# Time-invariant system question

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)\$$.

• Thank you for your reply. Could you please explain why y1(t-T) becomes x1(2(t-T)-5)? – user207787 Sep 15 '19 at 15:00
• What is the difference between transformation and substitution? I thought it was the same, but then, the answer given above is wrong. – Huisman Sep 15 '19 at 21:55
• @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. – The Photon Sep 15 '19 at 23:40
• @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. – The Photon Sep 15 '19 at 23:43
• @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 $x_2(t) = x_1(t-T)$ – 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 $$y_1(r)|_{r=t-T} = x_1(2r-5)|_{r=t-T}$$ $$y_1(t-T) = x_1(2t-5-2T)$$

• 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)? – user207787 Sep 15 '19 at 15:29
• @user207787 Please find updated answer. I hope it becomes more clear why which substitution is done. – Huisman Sep 15 '19 at 21:39