \int_{-\infty}^{\infty} Bh(t)\, dt </math> |
| + | |||
| + | <math> \Rightarrow ~y(t) \le B\int_{-\infty}^{\infty} e^t[u(t-2) - u(t-5)]\, dt </math> | ||
| + | |||
| + | <math> \Rightarrow ~y(t) \le B\int_{2}^{5} e^t\, dt </math> | ||
| + | |||
| + | <math> \Rightarrow ~y(t) \le B*(e^5 - e^2) </math> | ||
| + | |||
| + | Hence <math> y(t) \le B*c \le C~, ~~~ (where~c = e^5 - e^2 ~and ~~C = B*c) </math> | ||
| + | |||
| + | <math> \therefore y(t) ~is ~bounded </math> | ||
Latest 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 $
