Question

Problem 8.36 requires reference to 8.13, I only need 8.36 solved

8.36 High-pass filter** Consider the setup in Problem 8.13, but with the capacor replaced by an inductor. Calculate the ratio Ivl2/ Choose values for R and L to make l Vil2 = 0.1 for a 100 Hz signal. This circuit is the most primitive of "high-pass" filters, providing attenuation that increases with decreasing frequency. Show that, for sufficiently low frequencies, the signal power is reduced by a factor 1/4 for every halving of the frequency. 8.13 Low-pass filter In Fig. 8.29 an alternating voltage Vo cos ot is applied to the termi- nals at A. The terminals at B are connected to an audio amplifier of very high input impedance. (That is, current flow into the amplifier is negligible.) Calculate the ratio VVHere s the abso- lute value of the complex voltage amplitude at terminals B. Choose values for R and C to make l V12/V0 = 0.1 for a 5000 Hz signal. This circuit is the most primitive of "low-pass" filters, providing attenuation that increases with increasing frequency. Show that, for sufficiently high frequencies, the signal power is reduced by a factor 1/4 for every doubling of the frequency. Can you devise a filter with a more drastic cutoff -such as a factor 1/16 per octave? Figure 8.29.

Answer 1

2. 8 nfLV Rt i2nFL VnfL 2nfL 2ntL iv I0

For low frequency, i.e. when signal does not pass

then
R2
since at low frequencies $1 + \frac{R^{2}}{4\pi^{2}f^{2}L^{2}} \approx \frac{R^{2}}{4\pi^{2}f^{2}L^{2}}$

and thus when f goes to f/2, power goes as square of f = 1/4 of initial power