Frequency Response. Re ve jφ e jωt ( ) where v is the amplitude and φ is the phase of the sinusoidal signal v(t). ve jφ

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27 Frequency Response Before starting, review phasor analysis, Bode plots... Key concept: small-signal models for amplifiers are linear and therefore, cosines and sines are solutions of the linear differential equations which arise from R, C, and controlled source (e.g., G m ) networks. It is much more efficient to work with imaginary exponentials rather than cosine and sine functions; the measured function v(t) is considered (by convention) to be the real part of this imaginary exponential vt () vcos( ωt + φ) Re( ve ( jωt + φ) ) Re ve jφ e jωt ( ) where v is the amplitude and φ is the phase of the sinusoidal signal v(t). The phasor V is defined as the complex number V ve jφ Therefore, the measured function is related to the phasor by vt () Re( Ve jωt )

Circuit Analysis with Phasors The current through a capacitor is proportional to the derivative of the voltage: d it () C vt () dt We assume that all signals in the circuit are represented by sinusoids. Substitution of the phasor expression for voltage leads to: vt () Ve jωt Ie jωt d jωt C ( Ve ) jωcve jωt dt which implies that the ratio of the phasor voltage to the phasor current through a capacitor (the impedance) is Zjω ( ) V -- I --------- jωc Implication: the phasor current is linearly proportional to the phasor voltage, making it possible to solve circuits involving capacitors and inductors as rapidly as resistive networks... as long as all signals are sinusoidal.

Phasor Analysis of the Low-Pass Filter Voltage divider with impedances -- R + + C Replacing the capacitor by its impedance, / (jωc), we can solve for the ratio of the phasors / ---------- /jωc ------------------------ R + /jωc multiplying by jωc/jωc leads to ---------- ----------------------- + jω

Frequency Response of LPF Circuits Bode plots: magnitude and phase of the phasor ratio: / the range of frequencies is very wide (DC --> 0 8 Hz, for example) --> plot frequency axis on log scale the range of magnitudes is also very wide (and we care about ratios of 0.00 in some applications): --> plot magnitude on log scale define magnitude in decibels db by ---------- db 20log ---------- phase is usually expressed in degrees (rather than radians): ---------- atan Im( ) ---------------------------------- Re( )

Complex Algrebra Review * Magnitudes: Z ----- Z 2 Z -------- Z 2 2 2 X + Y ---------------------- 2 2 X 2 + Y 2, where Z X + jy Z 2 X 2 + Y 2 * Phases: Z ----- Z Z Z 2 2 Y atan ----- X Y 2 atan----- X 2 * Examples:

Magnitude and Phase Plots of the Low Pass Filter / --> for low frequencies; / --> 0 for high frequencies log scale Break point 0 db scale 0. 0.0 0.00 3dB 2 0.707 /ω db 20 decade 20 40 60 0.000 0.0 0. 0 (a) 00 000 80 ω log scale V out 0 45 90 35 80 0.0 0. Break point (b) 0 00 000 ω log scale The break point is when the frequency is equal to ω ο /, at which the ratio of phasors has a magnitude of - 3 db and the phase is -45 o. The break frequency defines low and high frequencies.

Finding the Waveform from the Bode Plot Suppose that v in (t) 00 mv cos (ω ο t + 0 o ) note that the input signal frequency is equal to the break frequency and that the phase is 0 o... the input signal phase is arbitrary and is generally selected to be 0. the output phasor is: V --------------------------------- in + j( ω o ω o ) V ---------- in + j magnitude: ---------- db 00mV 3dB ---------- ----------------- 7mV 2 2 phase: ---------- ( + j) 0 45 V 45 in output waveform v out (t) is given by: ( 7mV)e j45 v out () t Re e jω ot Re 7mVe j45 e jω ot ( ) v out () t 7 mvcos( ω o t 45 o )

Bode Plots of General Transfer Functions Procedure is to identify standard forms in the transfer functions, apply asymptotic techniques to sketch each form, and then combine the sketches graphically Hjω ( ) Ajω( + jωτ 2 )( + jωτ 4. )...( + jωτ n ) ----------------------------------------------------------------------------------------------- ( + jωτ )( + jωτ 3 )...( + jωτ n ) where the τ i are time constants -- (/τ i ) are the break frequencies, which are called poles when in the demoninator and zeroes when in the numerator From complex algebra, the factors can be dealt with separately in the magnitude and in the phase and the results added up to find H(jω) and phase (H(jω)) Three types of factors:. poles (binomial factors in the denominator) 2. zeroes (binomial factors in the numerator) 3. jω in the numerator (or denominator)

Rapid Sketching of Bode Plots Poles: - 3 db and -45 o at break frequency 0 db below and -20 db/decade above 0 o for low frequencies and -90 o for high frequencies; width of transition is 0 and (/0) break frequency Zeros: +3 db and +45 o at break frequency 0 db below and + 20 db/decade above 0 o for low frequencies and +90 o for high frequencies; width of transition is 0 and (/0) break frequency * jω: +20 db/decade (0 db at ω rad/s) and +90 o contribution to phase Example: