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Investigating power levels at load through a bandpass filter

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A bandpass filter has a constant attenuation of 2 dB in its pass band between 1000 Hz and 2000 Hz. The filter is designed to operate with a 50 Ohm source and load. Below 1000 Hz and above 2000 Hz the filter attenuates signals. The slope of the filter frequency response outside the pass band is 18 dB per octave. You may assume that the frequency response of this filter is made up of three straight lines when plotted on a dB vs log frequency graph. The filter is placed between a signal generator and a 50 Ohm load resistor.

a. Make a sketch of the response of the bandpass filter from 500 Hz to 10 kHz. Use a dB scale on the vertical axis and a log frequency scale on the horizontal axis from 500 Hz to 2000 Hz. Insert numbers on the scale on each axis.

b. The output of the signal generator is a sine wave at a frequency of 1500 Hz. With the filter removed from the circuit the power dissipated in the resistor is 1.0 W = 0 dBW. What is the power (in dBW) dissipated in the resistor with the filter in place between the signal generator and load resistor?

c. The output of the signal generator is changed to a sine wave at a frequency of 500 Hz. The power level is not changed. What is the power (in dBW) dissipated in the resistor with the filter in place between the signal generator and load resistor?

d. The output of the signal generator is changed to a sine wave at a frequency of 2000 Hz. The power level is not changed. What is the power (in dBW) dissipated in the resistor with the filter in place between the signal generator and load resistor?

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Solution Summary

Details of a Bandpass filter attenuation and frequency response are presented and a solution developed to determine the power level dissipated in a known load resistor of 50 Ohms for known source drive voltage, load directly connected to source. The filter is put in place and the solution shows how to determine the power dissipated in the load resistor when operating out of band at a frequency 500 Hz and in band at a frequency of 2 kHz for the same input power. The filter response is drawn using a Bode Plot (log frequency vs dB output) using the frequency and attenuation data presented in the Question.

Solution Preview

(a) Please see the attachment for the drawing of the response based on the information about the passband of the filter given.

An 18 dB/Octave roll-off out of passband means that there is a drop of -18 dB for every doubling or halving of frequency out of band.

A doubling of frequency represents a change of Log(2f/f) = Log(2) = +0.3 .

A halving of frequency represents a change of Log(f/2f) = Log(0.5) = -0.3 .

We have therefore tried to show this out of ...

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