1. System Dynamics Dr. Mohammad Kilani
Class 5
Response of Linear Time Invariant Systems

2. System Dynamics Dr. Mohammad Kilani
Midterm Exam:
Sunday, August 2
12:30 – 2:00

3. Forced Response of Linear Time Invariant Systems
A linear time invariant system of order n has n roots and is written as:
The Laplace transformed equation of the system is:
Some of the roots are caused by F(s), while the others are caused by the system. Those caused by the system be 2m complex roots and r = n – 2m real roots:
The roots may be written as:
The overall solution is a linear combination of the following solutions
complex
real

4. Standard Forcing Functions
Impulse: No roots
Step: One root at s = 0
Ramp: Two roots at s = 0
Harmonic: Two roots at s = ±jω
step
Ramp
Parabolic
Pulse
Impulse
Harmonic

5. Transient Response of a Linear Time Invariant Systems

6. First Order Systems

7. Impulse Response of 1st Order System
τ = 0.5
τ = 1
τ = 2
A 1st order system subject to a unit impulse input may be written as:
Setting x(0) = 0 and transforming while noting that L[δ(t)] = 1, we obtain:
which gives the following expression for x(t):
t/τ xτ

8. Impulse Response of 1st Order System
The free response of a 1st order system with the initial condition x(0) = 1/τ has the transformed form:
The response of a 1st order system to a unit impulse is equivalent to the free response of the same system to a initial condition x(0) = 1/τ

9. Step Response of 1st Order System
A 1st order system subject to step input may be written as:
Setting x(0) = 0 and transforming while noting that L[kt] = 1/s, we obtain:
Using partial fraction expansion we get:
Solving for the constants, we obtain
Which gives the following expressions for X(s) and x(t)
t/τ x

10. Ramp Response of 1st Order System
A 1st order system subject to ramp input may be written as:
Setting x(0) = 0 and transforming while noting that L[kt] = 1/s2, we obtain:
Using partial fraction expansion and noting the repeated root at s = 0, we get:
Solving for the constants, we obtain
Which gives the following expressions for X(s) and x(t)

11. 1st Order Systems with Ramp Input Relative Input and Relative Excitation
x(t)
f(t) = kt kτ τ
The response is in steady state after approximately t =4τ . At steady state, x(t)= k(t − τ), so the response is parallel to the input but lags behind it by a time τ
The state error between the input and the output, e(t) is seen to approach a steady state value of kτ

12. Second Order Systems

13. Second Order Systems

14. Effect of Root Location
The figure shows how the location of the characteristic roots in the complex plane affects the free response.
The real part of the root is plotted on the horizontal axis, and the imaginary part is plotted on the vertical axis.

15. Effect of Root Location
Unstable behavior occurs when the root lies to the right of the imaginary axis.
Neutrally stable behavior occurs when the root lies on the imaginary axis.
The response oscillates only when the root has a nonzero imaginary part.
The greater the imaginary part, the higher the frequency of the oscillation.
The farther to the left the root lies, the faster the response decays.

16. Undamped Free Response of 2nd Order Systems
Consider the undamped systems shown. They all have the same model form: mẍ + kx = f(t). Suppose f(t)=0, x(0) = x0 and ẋ(0) = ẋ0.
The response has the form:
which is found to be . m k c x

17. Undamped Response of 2nd Order Systemsk c x
This solution shows that the mass oscillates about the rest position x = 0 with a frequency of ωn =√(k/m) radians per unit time. The period of the oscillation is 2π/ωn.
The frequency of oscillation ωn is called the natural frequency. The natural frequency is greater for stiffer springs (larger k values). The amplitude of the oscillation and the time shift depend on the initial conditions.

18. Damped Free Response of 2nd Order Systemsk c x
The figures shows the free response of the system mẍ + cẋ + kx = f(t) for four values of c, with m = 1, k = 16, x(0) = 1, and ẋ(0) = 0. For no damping, the system is neutrally stable and the mass oscillates with a constant amplitude and a radian frequency of √(k/m) = 4, which is the natural frequency ωn.
As the damping is increased slightly to c = 4, the system becomes stable and the mass still oscillates but with a smaller radian frequency (√12 = 3.464). The oscillations die out eventually as the mass returns to rest at the equilibrium position x = 0.

