SECTION 4: FIRST- AND SECOND-ORDER SYSTEMS ESE 499 – Feedback Control Systems First- and Second-Order Systems 2 All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 𝑠𝑠 𝑠𝑠 − 𝑝𝑝1 ⋯ 𝑠𝑠 − 𝑝𝑝𝑛𝑛 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 𝑠𝑠 2 + 𝑎𝑎11 𝑠𝑠 + 𝑎𝑎10 ⋯ 𝑠𝑠 2 + 𝑎𝑎𝑚𝑚1 𝑠𝑠 + 𝑎𝑎𝑚𝑚0 Real poles – 1st-order terms Complex-conjugate poles – 2nd-order terms These terms and, therefore, the poles determine the nature of the timedomain response 𝐺𝐺 𝑠𝑠 = Real poles – decaying exponentials Complex-conjugate poles - decaying sinusoids All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids K. Webb These are the natural modes or eigenmodes of the system ESE 499 First- and Second-Order Systems 3 Most real-world systems are higher than 1st or 2nd order But, many higher-order systems can reasonably be approximated as 1st or 2nd order If they have a dominant pole or dominant pair of poles Greatly simplifies control system design Instructive to examine the responses of 1st- or 2nd-order systems K. Webb Gain insight into relationships between pole locations and dynamic response Next, we’ll look at 1st- and 2nd-order impulse and step responses ESE 499 4 K. Webb Response of First-Order Systems ESE 499 First-Order System – Impulse Response 5 First-order transfer function: 𝐺𝐺 𝑠𝑠 = Single real pole at 𝐴𝐴 𝑠𝑠+𝜎𝜎 1 𝑠𝑠 = −𝜎𝜎 = − 𝜏𝜏 where 𝜏𝜏 is the system time constant Impulse response: 𝑔𝑔 𝑡𝑡 = ℒ −1 𝐺𝐺 𝑠𝑠 K. Webb 𝑔𝑔 𝑡𝑡 = 𝐴𝐴𝑒𝑒 𝑡𝑡 −𝜏𝜏 = 𝐴𝐴𝑒𝑒 −𝜎𝜎𝜎𝜎 = 𝐴𝐴𝑒𝑒 𝑡𝑡 −𝜏𝜏 ESE 499 First-Order System – Impulse Response 6 Initial slope is inversely proportional to time constant Response completes 63% of transition after one time constant Decays to zero as long as the pole is negative K. Webb ESE 499 Impulse Response vs. Pole Location 7 Increasing 𝜎𝜎 corresponds to decreasing 𝜏𝜏 and a faster response K. Webb ESE 499 First-Order System – Step Response 8 Step response in the Laplace domain 𝑌𝑌 𝑠𝑠 = � 𝐺𝐺 𝑠𝑠 = 1 𝑠𝑠 𝐴𝐴 𝑠𝑠 𝑠𝑠+𝜎𝜎 Inverse transform back to time domain via partial fraction expansion 𝑌𝑌 𝑠𝑠 = 𝐴𝐴 𝑠𝑠 𝑠𝑠+𝜎𝜎 = 𝑟𝑟1 𝑠𝑠 + 𝑟𝑟2 𝑠𝑠+𝜎𝜎 𝐴𝐴 = 𝑟𝑟1 + 𝑟𝑟2 𝑠𝑠 + 𝜎𝜎𝑟𝑟1 𝑠𝑠 0 : 𝜎𝜎𝑟𝑟1 = 𝐴𝐴 → 𝑟𝑟1 = 𝑌𝑌 𝑠𝑠 = 𝑠𝑠1 : 𝑟𝑟1 + 𝑟𝑟2 = 0 → 𝑟𝑟2 = − 𝐴𝐴/𝜎𝜎 𝑠𝑠 Time-domain step response K. Webb 𝐴𝐴 𝜎𝜎 − 𝐴𝐴/𝜎𝜎 𝑠𝑠+𝜎𝜎 𝐴𝐴 𝜎𝜎 𝑡𝑡 𝐴𝐴 𝐴𝐴 −𝜎𝜎𝜎𝜎 − 𝑦𝑦 𝑡𝑡 = − 𝑒𝑒 = 𝐵𝐵 − 𝐵𝐵𝑒𝑒 𝜏𝜏 𝜎𝜎 𝜎𝜎 ESE 499 First-Order System – Step Response 9 Initial slope is inversely proportional to time constant Response completes 63% of transition after one time constant Almost completely settled after 7𝜏𝜏 K. Webb ESE 499 Step Response vs. Pole Location 10 Increasing 𝜎𝜎 corresponds to decreasing 𝜏𝜏 and a faster response K. Webb ESE 499 Pole Location and Stability 11 First-order transfer function 𝐴𝐴 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 − 𝑝𝑝 where 𝑝𝑝 is the system pole Impulse response is 𝑔𝑔 𝑡𝑡 = 𝐴𝐴𝑒𝑒 𝑝𝑝𝑝𝑝 If 𝑝𝑝 < 0, 𝑔𝑔 𝑡𝑡 decays to zero Pole in the left half-plane System is stable If 𝑝𝑝 > 0, 𝑔𝑔 𝑡𝑡 grows without bound K. Webb Pole in the right half-plane System is unstable ESE 499 12 K. Webb Response of Second-Order Systems ESE 499 Second-Order Systems 13 Second-order transfer function 𝐺𝐺 𝑠𝑠 = 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 𝑠𝑠 2 +𝑎𝑎1 𝑠𝑠+𝑎𝑎0 = 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 2 𝑠𝑠+𝜎𝜎 2 +𝜔𝜔𝑑𝑑 (1) where 𝜔𝜔𝑑𝑑 is the damped natural frequency Can also express the 2nd-order transfer function as 𝐺𝐺 𝑠𝑠 = 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 2 𝑠𝑠 2 +2𝜁𝜁𝜔𝜔𝑛𝑛 𝑠𝑠+𝜔𝜔𝑛𝑛 (2) where 𝜔𝜔𝑛𝑛 is the un-damped natural frequency, and 𝜁𝜁 is the damping ratio 𝜔𝜔𝑑𝑑 = 𝜔𝜔𝑛𝑛 1 − 𝜁𝜁 2 𝜁𝜁 = 𝜎𝜎 𝜔𝜔𝑛𝑛 Two poles at 𝑠𝑠1,2 = −𝜎𝜎 ± 𝜎𝜎 2 − 𝜔𝜔𝑛𝑛2 = −𝜁𝜁𝜔𝜔𝑛𝑛 ± 𝜔𝜔𝑛𝑛 𝜁𝜁 2 − 1 K. Webb ESE 499 Categories of Second-Order Systems 14 The 2nd-order system poles are Value of 𝜁𝜁 determines the nature of the poles and, therefore, the response 𝜻𝜻 > 𝟏𝟏: Over-damped Two identical, real poles – time-scaled decaying exponentials 𝑠𝑠1,2 = −𝜎𝜎 = −𝜁𝜁𝜔𝜔𝑛𝑛 = −𝜔𝜔𝑛𝑛 𝟎𝟎 < 𝜻𝜻 < 𝟏𝟏: Under-damped Two distinct, real poles – sum of decaying exponentials – treat as two first-order terms 𝑠𝑠1 = −𝜎𝜎1 , 𝑠𝑠2 = −𝜎𝜎2 𝜻𝜻 = 𝟏𝟏: Critically-damped 𝑠𝑠1,2 = −𝜁𝜁𝜔𝜔𝑛𝑛 ± 𝜔𝜔𝑛𝑛 𝜁𝜁 2 − 1 Complex-conjugate pair of poles – sum of decaying sinusoids 𝑠𝑠1,2 = −𝜎𝜎 ± 𝑗𝑗𝜔𝜔𝑑𝑑 = −𝜁𝜁𝜔𝜔𝑛𝑛 ± 𝑗𝑗𝜔𝜔𝑛𝑛 1 − 𝜁𝜁 2 𝜻𝜻 = 𝟎𝟎: Un-damped K. Webb Purely-imaginary, conjugate pair of poles – sum of non-decaying sinusoids 𝑠𝑠1,2 = ±𝑗𝑗𝜔𝜔𝑛𝑛 ESE 499 2nd-Order Pole Locations and Damping 15 K. Webb ESE 499 16 Second-Order Poles - 0 ≤ 𝜁𝜁 ≤ 1 Can relate 𝜎𝜎, 𝜔𝜔𝑑𝑑 , 𝜔𝜔𝑛𝑛 , and 𝜁𝜁 to pole location geometry 𝜎𝜎 is the real part 𝜔𝜔𝑑𝑑 K. Webb 𝜔𝜔𝑛𝑛 is the imaginary part is the pole magnitude 𝜁𝜁 is a measure of system damping 𝜁𝜁 = 𝜎𝜎 𝜔𝜔𝑛𝑛 = sin 𝜃𝜃 ESE 499 Impulse Response – Critically-Damped 17 For 𝜁𝜁 = 1, the transfer function reduces to 𝐴𝐴 𝐴𝐴 𝐺𝐺 𝑠𝑠 = 2 2 = 𝑠𝑠 + 𝜔𝜔 𝑠𝑠 + 2𝜔𝜔𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛 𝑛𝑛 2 𝐴𝐴 = 𝑠𝑠 + 𝜎𝜎 2 Impulse response 𝑔𝑔 𝑡𝑡 = ℒ −1 𝐺𝐺 𝑠𝑠 𝑔𝑔 𝑡𝑡 = 𝐴𝐴𝐴𝐴𝑒𝑒 −𝜎𝜎𝜎𝜎 K. Webb ESE 499 Impulse Response – Critically-Damped 18 Speed of response is proportional to 𝜎𝜎 K. Webb ESE 499 Impulse Response – Under-Damped 19 For 0 < 𝜁𝜁 < 1, the transfer function is 𝐴𝐴 𝐺𝐺 𝑠𝑠 = 2 𝑠𝑠 + 2𝜁𝜁𝜔𝜔𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛2 Complete the square on the denominator 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 + 𝜁𝜁𝜔𝜔𝑛𝑛 2 𝐴𝐴 + 𝜔𝜔𝑛𝑛 1 − 𝜁𝜁 2 Rewrite in the form of a damped sinusoid 𝐺𝐺 𝑠𝑠 = 2 = 𝐴𝐴 𝑠𝑠 + 𝜁𝜁𝜔𝜔𝑛𝑛 2 + 𝜔𝜔𝑑𝑑2 𝐴𝐴 𝜔𝜔𝑑𝑑 𝐴𝐴 𝜔𝜔𝑑𝑑 = 𝜔𝜔𝑑𝑑 𝑠𝑠 + 𝜁𝜁𝜔𝜔𝑛𝑛 2 + 𝜔𝜔𝑑𝑑2 𝜔𝜔𝑑𝑑 𝑠𝑠 + 𝜎𝜎 2 + 𝜔𝜔𝑑𝑑2 Inverse Laplace transform for the time-domain impulse response K. Webb 𝐴𝐴 −𝜎𝜎𝜎𝜎 𝑔𝑔 𝑡𝑡 = 𝑒𝑒 sin(𝜔𝜔𝑑𝑑 𝑡𝑡) 𝜔𝜔𝑑𝑑 ESE 499 20 Under-Damped Impulse Response vs. 