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 + + + + 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 𝐺𝐺 𝑠𝑠 2 2 Real poles𝑠𝑠 –−1𝑝𝑝st1-order⋯ 𝑠𝑠 terms− 𝑝𝑝𝑛𝑛 𝑠𝑠 𝑎𝑎11𝑠𝑠 𝑎𝑎10 ⋯ 𝑠𝑠 𝑎𝑎𝑚𝑚1𝑠𝑠 𝑎𝑎𝑚𝑚0 Complex-conjugate poles – 2nd-order terms
These terms and, therefore, the poles determine the nature of the time- domain response Real poles – decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
K. Webb 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 Gain insight into relationships between pole locations and dynamic response Next, we’ll look at 1st- and 2nd-order impulse and step responses
K. Webb ESE 499 4 Response of First-Order Systems
K. Webb ESE 499 First-Order System – Impulse Response 5
First-order transfer function: = 𝐴𝐴 Single real pole𝐺𝐺 at𝑠𝑠 𝑠𝑠+𝜎𝜎 1 = =
where is the𝑠𝑠 system−𝜎𝜎 time− constant 𝜏𝜏 Impulse response: 𝜏𝜏 = = = 𝑡𝑡 −1 −𝜎𝜎𝜎𝜎 −𝜏𝜏 𝑔𝑔 𝑡𝑡 ℒ = 𝐺𝐺 𝑠𝑠 𝐴𝐴𝑒𝑒 𝐴𝐴𝑒𝑒 𝑡𝑡 −𝜏𝜏 K. Webb 𝑔𝑔 𝑡𝑡 𝐴𝐴𝑒𝑒 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 corresponds to decreasing and a faster response
𝜎𝜎 𝜏𝜏
K. Webb ESE 499 First-Order System – Step Response 8
Step response in the Laplace domain
= = 1 𝐴𝐴 𝑠𝑠 𝑠𝑠 𝑠𝑠+𝜎𝜎 Inverse transform back𝑌𝑌 𝑠𝑠 to time� 𝐺𝐺 domain𝑠𝑠 via partial fraction expansion
= = + 𝐴𝐴 𝑟𝑟1 𝑟𝑟2 𝑌𝑌 𝑠𝑠 =𝑠𝑠 𝑠𝑠+𝜎𝜎 + 𝑠𝑠 +𝑠𝑠+𝜎𝜎
𝐴𝐴 :𝑟𝑟1 𝑟𝑟=2 𝑠𝑠 𝜎𝜎𝑟𝑟1 = 0 𝐴𝐴 1 1 𝑠𝑠 : 𝜎𝜎𝑟𝑟+ 𝐴𝐴= →0 𝑟𝑟 𝜎𝜎= 1 𝐴𝐴 / 1 / 2 2 = 𝑠𝑠 𝑟𝑟 𝑟𝑟 → 𝑟𝑟 − 𝜎𝜎 𝐴𝐴 𝜎𝜎 𝐴𝐴 𝜎𝜎 Time-domain step 𝑌𝑌response𝑠𝑠 𝑠𝑠 − 𝑠𝑠+𝜎𝜎
= = 𝑡𝑡 − 𝐴𝐴 𝐴𝐴 −𝜎𝜎𝜎𝜎 𝜏𝜏 K. Webb 𝑦𝑦 𝑡𝑡 − 𝑒𝑒 𝐵𝐵 − 𝐵𝐵𝑒𝑒 ESE 499 𝜎𝜎 𝜎𝜎 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 corresponds to decreasing and a faster response
𝜎𝜎 𝜏𝜏
K. Webb ESE 499 Pole Location and Stability 11
First-order transfer function
= 𝐴𝐴 𝐺𝐺 𝑠𝑠 where is the system𝑠𝑠 − 𝑝𝑝 pole Impulse response is 𝑝𝑝 = 𝑝𝑝𝑝𝑝 If < 0,𝑔𝑔 𝑡𝑡 decays𝐴𝐴𝑒𝑒 to zero Pole in the left half-plane 𝑝𝑝System𝑔𝑔 is 𝑡𝑡stable
If > 0, grows without bound 𝑝𝑝Pole in𝑔𝑔 the𝑡𝑡 right half-plane System is unstable
K. Webb ESE 499 12 Response of Second-Order Systems
K. Webb ESE 499 Second-Order Systems 13
Second-order transfer function
= = (1) 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 2 2 2 where is the𝐺𝐺 damped𝑠𝑠 𝑠𝑠 +natural𝑎𝑎1𝑠𝑠+𝑎𝑎0 frequency𝑠𝑠+𝜎𝜎 +𝜔𝜔𝑑𝑑
Can also express the 2nd-order transfer function as 𝜔𝜔𝑑𝑑 = (2) 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 2 2 where is the𝐺𝐺 un𝑠𝑠 -damped𝑠𝑠 +2𝜁𝜁𝜔𝜔natural𝑛𝑛𝑠𝑠+𝜔𝜔𝑛𝑛 frequency, and is the damping ratio
𝜔𝜔𝑛𝑛 = 1 𝜁𝜁 2 𝜔𝜔𝑑𝑑= 𝜔𝜔𝑛𝑛 − 𝜁𝜁 𝜎𝜎 Two poles at 𝜁𝜁 𝜔𝜔𝑛𝑛 , = ± = ± 1 2 2 2 1 2 𝑛𝑛 𝑛𝑛 𝑛𝑛 K. Webb 𝑠𝑠 −𝜎𝜎 𝜎𝜎 − 𝜔𝜔 −𝜁𝜁𝜔𝜔 𝜔𝜔 𝜁𝜁 − ESE 499 Categories of Second-Order Systems 14
The 2nd-order system poles are
, = ± 1 2 Value of determines the nature𝑠𝑠1 2 −of𝜁𝜁 the𝜔𝜔𝑛𝑛 poles𝜔𝜔𝑛𝑛 and,𝜁𝜁 − therefore, the response > : Over𝜁𝜁 -damped Two distinct, real poles – sum of decaying exponentials – treat as two first-order terms 𝜻𝜻 𝟏𝟏= , =
1 1 2 2 =𝑠𝑠 : Critically−𝜎𝜎 𝑠𝑠 -damped−𝜎𝜎 Two identical, real poles – time-scaled decaying exponentials 𝜻𝜻 𝟏𝟏, = = =
1 2 𝑛𝑛 𝑛𝑛 <𝑠𝑠 < −:𝜎𝜎 Under−𝜁𝜁𝜔𝜔-damped−𝜔𝜔 Complex-conjugate pair of poles – sum of decaying sinusoids 𝟎𝟎 𝜻𝜻 𝟏𝟏 , = ± = ± 1 2 1 2 𝑑𝑑 𝑛𝑛 𝑛𝑛 =𝑠𝑠 : Un−-damped𝜎𝜎 𝑗𝑗𝜔𝜔 −𝜁𝜁𝜔𝜔 𝑗𝑗𝜔𝜔 − 𝜁𝜁 Purely-imaginary, conjugate pair of poles – sum of non-decaying sinusoids 𝜻𝜻 𝟎𝟎, = ±
1 2 𝑛𝑛 K. Webb 𝑠𝑠 𝑗𝑗𝜔𝜔 ESE 499 2nd-Order Pole Locations and Damping 15
K. Webb ESE 499 Second-Order Poles - 0 1 16 Can≤ relate𝜁𝜁 ≤ , , , and to pole location 𝑑𝑑 𝑛𝑛 geometry 𝜎𝜎 𝜔𝜔 𝜔𝜔 𝜁𝜁 is the real part
𝜎𝜎 is the imaginary part is the pole magnitude 𝜔𝜔𝑑𝑑 is𝜔𝜔 𝑛𝑛a measure of system damping 𝜁𝜁 = = sin 𝜎𝜎 K. Webb ESE 499 𝜁𝜁 𝜔𝜔𝑛𝑛 𝜃𝜃 Impulse Response – Critically-Damped 17
For = 1, the transfer function reduces to
= = = 𝜁𝜁 + 2 + + + 𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐺𝐺 𝑠𝑠 2 2 2 2 𝑠𝑠 𝜔𝜔𝑛𝑛𝑠𝑠 𝜔𝜔𝑛𝑛 𝑠𝑠 𝜔𝜔𝑛𝑛 𝑠𝑠 𝜎𝜎 Impulse response = −1 𝑔𝑔 𝑡𝑡 =ℒ 𝐺𝐺 𝑠𝑠 −𝜎𝜎𝜎𝜎 𝑔𝑔 𝑡𝑡 𝐴𝐴𝐴𝐴𝑒𝑒
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 < < 1, the transfer function is
𝜁𝜁 = + 2 + 𝐴𝐴 𝐺𝐺 𝑠𝑠 2 2 Complete the square on the denominator𝑠𝑠 𝜁𝜁𝜔𝜔𝑛𝑛𝑠𝑠 𝜔𝜔𝑛𝑛
= = + + + +𝐴𝐴 1 𝐴𝐴 2 2 2 𝐺𝐺 𝑠𝑠 2 2 𝑛𝑛 𝑑𝑑 𝑛𝑛 𝑛𝑛 𝑠𝑠 𝜁𝜁𝜔𝜔 𝜔𝜔 Rewrite in the form of 𝑠𝑠a damped𝜁𝜁𝜔𝜔 sinusoid𝜔𝜔 − 𝜁𝜁
= = + + + + 𝐴𝐴 𝜔𝜔𝑑𝑑 𝐴𝐴 𝜔𝜔𝑑𝑑 2 2 2 2 𝐺𝐺 𝑠𝑠 𝑑𝑑 𝑛𝑛 𝑑𝑑 𝑑𝑑 𝑑𝑑 Inverse Laplace transform𝜔𝜔 for𝑠𝑠 the𝜁𝜁 time𝜔𝜔 -domain𝜔𝜔 𝜔𝜔impulse𝑠𝑠 𝜎𝜎 response𝜔𝜔
= sin( ) −𝜎𝜎𝜎𝜎 𝐴𝐴 𝑑𝑑 𝑔𝑔 𝑡𝑡 𝑑𝑑 𝑒𝑒 𝜔𝜔 𝑡𝑡 K. Webb 𝜔𝜔 ESE 499 Under-Damped Impulse Response vs.