19. Damped Response of 2nd Order Systems
As the damping is increased further to c = 8, the mass no longer oscillates because the damping force is large enough to limit the velocity and thus the momentum of the mass to a value that prevents the mass from overshooting the equilibrium position.
For a larger value of c (c = 10), the mass takes longer to return to equilibrium because the damping force greatly slows down the mass.

20. Critical Damping
If the damping coefficient is equal to the critical value cc defined as:
then the radical in the roots formula is zero, and the two roots are equal. The system will have two repeated roots s1 = s2 = -c/2m. The system is referred to as critically damped
If c > cc there are two real distinct roots, and the system is overdamped.
If c < cc = 2√mk, complex roots occur, and the system is underdamped.

21. The Damping Ratio
The damping ratio is defined as the ratio of the actual damping constant c to the critical value cc:
For a critically damped system, ζ = 1.
Exponential behavior occurs if ζ >1 (the overdamped case).
Oscillations exist when ζ <1 (the underdamped case).

22. Canonical Form of a 2nd Order System
The roots of the characteristic equation ms2 + cs + k = 0 are purely imaginary when there is no damping. The imaginary part and the frequency of oscillation for this case is the undamped natural frequency ωn = √ (k/m).
We can write the characteristic equation in terms of the parameters ζ and ωn. First divide the equation by m and use the fact that ωn2= k/m and that
The characteristic equation becomes and its roots become
The form above is known as the canonical form of the characteristic equation. It is used to immediately identify the values of ωn and ζ for a system. Since oscillatory response occurs if ζ < 1, the canonical form and can be used as a quick check for the presence of an oscillatory behavior.

23. Canonical Form of a 2nd Order System
Example: Check if the system
is stable and find if it has an oscillatory behavior
Solution: The roots are
The system is stable for d > 0. Comparing the system with the conical form, we find the natural frequency and the damping ratio to be:
Since ζ > 1, the system does not exhibit oscillatory behavior for any value of d.

24. The Damped Frequency of an Underdamped 2nd Order System
The imaginary parts of the roots of an undamped 2nd order system is denoted by ωd and is given by
This corresponds to the damped frequency of oscillation and called the damped frequency, to distinguish it from the undamped natural frequency ωn.
The frequencies ωn and ωd have meaning only for the underdamped case (ζ < 1). The equations above show that ωd < ωn . Thus the damped frequency is always less than the undamped frequency

25. The Time Constant of Underdamped 2nd Order System
The real part of the roots of an underdamped 2nd order system is r = -ζωn . This is the exponent of the exponential term of the response. The time constant is thus equal to:
The tables in the two following slides summarize the free response of the stable, linear, second-order model in terms of the parameters ζ, ωn, and ωd, and the formulas used.

28. Graphical Interpretation of the Root Location of an Underdamped 2nd Order System
The complex conjugate root location of an underdamped 2nd order system can be plotted on the s plane as
The lengths of two sides of the right triangle shown in are ζωn and ωd. Thus the hypotenuse is of length
It makes an angle θ with the negative real axis, where

29. Graphical Interpretation of the Root Location of an Underdamped 2nd Order System
Therefore all roots lying on the circumference of a given circle centered on the origin are associated with the same undamped natural frequency ωn. All roots lying on the same line passing through the origin are associated with the same damping ratio. The limiting values of ζ correspond to the imaginary axis (ζ =0) and the negative real axis (ζ =1).
Roots lying on a given line parallel to the real axis all give the same damped natural frequency ωd. All roots lying on a line parallel to the imaginary axis have the same time constant. The farther to the left this line is, the smaller the time constant.

30. Transient Response of Linear Time-Invariant Systems
Consider a system defined by the differential equation:
where the coefficients a1, a2, … , an are real constants, x(t), is the dependent variable, and p(t) is the input function.
The differential equation has a solution of the form
where xc(t) is the solution of the homogeneous equation:
and xp(t) is a particular solution that satisfies the original equation, and depends on the input function p(t).

31. Transient Response of Linear Time-Invariant Systems
The complimentary part of the solution may be found analytically by assuming an exponential solution of the form shown, which produces
Substituting this solution into the homogeneous equation, we obtain the characteristic equation
If λi is a root of the characteristic equation then xi(t) = e (λ i)t is a solution of the associated homogenous equation. There are generally n roots for the characteristic equation resulting in n solutions.

