Introduction
Chris Winstead
January 19, 2015
1. Time-varying signals
(a) Thevenin (voltage source w/series resistor)
(b) Norton (current source w/parallel resistor)
2. Units
(a) ω = 2πf
(b) f is in Hz (cycles per second)
(c) ω is in radians per second.
(d) ω is “omega”, not w.
(e) A magnitude V in dB is 20 log10 (V )
(f) A power P in dB is 10 log10 (P ).
3. Spectrum of signals
(a) Time-domain = integral or summation over sinusoidal components
va (t) = VA sin (ωt + φ)
(b) Magnitude spectrum
i. Spectrum of a pure sinusoid
⇒
40
VS (s) (dB)
vs (t) = 100 sin (2πf t + φ)
A sinusoid has a single
Fourier component that
appears as an impulse
function in the spectral
representation.
20
0
102
104
ω (rad/sec)
ii. High-pass and low-pass RC circuits
iii. Laplace-domain analysis, transfer functions
iv. Magnitude of transfer function
|H(jω)|2 = H(jω) × H ∗ (jω)
v. Short-cut analysis: for analyzing the steady-state magnitude response for sinusoids at a
specific frequency ω, the cap C behaves like a resistor:
1
1
≈
sC
ωC
(Note that this doesn’t work for analyzing the phase response).
vi. Bode plots (see book appendix)
1
|H(f )| (dB)
vii. Unity gain frequency wt .
60
40
20
ft
0
101 102 103 104 105 106
f (Hz)
(c) Periodic signals have discrete-valued harmonics
VS (s) (dB)
i. Aliasing in oscilliscope FFT display
• The Sec/Div knob sets the sampling rate fS .
• If the signal frequency f > fS /2, then the scope will show an image at f − fS /2.
So if you increase f beyond fS /2, the signal peak on the FFT will appear to move
backwards.
ii. Spurious harmonics arising from distortion. A sinusoid has a single frequency component
at f0 ; a distorted sinusoid has harmonic components spaced at integer multiples fk = kf0 .
40
20
0
103
ω (rad/sec)
104
4. Amplifiers
(a) Gain: voltage, current
(b) Power gain: load power / input power
vo io
vi ii
= Av Ai
AP =
(c) Small-signal concepts, signal convention
iC (t) = IC + ic (t)
(d) Circuit models for amplifiers
i.
ii.
iii.
iv.
Voltage amp
Current amp
Transconductance amp (v-in, i-out)
Transresistance amp (i-in, v-out
(e) Cascades of amplifiers – resistive coupling between Rsig , Rin and between Rout , RL .
EXAMPLE: Linearization
A temperature sensor provides a change of 2mV per ◦ C, connected to a load of 10kΩ. The output changes
by 10mV when T is changed by 10◦C. What is the source resistance of the sensor?
2
RS
+
−
+
vs
RL
−
+
VS
vOUT
−
The sensor model is linearized:
dvS ∆T
vs = VS +
dT T0
where T0 is the reference temperature and ∆T is the variation from that temperature. To consider
only the variation in vOUT , we isolate the small signal portion:
dvS vout =
∆T
dT T0
The problem statement tells us that
dvS = 2mV/ ◦C
dT T0
It also tells us that vout = 10 mV, so we can solve for RS :
vout = vs
RL
RL + RS
RL
RL + RS
(2 mV/ ◦C) (10 ◦C) RL
→ RS =
− RL
vout
= 10 kΩ
= (2 mV/ ◦C) (10 ◦C)
EXAMPLE: Linearization
A thermistor is modeled by the Steinhart-Hart equation:
R = R0 e
−B
1
− T1
T0
where R0 and T0 are reference measurements, T and T0 are in Kelvin, and B is a device-specific parameter. For small temperature changes (e.g. changes in a room’s temperature), we can approximate
this using a linearized model centered around T0 :
d
−B T1 − T1
0
R ≈ R0 + ∆T
R0 e
dT
T =T
0
1
1
d
1
1
−B T − T
0
= R0 + ∆T R0 e
−B
−
dT
T0 T
T =T0
B
= R0 − ∆T R0
T02
3
So if T0 = 300 K , R0 = 10 kΩ and B = 50 K−1 , then for temperatures near 300 K we have
R ≈ 10 kΩ − ∆T × 5.5 Ω
So we should see a difference of about 5.5 Ω/K. To check the accuracy of this approximation, we can
compare the actual (nonlinear) equation to the linearized result:
1.02
·104
Actual
Linearized
R (Ω)
1.01
1
0.99
280
290
300
310
T (Kelvin)
320
Note that the accuracy is best for very small ∆T , and the error begins to grow as |∆T | increases.
4

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