I am confusing reading book by Chaparro (Signal System with Matlab) in example 2.7. He gives a question, a RL circuit with unit-step source $ v(t) = u(t) $ will result a response,

$$ i(t) = (1- e^{-t}) u(t) $$

If the input changed $ v(t) = u(t) - u(t-2) $, we are asked to the new response of RL system.

In the given solution (see the figure), he only use superposition and time invariance principle to bring the following,

$$ i(t) − i(t − 2) = 2(1 − e ^ {−t} ) u(t) − 2(1 − e ^ {(t−2)} ) u(t − 2) $$

Based on my understanding on superposition and time invariance, it should be,

$$ i(t) − i(t − 2) = (1 − e ^ {−t} )u(t) − (1 − e ^ {(-t+2)} ) u(t − 2) $$

But, I've checked it with graphical method (convolution) and Matlab script, the given answer (plot of $i(t)$, $-i(t-2)$ and $ i(t)-i(t-2)$) is true. Here is how,

First I obtain $h(t)$ from $i(t)$ by derrivation,

$$i(t) = v(t) * h(t)$$

I got $h(t)=e^{-t}u(t)$. By convolving $h(t)$ with $i(t)-i(t-2)$ I got the same plot on the book (Using matlab conv).

Anyone can explain, how superposition and time-invariant principle can bring the answer?

Thanks in advance.

Screenshot of the book (from Google Book)

enter image description here

Plot of $i(t)$, $-i(t-2)$ and $ i(t)-i(t-2)$

enter image description here


1 Answer 1


The answer to your last question is no. Nobody can explain how that answer was obtained because it's wrong. The fact that it is wrong is obvious, because due the term $e^{(t-2)}$ (with a positive power of $t$!), the output will keep increasing indefinitely, which is impossible for a stable system (such as the given RL-circuit).

Your answer is the correct one, and you must have made a mistake while checking the book's answer.

  • $\begingroup$ Thanks for your answer Matt. I updated my question by adding screenshot of the book to prove I didn't made mistake checking the book. $\endgroup$
    – sugab
    Oct 26, 2015 at 1:19
  • $\begingroup$ @bagustris: But the given equation doesn't match the plot. The plot is correct, the equation is not. As mentioned in my answer, how should $e^{t-2}$ ever go to zero with increasing $t$? $\endgroup$
    – Matt L.
    Oct 26, 2015 at 9:09
  • $\begingroup$ @bagustris: Just to avoid any misunderstanding: I didn't believe that you made a mistake copying the book's answer, I thought that you made a mistake verifying the book's answer, because it is obviously wrong, so it can't be verified. $\endgroup$
    – Matt L.
    Oct 26, 2015 at 9:18
  • $\begingroup$ Thanks matt. So far, it's clear for me. The problem is equation as solution of the problem that I can't understood and it can't be verified. Solved. $\endgroup$
    – sugab
    Oct 26, 2015 at 11:08
  • $\begingroup$ @bagustris: if Matt's answer is the solution, please give it a green tick! $\endgroup$
    – Peter K.
    Oct 26, 2015 at 11:10

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