20 𝑛𝑛 = sin = sin 1 𝜔𝜔 𝐴𝐴 −𝜎𝜎𝜎𝜎 −𝜁𝜁𝜔𝜔𝑛𝑛𝑡𝑡 2 𝑔𝑔 𝑡𝑡 𝑒𝑒 𝜔𝜔𝑑𝑑𝑡𝑡 𝐵𝐵𝑒𝑒 𝜔𝜔𝑛𝑛 − 𝜁𝜁 𝑡𝑡 𝜔𝜔𝑑𝑑
K. Webb ESE 499 Under-Damped Impulse Response vs.
21
= sin = sin 1 𝜁𝜁 𝐴𝐴 −𝜎𝜎𝜎𝜎 −𝜁𝜁𝜔𝜔𝑛𝑛𝑡𝑡 2 𝑔𝑔 𝑡𝑡 𝑒𝑒 𝜔𝜔𝑑𝑑𝑡𝑡 𝐵𝐵𝑒𝑒 𝜔𝜔𝑛𝑛 − 𝜁𝜁 𝑡𝑡 𝜔𝜔𝑑𝑑
K. Webb ESE 499 Impulse Response – Un-Damped 22
For = 0, the transfer function reduces to
𝜁𝜁 = + 𝐴𝐴 𝐺𝐺 𝑠𝑠 2 2 Putting into the form of a sinusoid𝑠𝑠 𝜔𝜔𝑛𝑛 = + 𝐴𝐴 𝜔𝜔𝑛𝑛 𝐺𝐺 𝑠𝑠 2 2 Inverse transform to get the 𝜔𝜔time𝑛𝑛 𝑠𝑠-domain𝜔𝜔𝑛𝑛 impulse response = −1 An un-damped sinusoid𝑔𝑔 𝑡𝑡 ℒ 𝐺𝐺 𝑠𝑠 = sin
𝐴𝐴 𝑛𝑛 𝑔𝑔 𝑡𝑡 𝑛𝑛 𝜔𝜔 𝑡𝑡 K. Webb 𝜔𝜔 ESE 499 Un-Damped Impulse Response vs.
23 𝑛𝑛 = sin 𝜔𝜔 𝐴𝐴 𝑔𝑔 𝑡𝑡 𝜔𝜔𝑛𝑛𝑡𝑡 𝜔𝜔𝑛𝑛
K. Webb ESE 499 Second-Order Step Response 24
The Laplace transform of the step response is 1 =
𝑌𝑌 𝑠𝑠 𝐺𝐺 𝑠𝑠 The time-domain step response𝑠𝑠 for each damping case can be derived as the the inverse transform of
= 𝑌𝑌 𝑠𝑠 −1 or as the integral of𝑦𝑦 the𝑡𝑡 correspondingℒ 𝑌𝑌 𝑠𝑠 impulse response
= 𝑡𝑡
K. Webb 𝑦𝑦 𝑡𝑡 �0 𝑔𝑔 𝜏𝜏 𝑑𝑑𝑑𝑑 ESE 499 Critically-Damped Step Response vs.
25 1 = = 1 𝜎𝜎 −1 𝐴𝐴 −𝜎𝜎𝜎𝜎 −𝜎𝜎𝜎𝜎 𝑦𝑦 𝑡𝑡 ℒ 𝐺𝐺 𝑠𝑠 2 − 𝑒𝑒 − 𝜎𝜎𝜎𝜎𝑒𝑒 𝑠𝑠 𝜎𝜎
K. Webb ESE 499 Under-Damped Step Response vs.
26 1 𝑛𝑛 = = 1 cos 𝜔𝜔sin −1 𝐴𝐴 −𝜎𝜎𝜎𝜎 𝜎𝜎 −𝜎𝜎𝜎𝜎 𝑦𝑦 𝑡𝑡 ℒ 𝐺𝐺 𝑠𝑠 2 − 𝑒𝑒 𝜔𝜔𝑑𝑑𝑡𝑡 − 𝑒𝑒 𝜔𝜔𝑑𝑑𝑡𝑡 𝑠𝑠 𝜔𝜔𝑛𝑛 𝜔𝜔𝑑𝑑
K. Webb ESE 499 Under-Damped Step Response vs.
27 1 = = 1 cos 𝜁𝜁sin −1 𝐴𝐴 −𝜎𝜎𝜎𝜎 𝜎𝜎 −𝜎𝜎𝜎𝜎 𝑦𝑦 𝑡𝑡 ℒ 𝐺𝐺 𝑠𝑠 2 − 𝑒𝑒 𝜔𝜔𝑑𝑑𝑡𝑡 − 𝑒𝑒 𝜔𝜔𝑑𝑑𝑡𝑡 𝑠𝑠 𝜔𝜔𝑛𝑛 𝜔𝜔𝑑𝑑
