+ - Hmewrk 0 Slutin ) In the circuit belw: a. Find the magnitude and phase respnse. b. What kind f filter is it? c. At what frequency is the respnse 0.707 if the generatr has a ltage f? d. What is the phase f the respnse at this frequency? e. Pick a few pints abve and belw this frequency and sketch the magnitude and phase respnses. 0 cs ωt 0.05 V Ans: Part a: Writing the ndal equatins and slving fr v gives the magnitude and phase respnse: 0 j 0.05 v j0.5 v v j0.5 (0.5 ) tan ( 0.5 ) Part b: Examining the equatin gives a respnse f at zer frequency and a respnse f 0 at infinite frequency, making this a lw pass filter. Part c: Setting the magnitude f the respnse equal t and slving fr ω :
(0.5 ) (0.5 ) (0.5 ) 0.5 4 Part d: Putting the previus answer int the phase respnse gives: tan ( 0.5 ) tan ( 0.5(4)) tan () 45 r - 4
+ - ) In the circuit belw: a. Find the magnitude and phase respnse. b. What kind f filter is it? c. At what frequency is the respnse 0.707 if the generatr has a ltage f? d. What is the phase f the respnse at this frequency? e. Pick a few pints abve and belw this frequency and sketch the magnitude and phase respnses. 0. cs ωt V Ans: Part a: Writing the ndal equatin and slving fr v gives the respnse: j 0. j 0. j 0. j 0. tan 0. 0. Part b: Lking at the magnitude equatin gives a zer respnse at zer frequency and a respnse f at infinite frequency, making this a high pass filter. slve fr ω: Part c: As in prblem, t find the frequency, set the magnitude respnse t and
v 0. 0. 0. 0. 5 angle: Part d: Plugging in the result frm part C int the phase part f the respnse gives the 0.(5) 45 r 4 tan tan ()
+ - 3) In the circuit belw: a. Find the magnitude and phase respnse. b. What kind f filter is it? c. At what frequency is the respnse 0.707 if the generatr has a ltage f? d. What is the phase f the respnse at this frequency? e. Pick a few pints abve and belw this frequency and sketch the magnitude and phase respnses. H cs ωt 6 V Ans: Part a: Writing the ndal equatin and slving fr v gives the respnse: j 6 j 3 j 3 j 3 v 3 tan 3 Part b: Lking at the magnitude equatin gives a respnse f at zer frequency and a respnse f 0 at infinite frequency, making this a lw pass filter. slve fr ω: Part c: As in prblem, t find the frequency, set the magnitude respnse t and
v 3 3 3 3 3 angle: Part d: Plugging in the result frm part C int the phase part f the respnse gives the 3 3 45 r - 4 tan tan ( )
+ - 4) In the circuit belw: a. Find the magnitude and phase respnse. b. What kind f filter is it? c. At what frequency is the respnse 0.707 if the generatr has a ltage f? d. What is the phase f the respnse at this frequency? e. Pick a few pints abve and belw this frequency and sketch the magnitude and phase respnses. 0 cs ωt H V Part a: Writing the ndal equatin and slving fr v gives the respnse: v 0 j 0 j 0 j 0 j 0 tan 0 Part b: Lking at the magnitude equatin gives a zer respnse at zer frequency and a respnse f at infinite frequency, making this a high pass filter. slve fr ω: Part c: As in prblem, t find the frequency, set the magnitude respnse t and
v 0 0 0 0 0 angle: Part d: Plugging in the result frm part C int the phase part f the respnse gives the 0 0 45 r 4 tan tan ()
+ - 5) In the circuit belw: a. Find the magnitude and phase respnse. b. What kind f filter is it? c. At what frequency is the respnse maximum? d. What is the phase f the respnse at this frequency? e. Pick a few pints abve and belw this frequency and sketch the magnitude and phase respnses. 0 cs ωt /4 /4H V Ans: Part a: Writing the ndal equatin and slving fr v gives the magnitude and phase respnse: 0 j j 4 4 4 j0 v 4 4 j0 v 4 4 j0 4 v 4 tan 0 4 4 0 4 Part b: Lking at the equatin, we see that the respnse at zer frequency wuld be 0 and the respnse at infinite frequency wuld als be 0. This means that there will be sme pint where there is a peak respnse, making this a band pass filter. Part c: Again, lking at the equatin, t have a peak respnse, the denminatr f the magnitude wuld have t be a minimum. At sme pint:
4 0 4 Slving fr : 4 4 6 4 Part d: Inserting the result frm part C int the phase equatin f the respnse gives: 4 4 4 4 0 tan tan 0
+ - 6) In the circuit belw: a. Find the magnitude and phase respnse. b. What kind f filter is it? c. At what frequency is the respnse minimum? d. What is the phase f the respnse at this frequency? e. Pick a few pints abve and belw this frequency and sketch the magnitude and phase respnses. /H cs ωt / V Ans: Part a: Writing the ndal equatin and slving fr v gives the respnse: j j j j j j j v j j v 4 90 tan 90 tan fr < fr >
Ntice the way yu must reslve the phase f the respnse. The numeratr f the cmplex respnse is an imaginary term. This crrespnds t an angle f 90 degrees if the imaginary part is psitive and -90 degrees if the imaginary part is negative. Since the numeratr is simply the cmplex term, we take its abslute value t get the magnitude. After dealing with the denminatr using Euler s relatin, we can deal with the entire respnse as we wuld any phasr fractin, yielding the result shwn abve. Part b: Lking at the respnse, we see that it wuld have a respnse f infinity ver infinity at zer frequency, which we can reslve as. Similarly, it will have a respnse f infinity ver infinity at infinite frequency, which we can reslve als as. At sme pint in the middle, the respnse will be a minimum which makes this a band stp r ntch filter. Part c: Inspectin f the equatin shws that the respnse will be a minimum if: 0 Part d: 0 Nte: the phase functin is nt cntinuus at ω=. Ging back t the riginal cmplex respnse, we see the imaginary term in the numeratr vanishes at ω=. Therefre its phase is 0 degrees. The imaginary term in the denminatr als vanishes, making its phase als equal t 0. Reslving this using phasrs makes the phase respnse here 0 degrees. If yu were t build a ntch filter and watch the utput as yu swept the signal generatr thrugh the ntch frequency (i.e. the frequency with minimum respnse), yu wuld bserve the utput ging t zer in ne phase and reappearing again as yu passed the ntch frequency 80 degrees ut f phase with the respnse it had belw the ntch frequency.