Feedback Negative feedback Positive Feedback Vin = +A If output slightly higher Vo=+A+ε = V-­‐ V+ -­‐V-­‐=-­‐ε è Vo =-­‐gain ε lowers Vo Vo=A-­‐β V+ -­‐V-­‐=+β è Vo =gain β increases Vo Vin = +A If output slightly higher Vo=+A+ε =V+ V+ -­‐V-­‐=ε è Vo =gain ε V+ è large GO TO RAIL + Vo=+A-­‐ε V+=A-­‐ε V+ -­‐V-­‐=-­‐ε V+ è -­‐gε GO TO RAIL -­‐ Tends to draw the two terminals toghether Ball in a valley negative feedback ball on top of a hill positive feed back There are many good resources on line to review the basic ideas of passive circuits and approaches to understanding. Our goal is to have a basic understanding of the role of impedance in circuit analysis. There are three categories of analysis that we will consider: 1. phasors 2. Fourier Transforms 3. LaPlace Transforms Of course they are all related. For example, my view is that the phasor approach is the technique used to employ the results of Fourier Transform. If you analyze your circuit in terms of frequency response then you can characterize the voltage current relationship between components in terms of vectors in the complex plane. The algebra-­‐trigonometry is then simpler in terms of the phasors. This work well when analyzing certain circuits such as the RLC circuit. One must apply the more complete complex analysis for circuits where impedances appear, for example, in the denominator. For many problems the spectral response of the circuit i.e. the Fourier components are not the optimal approach. The underlying justification of the Fourier approach relies on examining your systems response to an applied frequency that in principle began at -­‐∞ and extend to ∞. This makes this analysis less useful when you have initial conditions. The charging and discharging of a capacitor is an example of such a problem. There are also some functions that grow in time in such a manner that they do not have well behaved Fourier transforms but still can be treated using the LaPlace approach. Both Laplace transforms and Fourier transforms are essential tools to be used in analyzing circuits. Complex numbers and their relationships are necessary to understand both methods with phasors as nice tool to see some of the relationships clearer. Here are a few links that I found useful: http://www.stanford.edu/~boyd/ee102/ General treatment è http://fourier.eng.hmc.edu/e102/lectures/Laplace_Transform/Laplace_Tr ansform.html Derivation of Laplace & Fourier relationship èhttp://www.cambridge.org/us/features/chau/webnotes/chap2laplace.p df -2pager on 2nd order linear DIF. EQ. è http://www.math.psu.edu/tseng/class/Math251/Notes2nd%20order%20ODE%20pt1.pdf basic spice information for LTspice but perhaps usefule in general http://ltwiki.org/?title=Main_Page Block diagrams Boyd EE102 y=Su systems often denoted by block diagram: y output of a system S when u is applied as an input PSfrag replacements u S y • lines with arrows denote signals (not wires) • boxes denote systems; arrows show inputs & outputs • special symbols for some systems To view the problems that we have been dealing with consider the RLC analysis. The input can be taken as the AC supply and the output is the current through or voltage across a component. Systems 2–3 The key in analyzing circuits is to remember that all circuits represent coupled linear diff. eq. obtained by nodal analysis or Kickoff’s Laws. For example some set of components might require two coupled equations that depend upon the voltages generated by R,L,C components in a network where there are power supplies that generate voltages 𝑓! 𝑡 , 𝑓! (𝑡). 𝑑 𝑑 𝛾! 𝐼! 𝑡 + 𝛼! 𝐼! 𝑡 + 𝛿! 𝐼! 𝑡 𝑑𝑡 + 𝛾! 𝐼! 𝑡 + 𝛼! 𝐼! 𝑡 + 𝛿! 𝐼! 𝑡 𝑑𝑡 = 𝑓! (𝑡) 𝑑𝑡 𝑑𝑡 ! ! 𝛾! !" 𝐼! 𝑡 + 𝛼! 𝐼! 𝑡 + 𝛿! 𝐼! 𝑡 𝑑𝑡 + 𝛾! !! 𝐼! 𝑡 + 𝛼! 𝐼! 𝑡 + 𝛿! 𝐼! 𝑡 𝑑𝑡 = 𝑓! (𝑡) In considering this problem as a Fourier or spectral analysis problem my approach is to remember that in projectile motion problems we decouple the horizontal and vertical parts of the projectiles trajectory. We simply find the horizontal motion by propagating the projectile horizontally using the horizontal velocity. We treat the vertical motion by considering the gravitational acceleration and the initial vertical velocity. The simplicity and correctness of this approach relies on the fact that the motion is described in terms of orthogonal directions. The horizontal motion is completely independent of the vertical. This very same feature is present in the spectral analysis. The amount that a circuit oscillates at any frequency ω1 is completely independent of its oscillation at frequency ω2. You are then free to find the way the circuit responds to each frequency separately. Given some arbitrary driving or input 𝑓! 𝑡 on finds the part of this signal that is oscillating at ω1 ,solve the problem with this much simpler function then proceed to find the response to all the frequencies present in 𝑓! 𝑡 and add them together. This topic was discussed briefly when reviewing test1 and some of these comments are available on the web site So far we have examined the response of some circuit networks to a fixed AC signal and recognized that networks and/or devices could be characterized by frequency response. The next step in our formal approach to networks will be using the LaPlace transform to solve the diff. eq. that govern the network behavior. Consider the derivative and integral as expressed using the Fourier and Laplace transforms Fourier ! = 𝜈 𝑖𝑠 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑖𝑛 𝐻𝑧 𝐹 𝑠 = ℒ 𝑓 𝑡 (𝑠) ! 𝑓 𝑡 𝑒 !!!!"# 𝑑𝑡 = 𝑓 𝑡 𝑒 !!" 𝑑𝑡 ! !! 𝑠 = 𝜎 + 𝑗𝜔 𝐻 𝜔 = ℱ ℎ 𝑡 (𝜔) ! 1 = ℎ 𝑡 𝑒 !!"# 𝑑𝑡 2𝜋 !! 𝑓 𝑡 = ℱ !! 𝐹 𝜈 (𝑡) notation: Let f, F be functions when using ν Let h, H “””” ω h(t) = f(t) 𝑑! 𝑑𝑡 ! Laplace 𝐹 𝜈 = ℱ 𝑓 𝑡 (𝜈) 𝜔 = 2𝜋𝜈 ! = 𝑓 𝑡 = ℒ !! 𝑓 𝑠 (𝑡) 1 !!!! = 𝑓 𝑠 𝑒 !" 𝑑𝑠 2𝜋𝑗 !!!! 𝐹 𝜈 𝑒 !!!"# 𝑑𝜈 !! ℎ 𝑡 = ℱ !! 𝐻 𝜔 (𝑡) ! 1 = 𝐻 𝜔 𝑒 !"# 𝑑𝜔 2𝜋 !! ℱ ℎ!! 