K. Webb ESE 499 Un-Damped Step Response vs.
28 1 𝑛𝑛 = = 1 cos 𝜔𝜔 −1 𝐴𝐴 𝑦𝑦 𝑡𝑡 ℒ 𝐺𝐺 𝑠𝑠 2 − 𝜔𝜔𝑛𝑛𝑡𝑡 𝑠𝑠 𝜔𝜔𝑛𝑛
K. Webb ESE 499 29 Step Response Characteristics
K. Webb 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:
. 1 8 𝑟𝑟 𝜔𝜔𝑛𝑛 K. Webb 𝑡𝑡 ≈ ESE 499 Step Response – Overshoot 31
Overshoot is the magnitude of a signal’s excursion beyond its final value Expressed as a percentage of full- scale swing
Overshoot increases as decreases
𝜁𝜁 %OS 0.45 20 𝜻𝜻 0.5 16 0.6 10 0.7 5 = 𝜁𝜁𝜁𝜁 100% − 2 K. Webb 1−𝜁𝜁 ESE 499 %𝑂𝑂𝑂𝑂 𝑒𝑒 ⋅ 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,
For ±1%𝜎𝜎 settling: . . = 4 6 4 6 𝑡𝑡𝑠𝑠 ≈ 𝜎𝜎 𝜁𝜁𝜔𝜔𝑛𝑛 K. Webb ESE 499 33 The Convolution Integral
In this sub-section, we’ll see that the time- domain 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 =
Recall that multiplication𝑌𝑌 𝑠𝑠 𝐺𝐺 𝑠𝑠 in� 𝑈𝑈 the𝑠𝑠 Laplace domain corresponds to convolution in the time domain = = −1 Time-domain𝑦𝑦 𝑡𝑡 ℒ output𝐺𝐺 𝑠𝑠 𝑈𝑈given𝑠𝑠 by𝑔𝑔 the𝑡𝑡 ∗convolution𝑢𝑢 𝑡𝑡 of the input signal and the impulse response = = 𝑡𝑡 𝑦𝑦 𝑡𝑡 𝑔𝑔 𝑡𝑡 ∗ 𝑢𝑢 𝑡𝑡 ∫0 𝑔𝑔 𝜏𝜏 𝑢𝑢 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝑑𝑑 K. Webb 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 in0 time and shifted by Multiply impulse response and flipped/shifted𝑡𝑡 input Integrate over = 0 …
Each time point𝜏𝜏 of 𝑡𝑡given by result of integral with shifted by 𝑦𝑦 𝑡𝑡
K. Webb ESE 499 𝑢𝑢 −𝜏𝜏 𝑡𝑡 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 tf.m – creates a transfer function model ss.m – creates a state-space model zpk.m – creates a zero-pole-gain model
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 + + + + = 𝑟𝑟 + 𝑟𝑟−1 + + + 𝑏𝑏1𝑠𝑠 𝑏𝑏2𝑠𝑠 ⋯ 𝑏𝑏𝑟𝑟𝑠𝑠 𝑏𝑏𝑟𝑟+1 𝑛𝑛 𝑛𝑛−1 𝐺𝐺 𝑠𝑠 1 2 𝑛𝑛 𝑛𝑛+1 Inputs to tf(…)𝑎𝑎are𝑠𝑠 𝑎𝑎 𝑠𝑠 ⋯ 𝑎𝑎 𝑠𝑠 𝑎𝑎 Num = [b1,b2,…,br+1]; Den = [a1,a2,…,an+1];
K. Webb ESE 499 Step Response Simulation – step(…) 41
[y,t] = step(sys,t) sys: system model t: optional time vector or final time value y: output step response t: output time vector If no outputs are specified, step response is automatically plotted
Time vector (or final value) input is optional If not specified, MATLAB will generate automatically
K. Webb ESE 499 Impulse Response Simulation – impulse(…) 42
[y,t] = impulse(sys,t) sys: system model t: optional time vector or final time value y: output impulse response t: output time vector If no outputs are specified, impulse response is automatically plotted
Time vector (or final value) input is optional If not specified, MATLAB will generate automatically
K. Webb 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:
= + cos + sin 2 ∞ 𝑎𝑎0 𝑓𝑓 𝑡𝑡 � 𝑎𝑎𝑛𝑛 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 𝑏𝑏𝑛𝑛 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 where and 𝑛𝑛=are1 given by the Fourier integrals
Sinusoids𝑛𝑛 are basis𝑛𝑛 signals from which all other periodic signals 𝑎𝑎can be constructed𝑏𝑏 Sinusoidal system response is of particular interest
K. Webb ESE 499 Fourier Series 47
K. Webb ESE 499 System Response to a Sinusoidal Input 48
Consider an -order system poles: , 𝑡𝑡푡, … Real or complex𝑛𝑛 1 2 𝑛𝑛 𝑛𝑛 Assume𝑝𝑝 all𝑝𝑝 are distinct𝑝𝑝 Transfer function is:
= (1) 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 Apply a sinusoidal𝐺𝐺 𝑠𝑠 input𝑠𝑠−𝑝𝑝1 𝑠𝑠 −to𝑝𝑝2 the⋯ 𝑠𝑠− system𝑝𝑝𝑛𝑛
= sin = ℒ 𝜔𝜔 2 2 Output is given𝑢𝑢 𝑡𝑡 by𝐴𝐴 𝜔𝜔𝜔𝜔 𝑈𝑈 𝑠𝑠 𝐴𝐴 𝑠𝑠 +𝜔𝜔
= = (2) 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 𝜔𝜔 2 2 𝑌𝑌 𝑠𝑠 𝐺𝐺 𝑠𝑠 𝑈𝑈 𝑠𝑠 𝑠𝑠−𝑝𝑝1 𝑠𝑠−𝑝𝑝2 ⋯ 𝑠𝑠−𝑝𝑝𝑛𝑛 � 𝐴𝐴 𝑠𝑠 +𝜔𝜔 K. Webb ESE 499 System Response to a Sinusoidal Input 49
Partial fraction expansion of (2) gives
= + + + + + (3) 𝑟𝑟1 𝑟𝑟2 𝑟𝑟𝑛𝑛 𝑟𝑟𝑛𝑛+1𝑠𝑠 𝑟𝑟𝑛𝑛+2𝜔𝜔 2 2 2 2 Inverse𝑌𝑌 𝑠𝑠 transform𝑠𝑠−𝑝𝑝1 𝑠𝑠− of𝑝𝑝2 (3)⋯ gives𝑠𝑠−𝑝𝑝 𝑛𝑛the𝑠𝑠 time+𝜔𝜔 -domain𝑠𝑠 +𝜔𝜔 output = + + + + cos + sin (4) 𝑝𝑝1𝑡𝑡 𝑝𝑝2𝑡𝑡 𝑝𝑝𝑛𝑛𝑡𝑡 𝑦𝑦 𝑡𝑡 𝑟𝑟1𝑒𝑒 𝑟𝑟2𝑒𝑒 ⋯ 𝑟𝑟𝑛𝑛𝑒𝑒 𝑟𝑟𝑛𝑛+1 𝜔𝜔𝜔𝜔 𝑟𝑟𝑛𝑛+2 𝜔𝜔𝜔𝜔 transient steady state
Two portions of the response: Transient Decaying exponentials or sinusoids – goes to zero in steady state Natural response to initial conditions Steady state Due to the input – sinusoidal in steady state
K. Webb ESE 499 Steady-State Sinusoidal Response 50
We are interested in the steady-state response
= cos + sin (5)
A trig. identity𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡provides𝑟𝑟𝑛𝑛+ 1insight𝜔𝜔 into𝜔𝜔 𝑟𝑟𝑛𝑛+2 : 𝜔𝜔𝜔𝜔
cos + sin =𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡 + sin + where 2 2 𝛼𝛼 𝜔𝜔𝜔𝜔 𝛽𝛽 𝜔𝜔𝜔𝜔 𝛼𝛼 𝛽𝛽 𝜔𝜔𝜔𝜔 𝜙𝜙 = tan −1 𝛼𝛼 Steady-state response𝜙𝜙 to a sinusoidal𝛽𝛽 input = sin
is a sinusoid of the𝑢𝑢 𝑡𝑡 same𝐴𝐴 frequency,𝜔𝜔𝜔𝜔 but, in general, different amplitude and phase = sin + Where (6) 𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡 𝐵𝐵 𝜔𝜔𝜔𝜔 𝜙𝜙 = + and = tan 2 2 −1 𝑟𝑟𝑛𝑛+1 K. Webb 𝐵𝐵 𝑟𝑟𝑛𝑛+1 𝑟𝑟𝑛𝑛+2 𝜙𝜙 𝑟𝑟𝑛𝑛+2 ESE 499 Steady-State Sinusoidal Response 51 = sin = sin +
Steady𝑢𝑢 -𝑡𝑡state𝐴𝐴 sinusoidal𝜔𝜔𝜔𝜔 → response𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡 𝐵𝐵is a scaled𝜔𝜔𝜔𝜔 𝜙𝜙 and phase-shifted sinusoid of the same frequency Equal frequency is a property of linear systems
Note the term in the numerator of (3) will affect the residues 𝜔𝜔 Residues determine amplitude and phase of the output 𝜔𝜔 Output amplitude and phase are frequency-dependent = sin +
𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡 𝐵𝐵 𝜔𝜔 𝜔𝜔𝜔𝜔 𝜙𝜙 𝜔𝜔 K. Webb ESE 499 Steady-State Sinusoidal Response 52
= sin + Linear System = sin +
𝑢𝑢 𝑡𝑡 𝐴𝐴 𝜔𝜔𝜔𝜔 𝜃𝜃 𝑦𝑦𝑠𝑠𝑠𝑠 𝑡𝑡 𝐵𝐵 𝜔𝜔𝜔𝜔 𝜙𝜙
Gain – the ratio of amplitudes𝐺𝐺 of𝑠𝑠 the output and input of the system
= 𝐵𝐵 Phase – phase difference𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 between system input and output 𝐴𝐴 =
Systems will, in general,𝑃𝑃푃𝑃𝑃𝑃𝑃𝑃𝑃 exhibit𝜙𝜙 −frequency𝜃𝜃 -dependent gain and phase We’d like to be able to determine these functions of frequency The system’s frequency response
K. Webb 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 Frequency response Function – 54
Frequency response function Substitute for in the transfer function 𝐺𝐺 𝑗𝑗𝑗𝑗
𝑗𝑗𝑗𝑗 𝑠𝑠 = | A complex-valued function of frequency 𝐺𝐺 𝑗𝑗𝑗𝑗 𝐺𝐺 𝑠𝑠 𝑠𝑠→𝑗𝑗𝑗𝑗 at each is the gain at that frequency Ratio of output amplitude to input amplitude 𝐺𝐺 𝑗𝑗𝑗𝑗 𝜔𝜔 at each is the phase at that frequency Phase shift between input and output sinusoids ∠𝐺𝐺 𝑗𝑗𝑗𝑗 𝜔𝜔 Another valid system model Typically represented graphically
K. Webb ESE 499 Plotting the Frequency Response Function
55
is a complex-valued function of frequency Has both magnitude and phase 𝐺𝐺 𝑗𝑗𝑗𝑗 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 plot phase are below degrees
K. Webb 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 10MHz For a 10KHz sinusoidal sinusoidal input, the input, the gain is -32dB gain is 0dB (1) (0.025), and and the phase the phase shift is 0°. shift is -176°.
