Doubt on time invariant system

Question

1. Now I am delaying the output of a system (which takes $x \left( t \right)$ as input and gives $t \cdot x \left( t \right)$ as output) by $T$ then final output is:

Let's denote the output of the system, $t \cdot x \left( t \right)$, as $y_{1} \left( t \right)$. So, we have:

$$ y_{1} \left( t \right) = t \cdot x \left( t \right) \tag{1} \label{1}$$

Now, if you want to delay this output by a time delay $T$, the output of the delayed system, $y_{2} \left( t \right)$, will be:

$$ y_{2} \left( t \right) = y_{1} \left( t - T \right) = \left( t - T \right) \cdot x \left( t - T \right) \tag{2} \label{2} $$

So, the output of the system with the delayed output is given by $\left( t - T \right) \cdot x \left( t - T \right)$. This represents the original output, but it has been delayed by a time delay of $T$ units.

(2)I am first delaying the $x \left( t \right)$ by $T$ and giving it as input to the above system then what is the final output?

• Delayed Input: Delay the input signal $x \left( t \right)$ by $T$ to obtain $x \left( t - T \right)$.

• Apply delayed input to the system: Feed the delayed input $x \left( t - T \right)$ into the system with the output equation $y_{1} \left( t \right) = t \cdot x \left( t \right)$.

• Calculate the Output: To find the final output $y_{2} \left( t \right)$, you need to evaluate $y_{1} \left( t \right)$ at $t - T$, as the system output is $y_{1} \left( t - T \right)$. So, substitute $t - T$ for $t$ in the expression for $y_{1} \left( t \right)$:

$$ y_{2} \left( t \right) = \left( t - T \right) \cdot x \left( t - T \right) \tag{3} \label{3}$$

From the above statements I can conclude the system is time invariant. But it is not true $y \left( t \right) = t \cdot x \left( t \right)$ is a time-varying system.

Where did I make the mistake?

I want to know why we use these steps (algorithm) to find the time invariance in the first place. What is the relation between this algorithm and the definition of time invariance?

Please give a clear explanation and explain why the system, which takes input $x \left( t \right)$ and gives output $x \left( a t \right)$, where $a$ is constant, is time-varying after explaining the above question.

Answer 1

An operator, a system, $\mathcal L$ represented by multiplication with a time function, $\ell (t)$, for example, $\ell(t)=t$, is not time invariant. To see this, recall the definition of time invariance.

The operator $\mathcal L$ is called time invariant if for $y(t)=\mathcal L [x(t)]$ and for any $T$: $y(t-T)=\mathcal L [x(t-T)]$. Less formally, the operator $\mathcal L$ may depend on $x$ explicitly but on the time variable may only depend implicitly through $x$.

But writing $y(t)=\ell(t) x(t)$ you do get $y(t-T)=\ell(t-T)x(t-T)$ that is not by the definition of being time variant represented by the multiplier $\ell(t)$. Were it time invariant only the input is delayed not the operator, that is, $y(t-T) = \ell(t)x(t-T)$, which is not the case here.

Regarding time scaling, $x(at)$, it is a time variant operation exactly because of its explicit time dependence that can be seen when writing $x(a(t-T)) = x(at-aT)$ in which the output is delayed not by $T$ but by $aT$.