AC BEHAVIOR OF COMPONENTS

AC BEHAVIOR OF COMPONENTS

AC Behavior of Capacitor Consider a capacitor driven by a sine wave voltage: I(t) 2 1 U(t) ~ C 0-1 -2 0 2 4 6 The current: is shifted by 90 o (sin cos)! 1.0 0.5 0.0-0.5-1.0 0 2 4 6 8 CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 2

Complex Impedance To simplify our calculations, we would like to extend the relation R= U/I to capacitors, using an impedance Z C. In order to get the phase right, we use complex quantities: for voltages and currents. By mixing complex and real parts, we can mix sin() and cos() components and therefore influence the phase. Note: Often j is used instead of i for the complex unit, because i is used as current symbol Often s is used for iω (or jω) CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 3

From Complex Values back to Real Quantities To find ( back ) the amplitude of such a complex signal, we calculate the length (magnitude) of the complex vector as Im z Re z * To get the phase, we use real and imaginary parts: z Im(z) Note: this simple formula works only in 2 quadrants. You may have to look at signs of Re(z) and Im(z) Re(z) CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 4

Hints for Mathematica Mathematica knows complex arithmetic Useful Functions are Abs[] and Arg[] Remember: Imaginary Unit is typed as ESC i i ESC If you want to simplify expression, M. has to know that expressions like ω, R, C, U are real. This can be done with Assumptions: Sometimes ComplexExpand[] can be used. It assumes all arguments are real (but not necessarily > 0): CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 5

Complex Impedance of the Capacitor We know that With we have Therefore Similar: The impedance of a capacitor becomes very small at high frequencies CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 6

Checking this for a Capacitor For an input voltage (sine wave of freq. ω) with phase = 0 we have The amplitude of I(t) is The phase is: We have dropped the time variant part and the constant U 0 CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 7

Simplifying even more As we have just seen, the propagates trivially to the output. We therefore drop this part and just use 1! CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 8

Recipe to Calculate Transfer Functions Replace all component by their complex impedances (1/(sC), sl, R) Assume a unit signal of 1 at the input (in reality it is ) Write down all node current equations or current equalities using Kirchhoff s Law (they depend on s) You need N equations for N unknowns Solve for the quantity you are interested in (most often V out ) Analyze the result (amplitude / phase / ) CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 9

Example: Low Pass Consider I R R v in C I C v out We have only one unknown: v out Current equality at node v out : Solve for v out : CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 10

Mathematica Hint Write down each node equation (here only 1): Solve them: v in R v out C Define a transfer function: CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 11

Low Pass as complex voltage divider R Z1 C Z2 This is an ac voltage divider with two impedances Z 1 = R and Z 2 = 1/sC Using the voltage divider formula, we get with ω 0 = 1/(RC), the corner frequency. CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 12

The HIGH Pass By exchanging R and C, low frequencies are blocked and high frequencies pass through. This is the High-Pass. C Z1 R Z2 We get CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 13

More Complicated Example R v 1 C v in R C vout We have now two unknowns: v 1, v out Eliminating v 1 gives: CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 14

Mathematica Steps Node equation (here 2): v in R v 1 C R v out Solve them: C Define a transfer function: CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 15

Mathematica Steps Replace s by i ω Calculate (squared) gain as absolute value To plot, convert to db (sqrt leads to factor 10 instead of 20) For phase, better use ArcTan[Re,Im] to get quadrant right CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 16

BODE PLOT

Transfer Function The transfer function of a linear, time invariant system visualizes how the amplitude and phase of a sine wave input signal of constant frequency ω appears at the output The frequency remains unchanged The transfer function H(ω) contains The phase change Φ(ω) The gain v(ω) = amp_in / amp_out (ω) in H(ω) out v(ω) Φ(ω) CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 18

Bode Diagram: Definition 20 db 0 db The Bode Plot shows gain (+ phase) of the transfer function The frequency (x-axis) is plotted logarithmically Gain is plotted (y-axis) logarithmically, often in decibel DB(x) = 20 log 10 (x): 10 +20 db 100 +40 db 2 6 db (not exactly!) 1 0 db / 2-6 db / 2-3 db -20 db -40 db dbs for multiplied quantities just add! CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 19

