The first assignment for AD Converters (ADC) was all about noise, aliasing and distortion. Corrections to the exercises will be processed and marked in red.

1. Sampling and Jitter

How much SNR can be obtained if a signal of is sampled with a sample rate of with jitter? What happens with the if the sample speed is increased to at the same jitter specification? A band-pass filter limits the bandwidth to a range from to . What is the in that bandwidth for both sample rates

To calculate the SNR for a 10 MHz sampled signal, we use the SNR formula:

We take our signal for to be a sine wave , giving us . Our equation for the jitter error power is:

Substituting Eq. (2) into Eq. (1) for with and we get dB. This is the maximum achievable SNR limited by the jitter.

If we change the sample rate, the SNR would be unaffected, since our equation is not influenced by the sample rate. It depends on the input frequency. Our full bandwidth jitter noise would just be spread across a larger Nyquist band, so the layer of jitter noise would become thinner. This means that the spectral noise density, however, does go down.

Chapter 7.6 of Marcel Pelgrom’s A/D Converters states that the jitter has a flat spectrum, similar to thermal noise. Thus, the jitter error power is inversely related to the sample rate; . If we define the power spectral density of the noise as we get the following equation for the jitter error power:

where and are the higher and lower frequencies of the filter. Let us first solve for , since for a fixed input frequency this will be constant:

Now that we have solved for the noise density it can be proven that for a BW equal to the Nyquist bandwidth, the jitter error power rate will remain the same resulting in equal SNR since the term and the scaling factor of 2 will cancel out. To now find the SNR after filtering we substitute Eq. (4) into the integral of Eq. (3):

Now we can calculate the :

Simplified and in dB:

For a sample rate of 80 MS/s and 310 MS/s we get that the SNR is roughly equal to 83 dB and 89 dB respectively.


2. Aliasing

A signal source delivers a signal that consists of three components: at , and . The signal is processed by a sampling system with an unknown sample rate. The output contains in the band only frequencies at , , and . What sampling frequency was used? Complete the spectrum till .

To find aliasing we can use:

where are the harmonics and are multiples of . No distortion is mentioned, so no harmonics. We assume .

Let us analyze the spacing of the absolute frequencies to determine the sampling frequency:

The relationship between input and output is given by:

Assuming the most direct folding case with we get MHz, giving a Nyquist boundary of 0.455 MHz. Using Eq. (7):

(MHz) (MHz)
330.27
340.64
440.36
450.55
550.45
560.46

3. RF Oscillator

An RF oscillator at contains harmonic distortion products at 2x and 3x time the oscillation frequency. The available spectrum analyzer can measure up to , but has a input bandwidth sampling circuit with variable sampling rate up to . Advice how to measure the harmonic distortion.

To view signals within a 10 MHz BW we use subsampling and intentional aliasing to map higher frequencies down to the lower end of the spectrum.

For a fundamental frequency GHz, the harmonics are GHz and GHz. To ensure all harmonics remain distinct and within the 10 MHz band, we choose such that aliases to , causing harmonics to alias to and .

Setting the target alias MHz with :

Fundamental (2.45 GHz):

2nd Harmonic (4.90 GHz):

3rd Harmonic (7.35 GHz):

All three components fall within the 10 MHz bandwidth and are spaced 1 MHz apart, allowing for clear identification.


4. Subsampling

A very fast ADC samples a MHz sinewave at a Gs/s sample rate. The fast data stream is subsampled: every 16th sample is passed to the output. Draw the resulting spectrum including a 2nd and 3rd order distortion component at dB and dB resp. Give a reason for using this subsample arrangement.

If the ADC samples at GHz, but only every 1 6th output is passed to the output, then the output sample rate becomes

Our Nyquist frequency is

As we can see, this is close to the input sine with frequency , lets use the aliasing formula again. We take to be from to for the second and third harmonic.

Fundamental (750.366 MHz):