y t ce t KA 1 2 2 sin t arctan Long time response purely sinusoidal linear system but Amplitude modified by quantity 2 2 What Happens if 2 2 1 What Happens if 2 2 1 How about the Phase The output follows the input The output goes to zero What if 4 What is the time delay Note 2 f 1 2 The period is T f 2 s The time delay is 1 2 40 at 10 rad s Input Output A sin t KA 1 2 2 Magnitude Ratio M Output Input 1 sin t dB 20log M 1 2 2 The frequency bandwidth often defined by the 3dB point 10 3 20 0 707 y t A sin t System 1 K1 M 1 System 1 K1 M 1 y t A sin t y AK1M 1 sin t System 2 K 2 M 2 2 y AK1K 2 M 1 M 2 sin t 1 2 SYSTEM MAG RESPONSE PRODUCT OF MAG RESPONSE OF COMPONENTS SYSTEM PHASE RESPONSE SUM OF PHASE RESPONSE OF COMPONENTS The input to our system may be some complex waveform that we can more easily represent as a fourier series F t a0 an cos nt bn sin nt n 1 or it may be multiple periodic waveforms Either way the output of the system is simply the linear sum of the separate output signals that would result from each input signal being applied separately Step function in 1st order system 0th order system K1 1 2 Output y t F t A H t Assuming that the initial condition for the first order system is y1 0 0 derive an expression for the output y t d y For a first order system we have y KF t dt For zero initial conditions we have that y t K1 A K1 Ae The zero order system follows this response with a static sensitivity of K2 What is the output t y t K 2 K1 A K1 Ae What is the steady state output K1K2A t Consider the following system 1st order system Step function in K1 1 1st order system 2 2 Output y t F t A H t Assuming that the initial conditions are y1 0 y2 0 0 1 Write a single first order ordinary differential equation for the output y t You do not need to solve this equation 2 What is the steady state response of this system The output of the first order system is given by y t K1 A K1 Ae What is the output of the second system t d y 1 2 y K 2 K1 A K1 Ae dt What is the steady state response of this system y K 2 K1 A t 1 f t a0 an cos n t bn sin n t n 1 Find the Fourier Series representation of the periodic displacement signal shown above Neglect frequency components beyond 0 2 Hz Find Fourier coefficients using the Euler formulas and then plug in Fourier Series 2 2 f t a0 an cos n t bn sin n t T T n 1 The period of the signal is 8 seconds so the first n 1 Fourier Component will be at 0 125 Hz and the second at 0 250 Hz Therefore we only have to go to n 1 1 a0 T T 2 2 1 f t dt dt 0 25 8 0 T 2 2 2 a1 cos 80 8 2 1 t dt 2 0 7853 T 2 2 b1 sin 80 8 2 1 t dt 1 a0 T T 2 2 1 f t dt dt 0 25 8 0 T 2 2 a1 cos 80 8 2 1 t dt 2 f t a0 a1 cos 0 7853 t b1 sin 0 7853 t 1 1 f t 0 25 cos 0 7853 t sin 0 7853 t 2 f 12 0 7853 f 1 5Hz 2 2 b1 sin 80 8 2 1 t dt 2 0 7853 T Using your expression from part b as the input function write a full expression for the signal output from the system in terms of K1 K2 1 and t2 these should be the only unknowns in your expression f t 0 25 1 cos 0 7853 t 1 sin 0 7853 t K1 1 f t 0 25K1 1 K1M 1 1 cos 0 7853 t 1 M 1 1 1 1 1 2 1 K1M 1 1 sin 0 7853 t 1 1 Arc tan 1 1 f t 0 25K1 1 K1M 1 1 cos 0 7853 t 1 K1M 1 1 sin 0 7853 t K 2 2 f t 0 25K1K 2 1 1 K1K 2 M 1 1 M 2 1 cos 0 7853 t 1 2 K1K 2 M 1 1 M 2 1 sin 0 7853 t 1 2 M 2 1 1 1 1 2 2 2 Arc tan 1 2 Assuming that the output of the system is 3V at a frequency of 0 rad s determine a numerical value for the product of the two static sensitivities K1 K 2 f t 0 25K1K 2 1 1 K1K 2 M 1 1 M 2 1 cos 0 7853 t 1 2 K1K 2 M 1 1 M 2 1 sin 0 7853 t 1 2 0 25K1K 2 3 K1K 2 12
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