Question 6.1: Using the Transfer Function to Determine the Output The tran...
Using the Transfer Function to Determine the Output
The transfer function H(f ) of a filter is shown in Figure 6.4. [Notice that the magnitude |H(f )| and phase ∠H(f ) are shown separately in the figure.] If the input signal is given by
vin(t) = 2 cos(2000πt + 40° )
find an expression (as a function of time) for the output of the filter.
By inspection, the frequency of the input signal is f = 1000 Hz. Referring to Figure 6.4, we see that the magnitude and phase of the transfer function are |H(1000)| = 3 and ∠H(1000) = 30° , respectively. Thus, we have
H(1000)=3\angle 30^\circ =\frac{\mathrm{V_{out}}}{\mathrm{V_{in}}}
The phasor for the input signal is Vin = 2∠40°, and we get
\mathrm{V_{out}} =H(1000)\times \mathrm{V_{in}}=3\angle 30^\circ \times 2\angle 40^\circ=6\angle 70^\circ
Thus, the output signal is
\mathrm{v_{out}}(t)=6\cos{(2000\pi t+70^\circ)}
In this case, the amplitude of the input is tripled by the filter. Furthermore, the signal is phase shifted by 30°. Of course, this is evident from the values shown in the plots of the transfer function at f = 1000.