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I seem to struggle with the notations and concepts used in signals & systems.

The book states, given that x2(t) = x1(t-T), x2(2t-5) is equal to x1(2t-5-T).

This might be a dumb question, but why wouldn't x2(2t-5) be x1(2t-5-2T)?


You were given

$$ x_2(t) = x_1(t-T)$$

If you apply the transformation \$t\to t-T\$ to the expression \$2t-5\$, you get \$(2t-5)-T=2t-5-T\$, not \$2t-5-2T\$.

Your expression for \$y(t)\$ involves multiplying the argument of \$x(t)\$ by 2, but that's irrelevant to the question of determining \$x_2(\cdot)\$ from \$x_1(\cdot)\$.

  • \$\begingroup\$ Thank you for your reply. Could you please explain why y1(t-T) becomes x1(2(t-T)-5)? \$\endgroup\$ – user207787 Sep 15 '19 at 15:00
  • \$\begingroup\$ What is the difference between transformation and substitution? I thought it was the same, but then, the answer given above is wrong. \$\endgroup\$ – Huisman Sep 15 '19 at 21:55
  • \$\begingroup\$ @Huisman, I didn't mean to make any distinction between transformation and substitution, and I don't know what the mathematical distinction is in any case. If substitution is a better fit, I can change it. \$\endgroup\$ – The Photon Sep 15 '19 at 23:40
  • \$\begingroup\$ @user207787, because you were given that for input \$x_1(t)\$, \$y_1(t)=x_1(2t-5)\$. I agree with the implication of Huisman's answer that the question could be asked much more clearly by using some dummy variables. \$\endgroup\$ – The Photon Sep 15 '19 at 23:43
  • \$\begingroup\$ @ThePhoton If you apply the substitution \$t = t-T\$ to the expression \$2t-5\$, you get \$2t-5-2T\$. So, if transformation and substitution would be the same, your answer is incorrect. I think you should rephrase it, but I'm not sure how. It took me time as well to clearly explain why in one case you need to subsitute \$t=t-T\$ and why in the other case substitute \$x_2(t) = x_1(t-T)\$ \$\endgroup\$ – Huisman Sep 16 '19 at 17:30

Note that (apart from the subscripts of x) the substitution of $$ x_2(2t-5) |_{x_2(t) = x_1(t-T)}$$ is not equal to the substitution of $$ x_1(2t-5) |_{ t = t-T }$$

The first substitutes \$x\$, the second equation substitutes \$t\$.

You could also write for the first equation:
$$ x_2(2t-5) |_{x_2(u) = x_1(u-T)}$$ where \$u\$ is just another letter chosen as dummy variable.

Now, $$ u= 2t-5 $$ and so

$$ x_2(2t-5) |_{x_2(u) = x_1(u-T)} = x(2t-5-T)$$

EDIT: more clarification

For a time-invariant system holds:
If you delay (or advance) the input, the output is similarly delayed (or advanced).

The textbook wants to prove whether the system is time-invariant or not, by using two input signals, \$x_1\$ and the time delayed version of \$x_1(t)\$ called \$x_2(t)\$. \$x_2(t)\$ is delayed by T, so: $$ x_2(t) = x_1(t-T) $$

Feeding these 2 inputs to the system, we get:
\$y_1(t)\$ is the output for the input \$x_1\$,
\$y_2(t)\$ is the output for the input \$x_2\$.

According above given definition, if \$y_2(t)\$ is a time delayed (by exactly T) version of \$y_1\$, the system is time-invariant. So, when $$ y_2(t) = y_1(t-T) $$ To prove the equation the textbook expresses both parts in terms of \$x_1(t)\$

Expressing \$y_2(t)\$
When we feed \$x_2(t)\$ to the system, the textbook's second equation states: $$ x_2(t) \rightarrow y_2(t) = x_2(2t-5) $$ Writing \$y_2(t)\$ in terms of \$x_1(t)\$ requires the substitution of \$ x_2(t) = x_1(t-T) \$.
We can also write this with another dummy variable \$u\$: $$ x_2(u) = x_1(u-T) $$

So, $$ y_2(t) = x_2(2t-5) |_{x_2(u) = x_1(u-T)} = x_1(2t-5-T) $$

Expressing \$y_1(t-T)\$
The textbook's first equation is: $$ y_1(t) = x_1(2t-5) $$ We can also write this with another dummy variable \$r\$: $$ y_1(r) = x_1(2r-5) $$ In order to get \$y_1(t-T)\$ we should substitute for \$r\$ $$ r = t-T $$ so $$ y_1(r)|_{r=t-T} = x_1(2r-5)|_{r=t-T} $$ $$ y_1(t-T) = x_1(2t-5-2T) $$

  • \$\begingroup\$ Thanks, that helped a lot. Could you please explain why y1(t-T) becomes x1(2(t-T)-5) instead of x1(2t-5-T)? \$\endgroup\$ – user207787 Sep 15 '19 at 15:29
  • \$\begingroup\$ @user207787 Please find updated answer. I hope it becomes more clear why which substitution is done. \$\endgroup\$ – Huisman Sep 15 '19 at 21:39

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