𝑡 𝜔 = 𝑗𝜔 ! ℱ ℎ 𝑡 = 𝑗𝜔 ! 𝐻(𝜔) 𝑑 → 𝑠 𝐹 𝑠 − 𝑓 0 𝑑𝑡 𝜔 require lim!→±! ℎ 𝑥 = 0 No initial conditions since the integration starts at -∞ Note the Fourier function form orthogonal basis 𝑒 !! !" ! → 𝑠 ! 𝐹 𝑠 − 𝑠𝑓 0 - 𝑓′ 0 𝑠 ! 𝐹 𝑠 − 𝑠 !!! 𝑓 0 𝑑𝑓 𝑑𝑓 !!! − 𝑠 !!! 0 … − 𝑠 ! !!! 0 𝑑𝑡 𝑑𝑡 Note that the LaPlace is similar but requires a knowledge of the driving function and its derivatives at t=0 LaPlace 𝑒 !!" = 𝑒 !(!!!")! !!!!"# ! 𝑑𝑡 𝑔 𝑡 = 𝐺 𝜔 = 1 𝐹(𝜔) 𝑗𝜔 𝑓 𝜏 𝑑𝜏 ! 𝐺 𝑠 = 1 𝐹(𝑠) 𝑠 f is assumed to be a differentiable function, and its derivative is assumed to be of exponential type. This can then be obtained by integration by parts evaluate at the point just before 0. Differentiation derivative formula for discontinuous functions disc. at t=0 if signal f is discontinuous at t = 0, then Second Differentiation L(f !) = sF (s) − f (0−) f is assumed twice differentiable and the second derivative to be of exponential type. Follows by applying the Differentiation property to f′(t). f is assumed to be n-­‐times differentiable, with nth derivative of exponential type. Follow by mathematical induction example: f is unit step, so f !(t) = δ(t) L(f !) = s General Differentiation ! " 1 −0=1 s Linearity convolutions f(t) and g(t) are extended by zero for t < 0 in the definition of the convolution. theIntegration Laplace transform is linear : if f and g are any signals, and a is any the Laplace u(t) is the Heaviside step function. Note (u scalar, we have ∗ f)(t) is the convolution of u(t) and f(t). scalar, we have linear The Laplace transform L(af ) = aF, L(f + g) = F + G Time scaling i.e., homogeneity homogeneity & superposition i.e., hold 3–22 transform of a function multiplied by a constant is the transform mult. by same constant transform of the sum of two functions is the sum of the transforms if you change scales the transform changes as shown time scaling define signal g by g(t) = f (at), where a > 0; then example: example: Time delay Exponential ! G(s) ="(1/a)F (s/a) scaling t ) L 3δ(t) − 2et = 3L(δ(t)) − 2L(e exponential at f bescaled a signalbyand a scalar, and define g(t) = e f (t); then makes sense: timesletare a,afrequencies by21/a scaling = 3− s−1 let’s check: G(s) = F (s − a) let f be a signal and T > 0; define the signal g as 3s − 5 ! ∞= translate ! in t∞ ime g is same function but T later ! −st − 0 < T1 dτ = (1/a)F (s/a) f (at)e dt = (1/a) f0(τ≤)est−(s/a)τ G(s) = let’s check: 0 g(t) = f (t − T0 ) t ≥ T ∞ e−(s−a)tf (t) dt = F (s − a) where τ= athave G(s) = e then we F (s) (g is f , delayed by T seconds &0 ‘zero-padded’ up 0to T ) The Laplace Laplace transform transform t The L(e ) = 1/(s − 1) so example: derivation: ! ∞ G(s) = −sT e−steatf (t) dt = f (t) PSfrag replacements example: L(cos t) = s/(s2 + 1), and hence g(t) ! ! ∞ G(s) = −t t The Laplace transform (s + 1)T + 1 t= =T ! = e The Laplace transform 3–10 3–10 1 ∞1 = s+1 s + 1 e L(eg(t) =1 s2 −ea−st (s/a) = f2 (t − T ) dt cosdt t) =− L(eat) = −st (1/a) 0 The Laplace transform ! ∞ 0 −sT s + 2s + 2 t e−s(τ +T )f (τ ) dτ F (s) 3–16 3–14 3–15 SUMMARY OF LAPLACE TRANSFORMS 3 where (of course) g(t) does not depend of f (t). Short Table of Laplace Transforms f (t) 1 s 1 s2 n! 1 t tn eat tn eat cos(!t) sin(!t) eat cos(!t) eat sin(!t) t cos(!t) t sin(!t) sin(!t) F (s) !t cos(!t) sn+1 1 s a n! (s a)n+1 s 2 s + !2 ! 