Signals and Systems: Response of LTI systems to Complex Exponentials

[mayan]

How will we evaluate the integral for calculation of Amplitude Factor for an LTI system for which the input and output are related by a time shift of 3, i.e.,

y(t) = x(t - 3)


The answer is: H(s) = e^(-3s)

I want to understand the Mathematics behind the evaluation of the integral.

Thanks.

 

[rbj]

i'm not sure how the title of your post is related to the content.

by definition, on the left side you have,

$$Y(s) = \mathcal{L} \left\{y(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} y(t) \,dt$$

and, also by definition,

$$X(s) = \mathcal{L} \left\{x(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} x(t) \,dt$$

and, it turns out that for LTI systems, after you show that the convolution operator and some impulse response is what relates the input x(t) to the output y(t), then

$$Y(s) = H(s) X(s)$$

where

$$H(s) = \mathcal{L} \left\{h(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} h(t) \,dt$$

and h(t) is the impulse response of the LTI system and completely characterizes the input/output relationship of the LTI system.

now, in your specific case,

$$Y(s) = \mathcal{L} \left\{y(t)\right\} = \mathcal{L} \left\{x(t-3)\right\} = \int_{-\infty}^{+\infty} e^{-st} x(t-3) \,dt$$

which, after you do a trivial substitution of variable of integration is

$$Y(s) = \int_{-\infty}^{+\infty} e^{-s(t+3)} x(t) \,dt = e^{-s \cdot 3} \int_{-\infty}^{+\infty} e^{-st} x(t) \,dt = e^{-s \cdot 3} \cdot X(s)$$

which means that

$$H(s) = e^{-s \cdot 3}$$.