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Systems and Solutions Class 4 1st Order Systems 1st Order Systems dx dt x f t x 0 x o time constant (s) x Instrument response (output) f t Quantity to be measured (input) 1st Order Systems dx dt x f t x 0 x o Step Input f ( t ) K s u s t Ramp Input f ( t ) K r tu s t Harmonic Input f ( t ) K h cos t 1st Order Systems with Step Input dx dt x K s u t x 0 x o 0, t 0 u t 1, t 0 2 u(t) 1 0 -1 0 Time, t 1 2 1st Order Systems with Step Input Solution by Integration dx x Ks dt x K xo x 0 x o dx s Ks x dt dx Ks x x K xo dt dx s x dt x 0 u K s x , du dx u dx t t dt 0 uo du t u dt , 0 u ln uo u t , ln u o Ks x x Ks K s xo xo K s t e x t K s x o K s e t t Error Ratio 1st Order Systems with Step Input Error Ratio and Excitation Ratio 1 Error: Output deviation from input 0.8 Excitation: Output deviation from its initial value 0.4 Excitation Ratio 0.6 Error Ratio 0.2 time, t 0 Error Ratio rerr rerr rerr Current Error Starting Error Current deviation from input value 0 K s xo x Ks xo K s e t 2 Excitation Ratio rex rex Starting deviation from input value Ks x 1 rex 3 4 5 Current Excitation Desired Excitation Output deviation from its initial value Input deviation from initial output x xo K s xo 1 K s xo K s xo K xo x xo rex 1 s K s xo K s xo Ks x rex 1 K s xo x xo K s xo t 1 rerr 1 e 1st Order Systems with Step Input Solution by Superposition Assume a solution of the form dx dt x t x h t x p t x Ks where x 0 x o x h x h 0 and x p x p K s Solution of the particular equation x p x p K s For the homogenoeu s ODE is found by observatio n x h x h 0 x p t K s Assume The complete solution is thereofore x h t Ce rt x h t rCe rt x t x p t x h t K s Ce Substituti ng into homogeneou s ODE x h x h 0 , rCe rt Ce rt 0 , r 1 Ce r 1, r 1 x h t Ce t t To satisfy the initial condition, rt 0 x 0 K s C x o , C x o K s x t K s x o K s e t 1st Order Systems with Step Input Solution by Laplace Transform dx dt X ( s ) xo x Ks sX ( s ) x o X ( s ) sX ( s ) X ( s ) Ks s X ( s ) s 1 X (s) s s 1 A B X ( s ) x o s s 1 x 0 x o X (s) s K s xo s xo K s sxo s K s sxo s s 1 xo s K s xo s s 1 Ks s K s xo A s 0 s s 1 K s xo s 0 s K s xo B s 1 s s 1 B 1 K s xo 1 s 1 1 K s xo K x 1 K s xo X ( s ) x o s o s s 1 X (s) Ks xo K s s 1 t x o K s e s x t K s 1st Order Systems with Unit Step Input and Unit Time Constant MATLAB Simulation by Transfer Function dx dt >> num=1; >> den=[1 1]; >> sys = tf(num,den); >> step(sys) >> grid x f t x 0 0 sX ( s ) X ( s ) F s Step Response 1 X ( s ) s 1 F s F s 1 s 1 0.8 0.7 Transfer Function 0.6 Amplitude X (s) 0.9 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 Time (sec) 4 5 6 1st Order Systems with Unit Step Input and Unit Time Constant MATLAB Simulation by Simulink dx dt dx dt x u t 1 u t x 1st Order Systems with Ramp Input dx dt x x o K r tu t x 0 x o 0, t 0 u t 1, t 0 tu t 1st Order Systems with Ramp Input Solution by Superposition Solution of the particular equation dx dt x x o K rt x 0 x o Assume a solution of the form x p x p x o K r t x t x h t x p t Assume a solution of the form where x p t At B , x p t A x h x h 0 Substitute into the ODE and A At B x o K r t x p x p x o K r t A Kr A B xo , For the homogenoeu s ODE B xo A xo K r x h x h 0 x p t K r t x o Assume x h t Ce rt x h t rCe The complete solution is thereofore rt x t x p t x h t x o K r t Ce Substituti ng into homogeneou s ODE x h x h 0 , rCe rt Ce rt 0 , r 1 Ce r 1, r 1 x h t Ce t To satisfy the initial condition, rt 0 x 0 x o K r C x o , C K r x t x o K r t K r e x t x o K r t K r 1 e t t t 1st Order Systems with Ramp Input Steady State Error and Relative Error dx dt x x o K r tu t x t x o K r t K r 1 e t x 0 x o 1 error Input - Output 0.9 error x o K r t x t error K r 1 e t Err r 0.8 0.7 Steady state error 0.6 e ss lim error 0.5 t e ss lim K r 1 e t t 0.4 0.3 e ss K r Relative error Err r K r 1 e K r Error Steady State Error t 1 e 0.2 0.1 0 t t 0 0.5 