2 s + !2 s a (s a)2 + ! 2 ! (s a)2 + ! 2 s2 ! 2 (s2 + ! 2 )2 2!s 2 (s + ! 2 )2 2! 3 (s2 + ! 2 )2 Department of Mathematics and Statistics, Texas Tech University, Lubbock, TX 79409-1042 E-mail address: lance.drager@ttu.edu Example Do some transforms-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐ let’s find Laplace transform of f (t) = et: ! ∞ ! ∞ t −st e e dt = e(1−s)t dt = F (s) = 0 0 "∞ 1 (1−s)t "" 1 e = " 1−s s−1 0 provided we can say e(1−s)t → 0 as t → ∞, which is true for $s > 1: " " " " " " " (1−s)t" " −j(#s)t" " (1−$s)t" e = e e " " " " " = e(1−$s)t " # $% & =1 • the integral defining F makes sense for all s ∈ C with $s > 1 (the ‘region of convergence’ of F ) • but the resulting formula for F makes sense for all s ∈ C except s = 1 More examples constant: (or unit step) f (t) = 1 (for t ≥ 0) F (s) = ! ∞ −st e 0 "∞ 1 −st "" 1 dt = − e " =s s 0 t→ provided wefirst can express say e−stf → sinusoid: (t) 0=ascos ωt ∞, as which is true for %s > 0 since " −st" "" −j(#s)t"" "" −($s)t"" −($s)t "e "f= "e + (1/2)e " = e−jωt " jωt (t)"e= (1/2)e # $% & =1 can find F as F makes sense for all s with %s > 0 •now the we integral defining ! ∞ • but the resulting formula for " F makes sense for all s# except s = 0 e−st (1/2)ejωt + (1/2)e−jωt dt F (s) = 0 = (1/2) ! ∞ Impulses at t = 0 (−s+jω)t dt + (1/2) e 0 ! ∞ e(−s−jω)t dt 0 if f contains impulses at t = 0 we choose to include them in the integral 1 1 The Laplace transform 3–6 defining F : +! (1/2) = (1/2) ∞ s − jω s + jω f (t)e−st dt s F (s) = = 2 0− s + ω2 (you can also choose to not include them, but this changes some formulas (validseefor&"s > 0; final formula OK for s #= ±jω) we’ll use) example: impulse function, f = δ ! ∞ " The Laplace transform δ(t)e−st dt = e−st"t=0 = 1 F (s) = 3–7 0− derivation of derivative formula: start from the defining integral Derivative transform similarly for f = δ (k) we have F (s) = ! ∞ 0− δ (k)(t)e−st ! ∞ " " d −st"" f " (t)ek −st dt G(s) k= dt = (−1) k e " = s e−st"t=0 = sk dt 0 t=0 k integration by parts yields The Laplace transform G(s) = = 3–9 "∞ e−stf (t)" − 0 ! ∞ f (t)(−se−st) dt 0 lim f (t)e−st − f (0) + sF (s) t→∞ for "s large enough the limit is zero, and we recover the formula G(s) = sF (s) − f (0) Example: RC circuit 1Ω PSfrag replacements y 1F u 𝑄 −𝑢 𝑡 + 𝐼𝑅 + = 0 apacitor is uncharged at t = 0, i.e., y(0) 𝐶 =0 (t) is a unit step m last lecture, !" ! 𝑢 𝑡 = 𝑅 !" + ! 𝑢 𝑡 = 𝑉! [𝑠𝑡𝑒𝑝 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛] !" (t)! [𝑠𝑡𝑒𝑝] + y(t)==𝑅𝐶 u(t) + Q y !