32. Transient Response of Linear Time-Invariant Systems
If all the n roots, λi, are real and different then the n individual solutions are
and the corresponding general solution is
The complementary solution is referred to as the transient response of the original system equation. If any of the roots, λi, is positive, the transient response grows exponentially with time with no bounds, and the system is unstable. If all the roots are negative, the response vanishes to zero as t approaches infinity. The system will have n time constants, τi = 1/(-λi). The highest time constant is τd = 1/(-λd) where λd is the root with the minimum negative value (root with the minimum absolute value). This root is referred to as the dominant root, λd . The settling time is ts = 4 τd.

33. Transient Response of Linear Time-Invariant Systems
If complex roots, λi = σi + jωi occur, they must occur in conjugate pairs since the coefficients of the original system equation are real. Thus λ’i = σi - jωi is the conjugate root . The two solutions corresponding to these roots are
If the real part of the complex roots, σi , is negative, the transient response vanishes to zero as t approaches infinity. σi is positive, it grows with no bound.

34. Step Response of a Linear Time Invariant Systems
Consider a system with a transfer function [Refer to Ogata page 399].
An analytical expression may be obtained by partial fraction expansion. If the input is a step input, U(s) = 1/s , The expression for X(s) can be written as:
where ai is the residue of the pole at s = -p;. [If the system involves multiple poles, then X(s) will have multiple-pole terms. The response x(t) is given by the inverse Laplace transform of the equation above as:

35. Step Response of a Linear Time Invariant Systems
Consider the case where the poles of X(s) consist of real poles and pairs of complex conjugate poles. A pair of complex poles yield a second-order term in s. The equation above can be written as:
The response is thus composed of a number of terms involving the simple functions found in the responses of first-order and second-order systems. The unit-step response x(t), is found from the inverse Laplace transform of X(s) to be

36. Step Response of a Linear Time Invariant Systems
Thus, the response curve is the sum of a number of exponential curves and sinusoidal curves. [If the poles of X(s) involve multiple poles, then X(I) must have the corresponding multiple-pole terms]
If all poles of X(s) lie in the left-half s-plane, then the exponential terms [including those terms multiplied by t, t2, etc., that occur when X(s) involves multiple poles] and the damped exponential terms in x(t) will approach zero as the time t increases. The steady-state output is then x(∞) = a.

37. Transient Response of a Linear Time Invariant Systems
These transient-response specifications are of an underdamped 2nd order system are illustrated.
Note that the response need not be that of a higher order system could also be described in terms of the same parameters.

38. Systems Response to Step Input: Maximum Peak or Overshoot
The maximum or peak overshoot Mp is the maximum deviation of the output x above its steady-state value xss . It is sometimes expressed as a percentage of the final value and denoted M%.
Because the maximum overshoot increases with decreasing ζ , it is sometimes used as an indicator of the relative stability of the system.

39. Systems Response to Step Input: Peak Time and Settling Time
The peak time tp is the time at which the maximum overshoot occurs.
The settling time ts is the time required for the oscillations to stay within some specified small percentage of the final value. The most common values used are 2% and 5%.

40. Systems Response to Step Input: Steady State Error
If the final value of the response differs from some desired value, a steady-state error exists.
The steady state error is the difference between the final value and the desired steady state value

41. Systems Response to Step Input: Rise Time and Delay Time
The rise time tr can be defined as the time required for the output to rise from 0% to 100% of its final value.
However, no agreement exists on this definition. Sometimes, the rise time is taken to be the time required for the response to rise from 10% to 90% of the final value.

42. Systems Response to Step Input: Delay Time
The delay time td is the time required for the response to reach 50% of its final value

43. Description of Underdamper Response of 2nd Order System
We can express the free response and the step response of the underdamped secondorder in terms of the parameters ζ and ωn as follows.
The form of the free response is as shown in the table for the overdamped, case, critically damped case, and underdamped case.

44. Description of Underdamper Response of 2nd Order System

45. 1st Order Systems with Impulse Input
eqn = 'D1x + x = 0';
init= 'x(0)=1';
x=dsolve(eqn,init,'t');
t=0:0.1:10;
xx = eval(vectorize(x));
plot(t,xx);
The free response of a 1st order system with initial conditions x(0) = xo is equal to that of the response of the system to an impulse of magnitude Ki = τ xo