High Pass Filter with pulses trains input

[Highlander-SP]

If i have the folloing transfer function:

H (ω) = u( ω - 2.pi.10³).exp(-jω10-³) + u[ - ( ω + 2.pi.10³)].exp(-jω10-³)

(where "u" is the Heaviside function, unity step)

and pulses trains as a input signal (in time domain), as picture attached, what should be the output signal and what's the explanation for that?


tks

 

[zorro]

Hi Highlander-SP,

First of all, note that the transfer function rejects all frequencies between –1000 Hz and 1000 Hz. All other frequencies pass with unity gain and a delay of 1 ms.
The pulse train is periodic with fundamental frequency 2.5 KHz.
The fundamental and all its harmonics pass across the transfer function. The only component that doesn’t pass is d.c. (frequency=0)
Then, the output is the same as the input but without the d.c., and delayed by 1 ms. It is evident that the d.c. component is 1V.
Thus, the output is y(t)=x(t-1ms)-1V.

Regards

Z