Gary Hecht
Nyquist Sampling Theorem
Homework #1 Answers
1. Assume a 2 kHz sinusoid is multiplied by a 20 kHz sinusoid. What sinusoid(s) will be present
in the result?
(2 kHz sinusoid) • (20 kHz sinusoid) = 18 kHz sinusoid and 22 kHz sinusoid
(Difference
(Sum
frequency)
frequency)
2. Assume a 1.0 volt DC signal is added to the 20 kHz sinusoid in problem 1 (before the
multiplication process). What sinusoid(s) will be present in the result?
[2 kHz sinusoid] • [1 volt (DC) + 20 kHz sinusoid] = (2 kHz sinusoid) • (1 volt (DC)) +
(2 kHz sinusoid) • (20 kHz sinusoid)
= (2 kHz sinusoid) +
(18 kHz sinusoid) +
(22 kHz sinusoid)
3. Assume a 3 kHz squarewave is applied to an ideal low-pass filter with a ‘cutoff’ frequency of 2
kHz. Sketch the waveform that will appear on the output of the low-pass filter.
(Output = 0v since all sinusoids of the squarewave
are rejected by the low-pass filter)
────────────────────────────────────────►
4. Assume a 3 kHz squarewave is applied to an ideal low-pass filter with a ‘cutoff’ frequency of 4
kHz. Sketch the waveform that will appear on the output of the low-pass filter.
(Output = 3 kHz sinusoid since only the first harmonic (sinusoid)
of the squarewave passes through the low-pass filter)
────────────────────────────────────────►
5. Assume a 3 kHz squarewave is applied to an ideal low-pass filter with a ‘cutoff’ frequency of 10
kHz. Sketch the approximate waveform that will appear on the output of the low-pass filter.
Only the first harmonic (3 kHz) and the third harmonic (9 kHz)
(sinusoids) of the square wave will pass through the low-pass
filter creating a “rippling” square wave on the output (the tops
────────────────────────────────────────►
and bottoms of the square wave will not be flat (they will have
ripples) and the ‘sides’ will not be perfectly vertical)
2
6. Assume a signal consists of three simultaneously present sinusoids at frequencies of 3 kHz, 7
kHz, and 11 kHz. Further assume that this signal is to be sampled. According to the Nyquist
Sampling Theorem, what is the minimum (lowest) sampling frequency that will allow these three
sinusoids to be correctly ‘reconstructed’ from the samples (i.e., from the ‘Sampled Signal’)?
fS (min) = 2•(Highest frequency sinusoid) = 2•(11 kHz) = 22 kHz
Minimum sampling frequency = fS (min) = ___22___ kHz
7. Assume a signal consists of two simultaneously present sinusoids at frequencies of 5 kHz and 8
kHz. Also assume that this signal is sampled at a rate of 20,000 samples per second (fS = 20
kHz). The ‘Sampled Signal’ (the signal created by the sampling process) will have numerous
sinusoids including sinusoids at frequencies of 5 kHz and 8 kHz (the original information).
Determine the specific frequency of the first sinusoid, in the ‘Sampled Signal’, that has a
frequency above 8 kHz. To determine the answer to this question, you may need to review the
specifics of how the sampling process creates various sinusoids.
Sampled Signal
Sampled Signal
Sampled Signal
Sampled Signal
=
=
=
=
[5 kHz + 8 kHz]•[(DC offset) + fS + 2•fS + 3•fS +…]
[5 kHz + 8 kHz]•[(DC offset) + 20 kHz + 2•(20 kHz) + 3•(20 kHz)…]
5 kHz + 8 kHz + (20 – 5)kHz + (20 + 5)kHz + (20 – 8)kHz + (20 + 8)kHz…
5 kHz + 8 kHz + (15)kHz + (25)kHz + (12)kHz + (28)kHz…
First sinusoidal frequency above 8 kHz in the ‘Sampled Signal’ = ___12___ kHz
8. Assume a perfect 3 kHz squarewave signal is to be sampled. According to the Nyquist
Sampling Theorem, at what frequency would the squarewave need to be sampled such that the
squarewave can be perfectly ‘reconstructed’ from the samples?
fS (min) = 2•(Highest frequency sinusoid) = 2•(∞ kHz) = ∞ kHz
Required Sampling frequency = fS = ____∞____ kHz
9. Assume a perfect 5 kHz squarewave (with 50% duty cycle) signal is to be sampled. Assume
that you will be satisfied if only the first three sinusoids that describe the squarewave are
correctly ‘reconstructed’ from the ‘Samples Signal’. According to the Nyquist Sampling
Theorem, what is the minimum sampling frequency that will meet the desired goal of correctly
reconstructing the first three sinusoids that express the squarewave?