𝜔𝜔𝑛𝑛 𝐴𝐴 −𝜎𝜎𝜎𝜎 𝑔𝑔 𝑡𝑡 = 𝑒𝑒 sin 𝜔𝜔𝑑𝑑 𝑡𝑡 = 𝐵𝐵𝑒𝑒 −𝜁𝜁𝜔𝜔𝑛𝑛 𝑡𝑡 sin 𝜔𝜔𝑛𝑛 1 − 𝜁𝜁 2 𝑡𝑡 𝜔𝜔𝑑𝑑 K. Webb ESE 499 21 Under-Damped Impulse Response vs. 𝜁𝜁 𝐴𝐴 −𝜎𝜎𝜎𝜎 𝑔𝑔 𝑡𝑡 = 𝑒𝑒 sin 𝜔𝜔𝑑𝑑 𝑡𝑡 = 𝐵𝐵𝑒𝑒 −𝜁𝜁𝜔𝜔𝑛𝑛 𝑡𝑡 sin 𝜔𝜔𝑛𝑛 1 − 𝜁𝜁 2 𝑡𝑡 𝜔𝜔𝑑𝑑 K. Webb ESE 499 Impulse Response – Un-Damped 22 For 𝜁𝜁 = 0, the transfer function reduces to 𝐴𝐴 𝐺𝐺 𝑠𝑠 = 2 𝑠𝑠 + 𝜔𝜔𝑛𝑛2 Putting into the form of a sinusoid 𝐴𝐴 𝜔𝜔𝑛𝑛 𝐺𝐺 𝑠𝑠 = 𝜔𝜔𝑛𝑛 𝑠𝑠 2 + 𝜔𝜔𝑛𝑛2 Inverse transform to get the time-domain impulse response 𝑔𝑔 𝑡𝑡 = ℒ −1 𝐺𝐺 𝑠𝑠 An un-damped sinusoid K. Webb 𝐴𝐴 𝑔𝑔 𝑡𝑡 = sin 𝜔𝜔𝑛𝑛 𝑡𝑡 𝜔𝜔𝑛𝑛 ESE 499 23 Un-Damped Impulse Response vs. 𝜔𝜔𝑛𝑛 𝐴𝐴 𝑔𝑔 𝑡𝑡 = sin 𝜔𝜔𝑛𝑛 𝑡𝑡 𝜔𝜔𝑛𝑛 K. Webb ESE 499 Second-Order Step Response 24 The Laplace transform of the step response is 1 𝑌𝑌 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝑠𝑠 The time-domain step response for each damping case can be derived as the the inverse transform of 𝑌𝑌 𝑠𝑠 𝑦𝑦 𝑡𝑡 = ℒ −1 𝑌𝑌 𝑠𝑠 or as the integral of the corresponding impulse response 𝑡𝑡 K. Webb 𝑦𝑦 𝑡𝑡 = � 𝑔𝑔 𝜏𝜏 𝑑𝑑𝑑𝑑 0 ESE 499 25 Critically-Damped Step Response vs. 𝜎𝜎 𝑦𝑦 𝑡𝑡 = K. Webb ℒ −1 1 𝐺𝐺 𝑠𝑠 𝑠𝑠 𝐴𝐴 = 2 1 − 𝑒𝑒 −𝜎𝜎𝜎𝜎 − 𝜎𝜎𝜎𝜎𝑒𝑒 −𝜎𝜎𝜎𝜎 𝜎𝜎 ESE 499 26 Under-Damped Step Response vs. 𝜔𝜔𝑛𝑛 𝑦𝑦 𝑡𝑡 = K. Webb ℒ −1 1 𝐺𝐺 𝑠𝑠 𝑠𝑠 𝐴𝐴 𝜎𝜎 −𝜎𝜎𝜎𝜎 −𝜎𝜎𝜎𝜎 = 2 1 − 𝑒𝑒 cos 𝜔𝜔𝑑𝑑 𝑡𝑡 − 𝑒𝑒 sin 𝜔𝜔𝑑𝑑 𝑡𝑡 𝜔𝜔𝑑𝑑 𝜔𝜔𝑛𝑛 ESE 499 27 Under-Damped Step Response vs. 𝜁𝜁 𝑦𝑦 𝑡𝑡 = K. Webb ℒ −1 1 𝐺𝐺 𝑠𝑠 𝑠𝑠 𝐴𝐴 𝜎𝜎 −𝜎𝜎𝜎𝜎 −𝜎𝜎𝜎𝜎 = 2 1 − 𝑒𝑒 cos 𝜔𝜔𝑑𝑑 𝑡𝑡 − 𝑒𝑒 sin 𝜔𝜔𝑑𝑑 𝑡𝑡 𝜔𝜔𝑑𝑑 𝜔𝜔𝑛𝑛 ESE 499 28 Un-Damped Step Response vs. 𝜔𝜔𝑛𝑛 𝑦𝑦 𝑡𝑡 = K. Webb ℒ −1 1 𝐺𝐺 𝑠𝑠 𝑠𝑠 𝐴𝐴 = 2 1 − cos 𝜔𝜔𝑛𝑛 𝑡𝑡 𝜔𝜔𝑛𝑛 ESE 499 29 K. Webb Step Response Characteristics ESE 499 Step Response – Risetime 30 Risetime is the time it takes a signal to transition between two set levels Typically 10% to 90% of full transition Sometimes 20% to 80% A measure of the speed of a response Very rough approximation: K. Webb 𝑡𝑡𝑟𝑟 ≈ 1.8 𝜔𝜔𝑛𝑛 ESE 499 Step Response – Overshoot 31 Overshoot is the magnitude of a signal’s excursion beyond its final value Expressed as a percentage of fullscale swing Overshoot increases as 𝜁𝜁 decreases K. Webb 𝜻𝜻 %OS 0.45 20 0.5 16 0.6 10 0.7 5 %𝑂𝑂𝑂𝑂 = 𝑒𝑒 − 𝜁𝜁𝜁𝜁 1−𝜁𝜁 2 ⋅ 100% ESE 499 Step Response –Settling Time 32 Settling time is the time it takes a signal to settle, finally, to within some percentage of its final value Typically ±1% or ±5% Inversely proportional to the real part of the poles, 𝜎𝜎 For ±1% settling: K. Webb 𝑡𝑡𝑠𝑠 ≈ 4.6 𝜎𝜎 = 4.6 𝜁𝜁𝜔𝜔𝑛𝑛 ESE 499 33 The Convolution Integral In this sub-section, we’ll see that the timedomain output of a system is given by the convolution of its time-domain input and its impulse response. K. Webb ESE 499 Convolution Integral 34 Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function 𝑌𝑌 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 � 𝑈𝑈 𝑠𝑠 Recall that multiplication in the Laplace domain corresponds to convolution in the time domain 𝑦𝑦 𝑡𝑡 = ℒ −1 𝐺𝐺 𝑠𝑠 𝑈𝑈 𝑠𝑠 = 𝑔𝑔 𝑡𝑡 ∗ 𝑢𝑢 𝑡𝑡 Time-domain output given by the convolution of the input signal and the impulse response K. Webb 𝑦𝑦 𝑡𝑡 = 𝑔𝑔 𝑡𝑡 ∗ 𝑢𝑢 𝑡𝑡 = 𝑡𝑡 ∫0 𝑔𝑔 𝜏𝜏 𝑢𝑢 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝑑𝑑 ESE 499 Convolution 35 Time-domain output is the input convolved with the impulse response 𝑡𝑡 𝑦𝑦 𝑡𝑡 = 𝑔𝑔 𝑡𝑡 ∗ 𝑢𝑢 𝑡𝑡 = � 𝑔𝑔 𝜏𝜏 𝑢𝑢 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝑑𝑑 Input is flipped in time and shifted by 𝑡𝑡 Multiply impulse response and flipped/shifted input Integrate over 𝜏𝜏 = 0 … 𝑡𝑡 0 Each time point of 𝑦𝑦 𝑡𝑡 given by result of integral with 𝑢𝑢 −𝜏𝜏 shifted by 𝑡𝑡 K. Webb ESE 499 Convolution 36 K. Webb ESE 499 Convolution 37 K. Webb ESE 499 38 Time-Domain Analysis in MATLAB A few of MATLAB’s many built-in functions that are useful for simulating linear systems are listed in the following sub-section. K. Webb ESE 499 System Objects 39 MATLAB has data types dedicated to linear system models Two primary system model objects: Transfer function model State-space model Objects created by calling MATLAB functions – creates a transfer function model ss.m – creates a state-space model zpk.m – creates a zero-pole-gain model tf.m K. Webb ESE 499 Transfer Function Model – tf(…) 40 sys = tf(Num,Den) Num: vector of numerator polynomial coefficients Den: vector of denominator polynomial coefficients sys: transfer function model