Chapter 0 Time Response of Systems 0. Some Standard Time Responses Let us try to get some impulse time responses just by inspection: Poles F (s) f(t) s-plane Time response p =0 s p =0,p 2 =0 s 2 t p = a, a > 0 s a e at p = a, a > 0 s+a e at 4
Lecture Notes on Control Systems/D. Ghose/202 42 Poles F (s) f(t) s-plane Time response p = jb, p 2 = jb b s 2 +b 2 sin bt p = jb, p 2 = jb s s 2 +b 2 cos bt p = a + jb p 2 = a jb a>0, b>0 b (s+a) 2 +b 2 e at sin bt p = a + jb p 2 = a jb a>0, b>0 s+a (s+a) 2 +b 2 e at cos bt p = a + jb p 2 = a jb a<0, b>0 b (s+a) 2 +b 2 e at sin bt p = a + jb p 2 = a jb a<0, b>0 s+a (s+a) 2 +b 2 e at cos bt What do we observe from these time responses?. Poles are real No oscillations. 2. Poles are positive (on the right hand side of the s-plane Exponential increase, or unstable. 3. Poles are negative (on the left hand side of the s-plane Exponential decrease, or stable.
Lecture Notes on Control Systems/D. Ghose/202 43 4. Poles are imaginary Oscillations with no damping. 5. Poles are complex Damped oscillations. These can be summarized in the following figure. PROBLEM SET 3 Figure 0.: Time response for different pole placements. Find the partial fraction expansions of the following transfer functions: (a) G(s) = s+2 s 2 +5s+2 (b) G(s) = s 2 + (s+) 2 (s+3) (c) G(s) = s 2 +3 s 3 +3s 2 +5s+6 2. Find the time response of the following systems, driven by unit step input signals, using Laplace transforms: (a) d2 x dt 2 +5 dx dt +3x =3u(t)
Lecture Notes on Control Systems/D. Ghose/202 44 (b) Mass-spring system. 0ẍ +3x =5u(t) (c) Spring-mass-damper system. 6ẍ +2ẋ +3x =2u(t) Note that the initial conditions can be non-zero. What is an oscillation? It is a result of energy transfer between two energy storage elements. For example, in a spring-mass-damper system: mass: kinetic energy spring: potential energy damper: energy dissipator Figure 0.2: Spring-mass-damper system What is damping? It is the result of energy dissipation. 0.2 Time Response Characteristics What are the main system time response characteristics? These can be classified in the following three categories.. Transient response (Output signal soon after the input signal is applied) 2. Steady-state response (Output signal long time after the input signal is applied) 3. Stability Our main objective is to find the behaviour of a system (in terms of the output y(t)) for various input signals r(t). Normally, the input r(t) is not known in advance since all we may know is the system transfer function G(s). So, it is customary to check the behaviour of a system for the following types of inputs:. Unit Step: r(t) =u(t) R(s) = s
Lecture Notes on Control Systems/D. Ghose/202 45 2. Unit Ramp: r(t) =tu(t) R(s) = s 2 3. Unit Impulse: r(t) = δ(t) R(s) = 4. Sinusoid: r(t) =u(t)sinωt R(s) = ω s 2 +ω 2 Usually designers assume that if the system behaviour is OK for these functions then most probably it will be ok for other signals too. In fact one usually looks at system behaviour against step functions only. Why? Because most piecewise continuous signals can be represented as a collection of pulses (which in turn can be created by algebraic manipulation of unit step functions). This we have already seen. Another Assumption. Any linear constant-coefficient system can be broken down into a cascade of first and second order systems (remember that we did this when we did the partial fraction expansion). Figure 0.3: A LTI system represented as a cascade of smaller systems So it is logical to first examine the time response of first and second order systems against step inputs. 0.3 Characteristics of First Order Systems Consider the first order system (assuming zero initial condition), a ẏ + a 0 y = b 0 r Taking Laplace transforms on both sides (a s + a 0 )Y (s) =b 0 R(s)
Lecture Notes on Control Systems/D. Ghose/202 46 So, the transfer function G(s) is given by, G(s) = Y (s) R(s) = b 0 a s + a 0 = b 0 a s + a 0 a The poles of G(s) are the roots of the denominator polynomial. Define, Define, So, p = a 0 a = a 0 a Time constant Bandwidth K = b 0 Open loop gain or DC gain a 0 G(s) = b 0 a s + a 0 = b 0 a s + a 0 a = b 0 a 0 a 0 a s + a 0 a = K s + Suppose, Figure 0.4: Position of pole and the impulse response of a first order system r(t) =δ(t)
Lecture Notes on Control Systems/D. Ghose/202 47 which is an impulse function. Then, R(s) = and Taking inverse Laplace transform, Y (s) =G(s)R(s) =K s + y(t) =K e t = K e t t y(t) 0 K K e =0.37 K. is the time required for y(t) to reach 37% of its initial value. 2. The DC gain term arises from the observation that for a unit step input, y(t) K as t. Check: Y (s) =G(s)R(s) = K s + [ = K Taking inverse Laplace transform on both sides, y(t) =K [ ] e t So, s s + + s lim = K t So the idea is: Send in and get K, after things have settled down! Try getting the same result using the final value theorem. y( ) = lim s 0 sy (s) = lim What does the response look like? s 0 K s + = K ]
Lecture Notes on Control Systems/D. Ghose/202 48 Figure 0.5: Unit step response of a first order system