Impulse response corresponding to differential equation always real?

An important class of linear and time-invariant systems are modeled by linear constant-coefficient differential equations (LCCDEs), i.e.
dnydtn+cn−1dn−1ydtn−1+…+c2d2ydt2+c1dydt=x(t),dnydtn+cn−1dn−1ydtn−1+…+c2d2ydt2+c1dydt=x(t),\frac{d^n y}{dt^n}+c_{n-1}\frac{d^{n-1} y}{dt^{n-1}}+…+c_2\frac{d^2 y}{dt^2}+c_1\frac{d y}{dt}=x(t),
where y(t)y(t) is the output and x(t)x(t) is the input.

When the input is a Dirac delta, the output is called the impulse response. If some of the coefficients cnc_n’s are complex, then it’s possible for the impulse response to be complex. But if coefficients are all real, is the impulse response of such a system always purely real? If not, please provide a counterexample.



1 Answer


For this system to represent an LTI, you have to specify that all initial conditions must be zero. If Y(s)Y(s) is the Laplace transform of y(t)y(t) and x(t)=δ(t−a)x(t) = \delta(t-a) then we have

Y(s)=e−assn+cn−1sn−1+⋯c1s. Y(s) = \frac{e^{-as}}{s^n + c_{n-1}s^{n-1} + \cdots c_1s}.

Now we can factor the denominator into terms that look like (s−r)(s-r) for real roots or (s−α)2+β2(s-\alpha)^2 + \beta^2 if α+iβ\alpha + i \beta is a complex root.

The inverse Laplace transforms of functions of the type

1s−r\frac{1}{s-r} and

1(s−α)2+β2 \frac{1}{(s-\alpha)^2+\beta^2}

are well known to be real functions (see any table of inverse transforms). So we can write Y(s)Y(s) as a product of functions with real inverse Laplace transforms, and response y(t)y(t) is a convolution of these functions, which is real-valued.