object Transfer function is assumed to be of the form 𝑏𝑏1 𝑠𝑠 𝑟𝑟 + 𝑏𝑏2 𝑠𝑠 𝑟𝑟−1 + ⋯ + 𝑏𝑏𝑟𝑟 𝑠𝑠 + 𝑏𝑏𝑟𝑟+1 𝐺𝐺 𝑠𝑠 = 𝑎𝑎1 𝑠𝑠 𝑛𝑛 + 𝑎𝑎2 𝑠𝑠 𝑛𝑛−1 + ⋯ + 𝑎𝑎𝑛𝑛 𝑠𝑠 + 𝑎𝑎𝑛𝑛+1 Inputs to tf(…) are Num = [b1,b2,…,br+1]; Den = [a1,a2,…,an+1]; K. Webb ESE 499 Step Response Simulation – step(…) 41 [y,t] = step(sys,t) system model t: optional time vector or final time value y: output step response t: output time vector sys: If no outputs are specified, step response is automatically plotted Time vector (or final value) input is optional If K. Webb not specified, MATLAB will generate automatically ESE 499 Impulse Response Simulation – impulse(…) 42 [y,t] = impulse(sys,t) system model t: optional time vector or final time value y: output impulse response t: output time vector sys: If no outputs are specified, impulse response is automatically plotted Time vector (or final value) input is optional If K. Webb not specified, MATLAB will generate automatically ESE 499 Linear System Simulation – lsim(…) 43 [y,t,x] = lsim(sys,u,t,x0) sys: system model u: input signal vector t: time vector corresponding to the input signal x0: optional initial conditions – (for ss model only) y: output response t: output time vector x: optional trajectory of all states – (for ss model only) If no outputs are specified, response is automatically plotted Input can be any arbitrary signal K. Webb ESE 499 More MATLAB Functions 44 A few more useful MATLAB functions Pole/zero analysis: pzmap(…) pole(…) zero(…) eig(…) Input signal generation: gensig(…) Refer to MATLAB help documentation for more information K. Webb ESE 499 45 Response to Sinusoidal Inputs In this subsection of notes, we’ll examine the response of linear systems to sinusoidal inputs K. Webb ESE 499 Frequency-Domain Analysis – Introduction 46 We’ve looked at system impulse and step responses Also interested in the response to periodic inputs Fourier theory tells us that any periodic signal can be represented as a sum of harmonically-related sinusoids The Fourier series: ∞ 𝑎𝑎0 + � 𝑎𝑎𝑛𝑛 cos 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 + 𝑏𝑏𝑛𝑛 sin 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 𝑓𝑓 𝑡𝑡 = 2 𝑛𝑛=1 where 𝑎𝑎𝑛𝑛 and 𝑏𝑏𝑛𝑛 are given by the Fourier integrals Sinusoids are basis signals from which all other periodic signals can be constructed K. Webb Sinusoidal system response is of particular interest ESE 499 Fourier Series 47 K. Webb ESE 499 System Response to a Sinusoidal Input 48 Consider an 𝑛𝑛𝑡𝑡𝑡 -order system 𝑛𝑛 poles: 𝑝𝑝1 , 𝑝𝑝2 , … 𝑝𝑝𝑛𝑛 Real or complex Assume all are distinct Transfer function is: 𝐺𝐺 𝑠𝑠 = 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 𝑠𝑠−𝑝𝑝1 𝑠𝑠−𝑝𝑝2 ⋯ 𝑠𝑠−𝑝𝑝𝑛𝑛 (1) Apply a sinusoidal input to the system 𝑢𝑢 𝑡𝑡 = 𝐴𝐴 sin 𝜔𝜔𝜔𝜔 Output is given by K. Webb 𝑌𝑌 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝑈𝑈 𝑠𝑠 = ℒ 𝑈𝑈 𝑠𝑠 = 𝐴𝐴 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 𝑠𝑠−𝑝𝑝1 𝑠𝑠−𝑝𝑝2 ⋯ 𝑠𝑠−𝑝𝑝𝑛𝑛 𝜔𝜔 𝑠𝑠 2 +𝜔𝜔2 � 𝐴𝐴 𝜔𝜔 𝑠𝑠 2 +𝜔𝜔2 (2) ESE 499 System Response to a Sinusoidal Input 49 Partial fraction expansion of (2) gives 𝑌𝑌 𝑠𝑠 = 𝑟𝑟1 𝑠𝑠−𝑝𝑝1 + 𝑟𝑟2 𝑠𝑠−𝑝𝑝2 + ⋯+ 𝑟𝑟𝑛𝑛 𝑠𝑠−𝑝𝑝𝑛𝑛 + 𝑟𝑟𝑛𝑛+1 𝑠𝑠 𝑠𝑠 2 +𝜔𝜔2 + 𝑟𝑟𝑛𝑛+2 𝜔𝜔 𝑠𝑠 2 +𝜔𝜔2 (3) Inverse transform of (3) gives the time-domain output (4) 𝑦𝑦 𝑡𝑡 = 𝑟𝑟1 𝑒𝑒 𝑝𝑝1𝑡𝑡 + 𝑟𝑟2 𝑒𝑒 𝑝𝑝2𝑡𝑡 + ⋯ + 𝑟𝑟𝑛𝑛 𝑒𝑒 𝑝𝑝𝑛𝑛𝑡𝑡 + 𝑟𝑟𝑛𝑛+1 cos 𝜔𝜔𝜔𝜔 + 𝑟𝑟𝑛𝑛+2 sin 𝜔𝜔𝜔𝜔 transient steady state Two portions of the response: Transient Steady state K. Webb Decaying exponentials or sinusoids – goes to zero in steady state Natural response to initial conditions Due to the input – sinusoidal in steady state ESE 499 Steady-State Sinusoidal Response 50 We are interested in the steady-state response A trig. identity provides insight into 𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡 : where (5) 𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡 = 𝑟𝑟𝑛𝑛+1 cos 𝜔𝜔𝜔𝜔 + 𝑟𝑟𝑛𝑛+2 sin 𝜔𝜔𝜔𝜔 𝛼𝛼 cos 𝜔𝜔𝜔𝜔 + 𝛽𝛽 sin 𝜔𝜔𝜔𝜔 = 𝜙𝜙 = tan−1 𝛼𝛼 𝛽𝛽 𝛼𝛼 2 + 𝛽𝛽 2 sin 𝜔𝜔𝜔𝜔 + 𝜙𝜙 Steady-state response to a sinusoidal input 𝑢𝑢 𝑡𝑡 = 𝐴𝐴 sin 𝜔𝜔𝜔𝜔 is a sinusoid of the same frequency, but, in general, different amplitude and phase Where K. Webb 𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡 = 𝐵𝐵 sin 𝜔𝜔𝜔𝜔 + 𝜙𝜙 𝐵𝐵 = 2 2 𝑟𝑟𝑛𝑛+1 + 𝑟𝑟𝑛𝑛+2 and (6) 𝜙𝜙 = tan−1 𝑟𝑟𝑛𝑛+1 𝑟𝑟𝑛𝑛+2 ESE 499 Steady-State Sinusoidal Response 51 𝑢𝑢 𝑡𝑡 = 𝐴𝐴 sin 𝜔𝜔𝜔𝜔 Steady-state sinusoidal response is a scaled and phase-shifted sinusoid of the same frequency Equal → 𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡 = 𝐵𝐵 sin 𝜔𝜔𝜔𝜔 + 𝜙𝜙 frequency is a property of linear systems Note the 𝜔𝜔 term in the numerator of (3) will affect the residues Residues determine amplitude and phase of the output Output amplitude and phase are frequency-dependent 𝜔𝜔 K. Webb 𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡 = 𝐵𝐵 𝜔𝜔 sin 𝜔𝜔𝜔𝜔 + 𝜙𝜙 𝜔𝜔 ESE 499 Steady-State Sinusoidal Response 52 𝑢𝑢 𝑡𝑡 = 𝐴𝐴 sin 𝜔𝜔𝜔𝜔 + 𝜃𝜃 Linear System 𝐺𝐺 𝑠𝑠 𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡 = 