4.5 First-Order System Dynamic Response
4.5.2 Response to Sinusoidal-Input Forcing
Now consider a first-order system that is subjected to an input that varies sinusoidally in time. The governing equation is
τ ˙y + y = KF (t) = KA sin(ωt), (4.23)
where K and A are arbitrary constants. The units of K would be those of y divided by those of A. The general solution is
y(t) = yh+ yp= c0e−
t
in which c0 through c3are constants, where the first term on the right side
of this equation is the homogeneous solution, yh, and the remaining terms
constitute the particular solution, yp.
The constants c1through c3can be found by substituting the expressions
for y(t) and its derivative into Equation 4.23. By comparing like terms in the resulting equation,
c1= 0, (4.25) c2= KA ω2τ2+ 1, (4.26) and c3= −ωτC2= −ωτKA ω2τ2+ 1. (4.27)
The constant c0can be found by applying the initial condition y(0) = y0 to
Equation 4.24, where
c0= y + 0 − c3=
ωτ KA
ω2τ2+ 1. (4.28)
Thus, the final solution becomes y(t) = (y0+ ωτ D)e− t τ + D sin(ωt) − ωτD cos(ωt), (4.29) where D = KA ω2τ2+ 1. (4.30)
Now Equation 4.29 can be simplified further. The sine and cosine terms can be combined in Equation 4.29 into a single sine term using the trigono- metric identity
α cos(ωt) + β sin(ωt) =pα2+ β2sin(ωt + φ), (4.31)
where
φ = tan−1(α/β). (4.32)
Equating this expression with the sine and cosine terms in Equation 4.29 gives α = −ωτD and β = D. Thus,
D sin(ωt) − ωτD cos(ωt) = Dpω2τ2+ 1 = √ KA
ω2τ2+ 1 (4.33)
φ = tan−1(−ωτ) = − tan−1(ωτ ), (4.34) or, in units of degrees,
φ◦= −(180/π) tan−1(ωτ ). (4.35)
The minus sign is present in Equations 4.34 and 4.35 by convention to denote that the output lags behind the input.
The final solution is
y(t) = y0+ ( ωτ KA
ω2τ2+ 1)e −t
τ +√ KA
ω2τ2+ 1sin(ωt + φ). (4.36)
The first term on the right side represents the transient response while the second term is the steady-state response. For ωτ << 1, the transient term becomes very small and the output follows the input. For ωτ >> 1, the output is droplets attenuated and its phase is shifted from the input by φ radians. The phase lag in seconds (lag time), β, is given by
β = φ/ω. (4.37)
Examine this response further in a dimensionless sense. The magnitude ratio for this input-forcing situation is the ratio of the magnitude of the steady-state output to that of the input. Thus,
M (ω) = KA/ √ ω2τ2+ 1 KA = 1 √ ω2τ2+ 1. (4.38)
The dynamic error, using its definition in Equation 4.22 and Equation 4.38, becomes
δ(ω) = 1 −√ 1
ω2τ2+ 1. (4.39)
Shown in Figures 4.3 and 4.4, respectively, are the magnitude ratio and the phase shift plotted versus the product ωτ . First examine Figure 4.3. For values of τ ω less than approximately 0.1, the magnitude ratio is very close to unity. This implies that the system’s output closely follows its input in this range. At ωτ equal to unity, the magnitude ratio equals 0.707, that is, the output amplitude is approximately 71 % of its input. Here, the dynamic error would be 1 − 0.707 = 0.293 or approximately 29 %. Now look at Figure 4.4. When ωτ is unity, the phase shift equals −45◦. That is, the output signal lags the input signal by 45◦ or 1/8th of a cycle.
The magnitude ratio often is expressed in units of decibels, abbreviated as dB. The decibel’s origin began with the introduction of the Bel, defined in terms of the ratio of output power, P2, to the input power, P1, as
FIGURE 4.3
The magnitude ratio of a first-order system responding to sinusoidal-input forcing.
To accommodate the large power gains (output/input) that many systems had, the decibel (equal to 10 Bels) was defined as
Decibel = 10 log10(P2/P1). (4.41)
Equation 4.41 is used to express sound intensity levels, where P2corresponds
to the sound intensity and P1to the reference intensity, 10−12W/m2, which
is the lowest intensity that humans can hear. The Saturn V on launch has a sound intensity of 172 dB; human hearing pain occurs at 130 dB; a soft whisper at a distance of 5 m is 30 dB.
There is one further refinement in this expression. Power is a squared quantity, P2 = Q22 and P1= Q12, where Q2 and Q1 are the base measur-
ands, such as volts for an electrical system. With this in mind, Equation 4.41 becomes
Decibel = 10 log10(Q2/Q1)2= 20 log10(Q2/Q1). (4.42)
Equation 4.42 is the basic definition of the decibel as used in measure- ment engineering. Finally, Equation 4.42 can be written in terms of the magnitude ratio
FIGURE 4.4
The phase shift of a first-order system responding to sinusoidal-input forcing. The point M (ω) = 0.707, which is a decrease in the system’s amplitude by a factor of 1/√2, corresponds to an attenuation of the system’s input by −3 dB. Sometimes, this is called the droplets half-power point because, at this point, the power is one-half the original power.
