# time res webpage - Chapter 5:Time Response Analysis Test...

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Test signals Impulse Step Ramp Sinusoidal (frequency response) Impulse δ (t) 1 step u(t) 1/s Ramp t u(t) 1/s 2 Laplace Transforms Chapter 5:Time Response Analysis:

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Examples of zeroth order systems : Potentiometer dc amplifier dc tachogenerator zeroth order system : C(s) / R(s) = K ; a constant (algebraic equation)
First order systems: C(s) / R(s) = K / (1+s τ ) K : steady state value of the Function τ : time constant, τ smaller, faster response. C(t) = K ( 1- e -t / τ ) ;for unit step Step Response

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0 50 100 150 200 250 300 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 t1 t2 Time constant : t2 < t1 dc/dt = 1/ τ t=0 K=1 Error e(t) = e -t / τ e ss = 0 c t If K 1 and input is not unity then e ss ?
Ramp response of first order systems r (t) = t u(t); R (s) = 1 / s 2 1 = K Let ; t s 1 K R(s) C(s) + = ) s 1 ( s 1 ) 2 τ + = c(s

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τ + τ - - = τ s 1 s 1 s 1 ) s ( C 2 ) e 1 ( - t = c(t) -t/ τ - Error e(t) = τ (1- e -t / τ ) e ss = τ If K 1 and input is not unity then e ss ?
0 1 2 3 4 5 6 7 8 9 10 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 9 9 τ τ c t τ - -t/ e 1 = dc/dt

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2 n n 2 2 n + s 2 s ) s ( R ) s ( C ω ζω + = Second order systems Second order example position control
For unit step input R ( s ) = 1 s C(t) t =1.0 Applying final value theorem Steady state error for step input = 0

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θ r θ L K p K A + - + - K T R a 1 s(Js+B) sK b position control Example
Taking K P = 0.1 V / rad K A = 10 V / V K T (Nm/A) = K b (V/rad/sec. ) = 0.5 J = 1 Kg m2 B = 0.5 Nm / rad / sec 1 + s s 1 ) s ( ) s ( 2 r L + = θ θ ∴ ω n = 1 rad / sec and ζ = 0. 5 s 2 + s + 1 = ( ) ( ) s + 0.5 2 3 2 2 + × × j 3 2 - j 3 2 - 0.5

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In general ( ) ( ) 2 2 n 2 n 1 + s ζ - ω + ζω = s 2 + 2 ζ ω n s + ω n 2 2 n n 1,2 1 j s ζ - ω ± ζω - = = - ζ ω n ± j ω d ; ω d = ω n 1 - ζ 2 ω d damped freq of oscillations ζ damping factor ω n undamped natural frequency
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time res webpage - Chapter 5:Time Response Analysis Test...

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