0
\$\begingroup\$

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)\$:

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

\$\endgroup\$
1
  • \$\begingroup\$ Thank you Jakub for making my equations more readable :) \$\endgroup\$ Mar 10, 2017 at 10:11

1 Answer 1

1
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.