Transfer Functions An alternative approach which is frequently used is the transfer function (3.19() () () () ()Ys KGsFs QsGs=+1, 3.222,3) where F(s) is the Laplace transform of F(t), G(s) is the transfer function of the system, and Q(s) describes the initial state. For a first order system, 12(3.20 , 3.23 )G(s)= K/( s+1) τ,3 TransferFunctions.doc - Page 1Notice that if s = i ω i( )G (i ) = 1 / ( i + 1)=M ( ) eφωωτωω which is the result we obtained previously. For a second order system; 22nn1G (s) = (1/ ) + (2 / ) s + 1sζωω TransferFunctions.doc - Page 2Complex Inputs Superposition As long as the system is linear, we can apply the principle of superposition. TransferFunctions.doc - Page 3 ,3,312IF sin (3.22 , 3.33 )noiii = 1 F(t)= + ( t) AAω∑ Then the steady-state output is given by: 12sin (3.23 , 3.34 )noiiiii=1y(t)= K[ + M( ) ( t+ ( )] AAφωωω∑ so, given the step response, we can compute the frequency response and from that the response to any "well-behaved" input.Example Instrument Selection Second-Order System Suppose we have a periodic input 1.First we use Fourier analysis to determine the frequency content of the signal 2.Next we determine the frequency of the highest harmonic which has an amplitude greater than, say, 5% of the fundamental harmonic. If we require that .95 < M(ω) < 1.05 over the frequency range of the signal, the best choice is an instrument with a damping ratio of about 0.7. We need to keep ω/ωn < 0.6 (see Example 3.10) TransferFunctions.doc - Page 4If our signal has a frequency of 2 Hz and the 8th harmonic is the highest with significant amplitude, the instrument must have a natural frequency of n > 2 * 8 / .6 = 26.7 Hf TransferFunctions.doc - Page 5Fig. 3-18 in 2nd and 3rd Edition Fig 3-16 in 2nd and 3rd Edition TransferFunctions.doc - Page 6Coupled Systems If our instrument has more than one component, we treat the output of one part as input to the next Figure 3.24 in 2nd and 3rd Edition TransferFunctions.doc - Page
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