Example 1

Let's apply these results to a final example, in which the input is a voltage source and the output is the inductor current. The source voltage equals V in =2cos2π60t+3 V in 2 2 60 t 3 . We want the circuit to pass constant (offset) voltage essentially unaltered (save for the fact that the output is a current rather than a voltage) and remove the 60 Hz term. Because the input is the sum of two sinusoids--a constant is a zero-frequency cosine--our approach is

Because the circuit is a series combination of elements, let's use voltage divider to find the transfer function between V in V in and V V, then use the v-i relation of the inductor to find its current. where voltage divider=ⅈ2πfLR+ⅈ2πfL voltage divider 2 f L R 2 f L and inductor admittance=1ⅈ2πfL inductor admittance 1 2 f L [Do the units check?] The form of this transfer function should be familiar; it is a lowpass filter, and it will perform our desired function once we choose element values properly.

The constant term is easiest to handle. The output is given by 3|H0|=3R 3 H 0 3 R . Thus, the value we choose for the resistance will determine the scaling factor of how voltage is converted into current. For the 60 Hz component signal, the output current is 2|H60|cos2π60t+∠H60 2 H 60 2 60 t H 60 . The total output due to our source is

The cutoff frequency for this filter occurs when the real and imaginary parts of the transfer function's denominator equal each other. Thus, 2π f c L=R 2 f c L R , which gives f c =R2πL f c R 2 L . We want this cutoff frequency to be much less than 60 Hz. Suppose we place it at, say, 10 Hz. This specification would require the component values to be related by RL=20π=62.8 R L 20 62.8 . The transfer function at 60 Hz would be which yields an attenuation (relative to the gain at zero frequency) of about 1/6 16, and result in an output amplitude of 0.3R 0.3 R relative to the constant term's amplitude of 3R 3 R . A factor of 10 relative size between the two components seems reasonable. Having a 100 mH inductor would require a 6.28 Ω resistor. An easily available resistor value is 6.8 Ω; thus, this choice results in cheaply and easily purchased parts. To make the resistance bigger would require a proportionally larger inductor. Unfortunately, even a 1 H inductor is physically large; consequently low cutoff frequencies require small-valued resistors and large-valued inductors. The choice made here represents only one compromise.

The phase of the 60 Hz component will very nearly be -π2 2 , leaving it to be 0.3Rcos2π60t−π2=0.3Rsin2π60t 0.3 R 2 60 t 2 0.3 R 2 60 t . The waveforms for the input and output are shown in Figure 2.