Why is the nyquist frequency at $\frac{N}{2}$ (or $\lfloor \frac{N}{2}\rfloor$) for the DFT and what is the value for $X[k_{N/2-1}]$

Question

For the definition of the DFT we have

• $X[k] = \sum\limits_{n=0}^{N-1}x[n]\exp(- \frac{2 \pi i \cdot n}{N} k)$

Let's say for simplification that $N$ is even.

Then

• $k_{N/2-1} = \frac{N}{2}$ is considered as being the nyquist frequency
• and $X[k_{N/2-1}] = 0$

What is the inuitve and mathematical reason for the two statements above?

I can see that for every value of $k$ the complex sinewaves are getting faster and at some point they are too fast and we get an aliasing effect. The sampling rate of the complex sine waves could be interpreted as $\frac{1}{N}$ and $N$ is mostly equal to the fundamental period of $x[n]$

Answer 1

The Nyquist bin $X[N/2]$ (for even $N$) is real-valued for real-valued sequences $x[n]$, but it is generally not equal to zero. It is given by

$$X[N/2]=\sum_{n=0}^{N-1}x[n]e^{-j\pi n}=\sum_{n=0}^{N-1}x[n](-1)^n$$

So you just change the sign of every other time domain sample and then you compute the sum. That value can be zero but it doesn't need to be zero.

The frequency index $N/2$ corresponds to Nyquist (for even $N$) because the related complex exponential has maximum frequency, i.e., it is an alternating sequence of constant amplitude:

$$e^{-\frac{j2\pi n}{N}\cdot\frac{N}{2}}=e^{-j\pi n}=(-1)^n$$

Answer 2

@OuttaSpaceTime To answer the question in your comment: For an even-N point DFT, the frequency associated with the k = N/2 DFT bin is Fs/2 where Fs is the data sample rate measured in hertz.

Answer 3

No, the 1/N factor in the definition of an inverse DFT is totally unrelated to an notion of "frequency" with regard to the DFT.

In signal processing the notion of "frequency" is the ratio $\text{angular change}\over\text{a time interval}$. In DSP, frequency is often measured in $\text{radians}\over\text{sample time}$, or just $\text{radians}\over\text{sample}$ where "sample" is the time duration between signal samples. In the DFT we specify a complex exponential as $e^{-j2{\pi}kn/N}$ measured in radians/sample. So..., what are the 'angular change' and the 'time interval' values (what is the "frequency") of our $e^{-j2{\pi}kn/N}$ complex exponential?

In our $e^{-j2{\pi}kn/N}$ complex exponential the angle changes by $2{\pi}k/N$ radians for each sample and the time duration between samples is $1/F_s$ seconds. So our complex exponential has a frequency of $\text{angular change}\over\text{a time interval}$=${2{\pi}k/N\over {1/F_s}}=2{\pi}kF_s/N$ radians per second. That frequency of our DFT's complex exponential measured in cycles per second is ${2{\pi}kF_s/N\over 2{\pi}}=kF_s/N$ Hz.

Answer 4

@OuttaSpaceTime, When we perform an N-point DFT we're simultaneously computing N correlations of the input signal with N different complex exponential sequences. So, ask yourself: What are the frequencies of those complex exponentials? That is, how many times do N samples of the kth complex exponential rotate around a circle (how many times do N samples of the kth complex exponential rotate through an angle of two pi radians)?

The k=1 complex exponential cycles around a circle one time over N samples. It’s ‘cyclic’ frequency is 1*Fs/N = Fs/N Hz where Fs is the data sample rate measured in hertz.

The k=2 complex exponential cycles around a circle two times over N samples. It’s ‘cyclic’ frequency is 2*Fs/N Hz

The final k=N-1 complex exponential cycles around a circle N-1 times over N samples. It’s ‘cyclic’ frequency is (N-1)*Fs/N Hz

So we can say, the frequency of the complex exponential for the kth DFT bin is k*Fs/N Hz.

To substantiate what I’m claiming here, the DFT can be viewed as a bank of complex-valued bandpass filters. That is, if you have a long sequence of x(n) input samples you can perform an N-point DFT of the x(0)-thru-x(N-1) input samples and retain the X(k) sample and assign it to be the first sample of an ‘R’ sequence. Next, perform an N-point DFT of the x(1)-thru-x(N) samples and retain the X(k) sample and assign it to be the second sample of the ‘R’ sequence. Then perform an N-point DFT of the x(2)-thru-x(N+1) samples and retain the X(k) sample and assign it to be the third sample of the ‘R’ sequence. And so on.

The ‘R’ sequence that you’ve computed is the output of a complex-valued bandpass filter whose center frequency is k*Fs/N Hz. (The frequency magnitude response of that bandpass filter is essentially sin(x)/x.)