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Sinusoidal Response
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To investigate the response of a first-order system to a sinusoidal forcing function, the example of the mercury thermometer will be considered again. Consider a thermometer to be in equilibrium with a temperature bath at temperature x S. At some time t = 0, the bath temperature begins to vary according to the relationship x = xs +Asinot
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(5.17)
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where x = temperature of bath xs = temperature of bath before sinusoidal disturbance is applied A = amplitude of variation in temperature w = radian frequency, radkime In anticipation of a simple result we shall introduce a deviation variable X which is defined as x = x - x , (5,. 18) Using this new variable in Eq. (5.17) gives X = A sin ot By referring to a table of transforms, the transform of Eq. (5.19) is (5.19)
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X(s) = -J$
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(5.20)
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RESPONSE OF FIRST-ORDER SYSTEMS
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Combining Eqs. (5.7) and (5.20) to eliminate X(S) yields AW l/7 Y ( s ) = - s* + co* s + l/T
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(5.21)
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This equation can be solved for Y(t) by means of a partial-fraction expansion, as described in Chap. 3. The result is Y(t) = Aore- AWT A sin ot cos wt + r*w* + 1 - 7*&J* + 1 TW + 1 (5.22)
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Equation (5.22) can be written in another form by using the trigonometric identity p cos A + q sin A = r sin (A + 0) (5.23) where tane=~ 4 Applying the identity of Eq. (5.23) to (5.22) gives AWT -t/7 + Y(t) = sin (wt + 4) r*W*+le where t$ = tan- (-on) As t + ~0, the first term on the right side of Eq. (5.24) vanishes and leaves only the ultimate periodic solution, which is sometimes called the steady-state solution Y ( t )s = J& sin W + 4) (5.25) r = Jpq
(5.24)
By comparing Eq. (5.19) for the input forcing function with Eq. (5.25) for the ultimate periodic response, we see that 1. The output is a sine wave with a frequency w equal to that of the input signal. 2. The ratio of output amplitude to input amplitude is l/ ,/T*w* + 1. This is always smaller than 1. We often state this by saying that the signal is attenuated. 3. The output lags behind the input by an angle 14 I. It is clear that lag occurs, for the sign of 4 is always negative.*
*By convention, the output sinusoid lags the input sinusoid if C#J in Eq. (5.25) is negative. In terms of a recording of input and output, this means that the input peak occurs before the output peak. If C$ is positive in E$. (5.29, the system exhibits phase lead, or the output leads the input. In this book we shall always use the term phase angle (4) and interpret whether there is lag or lead by the convention phase lag phase lead
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For a particular system for which the time constant r is a fixed quantity, it is seen from Eq. (5.25) that the attenuation of amplitude and the phase angle 4 depend only on the frequency o. The attenuation and phase lag increase with frequency, but the phase lag can never exceed 90 and approaches this value asymptotically. The sinusoidal response is interpreted in terms of the mercury thermometer by the following example. Example 5.2. A mercury thermometer having a time constant of 0.1 min is placed in a temperature bath at 100 F and allowed to come to equilibrium with the bath. At time t = 0, the temperature of the bath begins to vary sinusoidally about its average temperature of lOOoF with an amplitude of 2 F If the frequency of oscillation is 101~ cycles/min, plot the ultimate response of the thermometer reading as a function of time. What is the phase lag In terms of the symbols used in this chapter 7 = 0.1 x s = 100 F A = 2 F 10 f = - cycles/min 7iw = 27rf = 21r~ = 20 radlmin 7r From Eq. (5.25), the amplitude of the response and the phase angle am calculated; thus 2 = ___ = 0.896 F vGG-T-l JGT 4 = --tan- 2 = -63.5 or Phase lag = 63.5 The response of the thermometer is therefore Y(t) = 0.896 sin (20t - 63.5 ) or Y(G) = 100 + 0.896 sin (20t - 63.5 ) To obtain the lag in terms of time rather than angle, we proceed as follows: A frequency of lO/rr cycles/min means that a complete cycle (peak to peak) occurs in (10/r)- min. Since one cycle is equivalent to 360 and the lag is 63.5 , the time corresponding to this lag is 63.5 - X (time for 1 cycle) 360 A
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