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(16.4)
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where Y is the output variable and X(t) is the forcing function or input variable. For specific cases of Eq. (16.4), refer to Eq. (5.5) for a first-order system and Eq. (8.4) for a second-order system. If X(t) is sinusoidal
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X(t) = Asinwr the solution of Eq. (16.4) will consist of a complementary solution, and a particular solution of the form Yp(t) = Cl sinwt + C2cosmt
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If the system is stable, the roots of the characteristic equation of (16.4) all lie to the left of the imaginary axis and the complementary solution will vanish exponentially in time. Then YP is the quantity previously defined as the sinusoidal or frequency response. If the system is not stable, the complementary solution grows exponentially and the term frequency response has no physical significance because Yp(t) is inconsequential. The problem now is the evaluation of Cl and C2 in Eq. (16.5). Since we are interested in the amplitude and phase of Yp(t), Eq. (16.5) is rewritten as Yp = D1 sin
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(16.6)
as was done previously [compare to Eq. (5.23) and related equations]. It will be convenient to change X(t) and Yp(t) from trigonometric to exponential form, using the identity ,je - e-jO sin8 = 2j Thus,
xtt) = A(ejmr _ e-jy Li
and from Eq. (16.6)
ypct) = $[ejW+D2) _ e-jW+02)]
(16.7)
(16.8)
FREQUENCYRESPONSE
Substitution of Eqs. (16.7) and (16.8) into Eq. (16.4) yields:
~~ ej(ot+D2) [a,(jo)" + un-l(jco)n-l + *f. + al(ju) 2j Dle-j@r+D2) 2.i z-.--e jot _ e-jor) A( 2j + ao]
[a,(- jto)n + a,-I(- jm)n-l + *. . + a~(-jo) + a01
(16.9)
The coefficients of ejot on both sides of Eq. (16.9) must be equal. Hence,
DlejD2[un(jo) + un-l(jW) - + .** + ul(jw) + a~] = A
(16.10)
Equation (16.10) will be satisfied if and only if 1
u,(jw)n + a,-l(jw) -l + ... + ul(jo) + a0
Dl =A
(16.11)
u,(jo) + u,-l(jw)n-l + *** + ul(jo) + a0
_ = 02
But D1/A and 02 are the AR and phase angle of the response, respectively, as may be seen from Eq. (16.6) and the forcing function. Furthermore, from Eq. (16.4) the transfer function relating X and Y is 1 Y(s) - = X(s) unsn + un-lF1 + . ** + u1.s + a0 Equations (16.11) and (16.12)* establish the general result.
Example 16.2. Find the frequency response of the system with the general secondorder transfer function and compare the results with those of Chap. 8. The transfer function is 1 r2s-2 + 2lTS + 1 Putting s = jw yields 1 1 - +r2u2 + j2cwr which may be converted to the polar form 1
(16.12)
*In writing this equation, it is assumed that X and Yhave been written as deviation variables, so that initial conditions are zero.
INTRODUCTION TO FREQUENCY RESPONSE
Hence, Ju - w*,*y -I- (2507)* -25w7 Phase angle = tan- I _ w2,2 which agree with Eq. (8.40).
(16.13)
Tkansportation Lag
The response of a transportation lag is not described by Eq. (16.4). Rather, a transportation lag is described by the relation Y(t) = X(t - 7) (16.14) which states that the output Y lags the input X by an interval of time r. If X is sinusoidal, X = A sin wt then from Eq. (16.14) Y = A sin w(t - T) = A sin (ot - wr) It is apparent that the AR is unity and the phase angle is (-UT). To check the substitution rule of the previous section, recall that the transfer function is given by
-TS G(s) = Y(s) X(s)=e
Putting s = j,, G(jw) = e-joT Then, m =I edjoT (= 1 Phase angle = 4e-jwT = -wr and the validity of the rule is verified.
Example 16.3. The stirred-tank heater of Chap. 1 has a capacity of 15 gal. Water
(16.15)
is entering and leaving the tank at the constant rate of 600 lb/mm. The heated water that leaves the tank enters a well-insulated section of 6-in.-ID pipe. ILvo feet from the tank, a thermocouple is placed in this line for recording the tank temperature, as shown in Fig. 16.1. The electrical heat input is held constant at 1,000 kw. If the inlet temperature is varied according to the relation Ti = 75 + 5sin46t where Ti is in degrees Fahrenheit and t is in minutes, find the eventual behavior of the thermocouple reading Tm. Compare this with the behavior of the tank temperature
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