# Is it wrong to first calculate the whole system's Frequency response and then apply input to it?

Having a system like below, which (i think) is LTI, and $$H(e^{j\omega}) = \begin{cases} 1 & |\omega|\leq \frac{\pi}{2} \\ 0 & |\omega| \gt \frac{\pi}{2} \end{cases}$$

I've tried to find whole system's frequency response by shifting $$H(e^{j\omega})$$ with value of $$\pi$$ and summed it up with itself which then got $$H_{total}(e^{j\omega}) = 1$$ for all $$\omega$$ and if I'm correct with that I think the system is a wire which makes $$y[n] = x[n]$$

• Don't think so. Any tone between pi/2 and pi will be blocked – Stanley Pawlukiewicz Jan 7 at 3:31
• @StanleyPawlukiewicz would you provide any proof please? – no0ob Jan 7 at 4:02
• Everything outside of the pass band of the first filter is blocked. Frequency translation of zero is zero. Is this LTI? – Stanley Pawlukiewicz Jan 7 at 4:23
• I don't know if it is LTI or not, i assumed it is, because we are using Fourier transform here, if not i think we couldn't! – no0ob Jan 7 at 4:27
• @StanleyPawlukiewicz If it's not LTI what should we do? Is there any calculable Impulse response for whole System? – no0ob Jan 7 at 4:32

The system is not just a wire, it is in fact time-varying, so it has no frequency response in the conventional sense. The sequence $$w[n]$$ is of course just a low-pass filtered version of the input signal. The filtered signal is band-limited to half the Nyquist frequency. Now just write the output $$y[n]$$ in the time domain as the sum of $$w[n]$$ and the modulated version of $$w[n]$$. You should be able to see that effectively every other (odd) sample of $$w[n]$$ is replaced by a zero. But there occurs no aliasing due to the low pass filter at the input.