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As the question states, is it possible to predict the peak value of a time-domain signal given its frequency-domain spectrum?

Since the time-domain signal is just the sum of the individual sinusoids making up the frequency domain spectrum (i.e. inverse Fourier transform), I have been reading some answers on estimating the peak amplitude of a sum of sinusoids:

Max of sum of sinusoids with arbitrary frequencies

Sum of Sinusoids with Same Frequency = Sinusoid (proof)

Maintain constant total amplitude of sum of sines

Peak to peak amplitude of sum of sinusoidals (harmonic frequencies)

However, none of these questions really precisely ask the question I am asking (i.e. no specific reference to FT), and I wonder if there is some "magic" in the theory of Fourier transforms that allows this to be done. Note that I realize the trivial answer here is: take the inverse FT and find the peak! I also realize that it can be done analytically for certain functions (e.g. square wave, delta spike, etc).

To add a twist, I would also ask if there is a way to predict the peak value of the multiplication of two frequency-domain spectra? In this case, I wonder if there's any "magic" in convolution theory that would allow this to be done. If you convolve two time-domain signals, is there a way to predict the peak value of the convolved signal? This is equivalent to asking if you can find the peak time-domain value of the multiplication of two spectra.

Any info is appreciated.

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    $\begingroup$ Anything that exactly predicted the peak would be functionally identical to the Fourier transform. Without recreating the Fourier transform you could probably predict an upper bound on the peak value to energy ratio from some version of the system bandwidth, but it would only be an upper bound, not a firm prediction. $\endgroup$
    – TimWescott
    Jun 23, 2023 at 19:29
  • $\begingroup$ What's wrong with taking the inverse Fourier Transform? Yes, it's trivial, but it works! Why make it more complicated. Look at the opposite question: how you can find the peak frequency of a time domain signal. The answer here is also: take the FT. $\endgroup$
    – Hilmar
    Jun 23, 2023 at 19:40
  • $\begingroup$ If your application allows for some margin of error, there's something you can do. So, what is this for? $\endgroup$
    – Jazzmaniac
    Jun 23, 2023 at 19:51

2 Answers 2

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If you know of no special properties the signal might have, and if you don't want to compute an inverse Fourier transform, the only bound you have in terms of the Fourier transform is the following:

$$|x(t)|=\left|\int_{-\infty}^{\infty}X(f)e^{j2\pi ft}df\right|\le\int_{-\infty}^{\infty}|X(f)|df\tag{1}$$

Even if the integral on the right-hand side of $(1)$ is finite, for most signals this bound is not very tight.

However, there are some signals for which the bound $(1)$ is actually tight. One such example is the sinc function. Its maximum occurs at $t=0$, and its Fourier transform is real-valued and non-negative:

$$\max_t |x(t)|=x(0)=\int_{-\infty}^{\infty}X(f)df=\int_{-\infty}^{\infty}|X(f)|df$$

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So, let's try.

Suppose we have a Fourier representation of our signal, $X(\omega)$.

Suppose the time-domain signal, $x(t)$, has a unique peak at time $t_{\tt max}$.

We know that

$$ x(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} X(\omega) e^{j\omega t} d\omega $$

So, if we know the time of the peak, $t_{\tt max}$, then it's clear that the value of the peak is just

$$x(t_{\tt max}) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} X(\omega) e^{j\omega t_{\tt max}} d\omega $$

That doesn't help us much.

Suppose the peak isn't unique, for example when $x(t)$ is a single sinusoid. Then, it's just a matter of finding the phase of $x(t)$ from the peak of the Fourier transform:

$$ \cos(\omega_0 t + \theta) \leftrightarrow \pi[ e^{j\theta} \delta(\omega - \omega_0) + e^{-j\theta} \delta(\omega + \omega_0)]$$

Then the peak will be whenever

$$ (\omega_0 t + \theta)\ {\tt mod}\ 2\pi = 0 $$

But we then need to find the peak of the Fourier transform, to know what $\omega_0$ is so we can find $\theta$.

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  • $\begingroup$ Since the twin peaks of the Fourier transforms are impulses at $\pm \omega_0$, how do we determine $\theta$ from this information? $\endgroup$ Jun 24, 2023 at 3:03
  • $\begingroup$ @DilipSarwate Well, usually the positive frequency peak is the only one of interest, no? Or do you mean something different? $\endgroup$
    – Peter K.
    Jun 24, 2023 at 19:47

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