Question 6.1: Using the Transfer Function to Determine the Output , The tr...
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
v_{in}(t)=2\cos(2000 \pi t+40^\circ)
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 \text{ and }\underline{/H(1000)}=30^\circ, respectively. Thus, we have
H(1000)=3 \underline{/30^\circ}=\frac{\pmb{\text{V}}_{\text{out}}}{\pmb{\text{V}}_{\text{in}}}
The phasor for the input signal is \pmb{\text{V}}_{\text{in}}=2 \underline{/40^\circ}, and we get
\pmb{\text{V}}_{\text{out}}=H(1000) \times \pmb{\text{V}}_{\text{in}}= 3 \underline{/30^\circ} \times 2 \underline{/40^\circ}=6 \underline{/70^\circ}
Thus, the output signal is
v_{\text{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 ƒ = 1000.