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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# Introduction