K. Webb 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: = 20 log
Conversion from𝐺𝐺 𝑗𝑗𝑗𝑗 dB𝑑𝑑𝑑𝑑 to a linear⋅ scale:10 𝐺𝐺 𝑗𝑗𝑗𝑗
= 10 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑑𝑑𝑑𝑑 20
K. Webb 𝐺𝐺 𝑗𝑗𝑗𝑗 ESE 499 Decibels – dB 60
Multiplying two gain values corresponds to adding their values in dB E.g., the overall gain of cascaded systems = +
1 2 𝑑𝑑𝑑𝑑 1 𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑 Negative𝐺𝐺 dB𝑗𝑗𝑗𝑗 values⋅ 𝐺𝐺 𝑗𝑗𝑗𝑗 corresponds𝐺𝐺 𝑗𝑗𝑗𝑗 to sub𝐺𝐺-unity𝑗𝑗𝑗𝑗 gain Positive dB values are gains greater than one
dB Linear dB Linear 60 1000 6 2 40 100 -3 1/ 2 = 0.707 20 10 -6 0.5 √ 0 1 -20 0.1
K. Webb ESE 499 Value of Logarithmic Axes - dB 61
Gain axis is linear in dB A logarithmic scale Allows for displaying detail at very large and very small levels on the same plot
Gain plotted in dB Two resonant peaks clearly visible
Linear gain scale Smaller peak has disappeared
K. Webb ESE 499 Value of Logarithmic Axes - dB 62
Frequency axis is logarithmic Allows for displaying detail at very low and very high frequencies on the same plot
Log frequency axis Can resolve frequency of both resonant peaks
Linear frequency axis Lower resonant frequency is unclear
K. Webb ESE 499 Gain Response – Terminology 63
Corner frequency, cut off frequency, -3dB frequency: Frequency at which ~ of peaking gain is 3dB below its low-frequency value 5𝑑𝑑𝑑𝑑
= = 1.45 𝑐𝑐 𝑐𝑐 𝜔𝜔 𝑟𝑟𝑟𝑟𝑟𝑟 𝑓𝑓 𝜔𝜔𝑐𝑐 This is the2𝜋𝜋 bandwidth = = 𝑠𝑠0𝑠𝑠. 𝑠𝑠 of the system 𝜔𝜔𝑐𝑐 𝑓𝑓𝑐𝑐 23𝐻𝐻𝐻𝐻 2𝜋𝜋 Peaking Any increase in gain above the low frequency gain
K. Webb 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 first- and second-order terms
= + 2 + + 2 + 𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 2 2 2 2 𝐺𝐺 𝑠𝑠 1 2 1 𝑛𝑛� 𝑛𝑛� 2 𝑛𝑛2 𝑛𝑛2 The same is true𝑠𝑠 − of𝑝𝑝 the𝑠𝑠 −numerator𝑝𝑝 ⋯ 𝑠𝑠 𝜁𝜁 𝜔𝜔 𝑠𝑠 𝜔𝜔 𝑠𝑠 𝜁𝜁 𝜔𝜔 𝑠𝑠 𝜔𝜔 ⋯ + 2 + + 2 + = 2 + 2 + 2 2 + 2 + 2 𝑠𝑠 − 𝑧𝑧1 𝑠𝑠 − 𝑧𝑧2 ⋯ 𝑠𝑠 𝜁𝜁𝑎𝑎𝜔𝜔𝑛𝑛𝑎𝑎𝑠𝑠 𝜔𝜔𝑛𝑛𝑎𝑎 𝑠𝑠 𝜁𝜁2𝜔𝜔𝑛𝑛𝑏𝑏𝑠𝑠 𝜔𝜔𝑛𝑛𝑏𝑏 ⋯ 2 2 2 2 𝐺𝐺 𝑠𝑠 1 2 1 𝑛𝑛� 𝑛𝑛� 2 𝑛𝑛� 𝑛𝑛� Can think of the𝑠𝑠 − transfer𝑝𝑝 𝑠𝑠 − function𝑝𝑝 ⋯ 𝑠𝑠 as a 𝜁𝜁product𝜔𝜔 𝑠𝑠 of𝜔𝜔 the𝑠𝑠 individual𝜁𝜁 𝜔𝜔 factors𝑠𝑠 𝜔𝜔 ⋯ For example, consider the following system
= + 2 + 𝑠𝑠 − 𝑧𝑧1 2 2 𝐺𝐺 𝑠𝑠 1 1 𝑛𝑛� 𝑛𝑛� Can rewrite as 𝑠𝑠 − 𝑝𝑝 𝑠𝑠 𝜁𝜁 𝜔𝜔 𝑠𝑠 𝜔𝜔 1 1 = + 2 + 1 2 2 𝐺𝐺 𝑠𝑠 𝑠𝑠 − 𝑧𝑧 ⋅ 1 ⋅ 1 𝑛𝑛� 𝑛𝑛� K. Webb 𝑠𝑠 − 𝑝𝑝 𝑠𝑠 𝜁𝜁 𝜔𝜔 𝑠𝑠 𝜔𝜔 ESE 499 Transfer Function Factors 66 1 1 = + 2 + 1 2 2 𝐺𝐺 𝑠𝑠 𝑠𝑠 − 𝑧𝑧 ⋅ 1 ⋅ 1 𝑛𝑛� Think of this as three𝑠𝑠 cascaded− 𝑝𝑝 𝑠𝑠 transfer𝜁𝜁 𝜔𝜔 𝑠𝑠 𝜔𝜔 functions𝑛𝑛�
= , = , = 1 1 2 2 𝐺𝐺1 𝑠𝑠 𝑠𝑠 − 𝑧𝑧1 𝐺𝐺2 𝑠𝑠 𝑠𝑠−𝑝𝑝1 𝐺𝐺3 𝑠𝑠 𝑠𝑠 +2𝜁𝜁1𝜔𝜔𝑛𝑛�𝑠𝑠+𝜔𝜔𝑛𝑛�
𝑈𝑈 𝑠𝑠 𝑌𝑌1 𝑠𝑠 𝑌𝑌2 𝑠𝑠 𝑌𝑌 𝑠𝑠 or 𝐺𝐺1 𝑠𝑠 𝐺𝐺2 𝑠𝑠 𝐺𝐺3 𝑠𝑠
1 1 + 2 + 1 2 𝑈𝑈 𝑠𝑠 𝑠𝑠 − 𝑧𝑧1 𝑌𝑌 𝑠𝑠 𝑌𝑌 𝑠𝑠 2 2 𝑌𝑌 𝑠𝑠 𝑠𝑠 − 𝑝𝑝1 𝑠𝑠 𝜁𝜁1𝜔𝜔𝑛𝑛�𝑠𝑠 𝜔𝜔𝑛𝑛�
K. Webb 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 𝑌𝑌1 𝑗𝑗𝑗𝑗 𝑌𝑌2 𝑗𝑗𝑗𝑗 𝑌𝑌 𝑗𝑗𝑗𝑗 𝐺𝐺1 𝑗𝑗𝑗𝑗 𝐺𝐺1 𝑗𝑗𝑗𝑗 𝐺𝐺1 𝑗𝑗𝑗𝑗 Instructive, therefore, to understand the responses of the individual factors First- and second-order poles and zeros
K. Webb ESE 499 First-Order Factors 68
First-order factors Single, real poles or zeros In the Laplace domain: = , = , = + , = 1 1 In the frequency𝐺𝐺 𝑠𝑠 𝑠𝑠domain𝐺𝐺 𝑠𝑠 𝑠𝑠 𝐺𝐺 𝑠𝑠 𝑠𝑠 𝑎𝑎 𝐺𝐺 𝑠𝑠 𝑠𝑠+𝑎𝑎 = , = , = + , = 1 1 Pole/zero𝐺𝐺 plots:𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝑎𝑎 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗+𝑎𝑎
K. Webb ESE 499 First-Order Factors – Zero at the Origin 69
A differentiator =
𝐺𝐺 𝑠𝑠 =𝑠𝑠
Gain𝐺𝐺: 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 = =
𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝜔𝜔 Phase:
= +90°,
∠𝐺𝐺 𝑗𝑗𝑗𝑗 ∀𝜔𝜔
K. Webb ESE 499 First-Order Factors – Pole at the Origin 70
An integrator 1 =
𝐺𝐺 𝑠𝑠 𝑠𝑠 1 =
Gain𝐺𝐺: 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 1 1 = =
𝐺𝐺 𝑗𝑗𝑗𝑗 Phase: 𝑗𝑗𝑗𝑗 𝜔𝜔
= = 90°, 1
K. Webb∠𝐺𝐺 𝑗𝑗𝑗𝑗 ∠ − 𝑗𝑗 𝜔𝜔 − ∀𝜔𝜔 ESE 499 First-Order Factors – Single, Real Zero 71
Single, real zero at = = + 𝑠𝑠 −𝑎𝑎 Gain: 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗Phase𝑎𝑎 : = + = tan 2 2 −1 𝜔𝜔 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 for 𝐺𝐺 𝑗𝑗𝑗𝑗 𝜔𝜔 𝑎𝑎 for 𝑎𝑎 = 0° 𝜔𝜔 ≪ 𝑎𝑎 𝜔𝜔 ≪ 𝑎𝑎 for 𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ 𝑎𝑎 for ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠𝑎𝑎 = 90° 𝜔𝜔 ≫ 𝑎𝑎 𝜔𝜔 ≫ 𝑎𝑎 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠𝑗𝑗𝑗𝑗 K. Webb 𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ 𝜔𝜔 ESE 499 First-Order Factors – Single, Real Zero 72
Corner frequency: =
𝑐𝑐 𝜔𝜔 =𝑎𝑎 2 = 1.414
𝐺𝐺 𝑗𝑗𝜔𝜔𝑐𝑐 =𝑎𝑎 + ⋅ 𝑎𝑎
𝑐𝑐 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐺𝐺 𝑗𝑗𝜔𝜔 = +45°𝑎𝑎 3𝑑𝑑𝑑𝑑
𝑐𝑐 For∠𝐺𝐺 𝑗𝑗𝜔𝜔 , gain increases at: / 𝜔𝜔 ≫ 𝜔𝜔𝑐𝑐 / 20𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 From6𝑑𝑑 𝑑𝑑~0𝑜𝑜.𝑜𝑜𝑜𝑜1 to ~10 , phase increases at a rate of: 𝑐𝑐 𝑐𝑐 ~45°/ 𝜔𝜔 𝜔𝜔 Rough approximation 𝑑𝑑𝑑𝑑𝑑𝑑
K. Webb ESE 499 First-Order Factors – Single, Real Pole 73
Single, real pole at = 1 = 𝑠𝑠 −𝑎𝑎 + 𝐺𝐺 𝑗𝑗𝑗𝑗 Gain: 𝑗𝑗𝑗𝑗Phase𝑎𝑎 : 1 = = tan + −1 𝜔𝜔 𝐺𝐺 𝑗𝑗𝑗𝑗 2 2 for ∠𝐺𝐺 𝑗𝑗𝑗𝑗 − for 𝜔𝜔 𝑎𝑎 𝑎𝑎 1 1 𝜔𝜔 ≪ 𝑎𝑎 𝜔𝜔 ≪ 𝑎𝑎 = 0°
𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ for ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠ for 𝑎𝑎 𝑎𝑎 1 1 𝜔𝜔 ≫ 𝑎𝑎 𝜔𝜔 ≫ 𝑎𝑎 = 90°
𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠ − K. Webb 𝜔𝜔 𝑗𝑗𝑗𝑗 ESE 499 First-Order Factors – Single, Real Pole 74
Corner frequency: =
𝜔𝜔𝑐𝑐 =𝑎𝑎 = 0.707 1 1 𝑐𝑐 𝑎𝑎 2 𝑎𝑎 𝐺𝐺 𝑗𝑗𝜔𝜔 = ⋅ 1 𝑐𝑐 𝑑𝑑𝑑𝑑 𝐺𝐺 𝑗𝑗𝜔𝜔 = 45°𝑎𝑎 𝑑𝑑𝑑𝑑 − 3𝑑𝑑𝑑𝑑
𝑐𝑐 For∠𝐺𝐺 𝑗𝑗𝜔𝜔 , gain− decreases at: / 𝜔𝜔 ≫ 𝜔𝜔𝑐𝑐 / −20𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 From−6𝑑𝑑 ~𝑑𝑑0.1𝑜𝑜𝑜𝑜𝑜𝑜 to ~10 , phase decreases at a rate of: 𝑐𝑐 𝑐𝑐 ~ 45°/𝜔𝜔 𝜔𝜔 Rough approximation − 𝑑𝑑𝑑𝑑𝑑𝑑 K. Webb ESE 499 Second-Order Factors 75
Complex-conjugate zeros Complex-conjugate poles 1 = + + = 2 2 + + 𝑛𝑛 𝑛𝑛 𝐺𝐺 𝑠𝑠 𝑠𝑠 2𝜁𝜁𝜔𝜔 𝑠𝑠 𝜔𝜔 𝐺𝐺 𝑠𝑠 2 2 𝑠𝑠 2𝜁𝜁𝜔𝜔𝑛𝑛𝑠𝑠 𝜔𝜔𝑛𝑛
= , = 1 K. Webb ESE 499 2 𝜎𝜎 𝜁𝜁𝜔𝜔𝑛𝑛 𝜔𝜔𝑑𝑑 𝜔𝜔𝑛𝑛 − 𝜁𝜁 2nd-Order Factors – Complex-Conjugate Zeros
76
Complex-conjugate zeros at = ±
= + + 𝑑𝑑 2 𝑠𝑠 −𝜎𝜎 𝑗𝑗𝜔𝜔 2 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 2𝜁𝜁𝜔𝜔𝑛𝑛 𝑗𝑗𝑗𝑗 𝜔𝜔𝑛𝑛 Gain: Phase: for for 2 2 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛 = 0° 2 2 for 𝐺𝐺=𝑗𝑗𝑗𝑗 ≈ 𝜔𝜔𝑛𝑛 for ∠𝐺𝐺=𝑗𝑗𝑗𝑗 ≈ ∠𝜔𝜔𝑛𝑛 𝜔𝜔 𝜔𝜔𝑛𝑛 = 𝜔𝜔 𝜔𝜔𝑛𝑛 = = +90° 2 2 for 𝐺𝐺 𝑗𝑗𝑗𝑗 2𝜁𝜁𝜔𝜔𝑛𝑛 for ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ∠𝑗𝑗𝑗𝑗�𝜔𝜔𝑛𝑛 2 2 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛 = +180° 2 2
K. Webb 𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ 𝜔𝜔 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠ − 𝜔𝜔 ESE 499 2nd-Order Factors – Complex-Conjugate Zeros
77
Response may dip below low-freq. value near Peaking increases as decreases 𝜔𝜔𝑛𝑛 𝜁𝜁 Gain increases at + / or + / for 40𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 12𝑑𝑑𝑑𝑑 𝑜𝑜𝑜𝑜𝑜𝑜 𝜔𝜔 ≫ Corner𝑛𝑛 frequency depends on𝜔𝜔 damping ratio, increases as decreases 𝜁𝜁 At 𝑓𝑓𝑐𝑐 = , 𝜁𝜁 = 90° Phase transition abruptness depends𝜔𝜔 𝜔𝜔 on𝑐𝑐 ∠𝐺𝐺 𝑗𝑗𝑗𝑗
𝜁𝜁 K. Webb ESE 499 2nd-Order Factors – Complex-Conjugate Poles
78
