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Could someone help me understand how a simultaneous time shift on two separate functions is possible? I am having trouble linking a property to this solution. Given the function:

$$x(t) = \mathrm{e}^{-(t-3)}u(t-1)$$ $$ X(f)= \mathcal{F}\{\mathrm{e}^2\mathrm{e}^{-(t-1)}u(t-1)\}$$ $$ = \mathrm{e}^2\mathcal{F}\{\mathrm{e}^{-(t-1)}u(t-1)\}$$

This is the point that doesn't seem clear. The solution of this problem is to use the time shift property. However, the solution only shows one time shift. i.e.

$$ = \mathrm{e}^2\mathrm{e}^{-j2\pi f}\frac{1}{1+j2\pi f}$$

Is it ok to transform several functions that have the same time shift? I was under the impression that I need to shift each function. Will this always the case or does it only apply to a unit step function? Can someone please shine some light on this!? Thanks!!

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Think of the time shift as operation applied to the time variable (namely $t$) rather than the function. Time shift by $1$ means: wherever you see $t$, replace it with $t-1$. So, the shift of $$t^2\sin(2t)-\sqrt{3-t}$$ by $1$ is $$(t-1)^2\sin(2(t-1))-\sqrt{3-(t-1)}$$ which we would probably simplify to $$(t-1)^2\sin(2t-2)-\sqrt{4-t}$$

Same happens in your example: $e^{-(t-1)} u(t-1)$ is the time shift of $e^{-t} u(t)$ by $1$.

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  • $\begingroup$ Very well put. Thank you! $\endgroup$ Commented Mar 7, 2013 at 2:51

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