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

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 \text{ Hz}

The first component of the input signal is

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

For this component, the phasor is \pmb{V}_{\text{in1}}=5 \underline{/0^\circ}, and the angular frequency is ω = 20π. Therefore, ƒ = 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)}

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 \underline{/-5.71^\circ}

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

\begin{matrix} \pmb{\text{V}}_{\text{out1}}&=&H(10) \times \pmb{\text{V}}_{\text{in1}} \\ &=& (0.9950 \underline{/-5.71^\circ}) \times (5 \underline{/0^\circ})=4.975 \underline{/-5.71^\circ} \end{matrix}

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

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

Similarly, the second component of the input signal is

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

and we have

\pmb{\text{V}}_{\text{in2}}=5 \underline{/0^\circ}

The frequency of the second component is ƒ = 100:

\begin{matrix} H(100)&=&\frac{1}{1+j(100/100)}=0.7071 \underline{/-45^\circ} \\ \pmb{\text{V}}_{\text{out2}} &=& H(100) \times \pmb{\text{V}}_{\text{in2}} \\ &=& (0.7071 \underline{/-45^\circ}) \times (5 \underline{/0^\circ})=3.535 \underline{ /-45^\circ} \end{matrix}

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

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

Finally, for the third and last component, we have

\begin{matrix} v_{\text{in3}}(t)&=&5 \cos(2000 \pi t) \\ \pmb{\text{V}}_{\text{in3}}&=&5 \underline{/0^\circ} \\ H(1000)&=&\frac{1}{1+j(1000/100)}=0.0995 \underline{/-84.29^\circ} \\ \pmb{\text{V}}_{\text{out3}}&=&H(1000) \times \pmb{\text{V}}_{\text{in3}} \\ &=& (0.0995 \underline{/-84.29^\circ}) \times (5 \underline{/0^\circ})=0.4975 \underline{/-84.29^\circ} \end{matrix}

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

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

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