HW9_Soln

# HW9_Soln - Escuti/Masnari 2007 ECE200 Fall 2007 Homework 9...

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Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 9 Solutions Filters and Operational-Amplifiers Page 1 of 14 Problem 1 – [10 pts] The frequency response of a band- pass filter and the power spectrum of the input signal are shown below. Find and sketch the power spectrum of the output signal. Solution: Pout = Pin + G, when both Pin and Pout are given in dBW. P out, 1kHz = - 60+ 40 = -20 dBW P out, 2kHz = 0 + 40 = 40 dBW P out, 3kHz = 0 + 30 = 30 dBW P out, 4kHz = -20 + 20 = 0 dBW P out, 5kHz = -60 + 10 = -50 dBW Output dBW 1 2 3 4 5 kHz 0 - 60 - 20 - 40 60 20 40

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Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 9 Solutions Filters and Operational-Amplifiers Page 2 of 14 Problem 2 – [16 pts] The transfer function (frequency response) of a certain band-pass filter is given by (a) Calculate the voltage gain Av(f) from 0 to 6000 Hz in steps of 1000 Hz (i.e. a total of seven frequencies). (b) Find the 3-dB cut-off frequency of the filter using either G(f) or Av(f). (c) The following signal is passed through this filter (which has units of Volts). Find the output signal. (d) Find the total signal power at the filter output . a) Power gain is already given in dB. As equation (6.6.3) G = 20 log A v => G 20 Av = 10 Thus, Frequency (Hz) G(f) Av(f) 0 - 60 0.001 1000 - 20 0.1 2000 0 1 3000 0 1 4000 0 1 5000 - 20 0.1 6000 - 60 0.001 b) Power gain is already given in dB. Since the filter has a power gain of 0 dB in the passband, the cut-off frequencies of this band pass filter exist at - 3 dB. To do this accurately, we need to find the line equations for the frequency ranges 1000 < f < 2000 & 4000 < f < 5000 Setting up the line equation using two points on the straight line we obtain:
Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 9 Solutions Filters and Operational-Amplifiers Page 3 of 14 Simplifying, we obtain the following straight line equation: G = 0.02f - 40 We need to find the frequency when G is equal to - 3 dB: - 3 = 0.02f - 40 Solving for f, we obtain 1850 Hz, which 150 Hz to the left of the 0 dB line. Due to symmetry, the upper cut-off frequency should be 150 Hz to the right of the 0 dB line. Therefore, the upper cut-off should be at 4150 Hz. c) ( ) 10 (0) (1000) 8 cos(2 1000 ) (2000) 6cos(2 2000 - )- (3000) 4cos (2 3000 - ) 4 4 4 out v v v v v t A A t A t A t ! = " + " " + + " " " (V), thus,

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## This homework help was uploaded on 03/20/2008 for the course ECE 200 taught by Professor Escuti during the Fall '07 term at N.C. State.

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HW9_Soln - Escuti/Masnari 2007 ECE200 Fall 2007 Homework 9...

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