Complex-conjugate zeros at = ± 1 =𝑠𝑠 −𝜎𝜎 𝑗𝑗𝜔𝜔𝑑𝑑 + + 𝐺𝐺 𝑗𝑗𝑗𝑗 2 2 𝑗𝑗𝑗𝑗 2𝜁𝜁𝜔𝜔𝑛𝑛 𝑗𝑗𝑗𝑗 𝜔𝜔𝑛𝑛 Gain: Phase: for for 2 1 2 1 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛 = 0°
for 𝐺𝐺=𝑗𝑗𝑗𝑗 ≈ 2 for ∠𝐺𝐺=𝑗𝑗𝑗𝑗 ≈ ∠ 2 𝜔𝜔𝑛𝑛 𝜔𝜔𝑛𝑛 1 1 𝜔𝜔 𝜔𝜔𝑛𝑛 = 𝜔𝜔 𝜔𝜔𝑛𝑛 = = 90°
for 𝐺𝐺 𝑗𝑗𝑗𝑗 2 for ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ∠ 2 − 2𝜁𝜁𝜔𝜔𝑛𝑛 𝑗𝑗𝑗𝑗�𝜔𝜔𝑛𝑛 2 1 2 1 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛 = 180°
𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ 2 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 ≈ ∠ − 2 − K. Webb 𝜔𝜔 𝜔𝜔 ESE 499 2nd-Order Factors – Complex-Conjugate Poles
79
Response may peak above low-freq. value near Peaking increases as decreases 𝜔𝜔𝑛𝑛 𝜁𝜁 Gain decreases at / or / for −40𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 −12𝑑𝑑𝑑𝑑 𝑜𝑜𝑜𝑜𝑜𝑜 𝜔𝜔 ≫ Corner𝑛𝑛 frequency depends on𝜔𝜔 damping ratio, increases as decreases 𝜁𝜁 At 𝑓𝑓𝑐𝑐 = , 𝜁𝜁 = 90° Phase transition abruptness depends𝜔𝜔 𝜔𝜔 on𝑐𝑐 ∠𝐺𝐺 𝑗𝑗𝑗𝑗 −
𝜁𝜁 K. Webb ESE 499 Pole Location and Peaking 80
Peaking is dependent on – pole locations No peaking at all for 1/ 2 = 0.707 = 0.707 – maximally-flat𝜁𝜁 or Butterworth response 𝜁𝜁 ≥ 𝜁𝜁
K. Webb ESE 499 Frequency Response Components - Example
81
Consider the following system 20 + 20 = + 1 + 100 𝑠𝑠 𝐺𝐺 𝑠𝑠 The system’s frequency response𝑠𝑠 function𝑠𝑠 is 20 + 20 = + 1 + 100 𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 As we’ve seen we can consider this𝑗𝑗𝑗𝑗 a product𝑗𝑗𝑗𝑗 of individual frequency response factors 1 1 = 20 + 20 + 1 + 100 𝐺𝐺 𝑗𝑗𝑗𝑗 ⋅ 𝑗𝑗𝑗𝑗 ⋅ ⋅ Overall response is the composite of the𝑗𝑗𝑗𝑗 individual𝑗𝑗 𝑗𝑗responses Product of individual gain responses – sum in dB Sum of individual phase responses
K. Webb 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 second- order 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 + = 2 + 2 + 2 𝑠𝑠 − 𝑧𝑧1 𝑠𝑠 − 𝑧𝑧2 ⋯ 𝑠𝑠 𝜁𝜁𝑎𝑎𝜔𝜔𝑛𝑛𝑛𝑛𝑠𝑠 𝜔𝜔𝑛𝑛𝑛𝑛 ⋯ 𝐺𝐺 𝑠𝑠 𝐾𝐾 2 2 We first want to put this into𝑠𝑠 − Bode𝑝𝑝1 𝑠𝑠 form− 𝑝𝑝2 :⋯ 𝑠𝑠 𝜁𝜁1𝜔𝜔𝑛𝑛1𝑠𝑠 𝜔𝜔𝑛𝑛1 ⋯
2 + 1 + 1 + + 1 2 = 𝑠𝑠 𝑠𝑠 𝑠𝑠 𝜁𝜁𝑎𝑎 ⋯ 2 2 𝑠𝑠 ⋯ 𝜔𝜔𝑐𝑐𝑐𝑐 + 1 𝜔𝜔𝑐𝑐𝑏𝑏 + 1 𝜔𝜔𝑛𝑛𝑛𝑛 + 𝜔𝜔𝑛𝑛𝑛𝑛 + 1 𝐺𝐺 𝑠𝑠 𝐾𝐾0 2 1 𝑠𝑠 𝑠𝑠 𝑠𝑠2 𝜁𝜁 𝑐𝑐1 𝑐𝑐2 ⋯ 𝑛𝑛1 𝑠𝑠 ⋯ The corresponding frequency𝜔𝜔 response𝜔𝜔 function,𝜔𝜔𝑛𝑛 1in Bode𝜔𝜔 form, is
2 + 1 + 1 + + 1 2 = 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝜁𝜁𝑎𝑎 ⋯ 𝑗𝑗𝑗𝑗 ⋯ 𝜔𝜔𝑐𝑐𝑐𝑐 𝜔𝜔𝑐𝑐𝑏𝑏 𝜔𝜔𝑛𝑛𝑎𝑎 2𝜔𝜔𝑛𝑛𝑛𝑛 0 + 1 + 1 + + 1 𝐺𝐺 𝑗𝑗𝑗𝑗 𝐾𝐾 2 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝜁𝜁1 𝑐𝑐1 𝑐𝑐2 ⋯ 𝑛𝑛1 𝑛𝑛1 𝑗𝑗𝑗𝑗 ⋯ Putting into Bode form𝜔𝜔 requires𝜔𝜔 putting𝜔𝜔 each of𝜔𝜔 the first- and second-order factors into Bode form 𝐺𝐺 𝑗𝑗𝑗𝑗 K. Webb ESE 499 First-Order Factors in Bode Form 87
First-order frequency-response factors include:
= , = + , = 𝑛𝑛 1 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗+𝜎𝜎 For the first factor, 𝐺𝐺 𝑗𝑗𝑗𝑗= 𝑗𝑗𝑗𝑗 , 𝜎𝜎is a𝐺𝐺 positive𝑗𝑗𝑗𝑗 or negative integer Already in Bode form 𝑛𝑛 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝑛𝑛 For the second two, divide through by , giving
= + 1 and 𝜎𝜎 = 𝑗𝑗𝑗𝑗 1 𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 𝜎𝜎 𝜎𝜎 𝐺𝐺 𝑗𝑗𝑗𝑗 𝜎𝜎 𝜎𝜎 +1 Here, = , the corner frequency associated with that zero or pole, so
𝑐𝑐 𝜎𝜎 𝜔𝜔 = + 1 and = 𝑗𝑗𝑗𝑗 1 𝑐𝑐 𝑗𝑗𝑗𝑗 𝜔𝜔𝑐𝑐 𝑐𝑐 𝐺𝐺 𝑗𝑗𝑗𝑗 𝜔𝜔 𝐺𝐺 𝑗𝑗𝑗𝑗 𝜔𝜔 𝜔𝜔𝑐𝑐+1
K. Webb ESE 499 Second-Order Factors in Bode Form 88
Second-order frequency-response factors include:
= + + and = 2 2 1 2 2 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 2𝜁𝜁𝜔𝜔𝑛𝑛 𝑗𝑗𝑗𝑗 𝜔𝜔𝑛𝑛 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 +2𝜁𝜁𝜔𝜔𝑛𝑛 𝑗𝑗𝑗𝑗 +𝜔𝜔𝑛𝑛 Again, normalize the coefficient, giving 0 / = +𝑗𝑗𝑗𝑗 + 1 and = 2 𝑗𝑗𝜔𝜔 2 2𝜁𝜁 1 𝜔𝜔𝑛𝑛 2 2 𝐺𝐺 𝑗𝑗𝑗𝑗 𝜔𝜔𝑛𝑛 𝜔𝜔𝑛𝑛 𝜔𝜔𝑛𝑛 𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 2𝜁𝜁 𝜔𝜔𝑛𝑛 +𝜔𝜔𝑛𝑛 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,
