I got unfortunally stuck by performing a (quite simple?) convolution integral. Given are those functions:
$$f_1(t) = k_1\cdot e^{b_1\cdot t}$$ and $$f_2(t) = k_2\cdot t$$
where $k_1, k_2$ and $b_1$ are constants.
Now I should compute $g(t) = f_1(t)\star f_2(t)$ in $0 < \tau \leq t$. I did it that way:
$$g(t) = \int_0^t k_1\cdot e^{b_1\cdot\tau} \cdot k_2\cdot (t-\tau)\text{d}\tau$$
What gives me:
$$ g(t) = k_2\cdot\frac{k_1}{b_1}(-b_1\cdot t + e^{b_1\cdot t_1} - 1)$$
Is this correct? Or did I something wrong?
Now, let $$f_1(t) = k_1\cdot e^{b_1\cdot t}$$ $$f_2(t) = k_2+k_3\cdot(t-t_1)$$ and $k_1, k_2, k_3$ are constants and $t_1$ one instant in time (a constant).
If I want to get $g(t) = f_1(t)\star f_2(t)$ in the interval $t_1 < \tau \leq t_2$, can I do it the same way, like "just" by solving the following integral
$$g(t) = \int_{t_1}^{t_2} k_1\cdot e^{b_1\cdot \tau}\cdot [ k_2+k_3\cdot (t_2-t_1 -\tau)]\text{d}\tau$$
I hope someone could help me ... it's a bit hard for me to understand because I didn't really need it before ...
If $g(t)=f_1(t)\star f_2(t)$, then $$ g(t)=\int_0^t f_1(\tau)f_2(t-\tau)\mathrm d\tau= k_1k_2\int_0^t e^{b_1\tau}(t-\tau)\mathrm d\tau $$ Which we evaluate by parts to be $$ k_1k_2[\frac{t-\tau}{b_1}e^{b_1\tau}+\frac{e^{b_1\tau}}{b_1^2}\vert_0^t]=\\ k_1k_2[\frac{e^{b_1t}-1}{b_1^2}e^{b_1\tau}-\frac{t}{b_1}]=g(t) $$