𝐶𝑉 !" Laplace transform, term by term: 𝑈 𝑠 = 𝐶𝑉! 1 𝑠 sY𝑑𝑄 (s)→+𝑠𝑌 Y (s) 𝑠 −=𝑦 1/s 0 = 𝑠𝑌 𝑠 𝑑𝑡 𝑄 → 𝑌(𝑠) ng y(0) = 0 and U (s) = 1/s) place transform ! 𝐶𝑉! ! = 𝑅𝐶𝑠𝑌 𝑠 + Y(s)= Y(s)(1+RCs) 𝑌 𝑠 = 𝐶𝑉! 𝑅𝐶 3–23 𝐶𝑉! 1 = 𝐶𝑉! 𝑅𝐶 ; 𝐴 = 𝑅𝐶𝑠 𝑠(1 + 𝑅𝐶𝑠) 𝐴(1 + 𝐴) ! ! − !!! = 𝐶𝑉! 𝑅𝐶 ! 𝑌 𝑠 = !!! !" !"# − !!! !!! !" !!!"# = 𝐶𝑉! RULE af (t) è aF(s) 𝐶𝑉! 1 − ℒ !! ! − !(!!!) = 𝐶𝑉! 𝑅𝐶 !(!!!) !" (!!!"#) ; ! ! − ! !(!!!) !" (!!!"#) !" (!!!"#) = ! ( ! !!) !" = ! (!! ! ) !" 𝑅𝑈𝐿𝐸 scaling by a generates 1/a in the transform or change the form of the fraction as above and read off the transform 𝐶𝑉! 1 − ! 𝑅𝐶 ! ! 𝑒 !" = 𝑄! 1 − 𝑒 !!" 𝑅𝐶 Examples simple discharge RC circuit: PSfrag replacements R circuit equation: RCv v C solution: v(t) = v(0)e 𝑄 𝐼𝑅 + = 0 𝐶 population dynamics: 𝑑𝑄 → 𝑠𝑌 𝑠 − 𝑦 0 𝑑𝑡 𝑄 𝑌(𝑠) is population • →y(t) of some bacteria at time t 0•=growth 𝑅𝐶 𝑠𝑌 𝑠 − 𝑦 0decay + 𝑌 𝑠 if negative) (or 0 = 𝑅𝐶 𝑌 𝑠 𝑠 + 1/𝑅𝐶 − 𝑅𝐶𝑦(0) d is death 𝑦(0) rate 𝑌 𝑠 = 𝑠 + 1/𝑅𝐶 (b−d)t • y(t) = y(0)e 𝐼 0 𝑒 ! rate is y ! = by − dy (grows if b > d; decays if b < d ! !" Natural response of first and second order systems Second order systems a(s2Y (s) − sy(0) − y !(0)) + b(sY (s) − y(0)) + cY (s) = 0 ! "# $ L(y !! ) ! solve for Y (just!!algebra!)! ay + by + cy = 0 "# L(y ! ) $ αs + β (0)χ+ by(0) asy(0) + ay of Roots solution of ay !! + byY! (s) + cy = 2 = = 0 is2 as + bs + −1) c as + bs + c a > 0 (otherwise multiply equation (two) roots of characteristic polynomial by χ are !√ " ! (0) + by(0) where α = ay(0) and β = ay−b 2 ± b αs − 4ac+ β = by Laplace transform:y(t) =λ1,2 L−1 2a 2 as + bs c 4ac) Natural response of first and distinct second order systems Real roots (b2+> ! as roots dependent he ccY onstants −denominator y (0)) + hb(sY (s) − y(0)on ) t+ (s) = 0 a(s Y (s) − sy(0) cases: ! "# $ ! "# $ 2 three +• roots bs c realisand called characteristic polynomial of the χ(s) = as L(y !! ) + 2 > 4ac are distinct: roots χ(s) Real = a(s − λ1b)(s −L(y λ2!)) (λ1, λ2 real) 2 i.e., we have as ! + bs + c = a(s − λ1)(s − λ2) 4–7 2 λ1 = −b + √ b2 − 4ac , 2a λ2 = −b − system √ b2 − 4ac 2a form of y = L (Y ) depends on roots of characteristic polynomial χ mY page (just4-6, algebra!) 2 • roots are real and equal: b = 4ac αs + β coefficients of numerator αs + β come Y (s) = λ1 = λ2 = −b/(2a)from initial conditions a(s − λ1)(s −αs λ2+ ) β by(0) asy(0) + ay !(0) + = 2 Y (s) = 2 as + bs + c as + bs + c ere α, β depend on initial conditions −1 Natural response of first and second order systems press Y asand β = = ay(0) 4–9 write this as a product of {Partial fractions} 𝑥 𝑦 ! + (0) by(0) ay (𝑠 − 𝜆1)+ (𝑠 − 𝜆2) r2 r1 + Y (s) = Find the transforms s −from λ1 the table. s − λ2 Partial Fractions ere r1 and r2 are found from R(x) = ƒ(x)/g(x) se of first and second order systems ich yields 4–7 r1 + r2 = α/a, −λ 2 r1 − λ1 r2 = β/a λ α+β , r1 = √ 2 b − 4ac al response of first and second order systems 1 al response of first and second order systems −λ2α − β r2 = √ b2 − 4ac 4– 4–1