0 2 𝐾𝐾 = 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 2𝜁𝜁𝑎𝑎 𝜔𝜔𝑐𝑐𝑐𝑐+1 𝜔𝜔𝑐𝑐𝑐𝑐+1 ⋯ 𝜔𝜔𝑛𝑛𝑛𝑛 +𝜔𝜔𝑛𝑛𝑛𝑛𝑗𝑗𝑗𝑗+1 ⋯ 2 𝐺𝐺 𝑗𝑗𝑗𝑗 𝐾𝐾0 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 2𝜁𝜁1 𝜔𝜔𝑐𝑐�+1 𝜔𝜔𝑐𝑐�+1 ⋯ 𝜔𝜔𝑛𝑛� +𝜔𝜔𝑛𝑛�𝑗𝑗𝑗𝑗+1 ⋯ K. Webb ESE 499 Bode Plot Construction 89
Frequency response function in Bode form
2 = 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 2𝜁𝜁𝑎𝑎 𝜔𝜔𝑐𝑐𝑐𝑐+1 𝜔𝜔𝑐𝑐𝑐𝑐+1 ⋯ 𝜔𝜔𝑛𝑛𝑛𝑛 +𝜔𝜔𝑛𝑛𝑛𝑛𝑗𝑗𝑗𝑗+1 ⋯ 2 𝐺𝐺 𝑗𝑗𝑗𝑗 𝐾𝐾0 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 2𝜁𝜁1 𝜔𝜔𝑐𝑐�+1 𝜔𝜔𝑐𝑐�+1 ⋯ 𝜔𝜔𝑛𝑛� +𝜔𝜔𝑛𝑛�𝑗𝑗𝑗𝑗+1 ⋯ Product of a constant DC gain factor, , and first- and second-order factors 𝐾𝐾0 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. Put the sinusoidal transfer function into Bode form 2. Draw a straight-line asymptotic approximation for the gain and phase response of each individual factor 3. 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
=
Constant𝐺𝐺 𝑗𝑗𝑗𝑗 gain𝐾𝐾0 =
Constant𝐺𝐺 𝑗𝑗𝑗𝑗 Phase𝐾𝐾0 = 0°
∠𝐺𝐺 𝑗𝑗𝑗𝑗
K. Webb ESE 499 Bode Plot – Poles/Zeros at the Origin 92
= 𝑛𝑛 > 0:𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 zeros at the origin 𝑛𝑛 <𝑛𝑛 0: poles at the origin 𝑛𝑛 Gain𝑛𝑛 : Straight line Slope = 20 = 6 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 at 𝑛𝑛=⋅ 1 𝑑𝑑𝑑𝑑𝑑𝑑 𝑛𝑛 ⋅ 𝑜𝑜𝑜𝑜𝑜𝑜
Phase0𝑑𝑑𝑑𝑑: 𝜔𝜔 = 90°
K. Webb ∠𝐺𝐺 𝑗𝑗𝑗𝑗 𝑛𝑛 ⋅ ESE 499 Bode Plot – First-Order Zero 93
Single real zero at =
Gain: 𝑐𝑐 𝑠𝑠 −𝜔𝜔 = + 1 for < 𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 𝜔𝜔𝑐𝑐 +20 = +6 > 0𝑑𝑑𝑑𝑑 𝜔𝜔 𝜔𝜔𝑐𝑐 for 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 Straight-line asymptotes 𝑐𝑐 intersect𝑑𝑑𝑑𝑑𝑑𝑑 at 𝑜𝑜,𝑜𝑜𝑜𝑜 𝜔𝜔 𝜔𝜔 Phase: 𝜔𝜔𝑐𝑐 0𝑑𝑑𝑑𝑑 0° for 0.1 45° for = 𝜔𝜔 ≤ 𝜔𝜔𝑐𝑐 90° for 10 𝜔𝜔 𝜔𝜔𝑐𝑐 0.1 10 for 𝜔𝜔 ≥ 𝜔𝜔𝑐𝑐 +45° 𝑑𝑑𝑑𝑑𝑑𝑑 𝜔𝜔𝑐𝑐 ≤ 𝜔𝜔 ≤ 𝜔𝜔𝑐𝑐 K. Webb ESE 499 Bode Plot – First-Order Pole 94
Single real pole at =
Gain: 𝑠𝑠 −𝜔𝜔𝑐𝑐 for < 1 = 20 = 6 for > + 1 0𝑑𝑑𝑑𝑑 𝜔𝜔 𝜔𝜔𝑐𝑐 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑗𝑗𝑗𝑗 𝑐𝑐 Straight-line asymptotes 𝑐𝑐 𝜔𝜔 intersect− 𝑑𝑑𝑑𝑑𝑑𝑑 at − 𝑜𝑜,𝑜𝑜𝑜𝑜 𝜔𝜔 𝜔𝜔 Phase: 𝜔𝜔𝑐𝑐 0𝑑𝑑𝑑𝑑 0° for 0.1 45° for = 𝜔𝜔 ≤ 𝜔𝜔𝑐𝑐 90° for 10 − 𝜔𝜔 𝜔𝜔𝑐𝑐 0.1 10 − for 𝜔𝜔 ≥ 𝜔𝜔𝑐𝑐 −45° 𝑑𝑑𝑑𝑑𝑑𝑑 𝜔𝜔𝑐𝑐 ≤ 𝜔𝜔 ≤ 𝜔𝜔𝑐𝑐 K. Webb ESE 499 Bode Plot – Second-Order Zero 95
Complex-conjugate zeros:
, = ±
Gain: 𝑠𝑠1 2 −𝜎𝜎 𝑗𝑗𝜔𝜔𝑑𝑑 for < = 2 + + 1 +40 = +12 > 𝑗𝑗𝑗𝑗 2𝜁𝜁 𝑛𝑛 for 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 0𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝜔𝜔 𝜔𝜔 𝑑𝑑𝑑𝑑 𝜔𝜔𝑛𝑛 𝜔𝜔𝑛𝑛 Straight-line asymptotes intersect𝑛𝑛 at , 𝑑𝑑𝑑𝑑𝑑𝑑 𝑜𝑜𝑜𝑜𝑜𝑜 𝜔𝜔 𝜔𝜔 -dependent peaking around 𝜔𝜔𝑛𝑛 0𝑑𝑑𝑑𝑑 Phase𝜁𝜁 : 𝜔𝜔𝑛𝑛 0° for 90° for = 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛 180° for 𝜔𝜔 𝜔𝜔𝑛𝑛 -dependent slope through 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛 Step-change for low 𝜁𝜁 𝜔𝜔𝑛𝑛 +180°/ for high 𝜁𝜁 𝑑𝑑𝑑𝑑𝑑𝑑 𝜁𝜁 K. Webb ESE 499 Bode Plot – Second-Order Pole 96
Complex-conjugate poles:
, = ±
Gain: 𝑠𝑠1 2 −𝜎𝜎 𝑗𝑗𝜔𝜔𝑑𝑑 < 1 for = + + 1 40 = 12𝑛𝑛 for > 𝐺𝐺 𝑗𝑗𝑗𝑗 2 0𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝜔𝜔 𝜔𝜔 𝑑𝑑𝑑𝑑 𝑗𝑗𝑗𝑗 2𝜁𝜁 𝑛𝑛 𝑛𝑛 𝑗𝑗𝑗𝑗 Straight-line asymptotes intersect𝑛𝑛 at 𝜔𝜔 𝜔𝜔 − , 𝑑𝑑𝑑𝑑𝑑𝑑 − 𝑜𝑜𝑜𝑜𝑜𝑜 𝜔𝜔 𝜔𝜔 -dependent peaking around 𝜔𝜔𝑛𝑛 0𝑑𝑑𝑑𝑑 Phase𝜁𝜁 : 𝜔𝜔𝑛𝑛 0° for 90° for = 𝜔𝜔 ≪ 𝜔𝜔𝑛𝑛 180° for − 𝜔𝜔 𝜔𝜔𝑛𝑛 -dependent slope through − 𝜔𝜔 ≫ 𝜔𝜔𝑛𝑛 Step-change for low 𝜁𝜁 𝜔𝜔𝑛𝑛 180°/ for high 𝜁𝜁 − 𝑑𝑑𝑑𝑑𝑑𝑑 𝜁𝜁 K. Webb ESE 499 Bode Plot Construction – Example 97
