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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?

## SOLUTION This solution is FREE courtesy of BrainMass!

(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 passband response in the plot above by drawing a roll-off response for an -18 dB drop every Â±0.3 change on the Log(f) axis.

(b) The first thing to notice is that the frequency of 1500 Hz (1.5 KHz) lies in the middle of the passband where the attenuation is - 2 dB. Thus the power dissipated in the resistor will be - 2 dB down on the power dissipated when there is no filter
i.e. Power Dissipated in R (@ 1500 Hz) = 0 dBW - 2 dB = - 2 dBW

(c) This is similar to part (b) but we are working at 500 Hz which is out of the passband so we need to know what is the attenuation due to the filter at this frequency. As 500 Hz is Â½ of the lower edge frequency (1 kHz) of the passband (i.e. one octave below) we can deduce given the information provided of a roll-off in response of - 18 dB/octave that the attenuation at 500 Hz {A(500 Hz)} is

A(500 Hz) = Attenuation at passband - 18 dB = - 2 dB - 18 dB = - 20 dB

Alternatively we could have taken a look at our Bode plot and worked out that

Log(f = 500) = 2.7 and then draw a vertical line to find the attenuation at this frequency by reading directly off the plot.

So we can deduce that the

Power dissipated in R (@ 500 Hz) = 0 dBW - 20 dB = - 20 dBW.

(d) Again this is analogous to the situation in part (b) where at 2000 Hz (2 KHz) we are just in the passband so we can say

Power dissipated in R (@ 2000 Hz) = 0 dBW - 2 dB = - 2 dBW.

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