Question

Recall that the one-sided Laplace transform of x(t)is defined as x(s)-J x()e "atfor any complex numl 1-0 0C A special case of this is X(ia) x(t)e-ω'dt, which is called the one-sided Fourier transform (FT) of x( 1-0 transforms the time domain into the frequency domain; a domain often preferred by engineers, as it decom its various frequency components. Consider the following approximation of the unit impulse δφ : x(t)-[u, (t)-4(t-A)] / Δ , where Δ is the pulse width. (a) Use the definition of the Laplace transform to show that for anyy(t)<» Y(s), y(t-to)<>e s'°Y(s)

Answer 1

The laplace transform of y(t) is given by

$Y(s) = \int_{-\infty}^{\infty}y(t)e^{-st}dt$

To calculate Laplace transform of y(t-t0), let's use the definition of Laplace transform,

Let,

$y_1(t) = y(t-t_0)$

$Y_1(s) = \int_{-\infty}^{\infty}y_1(t)e^{-st}dt$

  $= \int_{-\infty}^{\infty}y(t-t_0)e^{-st}dt$

Let's substitute

$t - t_0 = p$

It gives,

$t = t_0 &plus; p \Rightarrow dt = dp$

The important thing is the limits of integration will remain same since both are infinity.

Put these values in above integration equation,

$Y_1(s)= \int_{-\infty}^{\infty}y(p)e^{-s(t_0&plus;p)}dp$

$= \int_{-\infty}^{\infty}y(p)e^{-st_0}e^{-sp} dp$

$= e^{-st_0}\int_{-\infty}^{\infty}y(p)e^{-sp} dp$

Again do substitution p = t in the integration,

$Y_1(s)= e^{-st_0}\int_{-\infty}^{\infty}y(t)e^{-st} dt$

$= e^{-st_0} Y(s)$

Hence, if

$y(t) \leftrightarrow Y(s)$

$y(t-t_0) \leftrightarrow e^{-st_0}Y(s)$