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Say we have an integrator $y(t)= \int_{- \infty}^{t} x(t) dt$

It is incorrect to use the following method to determine if it is time variant/invariant?

First, when we have input signal $x(t),$

$y_1(t)= \int_{- \infty}^{t} x(t) dt$

Then, we shift $y(t)$ by $t_0$

$y_1(t-t_0)=\int_{- \infty}^{t-t_0} x(\tau-t_0) d\tau$

If we have input signal $x(t-t_0),$ we get

$y_2(t)=\int_{- \infty}^{t} x(\tau-t_0) d\tau$

And since $y_1(t-t_0)\neq y_2(t)$

Then the system must be time variant.

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You have an answer right now, but if you want to see it using the time-shifting approach, you should do like this:

First consider the input-output relationship $$x_1(t)\mapsto y_1(t)=\int_{-\infty}^{t}x_1(\tau)d\tau$$

Then consider the output to the shifted input: $$x_2(t)=x_1(t-t_0)\mapsto y_2(t)=\int_{-\infty}^{t}x_2(\tau)d\tau=\int_{-\infty}^{t}x_1(\tau-t_0)d\tau$$

change of variable: $\tau'=\tau-t_0\Rightarrow\tau=\tau'+t_0,\,d\tau=d\tau'$. Also the integration bound for $\tau'$ will become $-\infty$ (for $\tau\to\infty$) and $t-t_0$ (for $\tau=t$). Hence,

$$y_2(t)=\int_{-\infty}^{t-t_0}x_1(\tau'+t_0-t_0)d\tau'=\int_{-\infty}^{t-t_0}x_1(\tau')d\tau'=y_1(t-t_0)$$

Hence $y_2(t)=y_1(t-t_0)$, and the system is TI.

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This is not correct. First, you have to write it like that:

$$ \begin{align} y(t) = \int_{-\infty}^t x(\tau)d\tau \end{align} $$

I.e. your variable in the integrand is different from the integral limits. Accordingly, your calculation will be different, as you will only replace $t$ with $t-t_0$ and not $\tau$.

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  • $\begingroup$ Thanks for the answer, but I'm still a little confused, the system is time variant but the method I used to prove it is incorrect, right? (sorry I'm relatively new to all this) $\endgroup$ – user281270 Nov 19 '16 at 19:23
  • $\begingroup$ No, the system is time-invariant. Once you dont replace $t$ with $t-t_0$ in the integrand, you will see it will be time-invariant. Another idea for the explanation is (if you are familiar with this): The integration can be understood as the convolution of a unit impulse with the signal. Since convolution is time-invariant, so is the integrator. $\endgroup$ – Maximilian Matthé Nov 19 '16 at 19:54

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