| (26 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
| − | A Signal for which the output signal is constant times the input is referred as an '''eigenfunction''' of the system,andthe amplitude is called the system's | + | A Signal for which the output signal is constant times the input is referred as |
| + | |||
| + | an '''eigenfunction''' of the system,andthe amplitude is called the | ||
| + | |||
| + | system's ''eigenvalue''' | ||
| + | |||
| + | |||
| + | '''PART A:''' | ||
| + | ---- | ||
| + | |||
| + | |||
let the input be | let the input be | ||
| − | <math>x(t)={e^st}</math> | + | <math>x(t)= e^{st}</math> |
| + | |||
| + | we can determine the output using convolution integral | ||
| + | |||
| + | <math>y(t)= \int_{-\infty}^{\infty}h(T)x(t-T)\, dt </math> | ||
| + | |||
| + | = <math> \int_{-\infty}^{\infty}h(T){e^{s(t-T)}}\, dt </math> | ||
| + | |||
| + | = <math>e^{st} \int_{-\infty}^{\infty}h(T){e^{-sT}}\, dt </math> | ||
| + | |||
| + | = <math>H(s) e^{st}</math> | ||
| + | |||
| + | h(t) is the impulse response of the LTI SYSTEM | ||
| + | H(s) is the system fuction | ||
| + | |||
| + | <math>H(s)= \int_{-\infty}^{\infty}h(T){e^{-sT}}\, dT</math> | ||
| + | lets assume that: | ||
| + | |||
| + | <math>y(t)= 3x(t)</math> | ||
| + | |||
| + | let <math>x(t)= \delta\,\!(t)</math> | ||
| + | |||
| + | <math>h(t)= 3\delta\,\!(t)</math> | ||
| + | |||
| + | '''system function:''' | ||
| + | |||
| + | <math>H(s)=\int_{-\infty}^{\infty}3\delta\,\!(T)){e^{-j\omega\,\!T}}\, dT</math> | ||
| + | |||
| + | <math>H(s)=3{e^{-j\omega\,\!0}}</math> | ||
| + | |||
| + | <math>H(s)=3</math> | ||
| + | |||
| + | '''PART B: | ||
| + | ---- | ||
| + | ''' | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | The coefficients are as follows: | ||
| + | |||
| + | <math>a_1 = \frac{-j}{2}</math> | ||
| + | |||
| + | <math>a_-1 = \frac{j}{2}</math> | ||
| + | |||
| + | <math>a_2 = 1 </math> | ||
| + | |||
| + | <math>a_-2 = 1 </math> | ||
| + | |||
| + | <math>a_k = 0</math> | ||
| + | |||
| + | <math>=\frac{-s}{2}+\frac{s}{2}+s+s</math> | ||
| + | |||
| + | <math>=2s</math> | ||
Latest revision as of 14:56, 26 September 2008
A Signal for which the output signal is constant times the input is referred as
an eigenfunction of the system,andthe amplitude is called the
system's eigenvalue'
PART A:
let the input be
$ x(t)= e^{st} $
we can determine the output using convolution integral
$ y(t)= \int_{-\infty}^{\infty}h(T)x(t-T)\, dt $
= $ \int_{-\infty}^{\infty}h(T){e^{s(t-T)}}\, dt $
= $ e^{st} \int_{-\infty}^{\infty}h(T){e^{-sT}}\, dt $
= $ H(s) e^{st} $
h(t) is the impulse response of the LTI SYSTEM H(s) is the system fuction
$ H(s)= \int_{-\infty}^{\infty}h(T){e^{-sT}}\, dT $ lets assume that:
$ y(t)= 3x(t) $
let $ x(t)= \delta\,\!(t) $
$ h(t)= 3\delta\,\!(t) $
system function:
$ H(s)=\int_{-\infty}^{\infty}3\delta\,\!(T)){e^{-j\omega\,\!T}}\, dT $
$ H(s)=3{e^{-j\omega\,\!0}} $
$ H(s)=3 $
PART B:
The coefficients are as follows:
$ a_1 = \frac{-j}{2} $
$ a_-1 = \frac{j}{2} $
$ a_2 = 1 $
$ a_-2 = 1 $
$ a_k = 0 $
$ =\frac{-s}{2}+\frac{s}{2}+s+s $
$ =2s $
