EE 313 Linear Systems and Signals - Lecture 2
Lecture by Prof. Brian L. Evans
Before Lecture
Announcements
- Lecture slides on Periodic Signals in
PowerPoint format.
- Fall 2025, Part 1.
Notes in
Word and
PDF formats.
- Review. The Sampling Theorem says choose the sampling rate
fs so that
fs > 2 fmax
where so that frequencies up to and including the
frequency of interest, fmax, are captured.
- Slide 1-20. Zero frequency means a constant signal.
Consider v(t) = cos(2 pi f0 t).
Let f0 = 0, v(t) = cos(2 pi (0 Hz) t) = 1, which is a constant.
A constant value (DC value) has a frequency of 0 Hz.
- Slide 1-20. The playback of the sinusoid at 440 Hz was much quieter
than the sinusoid of 440 Hz squared which has a DC component and a component
at 880 Hz. Part of the explanation is that the human auditory system
has a much stronger response to 880 Hz than 440 Hz. See the
A-weighted human hearing curves.
- Slide 2-4. Phase shift. A phase shift of pi/4 looks the same as a phase shift
of 2 pi + pi/4.
- Slide 2-6. The sampling rate fs is 1 / Ts.
- Fall 2025, Part 2.
Notes in
Word and
PDF formats.
- Review. Interpreting the Sampling Theorem.
We can divide both sides of the inequality
fs > 2 fmax
to obtain
fmax < (1/2) fs.
That is, sampling at a sampling rate of fs
captures frequencies up to, but not including, (1/2) fs.
- Slide 2-8. Inverse Euler's formulas. Convert sinusoidal signals in
expressions to complex sinusoids to aid in simplifying the expression.
- ej theta + e-j theta = 2 cos(theta)
- ej theta - e-j theta = 2 j sin(theta)
- Slide 2-10.
x(t) = cos(2 pi 440 t) which takes values on [-1, 1].
For x(t),
A0 = 0
A1 = 1, f1 = 440, phi1 = 0
Input x(t) into a squaring block to produce output
y(t) = x2(t) = cos2(2 pi 440 t) which takes values on [0, 1].
y(t) = (1/2) + (1/2) cos(2 pi (880) t) by using the trigonometric identity for
cos2(theta) = (1/2) + (1/2) cos(2 theta).
For y(t),
A0 = 1/2
A1 = 0, f1 = 440, phi1 = 0
A2 = 1/2, f2 = 880, phi2 = 0
Last updated 09/04/25.
Send comments to Prof. Evans at
bevans@ece.utexas.edu
