Frequency Response of High-Order Systems


A thief, sentenced to the gallows for poaching on the King's land, bargains for his life by offering to teach the King's horse to talk in a year. To a skeptical friend the thief confides his hopes to avoid hanging: "Before the year is up, the horse may die, or the King may die, or I may die... or the horse may talk".

                                                                                                                Tudor joke


Consider two variables x and y related by the ordinary differential equation



where the coefficients aj and bj are constants. The relation is symmetrical, but we may regard x as the independent (input) variable and y as the dependent (output) variable. To determine the frequency response, we stipulate that x(t) and y(t) are both sinusoidal functions with the same frequency ω, but with different amplitudes X and Y, and with phases that differ by 2θ. Thus we can write these functions as



Substituting these functions into the preceding equation, and making use of the trigonometric identities



we arrive at an equation of the form



which can be satisfied by setting F(ω,θ) = 0 and G(ω,θ) = 0. This leads to the two equations






(The subscripts and A and B denote even and odd.) Dividing through each of the preceding two equations by Xcos(θ) and solving each of then for tan(θ), we get



Equating these two expressions for tan(θ), we arrive at the following relation between the amplitude ratio and the frequency



If we define normalized A functions as



and similarly for the normalized B functions, we can substitute for Y/X into the previous expressions for tan(θ) to give



Recalling that the full phase shift is 2θ, we can make use of the identity tan(2θ) = 2tan(θ)/[1−tan(θ)2] to show that the tangent of the full phase shift is



For a simple example, consider the first-order lead-lag transfer function, for which the only non-zero coefficients of the original differential equation are a0, b0, a1, and b1. In that case we have Ae = a0, Ao = a1ω, Be = b0, and Bo = b1ω, so the amplitude ratio and phase shift are given by equations (1) and (2) as



consistent with the results in the note on Lead-Lag Frequency Response. (In that note θ was defined as the full phase angle shift, whereas here it denotes half of the shift.)


Equation (1) can also be used to determine the bandwidth of a system, defined as the frequency at which the squared amplitude ratio equals 1/2. (This corresponds to the half power condition, since power is proportional to the square of the amplitude.) For example, a second-order system has the functions



so according to equation (1) the bandwidth frequency ωbw satisfies the relation



Clearing fraction and collecting on the frequency, this can be written as



We can easily solve this quadratic equation for ωbw2. In the special case of a unitary transfer function with no numerator dynamics, we have b1 = b2 = 0, and a0 = b0 = 1, and we can also set a2 = 1/ωn2 and a1 = 2ζ/ωn where ωn is the natural frequency and ζ is the damping ratio. Making these substitutions, the condition on the bandwidth is



Solving this leads to the well-known expression for the bandwidth of such a system



For a higher-order example, consider the system defined by the original differential equation with the coefficients given by



where τa and τb are constants with units of time. In this case we have



and therefore Y/X = 1 and



Thus there is no attenuation of the amplitude, and the output phase lags the input by (τa – τb)ω, so conventionally we say the output is delayed in time by τa – τb. This is consistent with the fact that the transfer function represented by the original differential equation with these coefficients can be depicted in operator notation as



Recognizing that the numerator and denominator are just exponential functions, this is equivalent to the transfer function



which is a pure “time shift”, as discussed in the note on Transfer Functions and Causation.


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