Dynamic circuits: Frequency domain analysis


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1 Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1
2 System behaviour: overview 2
3 System behaviour : review solution = transient + steady state Mathematically, natural response (system characteristic, c.f. personality of a person) forced response (system altered by external force, c.f. personality changes due to peer influence) Transient = complementary function (solution of the system when the sources are zero Steady state = particular solution when the sources are applied 3
4 Natural response (free oscillation) The natural response is important because it gives the characteristic of the circuit. The characteristic (like stability, transient speed, etc.) of the system can be studied by looking at its natural response. Natural response has nothing to do with the external forces. Hence, we can just look at the circuit with ALL SOURCES reduced to zero. Short circuit all voltage sources Open circuit all current sources The resulting circuit is called the FREE OSCILLATING circuit. 4
5 Free oscillating version of a circuit Free oscillating circuit Natural response: 5
6 Stability Definition from my book: p
7 Stability test Test procedure: Inspect the free oscillating circuit Find its solution (i.e., natural response) Check if the solution goes to zero as time tends to. Suppose the system is: The solution is: Here, λ i are eigenvalues (natural frequencies) which can be real or complex. Clearly, stability requires that ALL λ i are negative or having negative real parts. 7
8 Example: firstorder response The freeoscillating response is: The system is stable if λ < 0; otherwise it is unstable. stable unstable 8
9 Example: secondorder response The freeoscillating response is: xt ()= Ae + Ae Case 1: eigenvalues are real Stable if λ 1 and λ 2 are both negative. λ 1 t 2 λ t 1 2 Case 2: eigenvalues are complex, i.e., Solution is α t Stable if α is negative xt () = e ( Asinωt + Bcos ωt) λ12, = α ± jω unstable stable 9
10 Stability and characteristic equation The stability issue is about examining λ. Since Ae st is a solution, we can put it in the original differential equation and get The roots of this equation are the λ s. This important equation is called CHARACTERISTIC EQUATION. 10
11 Example Suppose we wish to study the stability of this circuit. We look at the freeoscillating circuit: State equation is: or The characteristic equation is 11
12 Example Solving the characteristic equation gives Clearly, if R, L and C are positive, then all λ s are negative. 12
13 Natural frequencies Natural frequencies = eigenvalues, λ k Roots of characteristic equation complex For an nth order system, there are n natural frequencies. The system solution is dynamical modes Exponential decay Exponential growth (unstable) Oscillatory Oscillatory with decaying amp Oscillatory with growing amp (unstable) 13
14 The complex frequency plane Exponential decay splane jω Exponential growth t x x x t σ x x sine wave x 14
15 Sources Suppose a voltage has a complex frequency s. v(t) = V o e st If s = ±jω, then it is pure sinusoidal since cos ωt = e jωt jωt + e 2 Sine waves have pure imaginary frequencies, ±jω rad/sec. 15
16 Impedances We can also extend the concept of complex frequency to impedance. Recall: in last lecture, we said that the impedance of an inductor is jωl when it is driven by a sine source. the impedance of a capacitor is 1/jωC when it is driven by a sine source. Now, we imagine the driving frequency is s. Thus, we have Z L = sl Z c = 1/sC V L = sl I L I C = sc V C 16
17 Example Consider the impedance: Thus, we can write: The freeoscillating circuit will have The characteristic equation is The natural frequency (eigenvalue) is λ = 1/CR which is 1/τ. 17
18 Transfer functions Transfer function ratio of two quantities 1. Voltage gain: v 2 /v 1 2. Current gain: i 2 /i 1 3. Transadmittance: i 2 /v 1 4. Transimpedance: v 2 /i 1 V 1 driven at s + linear circuit I 2 F(s) = I 2 (s)/v 1 (s) in the s domain all voltages and currents are set to frequency s 18
19 Example Suppose the circuit is driven by V 1 at frequency s. So, the inductor impedance = sl the capacitor impedance = 1/sC We can redraw the circuit as The transfer function from V 1 to V 2 is 19
20 Getting characteristic equation from transfer function The transfer function from V 1 to V 2 is Clearly, the freeoscillating circuit can be formed by setting V 1 =0. = 0 In general, Thus, the characteristic equation is just F(s) = Char. Eqn. is D(s) = 0 20
