Deriving of Duhamel Theorem

Deriving of Duhamel-theorem will be executed in the followings.

Let’s start from the weak derivative of convolution

$$ y(t) = \frac{\delta }{\delta t} \left \[ f(t)\ast h(t) \right \]$

Apply the following denoting:
$$ f(t)=1(t)f(t) ;\quad h(t)=1(t)h(t) ;\quad f(t-\tau)=1(t-\tau)f(t-\tau) ;\quad h(t-\tau)=1(t-\tau)h(t-\tau)$

Let’s describe the convolution formula in the argument of weak derivative:

$$ y(t) = \frac{\delta }{\delta t}\int_0^tf(\tau)h(t-\tau)1(t-\tau)d\tau=\int_0^tf(\tau) \frac{\delta }{\delta t}\left [ h(t-\tau)1(t-\tau) \right ] d\tau$

$$ y(t)= \int_0^tf(\tau) \left [ h(t-\tau) \frac{\delta 1(t-\tau)}{\delta t}+ 1(t-\tau) \frac{dh(t-\tau)}{dt} \right ] d\tau$
$$ y(t)= \int_0^tf(\tau) \left [ \delta(t-\tau)h(t-\tau) +1(t-\tau) \frac{dh(t-\tau)}{dt} \right ] d\tau$

$$ y(t)= \int_0^tf(\tau) \delta(t-\tau)h(t-\tau) d\tau + \int_0^tf(\tau) 1(t-\tau) \frac{dh(t-\tau)}{dt} d\tau$
If $ t>0 $$ \int_0^tf(\tau) \delta(t-\tau)h(t-\tau) d\tau=f(t)h(0)$ , and
$$ \int_0^tf(\tau) 1(t-\tau) \frac{dh(t-\tau)}{dt} d\tau=\int_0^tf(\tau) \frac{dh(t-\tau)}{dt} d\tau$ , where $$ 0<\tau<t$
Now, it is possible to get the final expression of Duhamel-theorem
$$ y(t)= f(t)h(0) + \int_0^tf(\tau) \frac{dh(t-\tau)}{dt} d\tau = f(t)h(0) + \int_0^tf(\tau) \dot{h}(t-\tau) d\tau$ , where $$ \frac{d}{dt}$ means the conventional derivative, which is denoted by “ $$ \dot{ }$ ” as usual.

Return to the ’Linear shift invariant system description by step response function’ chapter

Export

Medical Imaging

Site Language: English

Deriving of Duhamel Theorem

Sidebar

Medical Imaging

Site Language: English