# Nyquist Theorem - Why unique frequencies upto Fs/2 and not Fs? f+Fs is start of Aliasing

If any frequency, f, displays an alias at f + Fs, This shows that unique frequencies have a range of Fs. Why does Nyquist theorem say that actually there is only half of this with unique frequencies below Fs/2?

Has it got something to do with negative frequencies?

The Fourier transform of any strictly real time domain signal will be conjugate symmetric (in the frequency domain). That means that a strictly real pure sinusoid with a frequency of just below Fs/2 will also have a conjugate symmetric component just above -Fs/2 in the frequency domain, thus spanning nearly the full range Fs. If you try to FT a real signal with a frequency above Fs/2, then it plus it’s negative frequency component would span a range wider than Fs, and thus the components would alias against other frequencies within that range.

If you are dealing with complex signals (IQ or quadrature data, for instance), then samples can represent (nearly) the full range from 0 to Fs, because the IQ input to an FFT can leave out (filter) any negative frequency components (possible with complex signals, not strictly real ones).

Since you stated you are still confused, the following graphics may help further understand the other two good answers:

First be sure that you understand what the sampling process (which is multiplying an analog signal by a stream of impulses in the time domain) looks like in both time and frequency; a stream of impulses in the time domain is a stream of impulses in the frequency domain. Multiplication in the time domain is convolution in the frequency domain, so if we sample by multiplying by an impulse every T seconds, we convolve our waveform with a frequency impulse spaced every 1/T Hz: An example of this is sampling a 3 Hz cosine (where a 3 Hz cosine is represented in the frequency domain by an impulse at + and - 3 Hz) with a sampling clock of 20 Hz as shown in the graphic below. The top spectrum represents the input (analog) spectrum of the cosine wave. The middle spectrum is the spectrum of our sampling process (an impulse every 20 Hz including DC), and the bottom spectrum is the digital spectrum resulting from convolving the previous two. Note that the analog spectrum was replicated in this process; so everything from $$-F_s/2$$ to $$+F_s/2$$, what we call the "first Nyquist zone", is duplicated centered on every multiple of $$F_s$$. (Where $$F_s$$ is the sampling rate). Thus the unique spectrum is completely defined by what is between $$-F_s/2$$ to $$+F_s/2$$, but we could easily show instead the spectrum for 0 to $$F_s$$, so it is somewhat arbitrary what slice of the spectrum we choose.

Note for real signals, the negative frequency components are the complex conjugate of the positive frequency components, so in that case not unique: they are completely defined by the positive frequency components so in that case that is all we need to present to provide all relevant information (the spectrum from DC to $$F_s/2$$).

That is not the case for a complex signal, as explained by hotpaw2 and shown in the graphic below. • Beautiful illustration! – Gilles Nov 20 '18 at 23:27
• Vielen Dank Gilles! – Dan Boschen Nov 21 '18 at 1:49

Let's take specific numbers instead of symbols.

Suppose that the signal $$x(t)$$ being sampled is a pure sinusoid (extending from $$-\infty$$ to $$\infty$$) at $$1200$$ Hz, say $$\cos(2\pi\cdot 1200 t)$$, $$-\infty < t < \infty$$. (For the record, the Fourier transform of this signal is a pair of impulses at $$\pm 1200$$ Hz but we won't be needing to think about this very much). Suppose that we are sampling at a rate of $$1000$$ Hz, one sample every millisecond. Then, \begin{alignat}{6} x &= x(0) &&= \cos(2\pi\cdot 1200 \cdot 0) &&= \cos(0) &&= \cos(0)\\ x &= x(0.001) &&= \cos(2\pi\cdot 1200 \cdot 0.001) &&= \cos(2\pi\cdot 1.2) &&= \cos(2\pi \cdot 0.2)\\ x &= x(0.002) &&= \cos(2\pi\cdot 1200 \cdot 0.002) &&= \cos(2\pi\cdot 2.4) &&= \cos(2\pi \cdot 0.4)\\ x &= x(0.003) &&= \cos(2\pi\cdot 1200 \cdot 0.003) &&= \cos(2\pi\cdot 3.6) &&= \cos(2\pi \cdot 0.6)\\ \end{alignat} and so on. More generally, we have that $$x[n] = \cos\left(2\pi \cdot 1200\frac{n}{1000}\right) = cos(2\pi\cdot 1.2n) = \cos(2\pi \cdot 0.2n).$$ But notice that if instead we had been sampling $$\cos(2\pi \cdot 200t)$$ at $$1000 Hz$$, we would have gotten the same set of sample values as shown above : the $$n$$-th sample (at $$1000 Hz$$ sampling rate) of $$\cos(2\pi \cdot 200t)$$ is precisely $$x[n] = \cos\left(2\pi \cdot 200\frac{n}{1000}\right) = \cos(2\pi \cdot 0.2n).$$ In short, from the samples alone, we cannot tell whether we are sampling $$\cos(2\pi\cdot 1200 t)$$ at $$1000$$ Hz or sampling $$\cos(2\pi \cdot 200t)$$ at $$1000 Hz$$. Thus, the sinusoid at $$1200$$ Hz has been aliased into the sinusoid at $$200$$ Hz. More generally, all sinusoids at frequencies $$1000m + 200$$ Hz will alias into the sinusoid at $$200$$ Hz. For the record, I note that $$0 \leq 200\leq 500$$ where $$500$$ is half the sampling frequency $$F_s$$ of $$1000$$ Hz.