21 Transfer function on complex plane What do we mean by transfer function? It is the ratio of V 2 to V 1 at complex frequency s. We can thus plot it as a surface on the complex plane. Note: at s = λ, the transfer function goes to, like a pole pointing up to the sky! 21
22 Meaning of transfer function So, what do we actually mean by transfer function? More precisely, it is the ratio of V 2 to V 1 at a given frequency s, in the steady state. V 2 /V 1 22
23 Frequency response In particular, if we put s = jω, the input is a sinusoidal driving source. In other words, we are looking at the cross section of the surface when it is cut along the imaginary axis (yaxis). V 2 /V 1 23
24 Frequency response vs. transient Consider a simple firstorder system. The frequency response has a pole at s = λ. The transient is A exp (λt). So, there is correspondence between frequency response and transient. V 2 /V 1 1 s + λ v 2 (t) = Ae λt t 24
25 More examples time domain frequency domain Recall: Laplace transform pairs 25
26 Firstorder lowpass frequency response The transfer function of a firstorder lowpass RC filter is Gs ( ) V ( s) = 2 = V ( s) 1 1/ sc 1 = R + 1/ sc 1 + scr Hence, 1. this transfer function has a pole at p = 1 (At this frequency, G(s) goes to.) CR 2. for large magnitudes of s (i.e., s\ ), G(s) goes to 0. From this, we can roughly sketch G(s) on the complex plane. 26
27 Rubber sheet analogy Pole: 1/CR magnitude goes to. All surroundings go to 0, for large s. 27
28 Firstorder lowpass frequency response 28
29 Rubber sheet analogy Bandwidth depends on 1/CR 29
30 Frequency response closeup Exact formula: G( jω) = jωcr = ω C R Observations: If the pole is further away from the origin, then the frequency response starts to drop off at a higher frequency. =1/CR 3dB corner frequency At ω = 1/CR rad/s, the response drops to of the dc value, which is exactly 3 db below the dc value. 30
31 Example Consider a first order circuit having the following transfer function: Y( s) X( s) = 10 s Y/X The eigenvalue is The time domain response of the freeoscillating system is: y(t) = A e 10000t In the frequency domain, the ratio Y/X is 10 at dc, and starts to drop at around rad/s (or 1.59 khz) which is the 3dB corner frequency rad/s or 1.59 khz 31
32 Secondorder lowpass frequency response with complex pole pair Consider a second order circuit having the following transfer function: 1 Gs ( ) = s s ς + ω 2 n ωn Suppose the eigenvalues are complex. s = ςω ± jω 1 ς n n 2 Here, ζ is the damping factor ω n is the resonant frequency complex poles The response depends on ζ. ζ < : light damping freq response shows a peaking ζ > : heavy damping freq response shows no peaking ω n is roughly where the corner frequency is. 32
33 Complex pole pair seen on the splane The complex frequency plane is often called the splane. 33
34 Secondorder lowpass frequency response: Rubber sheet analogy 34
35 Secondorder lowpass frequency response 35
36 Resonance: a special case where ζ is small Exact resonance: when driving frequency is exactly the natural frequency (eigenvalue), the response is infinitely large! But, if you can only have sine waves as driving sources, then you can t have exact resonance for systems with complex poles. G small ζ Near resonance: for systems with ζ < , the response has a peak around ω = ω n (actually 2 to be more ωn 1 ς accurate). For small ζ, we can say that the system nearly resonates at ω n. ω ω n 36 ω
37 LC resonant circuits (theoretical cases: ζ = 0) + V(s) I(s) L C Is ( ) sc Y ( s) = = V( s) s LC Poles are ±jω n Shortcircuit at ω = ω n. Y or Z ζ = 0 + I(s) V(s) L V( s) sl Zs ( ) = = Is ( ) s LC C Poles are ±jω n Opencircuit at ω = ω n. ω = ω n ω In these cases, we see exact resonance because the natural frequencies fall on the imaginary axis and we can find sine wave driving sources. But, the question is do we have perfect inductor and capacitor, without resistance? 37
38 More example on frequency response In the steadystate, we can apply conventional analysis (like nodal) to get G 1 V 1 + G + G sl = G2 G2 + G3 + sc V GV 1 in 0 Thus, the input to output transfer function is For s=0, V 2 /V in 0 For s j, V 2 /V in 0 So, it s a bandpass 38
