I’ve got the following question:
The following relation holds: $\hat{v}_F(\omega) = \hat{Y}(\omega)\hat{F}(\omega)$
With: $F(t)=\sum_k F_{k}\sin(\omega_k t - \phi_k)$
The transformation from frequency to time domain would be:
$v_F(t) = \sum_k F_{k}|\hat{Y}(\omega_k)| \sin(\omega_k t+\angle \hat{Y}(\omega_k) +\phi_k)$
How do you get to that solution?
This is how I got started:
$\hat{v}_F(\omega) = \hat{Y}(\omega)\hat{F}(\omega) \Leftrightarrow v_f(t)= y(t)*f(t)$
so, expressing the convolution:
$\int_0^\infty y(\tau)\sin(\omega_k(t-\tau)+\phi_k)d\tau$
$= \frac{1}{2i}\int y(\tau)[e^{i(\omega_k(t-\tau)+\phi_k}-e^{-i(\omega_k(t-\tau)+\phi_k}] =\frac{1}{2i} e^{i\omega_kt + \phi_k}\int y(\tau)e^{-i\omega_k \tau}d\tau- \frac{1}{2i} e^{-i\omega_kt + \phi_k}\int y(\tau) e^{+i\omega_k \tau}d\tau$
$= \frac{1}{2i} e^{i\omega_kt + \phi_k} \hat{Y}(i\omega) - \frac{1}{2i} e^{-i\omega_kt + \phi_k} \hat{Y}(-i\omega)$
and I do not see how to proceed.
We have that $F(t)=\sum_k F_{k}\sin(\omega_k t - \phi_k)$ so $$F(w) = \mathcal{F}(F(t)) = \sum_k F_{k} \mathcal{F}[\sin(\omega_k t - \phi_k)]$$ Note that $$\sin(\omega_k t - \phi_k) = \cos(\phi_k) \sin (\omega_k t) - \sin(\phi_k)\cos(\omega_k t) $$ which means that \begin{equation} \tag{0} \mathcal{F}[\sin(\omega_k t - \phi_k)] = \cos(\phi_k) \mathcal{F}[\sin (\omega_k t)] - \sin(\phi_k)\mathcal{F}[\cos(\omega_k t)] \end{equation} You could easily verify (using the definition of Fourier transform) that \begin{align} \mathcal{F}[\sin (\omega_k t)] &= j \pi [\delta(\omega + \omega_k) - \delta(\omega - \omega_k)]\\ \mathcal{F}[\cos (\omega_k t)] &= \pi [\delta(\omega - \omega_k) + \delta(\omega + \omega_k)] \end{align} So $$\mathcal{F}[\sin(\omega_k t - \phi_k)] =R_k(\omega) + j I_k(\omega)$$ where $$ R_k(\omega) = - \sin(\phi_k)\pi[\delta(\omega - \omega_k) + \delta(\omega + \omega_k)]$$ and $$ I_k(\omega) = \cos(\phi_k)\pi[\delta(\omega + \omega_k) - \delta(\omega - \omega_k)]$$ So \begin{equation} \tag{1} \begin{split} V_F(\omega) &= Y(\omega) F(\omega)\\ &= Y(\omega) \sum_k F_{k}(R_k(\omega) + jI(\omega) )\\ &= \sum_k F_{k}(Y(\omega)R_k(\omega) + jY(\omega)I(\omega) ) \end{split} \end{equation} Notice that $$Y(\omega)R_k(\omega)=- \sin(\phi_k)\pi[\delta(\omega - \omega_k)Y(\omega) + \delta(\omega + \omega_k)Y(\omega)]$$ $$Y(\omega)I_k(\omega)= \cos(\phi_k)\pi[\delta(\omega + \omega_k)Y(\omega) - \delta(\omega - \omega_k)Y(\omega)]$$ Notice that $\delta(\omega + \omega_k)Y(\omega)= Y(-\omega_k)\delta(\omega + \omega_k)$ and $\delta(\omega -\omega_k)Y(\omega)= Y(\omega_k)\delta(\omega -\omega_k)$ so $$Y(\omega)R_k(\omega)=- \sin(\phi_k)\pi[Y(\omega_k)\delta(\omega -\omega_k) + Y(-\omega_k)\delta(\omega + \omega_k)]$$ $$Y(\omega)I_k(\omega)= \cos(\phi_k)\pi[Y(-\omega_k)\delta(\omega + \omega_k) - Y(\omega_k)\delta(\omega -\omega_k)]$$ Use exponential representation of your $Y(\omega_k) = \vert Y(\omega_k) \vert \exp(j \theta_k)$ where $\theta_k= \angle Y(\omega_k)$. If $y(t)$ is real, notice the following symmetry and anti-symmetry in frequency domain: \begin{align} \vert Y(\omega_k) \vert &= \vert Y(-\omega_k) \vert \\ \theta_k &= - \theta_{-k} \end{align} With this in mind, we get \begin{equation} \tag{2} Y(\omega)R_k(\omega)=- \sin(\phi_k)\pi \vert Y(\omega_k) \vert[ \exp(j\theta_k)\delta(\omega -\omega_k) + \exp(-j\theta_k)\delta(\omega + \omega_k)] \end{equation} \begin{equation} \tag{3} Y(\omega)I_k(\omega)= \cos(\phi_k)\pi \vert Y(\omega_k) \vert[\exp(-j\theta_k)\delta(\omega + \omega_k) -\exp(j\theta_k)\delta(\omega -\omega_k)] \end{equation} Replacing (2) and (3) in (1) , and factoring common terms, we get \begin{equation} \begin{split} V_F(\omega) &= -\sum_k F_{k} \vert Y(\omega_k)\vert \pi \big[ \sin \phi_k \exp(j\theta_k)+j\cos \phi_k \exp(j\theta_k) \big]\delta(\omega - \omega_k) \\ &+ \sum_k F_{k} \vert Y(\omega_k)\vert \pi \big[- \sin \phi_k \exp(-j\theta_k) + j\cos \phi_k \exp(-j\theta_k) \big]\delta(\omega + \omega_k) \\ \end{split} \end{equation} Using Euler's formula gives the terms in brackets as \begin{equation} \sin \phi_k \exp(j\theta_k)+j\cos \phi_k \exp(j\theta_k) = \sin(\phi_k - \theta_k) + j\cos(\phi_k - \theta_k) \end{equation} and \begin{equation} - \sin \phi_k \exp(-j\theta_k) + j\cos \phi_k \exp(-j\theta_k) = -\sin(\phi_k - \theta_k) + j\cos(\phi_k - \theta_k) \end{equation} Replacing we get \begin{equation} \begin{split} V_F(\omega) &= -\sum_k F_{k} \vert Y(\omega_k)\vert \pi [\sin(\phi_k - \theta_k) + j\cos(\phi_k - \theta_k)]\delta(\omega - \omega_k) \\ &+ \sum_k F_{k} \vert Y(\omega_k)\vert \pi [-\sin(\phi_k - \theta_k) + j\cos(\phi_k - \theta_k)]\delta(\omega + \omega_k) \\ \end{split} \end{equation} Recollecting the real and imaginary terms, \begin{equation} \begin{split} V_F(\omega) &= -j \sum_k F_{k} \vert Y(\omega_k)\vert \pi \cos(\phi_k - \theta_k)[\delta(\omega - \omega_k) -\delta(\omega + \omega_k) ]\\ &- \sum_k F_{k} \vert Y(\omega_k)\vert \pi \sin(\phi_k - \theta_k)[\delta(\omega - \omega_k) +\delta(\omega + \omega_k) ] \\ \end{split} \end{equation} That is \begin{equation} \begin{split} V_F(\omega) &= -\sum_k F_{k} \vert Y(\omega_k)\vert \Big\lbrace j \pi \cos(\phi_k - \theta_k)[\delta(\omega - \omega_k) -\delta(\omega + \omega_k) ]\\ &+ \pi \sin(\phi_k - \theta_k)[\delta(\omega + \omega_k) +\delta(\omega - \omega_k) ] \Big\rbrace \\ \end{split} \end{equation} Note that the terms in brackets are in the same form of equation (0), hence you can now apply the inverse fourier transform and arrive at your formula.