Homework 4 solutions

Homework 4 solutions - Homework 4 Note The K in the equation y t = KA y0 KA)e t is not necessarily the same as the K for static sensitivity in K =

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Homework 4 Note : The K in the equation 0 t/ y(t) KA (y KA)e τ =+− is not necessarily the same as the K for static sensitivity in K = dy/dx . (It is in some cases or if the problem is handled in a certain way, which is not the way it has been done in the examples you’ve seen so far. Look at the solutions when they are posted.) You may find it easier to consider 11 0 y( t ) K A ( y K A )e =+ and to find K 1 A 1 . 1. Figure 1a shows a temperature sensor response when it is taken from temperature T 0 = 25.0 °C = 298 Kelvin at time t = 0 and put onto an oven at 28.2 °C. Figure 1b shows the static calibration curve for this sensor. 51,000 52,000 53,000 54,000 55,000 56,000 57,000 58,000 59,000 60,000 61,000 24 25 26 27 28 29 Temperature ( o C) R(T) 51,000 52,000 53,000 54,000 55,000 56,000 57,000 58,000 59,000 60,000 61,000 0 100 200 300 400 500 600 700 800 900 1000 time (msec) R(t) a) (70 pts) What is the steady-state response of the sensor to an input of T(t) = C 2 + A 2 sin( ω t ), where = 20 rad/sec, C 2 = 26 °C, A 2 = 1.5 °C? Show your equations, the values you use, and your calculations. (Explicitly state any assumptions that you make in solving this problem and justify the reasonableness or necessity of those assumptions.) This system will be modeled as a first-order system. First, get from Figure 1, then use that to obtain the sine response.
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This note was uploaded on 11/28/2010 for the course ENES enes100 taught by Professor Staff during the Spring '10 term at Maryland.

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Homework 4 solutions - Homework 4 Note The K in the equation y t = KA y0 KA)e t is not necessarily the same as the K for static sensitivity in K =

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