Question 6.2: Using the Transfer Function with Several Input Components Su...

We start by breaking the input signal into its components. The first component is

v_{\mathrm{in1}}(t)=3

and the second component is

v_{\mathrm{in2}}(t)=2\cos{(2000\pi t)}

and the third component is

v_{\mathrm{in3}}(t)= \cos{(4000\pi t-70^\circ)}

By inspection, we see that the frequencies of the components are 0, 1000, and 2000Hz, respectively. Referring to the transfer function shown in Figure 6.4, we find that

 H(0) = 4

H(1000)=3\angle 30^\circ

and

H(2000)=2\angle 60^\circ

The constant (dc) output term is simply H(0) times the dc input:

v_{\mathrm{out1}}=H(0)v_{\mathrm{in1}}=4\times 3=12

The phasor outputs for the two input sinusoids are

\mathrm{V_{out2}}=H(1000)\times \mathrm{V_{in2}}=3\angle 30^\circ \times 2\angle 0^\circ =6\angle 30^\circ

\mathrm{V_{out3}}=H(2000)\times \mathrm{V_{in3}}=2\angle 60^\circ \times 1\angle -70^\circ =2\angle-10^\circ

Next, we can write the output components as functions of time:

v_{\mathrm{out1}}(t)=12

v_{\mathrm{out2}}(t)=6\cos{(2000\pi t +30^\circ)}

and

v_{\mathrm{out3}}(t)=2\cos{(4000\pi t -10^\circ)}

Finally, we add the output components to find the output voltage:

v_{\mathrm{out}}(t)= v_{\mathrm{out1}}(t)+ v_{\mathrm{out2}}(t)+ v_{\mathrm{out3}}(t)

and

v_{\mathrm{out}}(t)=12+6\cos{(2000\pi t +30^\circ)}+2\cos{(4000\pi t-10^\circ)}