𝐵𝐵 sin 𝜔𝜔𝜔𝜔 + 𝜙𝜙 Gain – the ratio of amplitudes of the output and input of the system 𝐵𝐵 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 = 𝐴𝐴 Phase – phase difference between system input and output 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝜙𝜙 − 𝜃𝜃 Systems will, in general, exhibit frequency-dependent gain and phase We’d like to be able to determine these functions of frequency K. Webb The system’s frequency response ESE 499 53 Frequency Response A system’s frequency response, or sinusoidal transfer function, describes its gain and phase shift for sinusoidal inputs as a function of frequency. K. Webb ESE 499 54 Frequency response Function – 𝐺𝐺 𝑗𝑗𝑗𝑗 Frequency response function A complex-valued function of frequency Ratio of output amplitude to input amplitude ∠𝐺𝐺 𝑗𝑗𝑗𝑗 at each 𝜔𝜔 is the phase at that frequency 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝐺𝐺 𝑠𝑠 |𝑠𝑠→𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 at each 𝜔𝜔 is the gain at that frequency Substitute 𝑗𝑗𝑗𝑗 for 𝑠𝑠 in the transfer function Phase shift between input and output sinusoids Another valid system model K. Webb Typically represented graphically ESE 499 Plotting the Frequency Response Function 55 𝐺𝐺 𝑗𝑗𝑗𝑗 is a complex-valued function of frequency Has both magnitude and phase Plot gain and phase separately Frequency response plots formatted as Bode plots Two sets of axes: gain on top, phase below Identical, logarithmic frequency axes Gain axis is logarithmic – either explicitly or as units of decibels (dB) Phase axis is linear with units of degrees K. Webb ESE 499 Bode Plots 56 Units of magnitude are dB Magnitude plot on top Logarithmic frequency axes Units of phase are degrees K. Webb Phase plot below ESE 499 Interpreting Bode Plots 57 Bode plots tell you the gain and phase shift at all frequencies: choose a frequency, read gain and phase values from the plot For a 10KHz sinusoidal input, the gain is 0dB (1) and the phase shift is 0°. K. Webb For a 10MHz sinusoidal input, the gain is -32dB (0.025), and the phase shift is -176°. ESE 499 Interpreting Bode Plots 58 K. Webb ESE 499 Decibels - dB 59 Frequency response gain most often expressed and plotted with units of decibels (dB) A logarithmic scale Provides detail of very large and very small values on the same plot Commonly used for ratios of powers or amplitudes Conversion from a linear scale to dB: 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑑𝑑𝑑𝑑 = 20 ⋅ log10 𝐺𝐺 𝑗𝑗𝑗𝑗 Conversion from dB to a linear scale: K. Webb 𝐺𝐺 𝑗𝑗𝑗𝑗 = 10 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑑𝑑𝑑𝑑 20 ESE 499 Decibels – dB 60 Multiplying two gain values corresponds to adding their values in dB E.g., the overall gain of cascaded systems 𝐺𝐺1 𝑗𝑗𝑗𝑗 ⋅ 𝐺𝐺2 𝑗𝑗𝑗𝑗 𝑑𝑑𝑑𝑑 = 𝐺𝐺1 𝑗𝑗𝑗𝑗 𝑑𝑑𝑑𝑑 + 𝐺𝐺2 𝑗𝑗𝑗𝑗 𝑑𝑑𝑑𝑑 Negative dB values corresponds to sub-unity gain Positive dB values are gains greater than one dB K. Webb Linear dB Linear 60 1000 6 2 40 100 -3 20 10 -6 0 1 -20 1/√2 = 0.707 0.5 0.1 ESE 499 Value of Logarithmic Axes - dB 61 Gain axis is linear in dB Gain plotted in dB A logarithmic scale Allows for displaying detail at very large and very small levels on the same plot Two resonant peaks clearly visible Linear gain scale K. Webb Smaller peak has disappeared ESE 499 Value of Logarithmic Axes - dB 62 Frequency axis is logarithmic Log frequency axis Allows for displaying detail at very low and very high frequencies on the same plot Can resolve frequency of both resonant peaks Linear frequency axis K. Webb Lower resonant frequency is unclear ESE 499 Gain Response – Terminology 63 Corner frequency, cut off frequency, -3dB frequency: Frequency at which gain is 3dB below its low-frequency value 𝑓𝑓𝑐𝑐 = This is the bandwidth of the system Peaking K. Webb 𝜔𝜔𝑐𝑐 2𝜋𝜋 ~5𝑑𝑑𝑑𝑑 of peaking 𝜔𝜔𝑐𝑐 = 1.45 𝑓𝑓𝑐𝑐 = 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 𝜔𝜔𝑐𝑐 = 0.23𝐻𝐻𝐻𝐻 2𝜋𝜋 Any increase in gain above the low frequency gain ESE 499 64 Response of 1st- and 2nd-Order Factors This section examines the frequency responses of first- and second-order transfer function factors. K. Webb ESE 499 Transfer Function Factors 65 We’ve already seen that a transfer function denominator can be factored into firstand second-order terms 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 − 𝑝𝑝1 𝑠𝑠 − 𝑝𝑝2 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 2 2 ⋯ 𝑠𝑠 2 + 2𝜁𝜁1 𝜔𝜔𝑛𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛𝑛 𝑠𝑠 2 + 2𝜁𝜁2 𝜔𝜔𝑛𝑛2 𝑠𝑠 + 𝜔𝜔𝑛𝑛2 ⋯ The same is true of the numerator 2 2 𝑠𝑠 − 𝑧𝑧1 𝑠𝑠 − 𝑧𝑧2 ⋯ 𝑠𝑠 2 + 2𝜁𝜁𝑎𝑎 𝜔𝜔𝑛𝑛𝑎𝑎 𝑠𝑠 + 𝜔𝜔𝑛𝑛𝑎𝑎 𝑠𝑠 2 + 2𝜁𝜁2 𝜔𝜔𝑛𝑛𝑏𝑏 𝑠𝑠 + 𝜔𝜔𝑛𝑛𝑏𝑏 ⋯ 𝐺𝐺 𝑠𝑠 = 2 2 𝑠𝑠 − 𝑝𝑝1 𝑠𝑠 − 𝑝𝑝2 ⋯ 𝑠𝑠 2 + 2𝜁𝜁1 𝜔𝜔𝑛𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛𝑛 𝑠𝑠 2 + 2𝜁𝜁2 𝜔𝜔𝑛𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛𝑛 ⋯ Can think of the transfer function as a product of the individual factors For example, consider