ECE 4429a – Advanced Digital Signal Processing Fall
2023
Sampling & Quantization MATLAB experiments
September 27 and October 13, 2023
NOTES:
1. Lab reports must be submitted individually before the deadline: One week after the lab
date (i.e. October 4 and October 20, 11:30 pm). Lab reports must be uploaded to
OWL. Email or locker submissions will not be accepted.
2. Get your attendance marked by TA. Lab reports will not be graded without
attendance.
3. Create a folder labeled “Lab1” on your Drop Box space in the course OWL site and
upload your report and associated code to that directory.
4. Create a Word document that contains MATLAB figures, results, and your comments for
each problem. Upload this document to the Lab1 folder.
5. Check the volume control before playing the signals through the headphones!
Sampling Experiments
1. (5 marks) Download the files “Sampling1.m” and “mySpectrogram.m” from the “Course
Content” folder. Go through the “Sampling1.m” file and add comments to each line
indicating its function. Upload the commented file to the Lab1 folder.
Uploaded
What happens when you run this MATLAB script? Include in your report sample
spectrograms when the sinusoidal frequency is 1000 Hz, 4000 Hz, and 7000 Hz. Discuss
these results within the context of sampling theorem.
F= 1kHz
F= 4kHz
F= 7kHz
When you run this MATLAB script you find that the frequency as a function of
time is plotted for the frequencies from 1kHz to 7kHz in 1kHz interval steps, for the given
sine function. At 1kHz, we can observe that the peak amplitude occurs at 1kHz. This means
that the function was sampled accurately as 1kHz<4kHz. At 4kHz we have no signal. This
can be attributed to:
𝑛
0.25 sin (2𝜋(4000) ∗ 8000) = 0.25 sin(𝜋𝑛) = 0
At 7kHz, we have a max amplitude at 1kHz, and this is due to aliasing as we’re not
sampling at the required 14kHz, rather 8kHz.
𝑓𝑎 = 𝑓𝑠 − 𝑓𝑜 = 8𝑘𝐻𝑧 − 7𝑘𝐻𝑧 = 1𝑘𝐻𝑧
2. (5 marks) Write a MATLAB script to generate 100 samples of the two sets of analog
signals:
a. 𝑥1(𝑡) = sin2(2000 𝜋 𝑡) cos2(2000 𝜋 𝑡), and 𝑦1(𝑡) = 18 − 18 cos(4000 𝜋 𝑡),when the
sampling rate is 6000 Hz.
b. 𝑥2(𝑡) = 2 sin(4000 𝜋 𝑡) cos3(2000 𝜋 𝑡), and 𝑦2(𝑡) = sin(2000 𝜋 𝑡) +
𝜋 𝑡), when the sample rate is 8000 Hz.
sin(6000
Properly name this script file and upload it to the Lab1 folder.
Show that the sampled versions of 𝑥1(𝑡) and 𝑦1(𝑡), and 𝑥2(𝑡) and 𝑦2(𝑡) are identical.
Explain the result using the sampling theory.
Through MATLAB’s command window we can now prove that the signals are
identical. Despite starting with two different equations, via introducing an appropriate
sampling frequency, we have in turn created a ‘new’ signal. This new signal is identical to
its corresponding pair. This is because, we are sampling at an insufficiently high sampling
frequency, and are thus introducing aliasing. It is this aliasing that makes the functions
appear as equal.
Quantization Experiments
1. (5 marks) Write a MATLAB function to return a quantized signal, yq, the quantization
error, eq, and the signal-to-quantization noise ratio (SNR in dB). The inputs to this function are
the discrete-time analog signal, y, the full-scale range of the quantizer, [xmax and xmin], and the
number of bits, m. Upload this file to the Lab1 folder.
Uploaded
2.
(5 marks) Use the above function to quantize a sinusoid 𝑦(𝑡) = 3 sin (600𝜋𝑡 + 𝜋 ),
4
when the full-scale range is [5 -5], number of bits is 6, and the sample rate is 4000 Hz.
Let the duration of the sinusoid be 0.02 seconds.
a. Include a plot of the original and quantized sinusoids.
b. Report the theoretical and measured SNRs for this example.
The difference between the
two graphs is hardly
noticeable from just visual
appearance. This is due to the
super low voltage steps of
(10/64)V/step. We also got a
reported SNR of 34.34dB.
This is somewhat similar to
the theoretical SNR of
30.00dB, and is within the
5% region of error.
Bonus Sampling Question
1. (5 marks) The following MATLAB experiment demonstrates aliasing with a chirp signal. A
chirp is a signal whose frequency varies as a function of time. In the analog domain, the
equation for generating a linear chirp signal is
x(t) = Acos(
where
•
t2+2 f1t)
is the sweep rate in Hz/sec and f1 is the starting frequency in Hz.
Write the discrete-time equivalent of the above equation when sampled at a
frequency, fs.
𝑥[𝑛] = 𝑥(𝑛𝑇𝑠 ) = 𝐴𝑐𝑜𝑠(𝜋𝜇(𝑛𝑇𝑠 )2 + 2𝜋𝑓1 (𝑛𝑇𝑠 )
1
𝑤ℎ𝑒𝑟𝑒 𝑛 = 𝑑𝑖𝑠𝑐𝑟𝑒𝑡𝑒 𝑡𝑖𝑚𝑒 𝑖𝑛𝑑𝑒𝑥, 𝑎𝑛𝑑 𝑇𝑠 =
𝑓𝑠
•
With f1 = 100 Hz, = 2000, A =0.25, and fs = 48 kHz, generate 8 seconds of sampled
chirp signal. Use the “spectrogram” command (see #2 in Sampling Experiments
section) to display the time-frequency structure of the sampled chirp signal. Include
this spectrogram in your report, along with your comments on what you see. Play
the sampled chirp signal using the “sound” command and describe what you heard.
The signal sounds like it
continuously gets higher pitched
over the 8 second interval. This
makes sense with what we are
seeing, as the frequency linearly
increases as time goes on,
attributing to the increase in pitch
that we hear.
•
Repeat the above step but this time using a sampling frequency of 16 kHz. Provide
an explanation based on the sampling theorem as to why the sampled chirp sounds
different and its spectrogram looks different with f s = 16 kHz.
The sampled chirp sounds different, as in
this case the pitch gradually increases to its
peak frequency, and then halfway through
gradually declines back to its starting
frequency of 0. This can be attributed to
aliasing of the signal in which a peak
frequency of 8kHz is obtained. This makes
sense as 2Fmax = 16kHz, upon which we
get an aliased signal.