5 kHz squarewave = 5 kHz sinusoid + 3•(5 kHz sinusoid) + 5•(5 kHz sinusoid) +…
5 kHz squarewave = 5 kHz sinusoid + 15 kHz sinusoid + 25 kHz sinusoid +…
fS (min) = 2•(Highest frequency sinusoid) = 2•(25 kHz) = 50 kHz
Required Sampling frequency = fS = ___50___ kHz
10. Assume a 5kHz sinusoid is sampled by a sampling system with a sampling frequency of 8
kHz. What are the frequencies of the first two sinusoids in the ‘Sampled Signal’?
Sampled Signal = [5 kHz]•[(DC offset) + fS + 2•fS + 3•fS +…]
Sampled Signal = [5 kHz]•[(DC offset) + 8 kHz + 2•(8 kHz) + 3•(8 kHz)…]
Sampled Signal = 5 kHz + (8 – 5)kHz + (8 + 5)kHz +…
Sampled Signal = 5 kHz + 3kHz + 13kHz +… [Note: The 3 kHz sinusoid is an ‘alias’
of the 5 kHz sinusoid]
First two sinusoidal frequencies in the ‘Sampled Signal’ = ____3____ kHz
and ____5____ kHz
Gary Hecht
Nyquist Sampling Theorem
Homework #2 Answers
1. Most sampling systems use a sampling rate that is somewhat higher than the minimum required
for sampling the signal (fS (min) = 2•(Highest frequency sinusoid in the signal to be sampled)).
This is done because the real-world low-pass filters that are used for the anti-aliasing and
reconstruction filters in a sampling system have noticeably non-ideal characteristics (especially
near their cutoff frequencies).
When a sampling system uses a sampling rate that is higher than the minimum required, some
amount of so-called ‘guardband’ is created. The amount of guardband is calculated as the
difference between the actual sampling frequency and the minimum required sampling
frequency:
Guardband = Sampling frequency (actual) – Minimum required sampling frequency
Sampling frequency (actual) = Minimum required sampling frequency + Guardband
Sampling frequency = fS = 2•(Highest frequency sinusoid in the signal to be sampled) +
Guardband
Assume a signal consists of three simultaneously present sinusoids at frequencies of 3 kHz, 7
kHz, and 11 kHz. Further assume that this signal is to be sampled by a sampling system that
has 5 kHz of guardband. What is the required sampling frequency that will allow these three
sinusoids to be correctly ‘reconstructed’ from the samples with a guardband of 5 kHz?
fS = 2•(Highest frequency sinusoid) + Guardband = 2•(11 kHz) + 5 kHz = 27 kHz
Sampling frequency = fS = ____27____ kHz
2. Land-line telephone service (known as POTS (plain-old telephone service)) samples each call at
a rate of 8,000 samples per second (in each direction) where each sample is converted into an
8-bit digital value by an ADC (analog-to-digital converter) and then transferred through the
telephone system as a digital value. At the receiving end for a call, each sample is converted
back into an analog voltage by a DAC (digital-to-analog converter) and then sent to the speaker
in the handset.
For POTS, what is the telephone company’s internal (digital) ‘data rate’ (in bits per second) for
each direction of a call? Note: ‘Data rate’ is also known as ‘bandwidth’.
Data rate = (8,000 samples/second)•(8 bits/sample) = 64,000 bits/second
Data rate (or bandwidth) for POTS (each direction) = ___64,000___ bits per second
CDs have a sampling rate of 44,100 samples per second where each sample consists of 16
bits. Also, CDs have two independent ‘channels’ (allowing for stereo) and, as such, there are
two samples stored on the CD for each 1/44,100 of a second. How many times larger is the
total data rate for CDs as compared to POTS (one direction)?
CD data rate = (44,100 samples/sec)•(16 bits/sample)•(2 channels) = 1,411,200 bits/sec
Data Rate Ratio = (1,411,200 bits/sec) / (64,000 bits/sec) = 22.05
The CD data rate is ___22.05___ times larger than the data rate for POTS
3. Assume it is desired to build a sampling system that samples an analog signal, digitizes the
samples, stores the digitized samples, and later plays back the digitized samples so as to recreate the original analog signal. The sampling system is to have the following general
characteristics:
Sampling rate = 12,000 samples/sec (fS = 12,000 Hz)
Cutoff frequency (fc) for the anti-aliasing and reconstruction filters = 5 kHz
(assume no sinusoids above 5 kHz get through the low-pass filters)
Resolution per sample = 8 bits
Questions:
a. How much ‘guardband’ exists in the sampling system?
