Solution to Homework on Low Pass Filter Frequency Response Analysis

Homework 10 Solution 1) In the circuit below: a. Find the magnitude and phase response. b. What kind of filter is it? c. At what frequency is the response 0.707 if the generator has a voltage of 1? d. What is the phase of the response at this frequency? e. Pick a few points above and below this frequency and sketch the magnitude and phase responses. + 10 0.025 Vo cos ωt Ans: Part a: Writing the nodal equations and solving for vo gives the magnitude and phase response: 1  vo vo  1 10 j 0.025 1  vo 1  j 0.25  vo  vo  1 1  j 0.25 1 1  (0.25 ) 2  tan 1 (0.25 ) Part b: Examining the equation gives a response of 1 at zero frequency and a response of 0 at infinite frequency, making this a low pass filter. Part c: Setting the magnitude of the response equal to 1 and solving for ω : 2 1 1  2 1  (0.25 ) 2 2  1  (0.25 ) 2 1  (0.25 ) 2 1  0.25 4 Part d: Putting the previous answer into the phase response gives:   tan 1 (0.25 )  tan 1 (0.25(4))  tan 1 (1)  45 or -  4 2) In the circuit below: a. Find the magnitude and phase response. b. What kind of filter is it? c. At what frequency is the response 0.707 if the generator has a voltage of 1? d. What is the phase of the response at this frequency? e. Pick a few points above and below this frequency and sketch the magnitude and phase responses. + 0.1 Vo 2 cos ωt Ans: Part a: Writing the nodal equation and solving for vo gives the response: 1  vo v  o 1 2 j 0.1 1 1  vo   j vo 0.2 1   1  1  j vo 0.2   1 vo  1 1 j 0.2 1  1  vo   tan 1   2  0.2   1  1    0.2  Part b: Looking at the magnitude equation gives a zero response at zero frequency and a response of 1 at infinite frequency, making this a high pass filter. Part c: As in problem 1, to find the frequency, set the magnitude response to solve for ω: 1 and 2 vo  1  2 1  1  1    0.2   1  2  1    0.2   1  1    0.2  1 1 0.2  5 2 2 2 Part d: Plugging in the result from part C into the phase part of the response gives the angle:  1  1   tan (1) 0.2(5)     tan 1    45 or  4 3) In the circuit below: a. Find the magnitude and phase response. b. What kind of filter is it? c. At what frequency is the response 0.707 if the generator has a voltage of 1? d. What is the phase of the response at this frequency? e. Pick a few points above and below this frequency and sketch the magnitude and phase responses. 1 2 + 2H Vo 6 cos ωt Ans: Part a: Writing the nodal equation and solving for vo gives the response: 1  vo vo  j 2 6 1  vo  j  3 vo   1  1  j  vo 3  1 vo   1 j 3 1   vo   tan 1    2  3   1   3 Part b: Looking at the magnitude equation gives a response of 1 at zero frequency and a response of 0 at infinite frequency, making this a low pass filter. Part c: As in problem 1, to find the frequency, set the magnitude response to solve for ω: 1 and 2 vo  1  2   2  1   3   1   3 1 1   1   3 2 2 2  3  3 Part d: Plugging in the result from part C into the phase part of the response gives the angle:  3     tan 1     tan 1 (1) 3   45 or -  4 4) In the circuit below: a. Find the magnitude and phase response. b. What kind of filter is it? c. At what frequency is the response 0.707 if the generator has a voltage of 1? d. What is the phase of the response at this frequency? e. Pick a few points above and below this frequency and sketch the magnitude and phase responses. + 10 1 1H Vo cos ωt 2 Part a: Writing the nodal equation and solving for vo gives the response: 1  vo vo  10 j 10 1  vo   j vo  10   1  1  j  vo   1 vo  10 1 j  vo   10   tan 1   2    10  1     1 Part b: Looking at the magnitude equation gives a zero response at zero frequency and a response of 1 at infinite frequency, making this a high pass filter. Part c: As in problem 1, to find the frequency, set the magnitude response to solve for ω: 1 and 2 vo  1  2 1  10  1      10  2  1      10  1     10 1 2 2 2    10 Part d: Plugging in the result from part C into the phase part of the response gives the angle:  10  1   tan (1)  10    tan 1    45 or  4 5) In the circuit below: a. Find the magnitude and phase response. b. What kind of filter is it? c. At what frequency is the response maximum? d. What is the phase of the response at this frequency? e. Pick a few points above and below this frequency and sketch the magnitude and phase responses. 1 + 20 1/4H cos ωt Vo 1/4 2 Ans: Part a: Writing the nodal equation and solving for vo gives the magnitude and phase response: 1  vo v v  o  o 1  20 j j  4 4  4  1  vo  j 20    vo 4    4  1  1  j 20     vo  4    vo  vo  1   4  1  j 20  4          4    tan 1  20     2  4      4   1   20       4   1 Part b: Looking at the equation, we see that the response at zero frequency would be 0 and the response at infinite frequency would also be 0. This means that there will be some point where there is a peak response, making this a band pass filter. Part c: Again, looking at the equation, to have a peak response, the denominator of the magnitude would have to be a minimum. At some point: 0   4 4  Solving for :  4  4   2  16 4 Part d: Inserting the result from part C into the phase equation of the response gives: 4  4   tan 1     tan 1  0  4 4   0  6) In the circuit below: a. Find the magnitude and phase response. b. What kind of filter is it? c. At what frequency is the response minimum? d. What is the phase of the response at this frequency? e. Pick a few points above and below this frequency and sketch the magnitude and phase responses. 1 + 2 1/2H Vo 2 cos ωt - 1/2 Ans: Part a: Writing the nodal equation and solving for vo gives the response: 1  vo vo vo    1 2  2  j j j    2 2  2  2   2  j     j    vo  2vo 2  2   2   j     2  2     2  j     vo  2    2  j   2  vo   2  2 j   2     2       90  tan 1  2   for  <2    2     2         2  vo    2  2      2   4    2     1 2  90  tan  2  for  >2          Notice the way you must resolve the phase of the response. The numerator of the complex response is an imaginary term. This corresponds to an angle of 90 degrees if the imaginary part is positive and -90 degrees if the imaginary part is negative. Since the numerator is simply the complex term, we take its absolute value to get the magnitude. After dealing with the denominator using Euler’s relation, we can deal with the entire response as we would any phasor fraction, yielding the result shown above. Part b: Looking at the response, we see that it would have a response of infinity over infinity at zero frequency, which we can resolve as 1. Similarly, it will have a response of infinity over infinity at infinite frequency, which we can resolve also as 1. At some point in the middle, the response will be a minimum which makes this a band stop or notch filter. Part c: Inspection of the equation shows that the response will be a minimum if: 0  2 2  2  Part d: 0 Note: the phase function is not continuous at ω=2. Going back to the original complex response, we see the imaginary term in the numerator vanishes at ω=2. Therefore its phase is 0 degrees. The imaginary term in the denominator also vanishes, making its phase also equal to 0. Resolving this using phasors makes the phase response here 0 degrees. If you were to build a notch filter and watch the output as you swept the signal generator through the notch frequency (i.e. the frequency with minimum response), you would observe the output going to zero in one phase and reappearing again as you passed the notch frequency 180 degrees out of phase with the response it had below the notch frequency.