Bode Diagram: Properties Power functions are straight lines: Plot[x 2,{x,0,10}] LogPlot[x 2,{x,0,10}] LogLogPlot[x 2,{x,1,10}] LogLogPlot[Table[x N,{N,-1,3}],{x,1,10}] 1000 100 10 1 0.1 1.0 1.5 2.0 3.0 5.0 7.0 10.0 y=x 3 y=x 2 y=x y=x 0 = 1 y=x -1 CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 20

Bode Diagram: Properties 1/x function has slope -1: Multiplied functions are added in plot: 2 f1=2+x;f2=x -1 ; LogLogPlot[{f1, 5 f1},{x,0.01,100}] LogLogPlot[{f1, f2,f1 * f2},{x,0.01,100}] 5 f1 f1+f2 f1=2+x f1=2+x f2=1/x CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 21

THE LOW PASS FILTER

Analysis of the Low Pass Transfer Function Transfer Function: Magnitude: Phase: (rad or degree) CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 23

Bode Plot of LowPass (Amplitude) 0 db -3 db -6 db -20 db v ω 0 = 10 1/Sqrt(2) = 0.707-3dB point Factor 10 decrease in amplitude for factor 10 increase in frequency = -20 db / decade or -6 db / octave -26 db 1 / 10 (-20 db) -40 db ω x 10 (Decade) CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 24

The same in db -3 db -3 db point Approximation with straight lines CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 25

Bode Plot of LowPass (Phase) Phase ω 0 = 10 Lin-Log Plot! -45 degree -90 degree ω CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 26

Series Connection of two Low Pass Filters Consider two identical LP filters. A unit gain buffer makes sure that the second LP does not load the first one: R R C C From the properties of the LogLog Plot, the TF of the 2 nd order LP is just the sum of two 1 st order LPs: -6 db -3 db 1 LP 1 LP: -6 db / octave 2 LPs: -12 db / octave 2 LPs CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 27

Why bother so much about the low pass? All circuits behave like low-passes (at some frequency)! CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 28

Caveat! So far, frequency is expressed with ω, i.e. in radian / second We have: ω = 2 π ν Therefore, the frequencies in Hertz are 2π lower!!! db ω 0 = 1 / (RC) = 1 / (1n x 1M) = 1 / 1m = 1 khz ν = 1000 / 6.28 Hz = 160 Hz CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 29

Low Pass and High Pass CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 30

A More Complex Example Consider a (High Pass) filter with an inductor: C R v in L v out The transfer function is It is of second order (s has exponent of 2 in denominator) Magnitude: L=C=1 R=0.1,0.5,1,2 H(s) = (C L s 2 )/(1+C R s+c L s 2 ) Inductive peaking 12 db / octave = 40 db / decade CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 31

Phase Phase For fun: When is filter steep & flat? Zoom to corner frequency: L=C=1, R={1,Sqrt[2],2} CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 32

CIRCUIT SIMPLIFICATIONS

Large and Small Values To roughly understand behavior of circuits, only keep the dominant components: R Large R Large R Small R Small R Large R Small R Large R Small Eliminate larger or the smaller part (depending on circuit!) Error ~ ratio of components CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 34

The same for Capacitors C Large C Large C Small C Small C Large C Small C Large C Small CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 35

Resistors AND Capacitors Behavior depends on frequency ( Z C = 1/(2πν C) ) C Z C >> R ν << 1/(2π RC) Low frequency R Z C << R ν >> 1/(2π RC) High frequency Low frequency High frequency CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 36

Fourier Decomposition Maybe later Also: Step / Impulse response via inverse Laplace Transform Later CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 37

To Do S. 27: Bild: du/dt ~ W S. 31: Wieso U0 exp(iwt) weg? Show that Z C = Z R at the corner frequency Phasensprung bei LCR checken! CCS - Basics P. Fischer, ZITI, Uni Heidelberg, Seite 38