39 Poles and zeros: review In general the transfer function is Poles roots of D(s) = 0 points on splane where G goes to. Zeros roots of N(s) = 0 points on splane where G goes to 0. For real systems, the degree of N(s) is always less than or equal to that of D(s), so that there cannot be more zeros than poles! WHY? Think about what happens when s. 39
40 Example Consider Poles : 2±j3 and 10 points on splane where G 1 goes to. Zeros : 0 and ±j10 points on splane where G 1 goes to 0. Also, lim G1( s) = 1 s 40
41 Rubber sheet analogy bandpass 41
42 Story half told! How about phase? Frequency response should characterized by  Magnitude  Phase shift For sinusoidal driving signals, we consider G(jω). Consider the simple RC circuit. The transfer function is V ( jω) G( jω) = 2 1 = V ( jω) 1 + jωcr The phase shift is given by 1 φ( j ω) = phase shift due to nominator phase shift due to denominator φ( jω) = 0 arctan( ωcr) 42
43 Full frequency response of RC circuit G( jω) V ( jω) = 2 1 = V ( jω) 1 + jωcr 1 Mag: G( jω) = jωcr = ω C R Phase: φ( jω) = 0 arctan( ωcr) 43
44 Phase response of secondorder system with a complex pole pair Gs ( ) = 1 s s ς + ω 2 ω n n G( jω) = 1 jως ω ω ω 2 1 = 2 ω 1 + 2jως 2 2 n n ω ω n n Phase: 2ςω ω φω ( ) = 0 arctan n 2 ω 1 2 ω n ςω arctan 2 = 2 2 ω ω n 44
45 Phase response of secondorder system with real poles Gs ( ) = 1 s ( 1+ s) G( jω) = 1 jω ( 1+ jω) Phase: ω ω φω ( ) = arctan arctan
46 Plotting frequency response Direct method: direct substitution of s = jω in the transfer function, and plot the magnitude and phase as a function of frequency. e.g., G( jω) 1 = 1 + jωcr Magnitude and phase are functions of ω: G( jω) = ω C R θω ( ) arctan ωcr = ( ) Then, we can plot them by any means, e.g., Matlab, mathematica, etc. 46
47 Plotting it with linear scale G( jω) = 1 + ω C R θω ( ) arctan ωcr = ( ) 47
48 Plotting frequency response Problem with scaling As seen before, the plots are usually nonlinear some square within a square root, etc. and some arctan function! The most difficult problem with linear scale is the limited range. LINEAR SCALE: where is 1000? 48
49 Log scale If the xaxis is plotted in log scale, then the range can be widened. LOG SCALE:
50 Bode technique : asymptotic approximation Consider the same firstorder transfer function G( jω) The square of the magnitude is: 1 = 1 + jωcr 2 1 G( ω) = ω C R Take the log of both sides and multiply by 20: log G( ω) = 10log( 1 + ω C R ) Define y = x = 20log G( ω) logω The unit of y is the decibel (db) 50
51 Bode technique for plotting frequency response In terms of x and y, we have y = 10log( 1 + ω C R ) For large ω and small, we can make some approximation. y = x = 20log G( ω) logω (a) ω >> 1/CR : y 10 log ω 2 C 2 R 2 (b) ω << 1/CR : y 10 log 1 (a) ω >> 1/CR : y 20 log ωcr = 20x + 20 log (1/CR) (b) ω << 1/CR : y 0 straight line of slope 20 horizontal line 51
52 The trick is: Bode approximation (magnitude) If we plot y versus x, then we get straight lines as asymptotes for large and small ω. (a) ω >> 1/CR : y 20x + 20 log (1/CR) (b) ω << 1/CR : y 0 Note: The approximation is poor for ω near 1/CR. 52
53 Bode approximation (phase) Same example: θω ( ) arctan ωcr = ( ) The approximation is 1 0 for ω << CR 1 θω ( ) = 45 for ω = CR 1 90 for ω >> CR one decade below 1/CR one decade above 1/CR 53
54 Bode plots: general technique Since Bode plots are logscale plots, we may plot any transfer function by adding together simpler transfer functions which make up the whole transfer function. Plot A Plot B Gs ( ) = s s s s s Plot C Plot D Plot E 54
55 Bode plots: standard forms 1. Simple pole 2. Simple zero 3. Integrating pole 4. Differentiating zero 5. Constant 1 Gs ( ) = 1 + s p Gs ( ) = 1 + sz Gs ( ) = 1 s / p s Gs ( ) = z Gs ( ) = A 6. Complex pole pair Gs ( ) = 1 s s ς + ω 2 ω n n 55
56 Bode plots: standard forms Simple pole: (logscale) (logscale) 56
57 Bode plots: standard forms Simple zero: (logscale) (logscale) 57
58 Bode plots: standard forms Integrating pole: (logscale) (logscale) 58
59 Bode plots: standard forms Differentiating zero: (logscale) (logscale) 59
60 Bode plots: standard forms Constant: (logscale) (logscale) 60
61 Bode plots: standard forms Complex pole pair (logscale) (logscale) 61
62 Example constant simple zero differentiating pole simple pole Note: ω = 2πf If ω = 200π rad/s, f = 100 Hz. 62
63 Example 63
64 Semilog graph paper allows compressed xaxis Effectively, the xaxis is being log ed usual label for practical convenience 64
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