advertisement

1. Time-varying signals (a) Thevenin (voltage source w/series resistor) (b) Norton (current source w/parallel resistor) 2. Units (a) ω = 2 πf (b) f is in Hz (cycles per second) (c) ω is in radians per second.

(d) ω is “omega”, not w .

(e) A magnitude (f) A power P V in dB is 20 log 10 ( V ) in dB is 10 log 10 ( P ).

3. Spectrum of signals (a) Time-domain = integral or summation over sinusoidal components v a ( t ) = V A sin ( ωt + φ ) (b) Magnitude spectrum i. Spectrum of a pure sinusoid v s ( t ) = 100 sin (2 πf t + φ ) ⇒ 40 20 A sinusoid has a single Fourier component that appears as an impulse function in the spectral representation.

0 10 2 ω 10 (rad/sec) 4 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 sC 1 ≈ ωC (Note that this doesn’t work for analyzing the phase response).

vi. Bode plots (see book appendix) 1

vii. Unity gain frequency w t .

60 40 20 0 10 1 10 2 10 3 10 4 f (Hz) 10 5 10 6 f t (c) Periodic signals have discrete-valued harmonics i. Aliasing in oscilliscope FFT display • • The Sec/Div knob sets the sampling rate f S .

If the signal frequency So if you increase f f > f S / 2, then the scope will show an beyond f S / image at f − f S / 2.

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 f 0 ; a distorted sinusoid has harmonic components spaced at integer multiples f k = kf 0 .

40 20 0 10 3 ω (rad/sec) 10 4 4. Amplifiers (a) Gain: voltage, current (b) Power gain: load power / input power A P = v o i o v i i i = A v A i (c) Small-signal concepts, signal convention i C ( t ) = I C + i c ( t ) (d) Circuit models for amplifiers i. Voltage amp ii. Current amp iii. Transconductance amp (v-in, i-out) iv. Transresistance amp (i-in, v-out (e) Cascades of amplifiers – resistive coupling between R sig , R in and between R out , R L .

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

R S + v s − + R L v OUT V S − + − The sensor model is linearized: v s = V S + dv S dT ∆ T T 0 where T 0 is the reference temperature and ∆ only the variation in v OUT T is the variation from that temperature. To consider , we isolate the small signal portion: v out = dv S dT T 0 ∆ T The problem statement tells us that dv S dT T 0 = 2mV/ ◦ C It also tells us that v out = 10 mV, so we can solve for R S : → v out R S = v s R L R L + R S = (2 mV/ = (2 mV/ ◦ C) (10 ◦ C) R L ◦ C) (10 ◦ C) R L R L + − R S R L v out = 10 kΩ

A thermistor is modeled by the Steinhart-Hart equation: R = R 0 e − B 1 T 0 − 1 T where R 0 rameter. For and T 0 are reference measurements, small temperature changes T and T 0 are in Kelvin, and B is a device-specific pa (e.g. changes in a room’s temperature), we can approximate this using a linearized model centered around T 0 : R ≈ R 0 + ∆ T d dT = R 0 + ∆ T R 0 e − B 1 T 0 − 1 T R 0 e − B 1 T 0 − 1 T d dT T = T 0 − B = R 0 − ∆ T R 0 B T 2 0 1 T 0 1 − T 3 T = T 0

So if T 0 = 300 K , R 0 = 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 · 10 4 Actual Linearized 1 .

01 1 0 .

99 280 290 300 T (Kelvin) 310 320 Note that the accuracy is best for very small ∆ T , and the error begins to grow as | ∆ T | increases.

4