the following system Can rewrite as 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 − 𝑝𝑝1 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 − 𝑧𝑧1 ⋅ K. Webb 𝑠𝑠 − 𝑧𝑧1 2 𝑠𝑠 2 + 2𝜁𝜁1 𝜔𝜔𝑛𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛𝑛 1 1 ⋅ 2 2 𝑠𝑠 − 𝑝𝑝1 𝑠𝑠 + 2𝜁𝜁1 𝜔𝜔𝑛𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛𝑛 ESE 499 Transfer Function Factors 66 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 − 𝑧𝑧1 ⋅ 1 1 ⋅ 2 2 𝑠𝑠 − 𝑝𝑝1 𝑠𝑠 + 2𝜁𝜁1 𝜔𝜔𝑛𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛𝑛 Think of this as three cascaded transfer functions 𝐺𝐺1 𝑠𝑠 = 𝑠𝑠 − 𝑧𝑧1 , or 𝑈𝑈 𝑠𝑠 K. Webb 𝑈𝑈 𝑠𝑠 𝐺𝐺1 𝑠𝑠 𝑠𝑠 − 𝑧𝑧1 𝑌𝑌1 𝑠𝑠 𝐺𝐺2 𝑠𝑠 = 𝑌𝑌1 𝑠𝑠 1 𝑠𝑠 − 𝑝𝑝1 1 𝑠𝑠−𝑝𝑝1 , 𝐺𝐺2 𝑠𝑠 𝑌𝑌2 𝑠𝑠 𝐺𝐺3 𝑠𝑠 = 𝑌𝑌2 𝑠𝑠 1 2 𝑠𝑠 2 +2𝜁𝜁1 𝜔𝜔𝑛𝑛𝑛 𝑠𝑠+𝜔𝜔𝑛𝑛𝑛 𝐺𝐺3 𝑠𝑠 𝑌𝑌 𝑠𝑠 1 2 𝑠𝑠 2 + 2𝜁𝜁1 𝜔𝜔𝑛𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛𝑛 𝑌𝑌 𝑠𝑠 ESE 499 Transfer Function Factors 67 In the Laplace domain, transfer function of a cascade of systems is the product of the individual transfer functions In the time domain, overall impulse response is the convolution of the individual impulse responses Same holds true in the frequency domain Frequency response of a cascade is the product of the individual frequency responses Or, the product of individual factors 𝑈𝑈 𝑗𝑗𝑗𝑗 𝐺𝐺1 𝑗𝑗𝑗𝑗 𝑌𝑌1 𝑗𝑗𝑗𝑗 𝐺𝐺1 𝑗𝑗𝑗𝑗 𝑌𝑌2 𝑗𝑗𝑗𝑗 𝐺𝐺1 𝑗𝑗𝑗𝑗 𝑌𝑌 𝑗𝑗𝑗𝑗 Instructive, therefore, to understand the responses of the individual factors K. Webb First- and second-order poles and zeros ESE 499 First-Order Factors 68 First-order factors Single, real poles or zeros In the Laplace domain: 1 𝑠𝑠 𝐺𝐺 𝑠𝑠 = 𝑠𝑠, 𝐺𝐺 𝑠𝑠 = , 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 + 𝑎𝑎, 𝐺𝐺 𝑠𝑠 = In the frequency domain 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗, 𝐺𝐺 𝑗𝑗𝑗𝑗 = Pole/zero plots: K. Webb 1 , 𝑗𝑗𝑗𝑗 1 𝑠𝑠+𝑎𝑎 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 + 𝑎𝑎, 𝐺𝐺 𝑗𝑗𝑗𝑗 = 1 𝑗𝑗𝑗𝑗+𝑎𝑎 ESE 499 First-Order Factors – Zero at the Origin 69 A differentiator 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 Gain: 𝐺𝐺 𝑗𝑗𝑗𝑗 Phase: = 𝑗𝑗𝑗𝑗 = 𝜔𝜔 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 = +90°, ∀𝜔𝜔 K. Webb ESE 499 First-Order Factors – Pole at the Origin 70 An integrator 1 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 1 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 Gain: 𝐺𝐺 𝑗𝑗𝑗𝑗 Phase: 1 1 = = 𝑗𝑗𝑗𝑗 𝜔𝜔 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 = ∠ − 𝑗𝑗 K. Webb 1 𝜔𝜔 = −90°, ∀𝜔𝜔 ESE 499 First-Order Factors – Single, Real Zero 71 Single, real zero at 𝑠𝑠 = −𝑎𝑎 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 + 𝑎𝑎 Gain: 𝐺𝐺 𝑗𝑗𝑗𝑗 = for 𝜔𝜔 ≪ 𝑎𝑎 + 𝑎𝑎2 𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ 𝑎𝑎 𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ 𝜔𝜔 for 𝜔𝜔 ≫ 𝑎𝑎 K. Webb 𝜔𝜔 2 Phase: ∠𝐺𝐺 𝑗𝑗𝑗𝑗 = for 𝜔𝜔 ≪ 𝑎𝑎 tan−1 𝜔𝜔 𝑎𝑎 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠𝑎𝑎 = 0° for 𝜔𝜔 ≫ 𝑎𝑎 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠𝑗𝑗𝑗𝑗 = 90° ESE 499 First-Order Factors – Single, Real Zero 72 Corner frequency: 𝐺𝐺 𝑗𝑗𝜔𝜔𝑐𝑐 𝐺𝐺 𝑗𝑗𝜔𝜔𝑐𝑐 = 𝑎𝑎 2 = 1.414 ⋅ 𝑎𝑎 𝑑𝑑𝑑𝑑 = 𝑎𝑎 ∠𝐺𝐺 𝑗𝑗𝜔𝜔𝑐𝑐 = +45° 𝑑𝑑𝑑𝑑 + 3𝑑𝑑𝑑𝑑 For 𝜔𝜔 ≫ 𝜔𝜔𝑐𝑐 , gain increases at: 𝜔𝜔𝑐𝑐 = 𝑎𝑎 20𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑𝑑𝑑 6𝑑𝑑𝑑𝑑/𝑜𝑜𝑜𝑜𝑜𝑜 From ~0.1𝜔𝜔𝑐𝑐 to ~10𝜔𝜔𝑐𝑐 , phase increases at a rate of: K. Webb ~45°/𝑑𝑑𝑑𝑑𝑑𝑑 Rough approximation ESE 499 First-Order Factors – Single, Real Pole 73 Single, real pole at 𝑠𝑠 = −𝑎𝑎 1 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 + 𝑎𝑎 Gain: 𝐺𝐺 𝑗𝑗𝑗𝑗 for 𝜔𝜔 ≪ 𝑎𝑎 𝜔𝜔 2 + 𝑎𝑎2 𝐺𝐺 𝑗𝑗𝑗𝑗 1 ≈ 𝑎𝑎 𝐺𝐺 𝑗𝑗𝑗𝑗 1 ≈ 𝜔𝜔 for 𝜔𝜔 ≫ 𝑎𝑎 K. Webb = 1 Phase: ∠𝐺𝐺 𝑗𝑗𝑗𝑗 = − tan−1 for 𝜔𝜔 ≪ 𝑎𝑎 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠ for 𝜔𝜔 ≫ 𝑎𝑎 𝜔𝜔 𝑎𝑎 1 = 0° 𝑎𝑎 1 = −90° ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠ 𝑗𝑗𝑗𝑗 ESE 499 First-Order Factors – Single, Real Pole 74 Corner frequency: 𝐺𝐺 𝑗𝑗𝜔𝜔𝑐𝑐 𝐺𝐺 𝑗𝑗𝜔𝜔𝑐𝑐 = 𝑑𝑑𝑑𝑑 𝑎𝑎 1 = = 0.707 ⋅ 2 1 𝑎𝑎 𝑑𝑑𝑑𝑑 ∠𝐺𝐺 𝑗𝑗𝜔𝜔𝑐𝑐 = −45° 1 𝑎𝑎 − 3𝑑𝑑𝑑𝑑 For 𝜔𝜔 ≫ 𝜔𝜔𝑐𝑐 , gain decreases at: 𝜔𝜔𝑐𝑐 = 𝑎𝑎 −20𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑𝑑𝑑 −6𝑑𝑑𝑑𝑑/𝑜𝑜𝑜𝑜𝑜𝑜 From ~0.1𝜔𝜔𝑐𝑐 to ~10𝜔𝜔𝑐𝑐 , phase decreases at a rate of: K. Webb ~ − 45°/𝑑𝑑𝑑𝑑𝑑𝑑 Rough approximation ESE 499 Second-Order Factors 75 Complex-conjugate zeros 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 2 + 2𝜁𝜁𝜔𝜔𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛2 K. Webb Complex-conjugate poles 1 𝐺𝐺 𝑠𝑠 = 2 𝑠𝑠 + 2𝜁𝜁𝜔𝜔𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛2 𝜎𝜎 = 𝜁𝜁𝜔𝜔𝑛𝑛 , 𝜔𝜔𝑑𝑑 = 𝜔𝜔𝑛𝑛 1 − 𝜁𝜁 2 ESE 499 2nd-Order Factors – Complex-Conjugate Zeros 76 Complex-conjugate zeros at 𝑠𝑠 = −𝜎𝜎 ± 𝑗𝑗𝜔𝜔𝑑𝑑 Gain: for 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛2 𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ 𝜔𝜔𝑛𝑛2 𝐺𝐺 𝑗𝑗𝑗𝑗 = 2𝜁𝜁𝜔𝜔𝑛𝑛2 𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ 𝜔𝜔2 for 𝜔𝜔 = 𝜔𝜔𝑛𝑛 for 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛2 K. Webb 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 2 + 2𝜁𝜁𝜔𝜔𝑛𝑛 𝑗𝑗𝑗𝑗 + 𝜔𝜔𝑛𝑛2 Phase: for 