I wish to find the maximum frequency of $\sin(\varphi(t))$, where $\varphi(t)=u(t)$. We have $\omega(t)=d\varphi(t)/dt=\delta(t)$. What is the the maximum instantaneous frequency $\omega_{\max}$?

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    $\begingroup$ Hint: $\sin(u(t)) = \begin{cases}\sin(0), & t < 0,\\\sin(1), & t > 0,\end{cases} = [\sin(1)]\cdot u(t)$. So we get .... $\endgroup$ – Dilip Sarwate Jun 4 '14 at 4:39
  • $\begingroup$ not meaning to pile on, but for continuous $t$, $\delta(t)$ is the Dirac delta, not the Kronecker delta, $$ \delta[n] \triangleq \begin{cases} 1 & n=0 \\ 0 & n \ne 0 \end{cases} $$ which is defined for only integer $n$. if the word "instantaneous" in the title question was replaced by "component", there could be a plausible answer to that question. $\endgroup$ – robert bristow-johnson Jun 4 '14 at 16:26

It's not very sensible to even try to assign an instantaneous frequency to signals like yours.

Your specific example fails for more fundamental reasons than usual however, namely for the fact that the Dirac delta is not a function but something mathematicians call a "distribution".

This for example implies that we cannot assign a value to $\delta(0)$ even though we know $\delta(x)=0$ for all $x\neq0$. The definition of the delta distribution is, that $\int_\mathbb{R} \delta(x) f(x) dx=f(0)$. From here it follows that $\delta(0)$ cannot have a finite value. However, stating that it is "infinite" there is also not enough. There's simply no way to think of any value at $0$ that makes sense.

So the answer to your question is that the instantaneous frequency is $0$ almost everywhere, and undefined at $t=0$. So the maximum (and minimum!) value of $\omega$ is $0$!

You can avoid this situation by replacing the unit step with a smooth approximation of a unit step. With this you can give your maximum instantaneous frequency any value you desire, even if you take a limit of a function sequence that results in the unit step itself. So the answer is absolutely arbitrary.


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