# Why can two different signals have the same periodogram?

I have just started trying to self-study an intro course on signal processing. I have just been introduced to the periodogram and have a hard time understanding why completely different signals can sometimes have the same periodogram. Take for example the two signals in the below image:

It is stated that these two signals have the same periodogram because they have the same magnitude function. I'm having trouble understanding what a magnitude function is and why these signals have the same periodogram.

Any help would be greatly appreciated as I have no idea how to move forward!

Good question, which I'll try and answer without complete mathematical rigorousness, but focus more on the intuition behind.

First, let's define the periodogram of a real discrete signal $$x[n]$$ as the Discrete Fourier Transform of its auto-correlation function: $$\mathcal{F}\{x[n]*x[-n]\} = \big|X[k]\big|^2$$ Here, $$n$$ is the sample number, $$*$$ the convolution operator, $$\mathcal{F}$$ the Discrete Fourier Transform, $$X$$ the Discrete Fourier Transform (I know, both the transform and the result share the same name), $$k$$ the frequency bin number and $$\big|\cdot\big|$$ the absolute value operator. $$\big|X[k]\big|$$ is called the magnitude spectrum.

In plain english, the periodogram of a signal $$x[n]$$ is an estimate of the spectral power content of $$x[n]$$, in other words, an estimate of the power at each frequency that make up $$x[n]$$.

The question is, for two signals $$x_1[n]\neq x_2[n]$$, how can $$\big|X_1[k]\big|^2 = \big|X_2[k]\big|^2$$

This can be answered using a combination of intuition and mathematics using the definition for the Discrete Fourier Transform. Let's call $$f_s$$ the sampling frequency with which $$x_1[n]$$ and $$x_2[n]$$ were sampled.

As I'm sure you're aware, the Discrete Fourier Transform can decompose a time-domain signal into its individual frequency components.

Ask yourself, what frequencies make up a chirp signal $$x_1[n]$$? That one is easy to answer intuitively, since a chirp signal is a signal in which the frequency increases (or decreases) with time, starting at frequency $$f_0$$ and ending at $$f_1$$. Since our signal is sampled at $$f_s$$, let's conveniently set $$f_0 = 0\,\text{Hz}$$ and $$f_1 = f_s/2\,\text{Hz}$$: the chirp is made-up of all frequencies between $$0$$ and $$f_s/2\,\text{Hz}$$ (discarding frequency resolution considerations here). Assuming the amplitude remains constant and arbitrarily setting it to $$1$$, that would give you, without the need for complicated mathematics: $$\big|X_1[k]\big|^2 = 1$$

Less intuitive question: what frequencies make up an impulse signal? Well, the answer is all of them. To show this, let's use the Discrete Fourier transform: $$X[k] = \sum_{n=0}^{N-1}x[n]e^{-j2\pi kn/N}$$ and compute it for $$x_2[n] = \delta[n-m] = \begin{cases}1 &n=m\\ 0 &n\neq m\end{cases}$$

$$X_2[k] = \sum_{n=0}^{N-1}\delta[n-m].e^{-j2\pi kn/N} = e^{-j2\pi k m/N}$$

The periodogram does not care about phase, it is only interested in the (squared) magnitude of $$X_2[k]$$: $$\big|X_2[k]\big|^2 = \big|e^{-j2\pi k m/N}\big|^2 = 1$$

So here you have it: $$\big|X_1[k]\big|^2 = \big|X_2[k]\big|^2$$

• Thanks alot for the indepth answer. While it made some things clearer i still can't seem to understand alot. At some point, for the impulse signal, we see that $x[n]$ is close to 15 from the graph above, how come then that the magnitude of $X_2[k]$ isnt 15 too? Commented Dec 16, 2022 at 12:43
• @maxi_knowledge95 That's a fair question. My answer deals with a perfect theoretical impulse signal of amplitude 1, whereas yours is a typical realizable impulse signal. But the bottom line is that although the periodograms won't be EXACTLY the same, they'll have the same spectral shape, which is what your textbook is really trying to convey.
– Jdip
Commented Dec 16, 2022 at 13:02

That's because the periodogram is expressed in terms of $$|X(f)|$$, thus it only cares about the magnitude of the spectrum.

A linear dependency between the phase and the frequency is a delay. In your case the chirp signal is an impulse with some dispersion. The periodogram is invariant to that, what is desirable. But it is also invariant to much more random things, the periodogram of a sample of a white random source would give roughly the same periodogram.

But the explanation to all of that is in the fact that periodogram ignores phase.

• The last line is the crux of the matter. If you know only the phases and not the amplitudes you can still reconstruct a significant part of the original data. A Golden Oldie paper on this is Alan V. Oppenheim, "The Importance of Phase in Signals", Proc. IEEE vol 69, May 1981. Download from rle.mit.edu/dspg/documents/ImportancePhaseSignals_1981.pdf Commented Dec 23, 2022 at 0:47
• @JonesTheAstronomer Was skeptical at first of this comment, but confirmed with a fairly complex image, and the reference is nice. Surprising considering I often hear it go the other way around, "most of it's in modulus", and I'd expect for images, which are concentrated in lows, to be greatly distorted by high freqs. Also tried 1D. It rather makes sense from information standpoint, phase takes twice the bits. Commented Dec 24, 2022 at 5:37

In view of the comment by @overlordgolddragon I thought that the following two slides (from a talk I gave about image data reconstruction in 2012) might add substance to remarks regarding the roles of amplitudes and phases in Fourier analysis.