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛2 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠𝜔𝜔𝑛𝑛2 = 0° for 𝜔𝜔 = 𝜔𝜔𝑛𝑛 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 = ∠𝑗𝑗𝑗𝑗𝑗𝜔𝜔𝑛𝑛2 = +90° for 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛2 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠ − 𝜔𝜔2 = +180° ESE 499 2nd-Order Factors – Complex-Conjugate Zeros 77 Response may dip below low-freq. value near 𝜔𝜔𝑛𝑛 Gain increases at +40𝑑𝑑𝑑𝑑/ 𝑑𝑑𝑑𝑑𝑑𝑑 or +12𝑑𝑑𝑑𝑑/𝑜𝑜𝑜𝑜𝑜𝑜 for 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛 Corner frequency depends on damping ratio, 𝜁𝜁 Peaking increases as 𝜁𝜁 decreases 𝑓𝑓𝑐𝑐 increases as 𝜁𝜁 decreases At 𝜔𝜔 = 𝜔𝜔𝑐𝑐 , ∠𝐺𝐺 𝑗𝑗𝑗𝑗 = 90° Phase transition abruptness depends on 𝜁𝜁 K. Webb ESE 499 2nd-Order Factors – Complex-Conjugate Poles 78 Complex-conjugate zeros at 𝑠𝑠 = −𝜎𝜎 ± 𝑗𝑗𝜔𝜔𝑑𝑑 𝐺𝐺 𝑗𝑗𝑗𝑗 = Gain: for 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛2 1 𝜔𝜔𝑛𝑛2 ≈ 𝐺𝐺 𝑗𝑗𝑗𝑗 1 = 2𝜁𝜁𝜔𝜔𝑛𝑛2 for 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛2 K. Webb 𝐺𝐺 𝑗𝑗𝑗𝑗 for 𝜔𝜔 = 𝜔𝜔𝑛𝑛 𝐺𝐺 𝑗𝑗𝑗𝑗 1 𝑗𝑗𝑗𝑗 2 + 2𝜁𝜁𝜔𝜔𝑛𝑛 𝑗𝑗𝑗𝑗 + 𝜔𝜔𝑛𝑛2 1 ≈ 2 𝜔𝜔 Phase: for 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛2 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠ for 𝜔𝜔 = 𝜔𝜔𝑛𝑛 1 2 = 0° 𝜔𝜔𝑛𝑛 1 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 = ∠ = −90° 𝑗𝑗𝑗𝑗𝑗𝜔𝜔𝑛𝑛2 for 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛2 1 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠ − 2 = −180° 𝜔𝜔 ESE 499 2nd-Order Factors – Complex-Conjugate Poles 79 Response may peak above low-freq. value near 𝜔𝜔𝑛𝑛 Gain decreases at −40𝑑𝑑𝑑𝑑/ 𝑑𝑑𝑑𝑑𝑑𝑑 or −12𝑑𝑑𝑑𝑑/𝑜𝑜𝑜𝑜𝑜𝑜 for 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛 Corner frequency depends on damping ratio, 𝜁𝜁 Peaking increases as 𝜁𝜁 decreases 𝑓𝑓𝑐𝑐 increases as 𝜁𝜁 decreases At 𝜔𝜔 = 𝜔𝜔𝑐𝑐 , ∠𝐺𝐺 𝑗𝑗𝑗𝑗 = −90° Phase transition abruptness depends on 𝜁𝜁 K. Webb ESE 499 Pole Location and Peaking 80 Peaking is dependent on 𝜁𝜁 – pole locations No peaking at all for 𝜁𝜁 ≥ 1/ 2 = 0.707 𝜁𝜁 = 0.707 – maximally-flat or Butterworth response K. Webb ESE 499 Frequency Response Components - Example 81 Consider the following system 20 𝑠𝑠 + 20 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 + 1 𝑠𝑠 + 100 The system’s frequency response function is 20 𝑗𝑗𝑗𝑗 + 20 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 + 1 𝑗𝑗𝑗𝑗 + 100 As we’ve seen we can consider this a product of individual frequency response factors 1 1 ⋅ 𝐺𝐺 𝑗𝑗𝑗𝑗 = 20 ⋅ 𝑗𝑗𝑗𝑗 + 20 ⋅ 𝑗𝑗𝑗𝑗 + 1 𝑗𝑗𝑗𝑗 + 100 Overall response is the composite of the individual responses K. Webb Product of individual gain responses – sum in dB Sum of individual phase responses ESE 499 Frequency Response Components - Example 82 Gain response K. Webb ESE 499 Frequency Response Components - Example 83 Phase response K. Webb ESE 499 84 Bode Plot Construction In this section, we’ll look at a method for sketching, by hand, a straight-line, asymptotic approximation for a Bode plot. K. Webb ESE 499 Bode Plot Construction 85 We’ve just seen that a system’s frequency response function can be factored into first- and secondorder terms Each factor contributes a component to the overall gain and phase responses Now, we’ll look at a technique for manually sketching a system’s Bode plot In practice, you’ll almost always plot with a computer But, learning to do it by hand provides valuable insight We’ll look at how to approximate Bode plots for each of the different factors K. Webb ESE 499 Bode Form of the Transfer function 86 Consider the general transfer function form: 2 𝑠𝑠 − 𝑧𝑧1 𝑠𝑠 − 𝑧𝑧2 ⋯ 𝑠𝑠 2 + 2𝜁𝜁𝑎𝑎 𝜔𝜔𝑛𝑛𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛𝑛𝑛 ⋯ 𝐺𝐺 𝑠𝑠 = 𝐾𝐾 2 𝑠𝑠 − 𝑝𝑝1 𝑠𝑠 − 𝑝𝑝2 ⋯ 𝑠𝑠 2 + 2𝜁𝜁1 𝜔𝜔𝑛𝑛1 𝑠𝑠 + 𝜔𝜔𝑛𝑛1 ⋯ We first want to put this into Bode form: 𝐺𝐺 𝑠𝑠 = 𝐾𝐾0 𝑠𝑠 +1 𝜔𝜔𝑐𝑐1 2𝜁𝜁 𝑠𝑠 𝑠𝑠 2 + 1 ⋯ 2 + 𝑎𝑎 𝑠𝑠 + 1 ⋯ 𝜔𝜔𝑐𝑐𝑏𝑏 𝜔𝜔𝑛𝑛𝑛𝑛 𝜔𝜔𝑛𝑛𝑛𝑛 𝑠𝑠 𝑠𝑠 2 2𝜁𝜁 + 1 ⋯ 2 + 1 𝑠𝑠 + 1 ⋯ 𝜔𝜔𝑐𝑐2 𝜔𝜔𝑛𝑛1 𝜔𝜔𝑛𝑛1 The corresponding frequency response function, in Bode form, is 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝐾𝐾0 𝑠𝑠 +1 𝜔𝜔𝑐𝑐𝑐𝑐 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐𝑐𝑐 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐1 𝑗𝑗𝑗𝑗 +1 ⋯ 𝜔𝜔𝑐𝑐𝑏𝑏 𝑗𝑗𝑗𝑗 +1 ⋯ 𝜔𝜔𝑐𝑐2 𝑗𝑗𝑗𝑗 𝜔𝜔𝑛𝑛𝑎𝑎 𝑗𝑗𝑗𝑗 𝜔𝜔𝑛𝑛1 2 2 + + 2𝜁𝜁𝑎𝑎 𝑗𝑗𝑗𝑗 + 1 ⋯ 𝜔𝜔𝑛𝑛𝑛𝑛 2𝜁𝜁1 𝑗𝑗𝑗𝑗 + 1 ⋯ 𝜔𝜔𝑛𝑛1 Putting 𝐺𝐺 𝑗𝑗𝑗𝑗 into Bode form requires putting each of the first- and second-order factors into Bode form K. Webb ESE 499 First-Order Factors in Bode Form 87 First-order frequency-response factors include: 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 𝑛𝑛 , 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 + 𝜎𝜎, 𝐺𝐺 𝑗𝑗𝑗𝑗 = For the first factor, 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 𝑛𝑛 , 𝑛𝑛 is a positive or negative integer 1 𝑗𝑗𝑗𝑗+𝜎𝜎 Already in Bode form For the second two, divide through by 𝜎𝜎, giving 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝜎𝜎 𝑗𝑗𝑗𝑗 𝜎𝜎 +1 and 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝜎𝜎 