Linking Fourier Coefficients of periodic functions

Let $\tau\in (0,1)$ and assume that we have a $\tau$-periodic function $$f_1(t) = \sum\limits_{k\in\mathbb{Z}} a^{(1)}_k e^{\frac{2\pi i k}{\tau}t},$$a $(1-\tau)$-periodic function$$f_2(t) = \sum\limits_{k\in\mathbb{Z}} a^{(2)}_k e^{\frac{2\pi i k}{1-\tau}t},$$and a $1$-periodic function $$f_3(t) = \sum\limits_{k\in\mathbb{Z}} a^{(3)}_k e^{2\pi i k t}$$

that are all smooth and connected by$$f_3(t)= \begin{cases} f_1(t) &: t\in[0,\tau]\\ f_2(t-\tau) &:t\in[\tau,1]\end{cases}.$$

My question would be: What relations are there between the Fourier coefficients of the series?

Clearly, for any $r\in\mathbb{Z}$, one at least has \begin{align}a^{(3)}_r &= \int\limits_0^1 f_3(t) e^{-2\pi i r t}dt \\&= \int\limits_0^\tau f_1(t) e^{-2\pi i rt}dt + \int\limits_\tau^1 f_2(t) e^{-2\pi i rt}dt\\&=\sum\limits_{k\in\mathbb{Z}} a^{(1)}_k\int\limits_0^\tau e^{2\pi i(\frac{k}{\tau}-r)t}dt + a^{(2)}_k\int\limits_\tau^1 e^{2\pi i(\frac{k}{1-\tau}(t-\tau)-rt)}dt \end{align}

The first integral is \begin{align}\int\limits_0^\tau e^{2\pi i(\frac{k}{\tau}-r)t}dt &= \frac{e^{-2\pi i r \tau}-1}{2\pi i( \frac{k}{\tau}-r)} \end{align} if $r\tau\not\in\mathbb{Z}$, and (essentially because of the same formula) else \begin{align}\int\limits_0^\tau e^{2\pi i(\frac{k}{\tau}-r)t}dt = \begin{cases}\tau &: k=r\tau \\ 0 &:\text{else}\end{cases}.\end{align}

The second integral can be treated in a similar way and yields\begin{align} \int\limits_\tau^1 e^{2\pi i(\frac{k}{1-\tau}(t-\tau)-rt)}dt &= \frac{1-e^{-2\pi ir \tau}}{2\pi i(\frac{k}{1-\tau}-r)}\end{align}if $r\tau\not\in\mathbb{Z}$, and else \begin{align} \int\limits_\tau^1 e^{2\pi i(\frac{k}{1-\tau}(t-\tau)-rt)}dt = \begin{cases}1-\tau &: k=r(1-\tau) \\ 0 &:\text{else}\end{cases}.\end{align}

Together, this means that $$a^{(3)}_r = \tau a^{(1)}_{r\tau}+(1-\tau)a^{(2)}_{r(1-\tau)}$$ if $r\tau\in\mathbb{Z}$, which seems to be quite nice. But if $r\tau\not\in\mathbb{Z}$, we only get$$a^{(3)}_r = \frac{e^{-2\pi i r\tau}-1}{2\pi i}\left(\sum\limits_{k\in\mathbb{Z}} a^{(1)}_k\frac{1}{\frac{k}{\tau}-r} - a^{(2)}_k\frac{1}{\frac{k}{1-\tau}-r}\right).$$

Intuitively, it would be nice if this turned out to be $0$, since $r\tau\not\in\mathbb{Z}$ (or equivalently $r(1-\tau)\not\in\mathbb{Z}$) means that the Fourier basis vectors of $f_3$ are "out of sync" with the ones of $f_1$ and $f_2$.