# What is the benefit of FFT when we're already sampling at an acceptable rate?

As I'm learning about DSP I can understand the process of taking a continuous signal into a discrete signal. I understand an oscillator producing different types of signals, including a periodic signals (especially sinusoids), phase shifts and amplitudes etc.,

I'm following a coursera class, and reading DSP theory.. Rather than blindly using the DSP libraries out there I want to know exactly why something is happening. My 1st roadblock is this:

1. If we already have enough samples of amplitudes collected at a reasonably sample rate (using Nyquist-Shannon) Why does it matter that a signal (Sinusoid) can be broken down into other sinusoids? Aren't the samples enough to reconstruct the original signal/sound?
2. What is advantageous by knowing that the Fast Fourier Transform can decompose a signal into multiple signals?
3. The sinusoids that make up a single sinusoid also have other sinusoids that compose them. If a signal has say 5 sinusoids, those 5 sinusoids, according to DFT if I'm correct, can also each have sinusoids.. Isn't this recursive or infinite? At what point can we say we have "enough" sinusoids to represent a signal.

Thanks in advance for the clarification & patience.

• Even as a beginner you should be able to formulate a meaningful title for your question. – Jazzmaniac Feb 2 '16 at 21:23
• @Jazzmaniac: Couldn't agree more... – jojek Feb 2 '16 at 21:39
• @Jazzmaniac Thnx for the feedback -- What is the benefit of FFT when we're already sampling at an acceptable rate? I'm still unclear. – sabzo Feb 3 '16 at 1:47
• The benefits of any particular representation or basis vector set depends on what you want to do. What do you want to do with your samples? – hotpaw2 Feb 3 '16 at 17:38

1. Not all periodic signals are sinusoids. Items 1 and 3 should not assume such. Not all sampling is done just for reconstruction. Look up other needs or uses for sampled signals.

2. A linear time-invariant system can behave differently to different frequency sinusoidal inputs (as provided by common solutions to certain low order differential equations), so it may be easier to figure how that system responds to other types of far more complicated inputs by breaking them down into the sum of simpler sinusoidal ones.

3. A DFT or FFT won't break down a pure (unmodulated) sinusoid that is integer periodic in aperture into other sinusoids (other than that a pure sinusoid of an arbitrary phase can be broken down into component cosine and sine sinusoids of that same frequency). But an FFT can "break down" other types of signals.

• I do understand not all periodic signals are linear, sorry If I made it seem otherwise. What I don't understand is why we NEED DFT or FFT? what about the FFT helps us transform a signal from analog to digital? If you could explain in "beginner speak", thnx. – sabzo Feb 3 '16 at 1:48
• That's a different question. You could try asking it (not in the comment section). – hotpaw2 Feb 3 '16 at 2:07

If we already have enough samples of amplitudes collected at a reasonably sample rate (using Nyquist-Shannon) Why does it matter that a signal (Sinusoid) can be broken down into other sinusoids? Aren't the samples enough to reconstruct the original signal/sound?

Why does it matter? This answer to another question is mostly about the continuous-time Fourier transform, but most of it applies to the DFT also.

Yes, the samples are enough to reconstruct the analog waveform if it was bandlimited enough when sampled. If the reason you need a discrete time signal does not involve the DFT, then there is no need to use it.

What is advantageous by knowing that the Fast Fourier Transform can decompose a signal into multiple signals?

The Fast Fourier Transform (FFT) is an algorithmic implementation of the Discrete Fourier Transform (DFT). The DFT's decomposition of a signal into complement (complex) sinusoids is useful for some things we want to do with the signal: filtering, for example, is the main reason we might want to transform to the Fourier domain.

The sinusoids that make up a single sinusoid also have other sinusoids that compose them. If a signal has say 5 sinusoids, those 5 sinusoids, according to DFT if I'm correct, can also each have sinusoids.. Isn't this recursive or infinite? At what point can we say we have "enough" sinusoids to represent a signal.

For a signal of length $N$ then $N$ (complex) sinusoids are enough to represent them.

The trick with the $N$ (complex) sinusoids used by the DFT is that they are periodic. That is the satisfy:

$$x[n] = x[n+P]$$

where $P$ is an integer and is the period.

No other discrete time (complex) sinusoids, apart from these $N$, satisfy this criterion.