0
\$\begingroup\$

enter image description here

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)?

\$\endgroup\$
2
\$\begingroup\$

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)\$.

\$\endgroup\$
  • \$\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
0
\$\begingroup\$

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 $$ y_1(r)|_{r=t-T} = x_1(2r-5)|_{r=t-T} $$ $$ y_1(t-T) = x_1(2t-5-2T) $$

\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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