But what about a sinusoid at frequency $$800$$ Hz that we sample at $$1000$$ Hz? (Note that this frequency is between $$F_s/2$$ and $$F_s$$.) Well, proceeding as above, we have that

\begin{alignat}{6} x &= x(0) &&= \cos(0) &&= \cos(0)\\ x &= x(0.001) &&= \cos(2\pi\cdot 0.8) &&= \cos(2\pi \cdot 0.8) &&= \cos(2\pi \cdot(-0.2))\\ x &= x(0.002) &&= \cos(2\pi\cdot 1.6) &&= \cos(2\pi \cdot 0.6)&&= \cos(2\pi \cdot(-0.4))\\ x &= x(0.003) &&= \cos(2\pi\cdot 2.4) &&= \cos(2\pi \cdot 0.4) &&= \cos(2\pi \cdot(-0.6))\\ \end{alignat} and so on. More generally, we have that $$x[n] = \cos\left(2\pi \cdot 800\frac{n}{1000}\right) = cos(2\pi\cdot 0.8n) = \cos(2\pi \cdot (-0.2)n).$$

In short, we cannot tell from the samples if we are sampling $$\cos(2\pi\cdot 800 t)$$ at $$1000$$ Hz or sampling $$\cos(2\pi \cdot (-200)t)$$ at $$1000 Hz$$. The sinusoid at $$800$$ Hz has been aliased to a sinusoid at $$-200$$ Hz.

"But, but, but," you exclaim, "I don't believe in negative frequencies, and neither should you. Do you have an explanation that doesn't involve negative frequencies at all?" Not to worry. Since $$\cos(-x) = \cos(x)$$, that sinusoid at $$-200$$ Hz is indistinguishable from a sinusoid at $$200$$ Hz. That is,

frequencies between $$F_s/2$$ and $$F_s$$ also alias into frequencies between $$0$$ and $$F_s/2$$.

The straightforward way of remembering what happens is to note that

Aliased frequency is the absolute difference between the actual signal frequency and the nearest integer multiple of the sampling frequency.

For a $$1200$$ Hz sinusoid, the frequency difference between $$1200$$ and $$1000$$ is $$200$$; for a $$3200$$ Hz sinusoid, the frequency difference between $$3200$$ and $$3\times 1000$$ is $$200$$; for the $$800$$ Hz sinusoid, the frequency difference is $$-200$$. All three frequencies alias to $$200$$ Hz (for those who refuse to believe in negative frequencies). But more generally, if you are broad-minded enough to accept negative frequencies as a concept, then note that for each nonzero integer $$k$$, the frequency band $$\left[kF_s - \frac 12 F_s, kF_s + \frac 12 F_s\right]$$ (centered at $$kF_s$$, an integer multiple of the sampling frequency) aliases to the low-pass frequency band $$\left[- \frac 12 F_s, \frac 12 F_s\right]$$.

Assume you have two continuous-time sinusoidal signals:

$$x_1(t) = e^{j2\pi f t} \\ x_2(t) = e^{j2\pi (f + f_s) t}$$

These are two pure tones whose frequencies differ by $$f_s$$. If we then sample each continuous-time signal at this rate $$f_s$$, then we have the discrete time signals:

$$x_1[n] = x_1(nT) = x_1\left(\frac{n}{f_s}\right)= e^{j2\pi f \frac{n}{f_s} } \\ x_2[n] = x_2(nT) = x_2\left(\frac{n}{f_s}\right)= e^{j2\pi (f + f_s) \frac{n}{f_s} }$$

We can do a bit more work to simplify them to:

$$x_1[n] = e^{j2\pi \frac{f}{f_s} n} \\ x_2[n] = e^{j\left(2\pi \frac{f}{f_s}n + 2 \pi n\right)} = e^{j2\pi \frac{f}{f_s}n}e^{j2\pi n}$$

Note that for all $$n$$, $$e^{j2\pi n} = 1$$, so:

$$x_2[n] = e^{j2\pi \frac{f}{f_s} n}$$

It is clear then, that the sampled signals $$x_1[n]$$ and $$x_2[n]$$ are identical, even though the continuous-time signals that were used to generate them were different. This is the essence of aliasing with respect to the Nyquist rate: if all of your frequency content is below the Nyquist rate, you will never have aliasing. Otherwise, you can end up with signals that are completely different in continuous time, but look identical once sampled.

Aside: This does not necessarily mean that if you're sampling a signal at rate $$f_s$$, all of your frequency content must be below $$f_s/2$$. There are more advanced techniques like bandpass sampling that can relax the constraint under certain conditions.

• I see this. But this has just shown that frequencies of difference F_s will alias but whats the proof that frequencies below F_s and above F_s/2 will alias too? @Jason – Natalie Johnson Nov 20 '18 at 16:03
• No, there are multiple ways to look at it, because, instead of using the band $[-f_s/2, f_s/2]$, it's perfectly acceptable to use $[f_s/2, 3f_s/2]$, for example. No two frequencies in that interval alias with one another, even though they are all positive. This is the notion behind bandpass sampling that I linked above. – Jason R Nov 20 '18 at 16:43
• That’s correct, there is no single region of unique frequencies; the concept is usually described as different Nyquist zones. It’s unclear exactly what you’re trying to ask. I’ve described the concept about as fully as I can. – Jason R Nov 20 '18 at 16:57
• The answer is that your assumption is not correct, as I’ve explained. – Jason R Nov 20 '18 at 17:00
• @NatalieJohnson If you flagged this answer (and another in a different post) as low-quality, please don't -- use the downvote button instead. An example of a low-quality answer would be one that only provides a link to an answer, or a single out-of-context equation, etc. – MBaz Nov 20 '18 at 17:04