1 𝑗𝑗𝑗𝑗 +1 𝜎𝜎 Here, 𝜎𝜎 = 𝜔𝜔𝑐𝑐 , the corner frequency associated with that zero or pole, so 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝜔𝜔𝑐𝑐 K. Webb 𝑗𝑗𝑗𝑗 𝜔𝜔𝑐𝑐 +1 and 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝜔𝜔𝑐𝑐 1 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐 ESE 499 Second-Order Factors in Bode Form 88 Second-order frequency-response factors include: 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 + 2𝜁𝜁𝜔𝜔𝑛𝑛 𝑗𝑗𝑗𝑗 + 𝜔𝜔𝑛𝑛2 and 𝐺𝐺 𝑗𝑗𝑗𝑗 = Again, normalize the 𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 = 2 𝜔𝜔𝑛𝑛2 𝑗𝑗𝜔𝜔 2 𝜔𝜔𝑛𝑛 + 2𝜁𝜁 𝜔𝜔𝑛𝑛 0 coefficient, giving 𝑗𝑗𝑗𝑗 + 1 1 2 𝑗𝑗𝑗𝑗 2 +2𝜁𝜁𝜔𝜔𝑛𝑛 𝑗𝑗𝑗𝑗 +𝜔𝜔𝑛𝑛 and 𝐺𝐺 𝑗𝑗𝑗𝑗 = 2 1/𝜔𝜔𝑛𝑛 𝑗𝑗𝑗𝑗 2 2𝜁𝜁 +𝜔𝜔 𝜔𝜔𝑛𝑛 𝑛𝑛 𝑗𝑗𝑗𝑗 +1 Putting each factor into its Bode form involves factoring out any DC gain component Lump all of DC gains together into a single gain constant, 𝐾𝐾0 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝐾𝐾0 K. Webb 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐𝑐𝑐 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐𝑐 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐𝑐𝑐 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐𝑐 ⋯ ⋯ 𝑗𝑗𝑗𝑗 2 2𝜁𝜁𝑎𝑎 + 𝑗𝑗𝑗𝑗+1 𝜔𝜔𝑛𝑛𝑛𝑛 𝜔𝜔𝑛𝑛𝑛𝑛 𝑗𝑗𝑗𝑗 2 2𝜁𝜁1 + 𝑗𝑗𝑗𝑗+1 𝜔𝜔𝑛𝑛𝑛 𝜔𝜔𝑛𝑛𝑛 ⋯ ⋯ ESE 499 Bode Plot Construction 89 Frequency response function in Bode form 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝐾𝐾0 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐𝑐𝑐 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐𝑐 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐𝑐𝑐 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐𝑐 ⋯ ⋯ 𝑗𝑗𝑗𝑗 2 2𝜁𝜁𝑎𝑎 + 𝑗𝑗𝑗𝑗+1 𝜔𝜔𝑛𝑛𝑛𝑛 𝜔𝜔𝑛𝑛𝑛𝑛 𝑗𝑗𝑗𝑗 2 2𝜁𝜁1 + 𝑗𝑗𝑗𝑗+1 𝜔𝜔𝑛𝑛𝑛 𝜔𝜔𝑛𝑛𝑛 ⋯ ⋯ Product of a constant DC gain factor,𝐾𝐾0 , and firstand second-order factors Plot the frequency response of each factor individually, then combine graphically Overall response is the product of individual factors Product of gain responses – sum on a dB scale Sum of phase responses K. Webb ESE 499 Bode Plot Construction 90 Bode plot construction procedure: 1. 2. 3. Put the sinusoidal transfer function into Bode form Draw a straight-line asymptotic approximation for the gain and phase response of each individual factor Graphically add all individual response components and sketch the result Next, we’ll look at the straight-line asymptotic approximations for the Bode plots for each of the transfer function factors K. Webb ESE 499 Bode Plot – Constant Gain Factor 91 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝐾𝐾0 Constant gain 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝐾𝐾0 Constant Phase ∠𝐺𝐺 𝑗𝑗𝑗𝑗 = 0° K. Webb ESE 499 Bode Plot – Poles/Zeros at the Origin 92 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 𝑛𝑛 > 0: 𝑛𝑛 poles at the origin Gain: 𝑛𝑛 zeros at the origin 𝑛𝑛 < 0: 𝑛𝑛 Straight line Slope = 𝑛𝑛 ⋅ 20 0𝑑𝑑𝑑𝑑 at 𝜔𝜔 = 1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑛𝑛 ⋅ 6 Phase: ∠𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑛𝑛 ⋅ 90° K. Webb 𝑑𝑑𝑑𝑑 𝑜𝑜𝑜𝑜𝑜𝑜 ESE 499 Bode Plot – First-Order Zero 93 Single real zero at 𝑠𝑠 = −𝜔𝜔𝑐𝑐 Gain: 0𝑑𝑑𝑑𝑑 for 𝜔𝜔 < 𝜔𝜔𝑐𝑐 +20 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 = +6 𝑑𝑑𝑑𝑑 𝑜𝑜𝑜𝑜𝑜𝑜 for 𝜔𝜔 > 𝜔𝜔𝑐𝑐 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐 Straight-line asymptotes intersect at 𝜔𝜔𝑐𝑐 , 0𝑑𝑑𝑑𝑑 Phase: 0° for 𝜔𝜔 ≤ 0.1𝜔𝜔𝑐𝑐 45° for 𝜔𝜔 = 𝜔𝜔𝑐𝑐 90° for 𝜔𝜔 ≥ 10𝜔𝜔𝑐𝑐 +45° 𝑑𝑑𝑑𝑑𝑑𝑑 K. Webb for 0.1𝜔𝜔𝑐𝑐 ≤ 𝜔𝜔 ≤ 10𝜔𝜔𝑐𝑐 ESE 499 Bode Plot – First-Order Pole 94 Single real pole at 𝑠𝑠 = −𝜔𝜔𝑐𝑐 Gain: 0𝑑𝑑𝑑𝑑 for 𝜔𝜔 < 𝜔𝜔𝑐𝑐 𝑑𝑑𝑑𝑑 −20 𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 −6 𝑜𝑜𝑜𝑜𝑜𝑜 for 𝜔𝜔 > 𝜔𝜔𝑐𝑐 Straight-line asymptotes intersect at 𝜔𝜔𝑐𝑐 , 0𝑑𝑑𝑑𝑑 𝐺𝐺 𝑗𝑗𝑗𝑗 = 1 𝑗𝑗𝑗𝑗 +1 𝜔𝜔𝑐𝑐 Phase: 0° for 𝜔𝜔 ≤ 0.1𝜔𝜔𝑐𝑐 −45° for 𝜔𝜔 = 𝜔𝜔𝑐𝑐 −90° for 𝜔𝜔 ≥ 10𝜔𝜔𝑐𝑐 −45° 𝑑𝑑𝑑𝑑𝑑𝑑 K. Webb for 0.1𝜔𝜔𝑐𝑐 ≤ 𝜔𝜔 ≤ 10𝜔𝜔𝑐𝑐 ESE 499 Bode Plot – Second-Order Zero 95 Complex-conjugate zeros: 𝑠𝑠1,2 = −𝜎𝜎 ± 𝑗𝑗𝜔𝜔𝑑𝑑 Gain: 0𝑑𝑑𝑑𝑑 for 𝜔𝜔 < 𝜔𝜔𝑛𝑛 +40 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 = +12 𝑑𝑑𝑑𝑑 𝑜𝑜𝑜𝑜𝑜𝑜 for 𝜔𝜔 > 𝜔𝜔𝑛𝑛 Straight-line asymptotes intersect at 𝜔𝜔𝑛𝑛 , 0𝑑𝑑𝑑𝑑 𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝜔𝜔𝑛𝑛 2 + 2𝜁𝜁 𝑗𝑗𝑗𝑗 + 1 𝜔𝜔𝑛𝑛 𝜁𝜁-dependent peaking around 𝜔𝜔𝑛𝑛 Phase: 0° for 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛 90° for 𝜔𝜔 = 𝜔𝜔𝑛𝑛 180° for 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛 𝜁𝜁-dependent slope through 𝜔𝜔𝑛𝑛 K. Webb Step-change for low 𝜁𝜁 +180°/𝑑𝑑𝑑𝑑𝑑𝑑 for high 𝜁𝜁 ESE 499 Bode Plot – Second-Order Pole 96 Complex-conjugate poles: 𝑠𝑠1,2 = −𝜎𝜎 ± 𝑗𝑗𝜔𝜔𝑑𝑑 Gain: 0𝑑𝑑𝑑𝑑 for 𝜔𝜔 < 𝜔𝜔𝑛𝑛 −40 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 = −12 𝑑𝑑𝑑𝑑 𝑜𝑜𝑜𝑜𝑜𝑜 for 𝜔𝜔 > 𝜔𝜔𝑛𝑛 Straight-line asymptotes intersect at 𝜔𝜔𝑛𝑛 , 0𝑑𝑑𝑑𝑑 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 𝜔𝜔𝑛𝑛 2 + 1 2𝜁𝜁 𝑗𝑗𝑗𝑗 + 1 𝜔𝜔𝑛𝑛 𝜁𝜁-dependent peaking around 𝜔𝜔𝑛𝑛 Phase: 0° for 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛 −90° for 𝜔𝜔 = 𝜔𝜔𝑛𝑛 −180° for 