Let $$x(t) = \frac{\sin t}{t} \qquad\text{and}\qquad y(t) = \frac{\sin 2t}{t}$$

  • Is it possible to find a LTI system such that $\mathcal{T}\{x(t)\} = y(t)$?
  • If not, what's the reason for that?

My try:

Assuming frequency response exists

$$Y(j\omega) = H(j\omega)X(j\omega) \tag{*}$$

Since $$\mathcal F\left\{\frac{\sin Wt}{\pi t}\right\} =\begin{cases} 1 &|\omega|\lt W \\0 &|\omega|\gt W\end{cases}$$ We have $$X(j\omega) =\begin{cases} \pi &|\omega|\lt 1 \\0 &|\omega|\gt 1\end{cases}$$ And $$Y(j\omega) =\begin{cases} \pi &|\omega|\lt 2 \\0 &|\omega|\gt 2\end{cases}$$ So it's not possible to find $H(j\omega)$ such that $(^*)$ holds. Because if $|\omega| \gt 1$ we have $X(j\omega) = 0$ which implies $Y(j\omega) = 0$ contradicting $Y(j\omega) = \pi$ for $1\lt |\omega| \lt 2$.

  • Is my argument right?
  • Also how we can prove the impossibility for the case frequency response doesn't exist? What's the intuition for this? I mean is it intuitively clear that we can't have $\mathcal{T}\{x(t)\} = y(t)$ for LTI system?
  • $\begingroup$ You are correct. For an argument as to why in the opposite direction (that is, $\mathcal T\{x(t)]=x\left(\frac t2\right)$ instead of $\mathcal T\{x(t)]=2x(2t)$ as in your question), a linear but not time-invariant system will work, see this recent question and its answers $\endgroup$ – Dilip Sarwate Jun 4 at 21:00
  • $\begingroup$ @DilipSarwate Thanks for your reply. How we can prove that when frequency response doesn't exist? $\endgroup$ – S.H.W Jun 4 at 21:52
  • 2
    $\begingroup$ Do you have an example of an LTI system for which the frequency response does not exist in mind? $\endgroup$ – Dilip Sarwate Jun 5 at 13:56
  • $\begingroup$ @DilipSarwate Yes, take a look: dsp.stackexchange.com/questions/67302/… I think we can construct other non-stable systems as well such that frequency response doesn't exist. $\endgroup$ – S.H.W Jun 5 at 16:55
  • $\begingroup$ No such LTI system can do this for you. You have a divide-by-zero problem. for $ 1 < \omega < 2$, your LTI system has to create non-zero energy at frequencies of where zero energy is input. $\endgroup$ – robert bristow-johnson Jun 5 at 17:54

Your Answer

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

Browse other questions tagged or ask your own question.