Signals and Systems: Response of LTI systems to Complex Exponentials

1. Jul 9, 2007

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.

2. Jul 9, 2007

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

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

Last edited: Jul 9, 2007
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