Frequency Response of HighOrder 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 a_{j} and b_{j} 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 



where 



(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 firstorder leadlag transfer function, for which the only nonzero coefficients of the original differential equation are a_{0}, b_{0}, a_{1}, and b_{1}. In that case we have A_{e} = a_{0}, A_{o} = a_{1}ω, Be = b_{0}, and B_{o} = b_{1}ω, so the amplitude ratio and phase shift are given by equations (1) and (2) as 



consistent with the results in the note on LeadLag 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 secondorder system has the functions 



so according to equation (1) the bandwidth frequency ω_{bw} satisfies the relation 



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



We can easily solve this quadratic equation for ω_{bw}^{2}. In the special case of a unitary transfer function with no numerator dynamics, we have b_{1} = b_{2} = 0, and a_{0} = b_{0} = 1, and we can also set a_{2} = 1/ω_{n}^{2} and a_{1} = 2ζ/ω_{n} where ω_{n} is the natural frequency and ζ is the damping ratio. Making these substitutions, the condition on the bandwidth is 



Solving this leads to the wellknown expression for the bandwidth of such a system 



For a higherorder example, consider the system defined by the original differential equation with the coefficients given by 



where τ_{a} and τ_{b} 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 (τ_{a} – τ_{b})ω, so conventionally we say the output is delayed in time by τ_{a} – τ_{b}. 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. 
