Let us say that there are some signals, and all of them are Fourier-transformable and are a unique, single-frequency sine wave.

Now, we combine (add) these signals into one signal. Then, we do Fourier transform into frequency contents.

Will Fourier transform show the frequency of each added signals?

Or will the transform show the different values as frequency contents?

What is the math/engineering behind these?


3 Answers 3


The Fourier transform is linear, which means that if you sum two signals, then also their spectrum will be the sum.

$$ \mathscr{F} [x(t) + y(t)] = \mathscr{F}[x(t)] + \mathscr{F}[y(t)] $$

If you consider the power, you can have two cases:

  1. The signals have different frequency
    Then their power just sums up, because: $$ P_{tot} = P_1 cos (\omega _1 t + \phi _1) + P_2 cos (\omega _2 t + \phi _2) $$

  2. The signals have the same frequency
    Then if they are in phase you have: $$ P_{tot} = P_1 cos (\omega _0 t + \phi _0) + P_2 cos (\omega _0 t + \phi _0) = (P_1 + P_2) \cdot cos(\omega _0 t + \phi_0) $$ If they're out of phase, the resulting power will be lower and precisely: $$ P_{tot} = P_1 cos (\omega _0 t + \phi _1) + P_2 cos (\omega _0 t + \phi _2) $$ and setting arbitrarily \$\phi_1 = 0\$ we obtain: $$ P_{tot} = P_1 cos (\omega _0 t) + P_2 cos (\omega _0 t + \Delta \phi) $$ which for \$\Delta \phi = \pi\$ gives the subtraction of the signals.

  • \$\begingroup\$ multiply signals in time => convolve signals in frequency domain (not sure what you mean about "reciprocal frequency and phase") \$\endgroup\$
    – Jason S
    Commented May 29, 2012 at 21:55
  • \$\begingroup\$ @JasonS nothing, I was confusing to something else: editing... \$\endgroup\$
    – clabacchio
    Commented May 29, 2012 at 22:06
  • \$\begingroup\$ One of the practical applications of this "feature" of the Fourier transom is the ability to detect the two DTMF tones coming from a touch-tone telephone keypad using a DSP. \$\endgroup\$
    – tcrosley
    Commented May 30, 2012 at 19:36

A fourier transform essentially shows the frequency spectrum of a signal. A sine wave is considered a pure frequency, so the fourier transform of a single sine would be a spike at its frequency. If the signal contains multiple sine waves, there will be a spike in the fourier transform for each one.

Keep in mind that such "spikes" in the fourier transform have some finite width and shape because the fourier integral is not performed for all time. The shorter the window the wider the spike has to be. Thinking of this another way, a sine wave changing in amplitude is no longer a pure frequency. The fourier transform's ability to resolve frequencies falls off with the reciprocal of the time window, so this represents a ambiguity in frequency space, hence even if just a single frequency is present the result will not be a infinitely thin spike.

  • 1
    \$\begingroup\$ What does it mean to say "the fourier integral is not performed for all time."? What is your definition of the Fourier integral? \$\endgroup\$ Commented May 29, 2012 at 12:43
  • \$\begingroup\$ The fourier integral is generally written for all time, but that is of course useless in any real case. Even when applying a fourier transform on a continuous signal, there is always some time limit outside which the signal is ignored. If not, we'd have to wait for the end of time to get a result. \$\endgroup\$ Commented May 29, 2012 at 13:04
  • \$\begingroup\$ Another way to see it is to assume that the signal is periodic, so infinite in time, and we integrate on the period to get the series; or, it's not necessarily periodic, but we integrate over an infinite time, like you say \$\endgroup\$
    – clabacchio
    Commented May 29, 2012 at 14:56

Think of linear systems as showing the results as independent signals but nonlinear sensing as a mixer with by-products. From Laplace transforms and Fourier analysis we get mixing, think of it as multiplying in the time domain and adding and subtracting in the frequency domain. I defer the math to my colleagues explanations or any book on Laplace transforms of a mixer output in the time domain or the analysis by Fourier.

So adding in time domain does not affect spectral content in spite of baseband apparent beat frequencies of the difference between spectral content of pure f1 and pure f2. Now our brains do a Fourier Analysis when we hear music instantaneously because of parallel processing of the signal. THe difference in beats of similar notes is an apparent result of nulling the signals then the phases cancel on the carrier rather than an intermodulation from mixing in a non-linear detector. Although our ears are also non-linear (See Fletcher-Munson curves) and logarithmic, they are pure linear systems with an adaptive detective of parallel nerve sensors and not actually a logarithmic detector in the passband which is non-linear. It is a logarithm detector in the Fourier transform of every continuous frequency.

Imagine a million Op Amp log detectors in parallel (See Analog computers) as our brains doing linear Fourier transforms with a 120 dB dynamic range. This may not match your application, but it shows how complex out brain's function with parallel processing and clearly distinguishing a pair of notes instantly.

I wish I could multiply as well as my calculator or do spectrogram of what I hear. Musicians can do this visually in their mind by hearing the sounds in a rather interesting way and have auditory memory like a strip chart with a "spectrogram" of Fourier content. The interest part is there is no conscious transform in our mind and we cannot imagine it in the time domain other than the envelope unless we digitize it onto music charts.

Engineers prefer Spectrum Analyzers. Many free tools on the web including Audacity's free Audio recording studio with a built in software Spectrum Analyzer.

I hope this helped even if I didn't have to use math.

I trust it raises more questions than you asked and answered at least one you intended. That's what learning is all about.

Thanks for listening ;)


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