# What is the instantaneous frequency of this signal

Suppose we have the signal

$x(t) = \cos(\pi t) + \cos(\pi t^2) + \cos(\pi \cos(\pi t))$

How can I find the formula which gives the instantaneous frequency $f(t)$?

• If you want frequency (say, in Hz), then probably t stands for time (in seconds). Then your expression makes no sense because you have units (of time and time squared) inside your cosine functions. The consine functions need a dimension-less argument. – Juancho Jan 20 '17 at 23:33

Concerning the comment from Juancho, we can assume that the time is normalized to e.g. 1s, meaning that $t=1$ corresponds to $1s$. Then, we dont have a problem with the units (this a very common operation and not specific to your problem).

Now, regarding the "instantaneous frequency" (also check Wikipedia and another question on math.se), it is not rigorously defined. Due to the uncertainty principle you normally cannot assign an accurate specific frequency to an infinitely short interval of time (which would the term "instantaneous" correspond to). So, in general, the question for an "instantaneous frequency" needs to be asked with care and you need to understand that its interpretation might not be as straight-forward as it seems.

However, in case of a simple sine wave, consider the simple expression

$$x(t)=\cos(2\pi \cdot f\cdot t),$$

where we know that its frequency is $f$. The argument to the $\cos$ is $\phi(t)=2\pi f t$. If we calculate $\frac{d\phi(t)}{dt}=2\pi f$, which tells us the phase of the cosine changes with rate $2\pi f$. This term can be considered as the instantaneous angular frequency (i.e. frequency is $f$, as we would expect) of the sine wave.

Let's extend this expression to the general one

$$x(t) = \cos(\phi(t))$$

where $\phi(t)$ is the phase of the cosine. We can now define the rate of change of the phase as the instantanous angular frequency: $\omega(t)=\frac{d\phi(t)}{dt}$.

With this idea, you can calculate the instantaneous frequencies (there are multiple) of your signal

$$\phi_1(t)=\pi t, \quad\phi_2(t)=\pi t^2,\quad \phi_3(t)=\pi\cos(\pi t).$$

So, the instantaneous angular frequencies are given by

$$\omega_1(t)=\pi,\quad\omega_2(t)=2\pi t,\quad\omega_3(t)=-\pi^2\sin(\pi t).$$

I.e. your signal consists of three different instantaneuos frequencies.