Homework 6 solutions 3

Homework 6 solutions 3 - Homework 6 40 pts 1. For a first...

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1 Homework 6 1. For a first order system, explain with the aid of one or more plots why the phase delay is 45° for ωτ = 1. 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 01234 ω t/ π Input Amplitude, 1 = A 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Output Amplitude, 1 = KA I will explain in terms of a temperature sensor. The left-hand axis is the normalized input amplitude (i.e., scaled so that 1 = A), and the right-hand axis is the normalized output amplitude (i.e., scaled so that 1 = KA). The amplitudes are then directly comparable. The time axis is ω t, but divided by π , so it shows the number of π (i.e., 2 corresponds to 2 π ). The gray curve shows the step response on the same time axis. Since ωτ = 1, t/ τ = 1 is at 0.32 on this axis. At ωτ = 1, the system has reached (1-e -1 ) = 63% of its final amplitude. This is indicated by the black lines. To allow comparison among the three curves, the sine waves were shifted π /4 to the left. Up until t = π , the ambient temperature (dark blue line) is higher than the sensor temperature (red line). The temperature of the sensor therefore rises – the input is driving the system (sensor) from where it is (cold) towards KA (= sensitivity * peak temperature). Times when the ambient is hotter than the sensor are indicated by the horizontal pink lines. The sensor temperature rises at a rate comparable to that of the gray line , but it is a little slower than the gray line because the sensor is not responding to a step input (constant temperature A), but to a more slowly rising ambient (sine wave going up).
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Homework 6 solutions 3 - Homework 6 40 pts 1. For a first...

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