Example Problem 4.2
Statement: Convert the sound intensity level of 30 dB to logeM (ω).
Solution: The relationship between logarithms of bases a and b is logbx = logax/ logab.
For this problem the bases are e and 10. So,
logeM (ω) = log10M (ω)/ log10e.
Now, log10e = 0.434 294. Also, loge10 = 2.302 585 and e = 2.718 282. Using Equation
4.43, log10M (ω) at 30 dB equals 1.500. Thus
logeM (ω) = 1.500/0.434 = 3.456.
The relationship between logarithms of two bases is used often when converting back and forth between base 10 and base e systems.
Systems often are characterized by their bandwidth and center frequency. Bandwidth is the range of frequencies over which the output amplitude of a system remains above 70.7 % of its input amplitude. Over this range,
M ω) ≥ 0.707 or −3 dB. The lower frequency at which Mω) < 0.707 is called the low cut-off frequency. The higher frequency at which M ω) < 0.707 is called the high cut-off frequency. The center frequency is the frequency equal to one-half the sum of the low and high cut-off frequencies. Thus, the bandwidth is the difference between the high and low cut-off frequencies. Sometimes bandwidth is defined as the range of frequencies that contain most of the system’s energy or over which the system’s gain is almost constant. However, the above quantitative definition is preferred and used most frequently.
Example Problem 4.3
Statement: Determine the low and high cut-off frequencies, center frequency, and the bandwidth in units of hertz of a first-order system having a time constant of 0.1 s that is subjected to sinusoidal-input forcing.
Solution: For a first-order system, M (ω) ≥ 0.707 from ωτ = 0 to ωτ = 1. Thus, the low cut-off frequency is 0 Hz and the high cut-off frequency is (1 rad/s s)/[(0.1 s)(2π rad/cycle)] = 5/π Hz. The bandwidth equals 5/π Hz − 0 Hz = 5/π Hz. The center frequency is 5/2π.
The following example illustrates how the time constant of a thermocou- ple affects its output.
Example Problem 4.4
Statement: Consider an experiment in which a thermocouple that is immersed in a fluid and connected to a reference junction/linearizer/amplifier micro-chip with a static sensitivity of 5 mv/◦C. Its output is E(t) in millivolts. The fluid temperature
varies sinusoidally in degrees Celsius as 115 + 12 sin(2t). The time constant τ of the thermocouple is 0.15 s. Determine E(t), the dynamic error δ(ω) and the time delay β(ω) for ω = 2. Assume that this system behaves as a first-order system.
Solution: It is known that
τ ˙E + E = KF (t). Substitution of the given values yields
0.15 ˙E + E = 5[115 + 12 sin 2t] (4.44)
with the initial condition of E(0) = (5 mv/◦C)(115◦C) = 575 mV.
To solve this linear, first-order differential equation with constant coefficients, a solution of the form E(t) = Eh+ Ep is assumed, where Eh = C0e−t/τ and Ep =
c1+ c2sin 2t + c3cos 2t. Substitution of this expression for E(t) into the left side and
grouping like terms gives
c1= 575, c2= 55.1, and c3= −16.5.
Equation 4.44 then can be rewritten as
E(t) = k0e−t/0.15+ 575 + 55.1 sin 2t − 16.5 cos 2t.
Using the initial condition,
Thus, the final solution for E(t) is
E(t) = 575 + 16.5e−t/0.15+ 55.1 sin 2t − 16.5 cos 2t
or, in units of◦C temperature
T (t) = 115 + 3.3e−t/0.15+ 11.0 sin 2t − 3.3 cos 2t.
The output (measured) temperature is plotted in Figure 4.5 along with the input (ac- tual) temperature. A careful comparison of the two signals reveals that the output lags the input in time and has a slightly attenuated amplitude. At t = 2 s, the actual temperature is ∼106 ◦C, which is less than the measured temperature of ∼109 ◦C.
Whereas, at t = 3 s, the actual temperature is ∼112◦C, which is greater than the
measured temperature of ∼109◦C. So, for this type of forcing, the measured temper-
ature can be greater or less than the actual temperature, depending upon the time at which the measurement is made.
The time lag and the percent reduction in magnitude can be found as follows. The dynamic error is
δ(ω = 2) = 1 − M(ω = 2) = 1 − 1
[1 + (2 × 0.15)2]1/2 = 0.04,
which is a 4 % reduction in magnitude. The time lag is
β(ω = 2) =φ(ω = 2) ω = − tan−1ωτ ω = (−16.7◦)(π rad/180◦) 2 rad/s = −0.15 s,
which implies that the output signal lags the input signal by 0.15 s. The last two terms in the temperature expression can be combined using a trigonometric identity (see Chapter 9), as
11.0sin2t − 3.3cos2t = 11.48sin(2t − 0.29), (4.45)
where 0.29 rad = 16.7◦is the phase lag found before.