1 1.5 2 2.5 3 3.5 4 1st Order Systems with Ramp Input Relative Input and Relative Excitation dx dt x x o K r tu t 4 x 0 x o 3.5 x t x o K r t K r 1 e t 3 Relative Input 2.5 1 2 Relative Input Input - Initial Value Relative Input K rt S.S.E K r K r t K r 1 e K r t Relative Excitation 1 t 0.5 Relative Excitation ex r ex r 1.5 Current Excitation Steady State Error t 1 e t t 0 0 0.5 1 1.5 2 2.5 3 3.5 4 1st Order Systems with Ramp Input Solution by Laplace Transform dx dt x x o K r t , x 0 x o xo sX ( s ) x o X ( s ) xo sX ( s ) X ( s ) X ( s ) s 1 X (s) s Kr s s 2 Kr s 2 xo xo s 2 xo s K r s 2 2 s s 2 X ( s ) xo s 2 1 1 s Kr xo X (s) s 1 A B X ( s ) x o 2 s s K 2 1 s s r d 2 xo K r xo B s 2 ds s s 1 xo s 0 Kr 2 1 s s x o C s 1 K r xo 2 s s 1 s 1 xo s 2 xo s K r s Kr 2 1 s s xo 2 A s K r xo s 2 s 1 s 0 Kr s s 1 C 2 x o K r s K r s 1 x t K r t x o K r K r e x t x o K r t K r 1 e t t 1st Order Systems with Unit Ramp Input and Unit Time Constant MATLAB Simulation by Transfer Function MATLAB does not have a ‘ramp’ command to plot the ramp response of the system. However, note that the response, Rst(s) of a system with transfer function G(s) to unit step input is Rst(s) = G(s)/s, and its response to a unit ramp input is Rrmp(s) = G(s)/s2 = (G(s)/s)/s . Thus, the response of G(s) to unit ramp is equal to the response of H(s) = G(s)/s to unit step. We may use MATLAB’s ‘step’ command to obtain the ramp response of a system G(s) simply by obtaining the step response of H(s) = G(s)/s to unit step. dx dt x f t >> num = [0 0 1]; >> den = [1 1 0]; >> t=0:0.1:5; >> sys = step(num,den,t); >> plot(t,sys,'o',t,t,'-') >> grid x 0 0 sX ( s ) X ( s ) F s X ( s ) s 1 F s G (s) X (s) H (s) G (s) F s s 1 s 1 1 s ( s 1) 1 s s 2 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0 1 2 3 4 5 1st Order Systems with Unit Step Input and Unit Time Constant MATLAB Simulation by Simulink dx dt dx dt x tu t 1 tu t x 1st Order Systems with Harmonic Input 1 dx dt x K h cos t x 0 0 0 -1 0 2 4 6 8 10 12 1st Order Systems with Harmonic Input Solution by Superposition sin 2 sin cos 2 cos sin 2 cos cos 2 cos cos 2 sin sin 2 sin Solution of the particular equation dx dt x p x p K h cos( t ) x K h cos t Assume a solution of the form x 0 0 x p t A cos( t ) x p t A sin( t ) Assume a solution of the form x t x h t x p t x p t A cos( t 2 ) where x h x h 0 and x p x p x o K r t Substitute into the ODE For the homogenoeu s ODE A cos( t 2 ) A cos( t ) K h cos( t ) x h x h 0 From the vector plot A A K h x h t Ce 2 2 Assume rt x h t rCe tan Substituti ng into homogeneou s ODE x h x h 0 , rCe rt Ce rt 0 , r 1 Ce r 1, r 1 x h t Ce 1 A Kh rt t rt 0 1 2 τAω 2 A ϕ Kh ωt 1st Order Systems with Harmonic Input Amplitude Ratio and Phase f(t) 1 dx dt x K h cos t x(t) ra 0 x 0 0 ϕ x t A cos( t ) 1 A Kh tan 1 2 -1 0 2 4 6 8 10 1.5 Amplitude Ratio ra ra A K h 1 Phase tan 1 1 Output Amplitude Input Amplitude 1 ϕ (τω) ra (τω) 2 0.5 0 -2 10 10 -1 10 0 τω 10 1 10 2 12 1st Order Systems with Harmonic Input Solution by Complex Exponential Euler’s Identity Multiplication & Division Rules y e i cos i sin z = Aeiωt z Ae z1 A1 e i t A z A cos t iA sin t i 1 i 2 z 2 A2 e ωt z x iy z1 z 2 A1 e x z1 z2 A1 e A2 e i 1 A2 e i 1 i 2 i 2 A1 A1 A2 e e i 1 2 A2 y Complex Conjugate z = Aeiωt z Ae i t z * Ae x Re z A cos t Power Rules A i t ωt 1 2 z z * x -ωt z Ae z Ae n z z* = i Ae-iωt 1 n i Ae i n A e n 1 n in 1 n A e i n i 1 2 Second Order Systems In the system shown, the input displacement, xi, will cause a deflection in the spring, and some time will be needed for the output displacement xo to reach the input displacement. xo xi c m k Second Order Systems F m xo k x i x o c x i x o m xo xi xo c m xo c x o kx o c x i kx i m k xo c k x o x o c x i kx i m k If m/k << 1 s2 and c/k << 1 s, the system may be approximated as a zero order system with unity gain. If, on the other hand, m/k << 1 s2 , but c/k is not, the system may be approximated