Consider a system with the following transfer function
10 + 20 = + 400 𝑠𝑠 𝐺𝐺 𝑠𝑠 The sinusoidal transfer function𝑠𝑠 :𝑠𝑠 10 + 20 = + 400 𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 Put it into Bode form 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗
10 20 + 1 0.5 + 1 = 20 = 20 400 𝑗𝑗𝑗𝑗 + 1 𝑗𝑗𝑗𝑗 + 1 ⋅ 400 400 𝐺𝐺 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗 Represent as a product of factors𝑗𝑗𝑗𝑗 ⋅ 𝑗𝑗𝑗𝑗 ⋅ 1 1 = 0.5 + 1 20 + 1 𝑗𝑗𝑗𝑗 400 𝐺𝐺 𝑗𝑗𝑗𝑗 ⋅ ⋅ ⋅ 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗
K. Webb ESE 499 Bode Plot Construction – Example 98
K. Webb ESE 499 Bode Plot Construction – Example 99
K. Webb ESE 499 Relationship between Pole/Zero Plots 100 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
= = 𝑠𝑠→𝑗𝑗𝑗𝑗 ∏𝑖𝑖 𝑠𝑠 − 𝑧𝑧𝑖𝑖 ∏𝑖𝑖 𝑗𝑗𝑗𝑗 − 𝑧𝑧𝑖𝑖 𝐺𝐺 𝑠𝑠 𝐺𝐺 𝑗𝑗𝑗𝑗 Numerator is a product∏𝑖𝑖 𝑠𝑠 −of𝑝𝑝 first𝑖𝑖 -order zero terms∏𝑖𝑖 𝑗𝑗𝑗𝑗 − 𝑝𝑝𝑖𝑖 Denominator is a product of first-order pole terms is a point on the imaginary axis represents a vector from to 𝑗𝑗𝑗𝑗 represents a vector from to 𝑗𝑗𝑗𝑗 − 𝑧𝑧𝑖𝑖 𝑧𝑧𝑖𝑖 𝑗𝑗𝑗𝑗 Gain is given by 𝑗𝑗𝑗𝑗 − 𝑝𝑝𝑖𝑖 𝑝𝑝𝑖𝑖 𝑗𝑗𝑗𝑗 = ∏𝑖𝑖 𝑗𝑗𝑗𝑗 − 𝑧𝑧𝑖𝑖 Phase can be calculated as𝐺𝐺 𝑗𝑗𝑗𝑗 ∏𝑖𝑖 𝑗𝑗𝑗𝑗 − 𝑝𝑝𝑖𝑖 =
Possible to evaluate∠𝐺𝐺 the𝑗𝑗 𝑗𝑗frequencyΣ∠ 𝑗𝑗𝑗𝑗 response− 𝑧𝑧𝑖𝑖 − Σ∠ graphically𝑗𝑗𝑗𝑗 − 𝑝𝑝𝑖𝑖 from a pole/zero diagram Not done in practice, but provides useful insight
K. Webb ESE 499 Bode Construction from Pole/Zero Plots
102
Consider the following system: + 3 = + 2 + .75 + 2 .75 𝑗𝑗𝑗𝑗 𝐺𝐺 𝑗𝑗𝑗𝑗 Evaluate at𝑗𝑗 𝑗𝑗 = 2. 𝑗𝑗� /𝑗𝑗𝑗𝑗 − 𝑗𝑗� Gain: 𝜔𝜔 5𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠𝑠𝑠.𝑠𝑠 .5 = . . 3+𝑗𝑗� 5 . 𝐺𝐺 𝑗𝑗� .5 = 2+𝑗𝑗� 75 2+𝑗𝑗� 25 . . 3 1 𝐺𝐺 𝑗𝑗� .5 =2 90⋅4.3897 8.
Phase: 𝐺𝐺 𝑗𝑗� → − 2𝑑𝑑𝑑𝑑 .5 = = 3 + .5 = 39.8° ∠𝐺𝐺 𝑗𝑗� 𝜃𝜃1 − 𝜃𝜃2 − 𝜃𝜃3 𝜃𝜃1 = ∠ 2 + 𝑗𝑗�.75 = 20.6° 𝜃𝜃2 = ∠ 2 + 𝑗𝑗�.25 = 64.8° 𝜃𝜃3 ∠.5 =𝑗𝑗� 45.5°
K. Webb ∠𝐺𝐺 𝑗𝑗� − ESE 499 103 Polar Frequency Response Plots
K. Webb ESE 499 Polar Frequency Response Plots 104
is a complex function of frequency Typically plot as Bode plots 𝐺𝐺 𝑗𝑗𝑗𝑗Magnitude and phase plotted separately Aids visualization of system behavior
A real and an imaginary part at each value of A point in the complex plane at each frequency Defines a curve in the complex plane 𝜔𝜔 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 A useful concept later, for the Nyquist stability criterion
K. Webb 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 Dependent on frequency rates of change of gain and phase
K. Webb 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 -order differential equation Time-domain Impulse𝑡𝑡푡 response 𝑛𝑛 representations Step response
Transfer function Frequency-domain Frequency response/Bode plot 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 . = . . 9 87 2 1 𝑠𝑠 +5 655𝑠𝑠+9 87 𝐺𝐺 𝑠𝑠 = . 987 2 𝐺𝐺2 𝑠𝑠 𝑠𝑠 +18 85𝑠𝑠+987
K. Webb 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) 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 no outputs are specified, bode response is automatically plotted – preferable to plot yourself Frequency vector input is optional 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 10 0 x1: last point in f is 10 𝑥𝑥 1 N: number of points in f𝑥𝑥 f: vector of logarithmically-spaced points Generates logarithmically-spaced points between 10 and 10 𝑥𝑥0 𝑁𝑁𝑥𝑥1 Useful for generating independent-variable vectors for log plots (e.g., frequency vectors for bode plots) Linearly spaced on a logarithmic axis
K. Webb ESE 499