<jupyter_start><jupyter_text>Shift-in-sShift-in-s and inversion can be applied in two ways (forwad and inverse).The first way (forward) is to transform a time function mutiplied by an exponential term, e.g.,$$y(t) = te^{3t}$$At first glance we do not know the transform of expression, but if we ignore the exponential term, we are left with t, and we know that the Laplace transform of t is,$$\mathcal{L}\left[t\right] = \frac{1}{s^2}$$from shift-in-s we also know that by multiplying t with $e^{3t}$ shifts the Laplace transform of $t$, i.e., $\frac{1}{s^2}$ by 3. Therefore, $$\mathcal{L}\left[e^{at}y(t)\right]=Y(s-a)$$, $$\mathcal{L}\left[e^{3t}t\right] = \frac{1}{(s-3)^2}$$The second way (inverse) is to identify the inverse of a shifted Laplace transform,$$\mathcal{L}^{-1}\left[\frac{1}{(s-3)^2}\right]$$First we need to workout the 'unshifted version' of the transform, which is $\frac{1}{s^2}$, and we know the inverse of this expression is $t$. Therefore,$$\mathcal{L}^{-1}\left[\frac{1}{(s-3)^2}\right] = te^{3t}$$Shift the slider below to show what happens to the time function as its Laplace transform is shifted.  Shift-in-tSimilar to shift-in-s, shift-in-t could be also be applied in two ways, the forward way,$$\mathcal{L}\left[y(t-a)u(t-a)\right] = Y(s)e^{-as}$$remember here $y(t-a)$ is multiplied by $u(t-a)$ because we want to make sure that the shifted function is 0 for $t<a$.For example,$$\mathcal{L}\left[e^{2t}\right] = \frac{1}{s-2}$$if $y(t)$ was shifted by $a$, then we would get the following,$$\mathcal{L}\left[e^{2(t-a)}u(t-a)\right] = \frac{e^{-as}}{s-2}$$This is the concept that we used in class to find the transform of shifted pulses. Though in the example covered in class we did not have to multiply a shifted unit step function to another shifted unit step function as they are by definition 0 for $t<a$.The inverse way involves finding the inverse of a Laplace transform that contains an exponential term, e.g., $$\mathcal{L}^{-1}\left[\frac{e^{-as}}{s-2}\right]$$here if we ignore the exponential term we get $\frac{1}{s-2}$, the inverse of which is $e^{2t}$. Now due to the presence of the exponential term in the Laplace transform we need to shift $e^{2t}$ by $a$, therefore the shifted time function is,$$e^{2(t-a)}$$,but remember this is not a unit step or dirac delta function, so we need to multiply this expression by a unit step function shifted by the same amount,$$\mathcal{L}^{-1}\left[\frac{e^{-as}}{s-2}\right] = e^{2(t-a)}u(t-a)$$Use the visual demonstration below to see the relationship between a function in the time domain, $f(t)$, and it's Laplace transform, $F(s)$ in the Laplace parameter domain. You can apply a shift-in-s or shift-in-s using the sliders below to see how these shifts affect either the original function or its Laplace transform.<jupyter_code>%matplotlib inline

from sourcecode_ode import Laplace
Laplace()

# Depending on the function chosen for f(t):
#     n is the order f the polynomial
#     w (omega) is the angular velocity
#     a is the exponential coefficient

# Some things to note:
#     omega/n/a slider changes when shifting t
#     need shift in t slider to go back to 0 when using shift in s slider and vice versa
#     use the t lim and s lim sliders to set the horizontal plotting limits
#     use the F(s) lim to set the vertical plotting limits for right-hand plot only<jupyter_output><empty_output>
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