by a first order system. Systems with a storage and dissipative capability but negligible inertial may be modeled using a first-order differential equation. Example – Automobile Accelerometer Consider the accelerometer used in seismic and vibration engineering to determine the motion of large bodies to which the accelerometer is attached. The acceleration of the large body places the piezoelectric crystal into compression or tension, causing a surface charge to develop on the crystal. The charge is proportional to the motion. As the large body moves, the mass of the accelerometer will move with an inertial response. The stiffness of the spring, k, provides a restoring force to move the accelerometer mass back to equilibrium while internal frictional damping, c, opposes any displacement away from equilibrium. Piezoelectric crystal xi xo c m k Zero-Order systems Can we model the system below as a zero-order system? If the mass, stiffness, and damping coefficient satisfy certain conditions, we may. xo xi c m k F m xo k x i x o c x i x o m xo k x i x o c x i x o m xi xo m xi k c m m xi First Order Systems Measurement systems that contain storage elements do not respond instantaneously to changes in input. The bulb thermometer is a good example. When the ambient temperature changes, the liquid inside the bulb will need to store a certain amount of energy in order for it to reach the temperature of the environment. The temperature of the bulb sensor changes with time until this equilibrium is reached, which accounts physically for its lag in response. In general, systems with a storage or dissipative capability but negligible inertial forces may be modeled using a first-order differential equation. 1st Order Systems with Step Input error ratio = excitation ratio = current error current deviation from final value = starting error starting deviation from final value current excitation current deviation from initial value = desired (input) excitation input deviation from initial value + = () 0 = 0 = + 0 − −/ Error Ratio − = −/ 0 − Excitation Ratio − 0 = 1 − −/ − 0 Excitation ratio may also be called response ratio = current response / desired response Note that the excitation ratio also represents the system response in case of x0=0 and K=1 Example 1 A bulb thermometer with a time constant τ =100 s. is subjected to a step change in the input temperature. Find the time needed for the response ratio to reach 90% Example 1 Solution A bulb thermometer with a time constant τ =100 s. is subjected to a step change in the input temperature. Find the time needed for the response ratio to reach 90% − 0 = 1 − −/ = 0.9 − 0 − = −/ = 0.1 0 − = ln 10 = 2.3 = 2.3 × = 230 . ≈ 4 minutes 1st Order Systems with Ramp Input + = 0 + () 0 = 0 excitation ratio = = 0 + − (1 − −/ ) Error = − = − (1 − −/ ) Steady State Error . . = lim () − () →∞ . . = = − 0 (1 − −/ ) = 1 − Note that using L’Hospital rule lim →0 = lim →0 1− (1 − −/ ) 1 − lim →0 1 − −/ −/ = 1 − lim = 1−1= 0 →0 1 excitation ratio = current excitation desired (input) excitation current deviation from initial value input deviation from initial value 1st Order Systems with Harmonic Input + = cos (ωt) = −/ + cos (ωt − φ) C depends on the initial conditions and the exponential term will vanish with time. We are interested in the particular steady solution cos (ωt − φ). Solving for and φ, we find = 1 + φ = tan−1 2 1st Order Systems with Harmonic Input Define the amplitude ratio = and the time ratio = where = 2 is the period of the excitation function, = = 1 1 + 2 = 1 1 + 4 2 2 φ = tan−1 = tan−1 2 1st Order Systems with Harmonic Input The amplitude ratio, Ar(ω), and the corresponding phase shift, ϕ, are plotted vs. ωτ. The effects of τ and ω on frequency response are shown. For those values of ωτ for which the system responds with Ar near unity, the measurement system transfers all or nearly all of the input signal amplitude to the output and with very little time delay; that is, X will be nearly equal to F in magnitude and ϕ will be near zero degrees. 