Show that the series RL circuit in Fig. 14.13 also acts like a high-pass filter:
a) Derive an expression for the circuit’s transfer function.
b) Use the result from (a) to determine an equation for the cutoff frequency in the series RL circuit.
c) Choose values for R and L that will yield a highpass filter with a cutoff frequency of 15 kHz.
a) Begin by constructing the s-domain equivalent of the series RL circuit, as shown in Fig. 14.14. Then use s-domain voltage division on the equivalent circuit to construct the transfer function:
H\left(s\right) =\frac{s}{s+R/ L}.
Making the substitution s=j\omega, we get
H\left(j\omega\right) =\frac{j\omega}{j\omega+R/ L}.
Notice that this equation has the same form as Eq.14.15 for the series RC high-pass filter.
H\left(j\omega\right) =\frac{j\omega}{j\omega+R/ L}.
b) To find an equation for the cutoff frequency, first compute the magnitude of H\left(j\omega\right):
\mid H\left(j\omega\right)\mid =\frac{\omega }{\sqrt{\omega ^{2}+\left(R/ L\right) ^{2} } } .
Then, as before, we set the left-hand side of this equation to \left(1/ \sqrt{2} \right) H_{max}, based on the definition of the cutoff frequency \omega _{c}. Remember that H_{max}=\mid H\left(j\infty \right)\mid for a high-pass filter, and for the series RL circuit, \mid H\left(j\infty \right)\mid=1. We solve the resulting equation for the cutoff frequency:
\frac{1}{\sqrt{2} } =\frac{\omega _{c} }{\sqrt{\omega _{c}^{2}+\left(R/ L\right) ^{2} } } , \omega _{c}=\frac{R}{L} .
This is the same cutoff frequency we computed for the series RL low-pass filter.
c) Using the equation for \omega _{c} computed in (b), we recognize that it is not possible to specify values for R and L independently. Therefore, let’s arbitrarily select a value of 500 \Omega for R. Remember to convert the cutoff frequency to radians per second:
L=\frac{R}{\omega _{c}}=\frac{500}{\left(2\pi \right)\left(15,000\right) } =5.31 mH .