Guardband = fs – 2•fc = 12 kHz – 2•(5 kHz) = 12 kHz – 10 kHz = 2 kHz
b. What is the ‘data rate’ (or ‘bandwidth’) of the sampled signal (in bits per second)?
Bandwidth = (12,000 samples/sec)•(8 bits/sample) = 96,000 bits/sec
c. Assume the anti-aliasing and reconstruction filters each use a 0.2 μF capacitor. What value
should be used for the resistor in these filters?
Note: fc = 1/(2πRC)
R = 1/[2π fc C] = 1/[2π(5•10+3)•(0.2•10-6)] = 0.159•10+3 =
159 ohms
d. If you wanted to store 5 minutes worth of samples, how many bits would need to be stored?
Eight bits are known as a ‘byte’ (i.e., 8 bits = 1 byte) – as such, how many bytes would
need to be stored for 5 minutes worth of samples?
5 minutes of samples = (96,000 bits/sec)•(60 secs/min)•(5 min) = 28,800,000 bits
5 minutes of samples = (28,800,000 bits) / (8 bits/byte) = 3,600,000 bytes
e. Assume the analog signal that is being sampled has a voltage range of –5v to +5v. This
analog signal, after being sampled, has each sample ‘digitized’ by an ADC (analog-to-digital
converter) that outputs an 8-bit numerical value for the voltage of each sample. Typically the
digital value of all 0’s corresponds to the lowest voltage possible for the analog signal (0000
0000 = –5v for this example). And, typically, the maximum digital value, plus 1, corresponds
to the highest voltage possible for the analog signal (1 0000 0000 (9 bits!) = +5v for this
example).
Assume the ADC divides the analog voltage range into evenly spaced ‘steps’ such that any
analog voltage within a given ‘step’ is assigned the same unique digital value. For this
example, how much voltage is there per step (known as the ‘step voltage’)?
0000 0000 (010) through 1 0000 0000 (25610) are 257 different numbers creating 256 steps
Step voltage = (Voltage range) / (Number of steps) = (10 volts) / (256 steps) = 39.1 mV
Gary Hecht
Nyquist Sampling Theorem
Homework #3 Answers
Assume it is desired to build a sampling system that samples an analog signal, digitizes the
samples, stores the digitized samples, and later plays back the digitized samples so as to re-create
the original analog signal. The sampling system is to have the following general characteristics:
Sampling rate = 10,000 samples/sec
Guardband = 2 kHz
Resolution per sample = 8 bits
Questions:
a. The anti-aliasing and reconstruction filters should be designed for what ‘cutoff’ frequency?
Guardband = fs – 2•fc
so
fc = [fs – Guardband]/2
fc = [fs – Guardband] / 2 = [10 kHz – 2 kHz] / 2 = 4 kHz
b. What is the ‘data rate’ (or ‘bandwidth’) of the sampled signal?
Bandwidth = (10,000 samples/sec)•(8 bits/sample) = 80,000 bits/sec
c. Assume the anti-aliasing and reconstruction filters each use a 0.2 μF capacitor. What value
should be used for the resistor in these filters?
Note: fc = 1/(2πRC)
R = 1/[2π fc C] = 1/[2π(4•10+3)•(0.2•10-6)] = 0.199•10+3 =
199 ohms
Further assume that 16 Mbit (total storage) memory ICs are used to store the samples where each
16 Mbit memory IC is "organized" "by 8" (i.e., 8 bits per memory location).
Note:
Total storage of
a memory IC (in bits) = (Number of locations)•(Number of bits per location)
d. How many samples can be stored in one 16 Mbit memory IC (“organized” “by 8”)? (Note that
one sample can be stored in each memory location of a “by 8” memory IC since our samples
consist of 8 bits per sample)
16 Mbits = (N locations)•(8 bits/location)
so N = [16 Mbits] / [8 bits/location] = 2M locations and, therefore, 2M samples
can be stored in one 16 Mbit memory IC
20
Note: 1 M = 2 = 1,048,576
e. How many 16 Mbit memory ICs are required to store one hour's worth of samples?
Number of samples in 1 hour = (10,000 samples/sec)•(60 sec/min)•(60 min/hour)
Number of samples in 1 hour = 36,000,000 samples
Number of ICs = (36,000,000 samples) / (2•1,048,576 samples/IC) = 17.2 ICs
So, 18 ICs would be needed