1st Order Systems with Harmonic Input At large values of ωτ the measurement system filters out any frequency information of the input signal by responding with very small amplitudes, which is seen by the small Ar(ω) , and by large time delays, as evidenced by increasingly nonzero ϕ. 1st Order Systems with Harmonic Input Any equal product of ω and τ produces the same results. If we wanted to measure signals with highfrequency content, then we would need a system having a small τ. On the other hand, systems of large τ may be adequate to measure signals of low-frequency content. Often the trade-offs compete available technology against cost. dB = 20 log Ar(ω) 1st Order Systems with Harmonic Input The dynamic error,δ(ω), of a system is defined as δ(ω) = (X(ω) – F)/F δ(ω) = Ar(ω) –1 It is a measure of the inability of a system to adequately reconstruct the amplitude of the input signal for a particular input frequency. We normally want measurement systems to have an amplitude ratio at or near unity over the anticipated frequency band of the input signal to minimize δ(ω) . As perfect reproduction of the input signal is not possible, some dynamic error is inevitable. We need some way to quantify this. For a first-order system, we define a frequency bandwidth as the frequency band over which Ar(ω) > 0.707; in terms of the decibel defined as dB = 20 log Ar(ω) This is the band of frequencies within which Ar(ω) remains above 3 dB Example 2 A temperature sensor is to be selected to measure temperature within a reaction vessel. It is suspected that the temperature will behave as a simple periodic waveform with a frequency somewhere between 1 and 5 Hz. Sensors of several sizes are available, each with a known time constant. Based on time constant, select a suitable sensor, assuming that a dynamic error of 2% is acceptable. Example 2. Solution A temperature sensor is to be selected to measure temperature within a reaction vessel. It is suspected that the temperature will behave as a simple periodic waveform with a frequency somewhere between 1 and 5 Hz. Sensors of several sizes are available, each with a known time constant. Based on time constant, select a suitable sensor, assuming that an absolute value for the dynamic error of 2% is acceptable. Accordingly, a sensor having a time constant of 6.4 ms or less will work. ω ≤ 0.02 −0.02 ≤ ω ≤ 0.02 −0.02 ≤ − 1 ≤ 0.02 0.98 ≤ ≤ 1.02 0.98 ≤ 1 1 + 2 ≤ 1.0 0 ≤ ≤ 0.2 The smallest value of will occur at the largest frequency = 2 = 2(5) 0 ≤ 2(5) ≤ 0.2 ≤ 6.4 × 10−3 s. Example 2. Solution A temperature sensor is to be selected to measure temperature within a reaction vessel. It is suspected that the temperature will behave as a simple periodic waveform with a frequency somewhere between 1 and 5 Hz. Sensors of several sizes are available, each with a known time constant. Based on time constant, select a suitable sensor, assuming that an absolute value for the dynamic error of 2% is acceptable. Accordingly, a sensor having a time constant of 6.4 ms or less will work. ω ≤ 0.02 −0.02 ≤ ω ≤ 0.02 −0.02 ≤ − 1 ≤ 0.02 0.98 ≤ ≤ 1.02 0.98 ≤ 1 1 + 2 ≤ 1.0 0 ≤ ≤ 0.2 The smallest value of will occur at the largest frequency = 2 = 2(5) 0 ≤ 2(5) ≤ 0.2 ≤ 6.4 × 10−3 s. 2nd Order Systems Example: Spring – mass damper RLC Circuits Accelerometers Mathematical Model: 2 + 2 + 2 = 2 Damping ratio (dimensionless) Natural frequency (1/s) : Input (quantity to be measured) : Output (instrument response) 2nd Order Systems with step input = () = 0 <0 1 ≥0 ds 2 2 + 2 + = 2 Damping ratio (dimensionless) Natural frequency (1/s) : Input (quantity to be measured) : Output (instrument response) : Arbitrary constant 2nd Order Systems with step input 2nd Order Systems with step input 2nd Order Systems with periodic input 2nd Order Systems with step input