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4. System Response • This module is concern with the response of LTI system. • L.T. is used to investigate the response of first and second order systems. Higher order systems can be considered to be the sum of the response of first and second order system. • Unit step, ramp, and sinusoidal signal play important role in control system analysis. It is therefore we will investigate this signals. 4. System Response Review of some LTI properties We will express system as in figure below, system input is r(t), output is c(t), and impulse response is h(t) r(t) h(t) c(t) 1. Impulse response Impulse response, denoted by h(t), is the output of the system when its input is impulse (t). h(t) is called the impulse response of the system or the weighting function 2. Convolution Output of LTI system is the convolution of its input and its impulse response: where r’(t) denotes the derivative of r(t) 4. Integral If the input is r(t)dt then the output is c(t)dt t c(t ) r (t ) * h(t ) r ()h(t )d t r (t )h()d Taking the L.T. of (1) yields C(s) = R(s)H(s) (2) Where C(s) is the L.T. of c(t) R(s) is the L.T. of r(t) H(s) is the L.T. of h(t) H(s) is called the transfer function (T.F) 3. Derivative If the input is r’(t) then the output is c’(t) (1) 5. Poles and Zero T.F. is usually rational and therefore can be expressed as N(s)/D(s). Poles is the values of s resulting T.F to be infinite. Zeroes is the values of s resulting T.F to be zero 4.1. Time Response of the First Order Systems. Here we will investigate the time response of the first order systems. The transfer function of a general first order system can be written as: G(s) C ( s) K R( s) τs 1 (1) We can found the differential equation first we write (1) as K 1 s C ( s) R( s) τ τ (3) (4) c(0) ( K / τ)R(s) s (1 / τ) s (1 / τ) (5) The eq. can be shown in the block diagram as shown in the figure bellow. R(s) K τ s c(0) Now we take the L.T of (3) and include the initial condition term to get 1 K sC ( s) c(0) C ( s) R( s) τ τ C ( s) (2) The diff. Eq. is the inverse L.T. of (2) 1 K c(t ) c(t ) r (t ) τ τ Solving for C(s) yields + 1 C(s) τ 1 s 1τ + Note that the initial condition as an input has a Laplace transform of c(0), which is constant. The inverse L.T of a constant is impulse (t). Hence the initial condition appears as the impulse function Here we can see that the impulse function has a practical meaning, even though it is not a realizable signal 4.1. Time Response of the First Order Systems. Since we usually ignore the initial condition in block diagram, we use the system block diagram as shown bellow. R(s) K C(s) τ s 1τ c(t) Suppose that the initial condition is zero then ( K / τ) C (s) R( s) s (1 / τ) (1) (2) Taking the inverse L.T of (2) yields tτ c(t ) K (1 e ) τ K (1 e τ ) t For unit step input R(s)=1/s, then ( K / τ) 1 K K ( s 1 / τ) s s s 1 / τ K t Unit step response C (s) The first term originates in the pole of input R(s) and is called the forced response or steady state response The second term originates in the pole of the transfer function G(s) and is called the natural response Figure below plot c(t) (3) The final value or the steady state value of c(t) is K that is lim c(t)= K t c(t) is considered to reach final value after reaching 98% of its final value. The parameter is called the time constant. The smaller the time constant the faster the system reaches the final value. 4.1. Time Response of the First Order Systems. A general procedure to find the steady state value is using final value theorem lim c(t) = lim sC(s) = lim sG(s)R(s) t s0 s0 For Unit step input then the final value is css(t) = lim G(s) s0 since R(s) = 1/s therefore c(t ) Kt K Ke τ ) t (2) This ramp response is composed of three term 1. a ramp 2. a constant 3. an exponential. System DC gain is the steady state gain to a constant input for the case that the output has a final value. Ramp Response c(t) r(t)=t c(t) For the input equal to unit ramp function r(t) = t and R(s) = 1/s2, C(s) is t C ( s) ( K / τ) 1 K K K (1) 2 2 ( s 1 / τ) s s s s 1/ τ 4.2. Time Response of Second Order System Case 3: =1 (real equal poles), c(t) is The standard form second order system is 2n C ( s) G( s ) 2 R(s) s 2n s 2n The poles of the TF is s = n c(t ) 1 k1et / k2tet / (1) This system is said to be critically damped jω(12) Case 4: =0 (imaginary poles), c(t) is Where = damping ratio n = natural frequency, or undamped frequency. c(t ) 1 sin(nt ) Consider the unit step response of this system C ( s ) G ( s ) R( s ) (s 2 2n 2n )s 2 n (2) For this system we have Time constant = = 1/n ; frequency = n =0 c(t) Case 1: 0<<1 (complex poles), c(t) is 0.2 (3) 0.7 This system is said to be underdamped Case 2: >1 (real unequal poles), c(t) is c(t ) 1 k1et / 1 k2et / 2 This system is said to be overdamped (3) This system is said to be undamped define 1 and tan1 ( ). c(t ) 1 1 ent sin(nt ) (5) 2 (4) n t