For the damped spring-mass system from above (eq. (95)) this gives us
u(t) =∫
−1
bexp(
−b(t−t′)
2 m)
sin(√
4 mk−b^2
2 m(t−t′))
f(t′)dt′.But keep in mind this is only possible if the response of the system is linear to the impulse.
Generalized susceptibility:
We now assume we have such a situation where we have a linear differential equationLu(t) = f(t)
with a driving forcef(t). We also assume that the driving force is sinusoidal,f(t) = Aexp(iωt).
Because the response of the system is linear, the response will also be sinusoidal with the same
frequencyωas the driving force. The general solution in terms of the Green’s function then is
u(t) =∫
g(t−t′)f(t′) =∫
g(t−t′)Aexp(iωt′)dt′We define the generalized susceptibility which is the response of the system divided by the driving
force
χ(ω) =
u
f=
∫
g(t−t′)Aexp(iωt′)dt′
Aexp(iωt)(97)
Since the integral is overt′, the factor withtcan be put in the integral.
χ(ω) =
u
f=
∫
g(t−t′)Aexp(−iω(t−t′))dt′This equation looks similar to a Fourier transformation. If the variables are changed toτ = t−t′
χ(ω) = A·∫
g(τ) exp(−iωτ)dτ (98)we see that the susceptibility is proportional to the Fourier transform of the Green’s function. This
is useful because if we now know the Green’s function we know the response of the system at all
frequencies. (Without loss of generality^9 the proportionality factorAcan be neglected.)
Any functiong(t)can be written as the sum of an even componentE(t)and an odd componentO(t)
g(t) = E(t) +O(t) (99)Ifg = 0fort < 0 , which is the case for the response function for a system (causality), this requires
that
O(t) = sgn(t)E(t). (100)Therefore the impulse response function is completely specified by either it’s even components or its
odd components.
The Green’s function written as the sum of its even and odd component and put in eq. (98) leads
to:
χ(ω) =∫
g(τ) exp(−iωτ)dτ =∫
E(τ) cos(ωτ)dτ+i∫
O(τ) sin(ωτ)dτ = χ′(ω) +iχ′′(ω) (101)(^9) ohne Beschränkung der Allgemeinheit