# what is the maximum instantaneous frequency of Dirac delta?

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 $\omega_{\max}$?

• Hint: $\sin(u(t)) = \begin{cases}\sin(0), & t < 0,\\\sin(1), & t > 0,\end{cases} = [\sin(1)]\cdot u(t)$. So we get .... – Dilip Sarwate Jun 4 '14 at 4:39
• 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. – robert bristow-johnson Jun 4 '14 at 16:26

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 $0$!