Coherent Sampling And The Distribution Of Quantization Noise

Question

I have a question concerning the distribution of noise in the frequency domain in case of Coherent Sampling. I read up about Coherent Sampling and understood that in order for a frequency $f_{in}$ to be located on one single bin of the DFT, the period or a multiple of it must fit exactly into the data window of the DFT, meaning that the following must be true:

$$ \frac{f_{in}}{f_s}=\frac{N_c}{N} $$ $N$ ... Record Length, $N_c$ ... Number of cycles in data window

Furthermore in the slides of a course about Mixed-Signal Processing I am attending it is mentioned that $N_c$ and $N$ need to be coprime in order for the quantization noise to be distributed uniformaly. Otherwise the quantization noise would show a discrete spectrum. Why is that the case?

Answer 1

This is most easily understood by examining the error between the true sinusoid and the quantized sinusoid. Let's start by defining a simple sinusoid. For simplicity, I will assume a cosine, but the same arguments apply for a sinusoid with an arbitrary phase offset. The true sinusoid is defined in continuous-time as

$$ x(t) = \cos\left(2 \pi f_{in} t\right). $$

Now, we need to sample this signal with a period of $T_s = \frac{1}{f_s}$ to obtain a discrete-valued (but unquantized) signal

$$ x_D(n T_s) = \cos\left(\frac{2 \pi f_{in} n}{f_s}\right).$$

Using the relationship that you provided in your question, we can rewrite this as

$$ x_D(n T_s) = \cos\left(\frac{2 \pi N_c n}{N}\right). $$

From this it should be clear that this signal lies in two bins, specifically the bins at $N_c$ and $N - N_c$ of an $N$ point FFT. However, this signal is not realizable with finite precision numbers. So, we need to quantize it. Let us assume that we are quantizing to $q+1$ bits using two's complement. A reasonable approximation to the quantized signal is given by

$$ x_Q(n T_s) = 2^{-q} \left\lfloor 2^{q} \cos\left(\frac{2 \pi N_c n}{N}\right) \right\rfloor.$$

Another way to represent this would be to represent the sinusoid added with an error term (i.e., the quantization noise $\epsilon_q(n T_s)$). This is captured in the following:

$$ x_{D}(n T_s) = x_Q(n T_s) + \epsilon_q(n T_s). $$

Now, let's consider what $\epsilon_q(n T_s)$ looks like for the two cases you mentioned. That is, the coprime case and the common factor case. Let's rearrange this slightly so that we can see things a little better,

$$ \epsilon_q(n T_s) = \cos\left(\frac{2 \pi N_c n}{N}\right) - 2^{-q} \left\lfloor 2^{q} \cos\left(\frac{2 \pi N_c n}{N}\right) \right\rfloor. $$

Now, consider the case where $N_c$ and $N$ share a common factor, $M$. Let's say $M$ is the greatest common factor. In this case, $N_c = M K_c$ and $N = M K$. Using this and the expression derived for $\epsilon_q(n T_s)$, we can show that

$$ \epsilon_q(n T_s) = \cos\left(\frac{2 \pi K_c n}{K}\right) - 2^{-q} \left\lfloor 2^{q} \cos\left(\frac{2 \pi K_c n}{K}\right) \right\rfloor, $$

and from this, we can see that

$$ \epsilon_q(n T_s) = \epsilon_q(n T_s + K T_s) $$

since

$$ \cos\left(\frac{2 \pi K_c n}{K}\right) = \cos\left(\frac{2 \pi K_c (n + K)}{K}\right).$$

In other words, the sequence is periodic with period $K < N$ since $K$ and $K_c$ are coprime. This means that all of the information in this signal can be captured with a $K$ point FFT. This, in turn, means that all of the energy from this signal is contained at the equivalent bins in the $N$ point FFT. These are the bins at $0, M, 2M, ... (K-1)M$. All of the other bins are zero-valued. This is why you see a discrete spectrum as shown in the slides you mentioned.

Now, if $N$ and $N_c$ are coprime, then the sequence is non-sinusoidal and aperiodic (or viewed another way, it is periodic with period $N$). There is no shorter period embedded in the sequence as there was with the common factor case. Therefore, all bins are required to represent the signal.

Answer 2

In addition to the good answer from hops, have a look at the following python code, which can illustrate this behavior:

Fs = 700.0
T = 1.
N = Fs*T
f0 = 5# change to something else, how you want
Nc = f0*T
t = np.arange(0, T, 1/Fs)
signal = np.sin(2*np.pi*f0*t)

print N, Nc


B = 2 # number of bits used for quantization
A = 1. # (one-sided) amplitude

delta = (2*A) / (2**B-1)
threshs = -A + delta * np.arange(2**B)

def quantize(signal, threshs):  
    # performs the quantization (for newer python version, you can have a look at numpy.digitize)
    T = threshs.reshape((1, -1))
    S = signal.reshape((-1,1))
    diff = S - T

    quantized = threshs[np.argmin(abs(diff), axis=1)]
    return quantized

quantized = quantize(signal, threshs)
qnoise = quantized - signal

plt.subplot(121)
plt.plot(t, signal)
plt.plot(t, quantized, 'r-')
plt.title('Fs=%.0f, f0=%.0f, N=%.1f Nc=%.1f' % (Fs, f0, N, Nc))
plt.plot(t, qnoise, 'g-')

plt.subplot(122)
f = np.linspace(-Fs/2, Fs/2, len(t), endpoint=False)
spec = lambda x: 20*np.log10(abs(np.fft.fftshift(np.fft.fft(x))))
plt.plot(f, spec(signal), label='signal')
plt.plot(f, spec(qnoise), label='noise')
plt.legend()
plt.xlim((0, Fs/2))
plt.ylim((-50, 100))

Here are two outputs, for different Fs:

As visible, the peak for the 5Hz signal is very clear, for both sampling frequencies. But, the noises look different. When N and Nc are not coprime, the noise is quite discrete (i.e. it shows strong peaks), but still, it is distributed everywhere in frequency domain. Looking at the coprime example (5 and 573), we see a bit more distributed noise, but still it exhibits strong peaks in the frequency domain.

Looking at the time-domain realization of both noises, it can be seen that both exhibit a decent periodicity. This periodicity creates the peaks in the frequency domain, regardless of whether the signal contains a full period of the noise or not.

So, with the above results, I would actually question the mentioned slides, which state the noise is uniformly distributed in the frequency domain. It is not, as any periodic time-domain signal will create periodic quantization noise, and hence create a (roughly) discrete noise spectrum.

Edit: This is the result with 16 bit quantization (i.e. B=16):

As visible, the effect is very different from the coarse quantization with 2bits above: For coprime numbers, the noise looks uniformly distributed. For matching numbers, the noise becomes discrete in frequency domain. This confirms the OPs claim. So, it depends strongly on the number of bits used for quantization.

Edit2: I have just increased the frequency of the signal to 250Hz. Now N=700 and Nc=250 (which are not coprime, both dividable by 50), the spectrum of the noise is still relatively uniform, but has some small peaks. So, the claim from the OP is really depending on the actual parameters. Only coprime-ness seems not to be sufficient. Anyway, it's an interesting thing to play around with.