i have a lowpass filter H(ω) which is
$ H(ω) = e^{-jω} $ on -2π≤ω≤2π, and $0$ elsewhere
and i have a function in fourier series y(t), i need to find the new signal (z(t))
after the application of the filter ($Z=H\cdot Y$),
i know that if H was H(ω)=1 on -2π≤ω≤2π, then i could easily accept only these sines, cosines which have frequency from -2pi to 2pi, but now, H is $ H(ω) = e^{-jω} $
how can i find z?
On
If $T$ is the period of your input signal $y(t)$, and $\omega_0=2\pi/T$ then the complex Fourier series of $y(t)$ is
$$y(t)=\sum_{n=-\infty}^{\infty}c_ne^{jn\omega_0t}$$
where $c_n$ are the Fourier coefficients of $y(t)$. For each complex exponential $e^{jn\omega_0t}$ the corresponding output of an LTI system with frequency response $H(\omega)$ is $H(n\omega_0)e^{jn\omega_0t}$. So the output signal is given by
$$z(t)=\sum_{n=-\infty}^{\infty}c_nH(n\omega_0)e^{jn\omega_0t}$$
which for the given frequency response becomes
$$z(t)=\sum_{|n|\le 2\pi/\omega_0}c_ne^{jn\omega_0(t-1)}$$
Hint: use the time-shifting property ("translation" here) of the Fourier transform.