Question 6.3: Calculation of RC Lowpass Output Suppose that an input signa...

The filter has the form of the lowpass filter analyzed in this section. The half-power frequency is given by

f_{B}=\frac{1}{2\pi RC}=\frac{1}{2\pi \times (1000/2\pi)\times 10\times 10^{-6}}=100 \mathrm{~Hz}

The first component of the input signal is

v_{\mathrm{in1}}(t)=5\cos{(20 \pi t)}

For this component, the phasor is V_in1 = 5∠0°, and the angular frequency is ω = 20π. Therefore, f = ω/2π = 10. The transfer function of the circuit is given by Equation 6.9, which is repeated here for convenience:

H(f)=\frac{1}{1+j(f/f_B)}        (6.9)

Evaluating the transfer function for the frequency of the first component ( f = 10), we have

H(10)=\frac{1}{1+j(10/100)}=0.9950\angle -5.71^\circ

The output phasor for the f = 10 component is simply the input phasor times the transfer function. Thus, we obtain

\mathrm{V_{out1}}=H(10)\times \mathrm{V_{in1}}=(0.9950\angle -5.71^\circ)\times (5\angle 0^\circ)=4.975\angle -5.71^\circ

Hence, the output for the first component of the input signal is

v_{\mathrm{out1}}(t)=4.975\cos{(20 \pi t -5.71^\circ)}

Similarly, the second component of the input signal is

v_{\mathrm{in2}}(t)=5\cos{(200 \pi t)}

and we have

\mathrm{V_{in2}}=5\angle 0^\circ

The frequency of the second component is f = 100:

H(100)=\frac{1}{1+j(100/100)}=0.7071\angle -45^\circ

\mathrm{V_{out2}}=H(100)\times \mathrm{V_{in2}}=(0.7071\angle -45^\circ)\times (5\angle 0^\circ)=3.535\angle -45^\circ

Therefore, the output for the second component of the input signal is

v_{\mathrm{out2}}(t)=3.535 \cos{(200 \pi t-45^\circ)}

Finally, for the third and last component, we have

v_{\mathrm{in3}}(t)=5\cos{(2000 \pi t)}

\mathrm{V_{in3}}=5\angle 0^\circ

H(1000)=1\frac{1}{1+j(1000/100)}=0.0995\angle -84.29^\circ

\mathrm{V_{out3}}=H(1000)\times \mathrm{V_{in3}}=(0.0995\angle -84.29^\circ)\times (5\angle 0^\circ)=0.4975\angle -84.29^\circ

Consequently, the output for the third component of the input signal is

v_{\mathrm{out3}}(t)=0.4975 \cos{(2000 \pi t -84.29^\circ)}

Now, we can write an expression for the output signal by adding the output components:

v_{\mathrm{out}}(t)=4.975\cos{(20 \pi t-5.71^\circ)}+3.535\cos{(200\pi t -45^\circ)}+0.4975\cos{(2000 \pi t -84.29^\circ)}

Notice that each component of the input signal v_in(t) is treated differently by this filter. The f = 10 component is nearly unaffected in amplitude and phase. The f = 100 component is reduced in amplitude by a factor of 0.7071 and phase shifted by -45°. The amplitude of the f = 1000 component is reduced by approximately an order of magnitude. Thus, the filter discriminates against the high-frequency components.