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Chapter 4. Continuous Time Signals Time Response. Poles and Zeros. Poles: the values of the Laplace transform variable, s, that cause the transfer function to become infinite Any roots of the denominator of the transfer function that are common to roots of the numerator. Zeros:
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Chapter 4 Continuous Time SignalsTime Response
Poles and Zeros Poles: • the values of the Laplace transform variable, s, that cause the transfer function to become infinite • Any roots of the denominator of the transfer function that are common to roots of the numerator. Zeros: • the values of the Laplace transform variable, s, that cause the transfer function to become zero • Any roots of the numerator of the transfer function that are common to roots of the denominator.
Figure 4.1a. System showinginput and output;b. pole-zero plotof the system;c. evolution of asystem response.Follow blue arrowsto see the evolutionof the responsecomponent generatedby the pole or zero.
Poles and Zeros • A pole of the input function generates the form of the forced response ( steady state response) • A pole of the transfer function generates the form of the natural response • A pole on the real axis generates an exponential response of the form e-at • The zeros and poles generate the amplitudes for both the forced and the natural responses.
Ex: Evaluating response using poles Problem: Given the system, write the output, c(t), in general terms. Specify the natural and forced parts of the solution. By inspection, each system pole generates an exponential as part of the natural response, and the input’s pole generates the s s response, thus Forced Natural Response Response
Steady State Step Response The steady state response (provided it exists) for a unit step is given by where G(s) is the transfer function of the system.
First Order Systems We define the following indicators: Steady state value, y:the final value of the step response (this is meaningless if the system has poles in the RHP). Time Constant, 1/a: The time it takes the step response to rise to 63% of its final value. Or the time for e-at to decay 37% of its initial value.
First Order Systems Rise time, Tr: The time elapsed for the waveform to go from 0.1 to 0.9 of its final value Settling time, Ts: the time elapsed until the step response enters (without leaving it afterwards) a specified deviation band, ±, around the final value. This deviation , is usually defined as a percentage of y, say 2% to 5%. Overshoot, Mp: The maximum instantaneous amount by which the step response exceeds its final value. It is usually expressed as a percentage of y
Figure 4.8Second-orderstep response componentsgenerated bycomplex poles
Figure 4.9System forExample 4.2 Factoring the denominator we get s1 = -5+j13.23 & s2 = -5-j13.23 The response is damped sinusoid
Figure 4.10Step responsesfor second-ordersystemdamping cases
The General Second Order system Natural Frequency Is the frequency of oscillation of the system without damping Damping Ratio General System
Figure 4.11Second-orderresponse as a function of damping ratio
Figure 4.12 Systems for Example 4.4 Since then Using values of a & b from each system we find for (a) overdamped for (b) critically damped for (c) underdamped
Figure 4.13Second-orderunderdampedresponses fordamping ratio values
Figure 4.16Normalized risetime vs. dampingratio for asecond-orderunderdampedresponse
Figure 4.18Lines of constantpeak time, Tp , settlingtime,Ts , and percentovershoot, %OSNote: Ts2 < Ts1 ;Tp2< Tp1; %OS1 <%OS2
Figure 4.19Step responsesof second-orderunderdamped systemsas poles move:a. with constant real part;b. with constant imaginary part;c. with constant damping ratio
Figure 4.20 Pole plot for Example 4.6 Find Solution:
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Figure 4.23Component responses of a three-pole system:a. pole plot;b. componentresponses: nondominant pole is neardominant second-order pair (Case I), far from the pair (Case II), andat infinity (Case III)
Figure 4.24 Step responses with additional poles Note that c2(t) is better approximate of c1(t) than c3(t) since the 3rd pole is farthest from dominant poles
Figure 4.25 Effect of adding a zero to a two-pole system The system has 2 poles at -1+j2.828 and -1-j2.828 Note: the farther the location of the zero from the dominant poles the lesser its effect on the response
Transfer Functions for Continuous Time State Space Models Taking Laplace transform in the state space model equations yields and hence G(s) is the system transfer function.
Laplace transform solution: Example Problem:Given the system represented in state space by the following equations: a) solve the state equation and find output b) find eigenvalues and poles Solution: First we find (sI-A)-1=
Laplace transform solution: Example Cont. Using U(s) = 1/(s+1), laplace transform of e-t, and B and x(0) from given equations, we sustitue in X(s) = (sI-A)-1[x(0) + BU(s)], we get The output is
Laplace transform solution: Example Cont. Taking Laplace transform of Y(s) we get b) We set det(sI-A) = 0 to find poles and eigenvalues which are -2, -3, and -4
Solution of Continuous Time State Space Models A key quantity in determining solutions to state equations is the matrixexponential defined as The explicit solution to the linear state equation is then given by
Time domain solution, Example: Problem:for the state equation and initial state vector shown, where u(t) is a unit step. Find the state-transition matrix and then solve for x(t). Solution: Find eigenvalues using det(sI-A) = 0. Hense s2+6s+8 = 0 and s1= -2 s2 = -4. Now we assume state transition matrix as Solve for the constants using
Time domain solution, Example: Cont. Also, Then,
Time domain solution, Example Cont. Final result is,
State Space via Laplace transform Example: Problem: Find the state transition matrix using laplace for the following system Solution: we find First find For which
State Space via Laplace transform Example Cont. Expanding each term in the matrix using partial fractions Finally, taking the inverse Laplace transform, we obtain
Step Response for Canonical Second Order Transfer Function On applying the inverse Laplace transform we finally obtain
Figure 4.5: Pole location and unit step response of a canonical second order system.
Summary • There are two key approaches to linear dynamic models: • the, so-called, time domain, and • the so-called, frequency domain • Although these two approaches are largely equivalent, they each have their own particular advantages and it is therefore important to have a good grasp of each.
Time domain • In the time domain, • systems are modeled by differential equations • systems are characterized by the evolution of their variables (output etc.) in time • the evolution of variables in time is computed by solving differential equations
Frequency domain • In the frequency domain, • modeling exploits the key linear system property that the steady state response to a sinusoid is again a sinusoid of the same frequency; the system only changes amplitude and phase of the input in a fashion uniquely determined by the system at that frequency, • systems are modeled by transfer functions, which capture this impact as a function of frequency.