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛 𝜁𝜁-dependent slope through 𝜔𝜔𝑛𝑛 K. Webb Step-change for low 𝜁𝜁 −180°/𝑑𝑑𝑑𝑑𝑑𝑑 for high 𝜁𝜁 ESE 499 Bode Plot Construction – Example 97 Consider a system with the following transfer function 𝐺𝐺 𝑠𝑠 = 10 𝑠𝑠 + 20 𝑠𝑠 𝑠𝑠 + 400 The sinusoidal transfer function: Put it into Bode form 𝐺𝐺 𝑗𝑗𝑗𝑗 = 10 𝑗𝑗𝑗𝑗 + 20 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 + 400 𝑗𝑗𝑗𝑗 +1 0.5 20 𝐺𝐺 𝑗𝑗𝑗𝑗 = = 𝑗𝑗𝑗𝑗 +1 𝑗𝑗𝑗𝑗 ⋅ 400 𝑗𝑗𝑗𝑗 ⋅ 400 Represent as a product of factors 10 ⋅ 20 𝐺𝐺 𝑗𝑗𝑗𝑗 = 0.5 ⋅ K. Webb 𝑗𝑗𝑗𝑗 +1 20 𝑗𝑗𝑗𝑗 +1 400 𝑗𝑗𝑗𝑗 1 1 +1 ⋅ ⋅ 𝑗𝑗𝑗𝑗 20 𝑗𝑗𝑗𝑗 +1 400 ESE 499 Bode Plot Construction – Example 98 K. Webb ESE 499 Bode Plot Construction – Example 99 K. Webb ESE 499 100 Relationship between Pole/Zero Plots and Bode Plots It is also possible to calculate a system’s frequency response directly from that system’s pole/zero plot. K. Webb ESE 499 Bode Construction from Pole/Zero Plots 101 Transfer function can be expressed as 𝐺𝐺 𝑠𝑠 = ∏𝑖𝑖 𝑠𝑠 − 𝑧𝑧𝑖𝑖 ∏𝑖𝑖 𝑠𝑠 − 𝑝𝑝𝑖𝑖 𝑠𝑠→𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 = ∏𝑖𝑖 𝑗𝑗𝑗𝑗 − 𝑧𝑧𝑖𝑖 ∏𝑖𝑖 𝑗𝑗𝑗𝑗 − 𝑝𝑝𝑖𝑖 Numerator is a product of first-order zero terms Denominator is a product of first-order pole terms 𝑗𝑗𝑗𝑗 is a point on the imaginary axis 𝑗𝑗𝑗𝑗 − 𝑧𝑧𝑖𝑖 represents a vector from 𝑧𝑧𝑖𝑖 to 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 − 𝑝𝑝𝑖𝑖 represents a vector from 𝑝𝑝𝑖𝑖 to 𝑗𝑗𝑗𝑗 Gain is given by 𝐺𝐺 𝑗𝑗𝑗𝑗 Phase can be calculated as = ∏𝑖𝑖 𝑗𝑗𝑗𝑗 − 𝑧𝑧𝑖𝑖 ∏𝑖𝑖 𝑗𝑗𝑗𝑗 − 𝑝𝑝𝑖𝑖 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 = Σ∠ 𝑗𝑗𝑗𝑗 − 𝑧𝑧𝑖𝑖 − Σ∠ 𝑗𝑗𝑗𝑗 − 𝑝𝑝𝑖𝑖 Possible to evaluate the frequency response graphically from a pole/zero diagram K. Webb Not done in practice, but provides useful insight ESE 499 Bode Construction from Pole/Zero Plots 102 Consider the following system: 𝐺𝐺 𝑗𝑗𝑗𝑗 = 𝑗𝑗𝑗𝑗 + 3 𝑗𝑗𝑗𝑗 + 2 + 𝑗𝑗𝑗.75 𝑗𝑗𝑗𝑗 + 2 − 𝑗𝑗𝑗.75 Evaluate at 𝜔𝜔 = 2.5𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠𝑠𝑠𝑠𝑠 Gain: 𝐺𝐺 𝑗𝑗𝑗.5 = 𝐺𝐺 𝑗𝑗𝑗.5 = Phase: 3+𝑗𝑗𝑗.5 2+𝑗𝑗𝑗.75 2+𝑗𝑗𝑗.25 3.1 2.9⋅4.7 𝐺𝐺 𝑗𝑗𝑗.5 = 0.389 → −8.2𝑑𝑑𝑑𝑑 ∠𝐺𝐺 𝑗𝑗𝑗.5 = 𝜃𝜃1 − 𝜃𝜃2 − 𝜃𝜃3 𝜃𝜃1 = ∠ 3 + 𝑗𝑗𝑗.5 = 39.8° 𝜃𝜃2 = ∠ 2 + 𝑗𝑗𝑗.75 = 20.6° 𝜃𝜃3 = ∠ 2 + 𝑗𝑗𝑗.25 = 64.8° K. Webb ∠𝐺𝐺 𝑗𝑗𝑗.5 = −45.5° ESE 499 103 K. Webb Polar Frequency Response Plots ESE 499 Polar Frequency Response Plots 104 𝐺𝐺 𝑗𝑗𝑗𝑗 is a complex function of frequency Typically plot as Bode plots A real and an imaginary part at each value of 𝜔𝜔 Magnitude and phase plotted separately Aids visualization of system behavior A point in the complex plane at each frequency Defines a curve in the complex plane A polar plot Parametrized by frequency – not as easy to distinguish frequency as on a Bode plot Polar plots are not terribly useful as a means of displaying a frequency response K. Webb A useful concept later, for the Nyquist stability criterion ESE 499 Polar Frequency Response Plots 105 Identical frequency responses plotted two ways: Bode plot and polar plot Note uneven frequency spacing along polar plot curve K. Webb Dependent on frequency rates of change of gain and phase ESE 499 106 Frequency and Time Domains A system’s frequency response and it’s various time-domain responses are simply different perspectives on the same dynamic behavior. K. Webb ESE 499 Frequency and Time Domains 107 We’ve seen many ways we can represent a system 𝑡𝑡𝑡 -order differential equation Impulse response Step response Transfer function Frequency response/Bode plot 𝑛𝑛 Time-domain representations Frequency-domain representations All are valid and complete models They all contain the same information in different forms Different ways of looking at the same thing K. Webb ESE 499 Time/Frequency Domain Correlation 108 𝐺𝐺1 𝑠𝑠 = 𝐺𝐺2 𝑠𝑠 = K. Webb 9.87 𝑠𝑠 2 +5.655𝑠𝑠+9.87 987 𝑠𝑠 2 +18.85𝑠𝑠+987 ESE 499 109 Frequency-Domain Analysis in MATLAB As was the case for time-domain simulation, MATLAB has some useful functions for simulating system behavior in the frequency domain as well. K. Webb ESE 499 Frequency Response Simulation – bode(…) 110 [mag,phase] = bode(sys,w) If no outputs are specified, bode response is automatically plotted – preferable to plot yourself Frequency vector input is optional sys: system model – state-space, transfer function, or other w: optional frequency vector – in rad/sec mag: system gain response vector phase: system phase response vector – in degrees If not specified, MATLAB will generate automatically May need to do: squeeze(mag) and squeeze(phase) to eliminate singleton dimensions of output matrices K. Webb ESE 499 Log-spaced Vectors – logspace(…) 111 f= logspace(x0,x1,N) x0: first point in f is 10𝑥𝑥0 𝑥𝑥 x1: last point in f is 10 1 N: number of points in f f: vector of logarithmically-spaced points Generates 𝑁𝑁 logarithmically-spaced points between 10𝑥𝑥0 and 10𝑥𝑥1 Useful for generating independent-variable vectors for log plots (e.g., frequency vectors for bode plots) K. Webb Linearly spaced on a logarithmic axis ESE 499
