| (One intermediate revision by the same user not shown) | |||
| Line 5: | Line 5: | ||
<math> y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau - t)h(\tau)\, d\tau </math> | <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(\tau)\, d\tau ~~~ | + | <math> \Rightarrow ~y(t) \le \int_{-\infty}^{\infty} Bh(\tau)\, d\tau ~~~\leftrightarrow~~~x(\tau - t)~is~stable~as~x(t)~is~stable.</math> |
<math> \Rightarrow ~y(t) \le B\int_{-\infty}^{\infty} e^\tau[u(\tau-2) - u(\tau-5)]\, d\tau </math> | <math> \Rightarrow ~y(t) \le B\int_{-\infty}^{\infty} e^\tau[u(\tau-2) - u(\tau-5)]\, d\tau </math> | ||
| Line 11: | Line 11: | ||
<math> \Rightarrow ~y(t) \le B\int_{2}^{5} e^\tau\, d\tau </math> | <math> \Rightarrow ~y(t) \le B\int_{2}^{5} e^\tau\, d\tau </math> | ||
| − | <math> \Rightarrow ~y(t) \le B | + | <math> \Rightarrow ~y(t) \le B(e^5 - e^2) </math> |
| − | Hence <math> y(t) \le | + | Hence <math> y(t) \le Bc \le C~, ~~~ (where~c = e^5 - e^2 ~and ~~C = B*c) </math> |
<math> \therefore y(t) ~is ~bounded </math> | <math> \therefore y(t) ~is ~bounded </math> | ||
Latest revision as of 19:37, 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(\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 $
