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I need help finding where I have made the error in this problem set by my control systems lecturer.

The Question: The output of a linear time invariant system for an input \$r(t)\$ equals \$c(t)\$. If the input signal is passed through a block with transfer function \$e^{-s}\$, and then applied to the system, what will the output, \$y(t)\$, be?

The answer should be \$y(t) = c(t) \cdot u(t-1)\$ but I get \$y(t) = u(t) \cdot c(t-1).\$

My working:

Transfer function one, \$H_1(s)\$, applies to the original system. Transfer function two, \$H_2(s)\$, applies to the system including the new block:

$$H_1(s) = \frac{C(s)}{R(s)}$$

$$H_2(s) = e^{-s} \cdot H_1(s) = \frac{e^{-s} \cdot C(s)}{R(s)} = \frac{Y(s)}{R(s)}$$

Therefore \$Y(s) = e^{-s} \cdot C(s)\$,

\$y(t)\$ is the inverse Laplace transform of \$Y(s)\$:

$$y(t) = u(t) \cdot c(t-1)$$

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  • \$\begingroup\$ Thank you Jakub for making my equations more readable :) \$\endgroup\$ – The Impossible Squish Mar 10 '17 at 10:11
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The Laplace property of time delay is: If L[f(t).u(t)] = F(s), then L[f(t-a).u(t-a) = e^(-as).F(s).
Both factors contain (t-1) in your case.

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