Calculation of RC Lowpass Output
Suppose that an input signal given by

v_in(t) = 5 cos(20πt) + 5 cos(200πt) + 5 cos(2000πt)

is applied to the lowpass RC filter shown in Figure 6.9. Find an expression for the output signal.

Question

Calculation of RC Lowpass Output
Suppose that an input signal given by

v_in(t) = 5 cos(20πt) + 5 cos(200πt) + 5 cos(2000πt)

is applied to the lowpass RC filter shown in Figure 6.9. Find an expression for the output signal.

Accepted Answer

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

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

The first component of the input signal is

[latex]v_{\mathrm{in1}}(t)=5\cos{(20 \pi t)}[/latex]

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:

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

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

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

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

[latex]\mathrm{V_{out1}}=H(10) imes \mathrm{V_{in1}}=(0.9950\angle -5.71^\circ) imes (5\angle 0^\circ)=4.975\angle -5.71^\circ[/latex]

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

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

Similarly, the second component of the input signal is

[latex]v_{\mathrm{in2}}(t)=5\cos{(200 \pi t)}[/latex]

and we have

[latex]\mathrm{V_{in2}}=5\angle 0^\circ[/latex]

The frequency of the second component is f = 100:

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

[latex]\mathrm{V_{out2}}=H(100) imes \mathrm{V_{in2}}=(0.7071\angle -45^\circ) imes (5\angle 0^\circ)=3.535\angle -45^\circ[/latex]

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

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

Finally, for the third and last component, we have

[latex]v_{\mathrm{in3}}(t)=5\cos{(2000 \pi t)}[/latex]

[latex]\mathrm{V_{in3}}=5\angle 0^\circ[/latex]

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

[latex]\mathrm{V_{out3}}=H(1000) imes \mathrm{V_{in3}}=(0.0995\angle -84.29^\circ) imes (5\angle 0^\circ)=0.4975\angle -84.29^\circ[/latex]

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

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

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

[latex]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)}[/latex]

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.

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

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