Response of an RC filter to a square wave:

Transcription

1 9/7/7 Miscellaneous exercises Filered square wave Response of an RC filer o a square wave: The experimener ofen uses a square wave for iming and iggering purposes. Wha happens o a square wave afer passing hrough a low-pass filer? In his exercise we will express he square wave as he sum of a sine wave a he fundamenal frequency and waves a he higher harmonics. Then we will use a circui response funcion ha aenuaes and phase shifs each harmonic o find he new se of aenuaed waves. The aenuaed waves are hen summed o find he waveform ha comes ou of he low-pass filer. The firs example is a simple low-pass RC filer and he second example is a RLC bandpass filer. I. The square wave The Fourier decomposiion of he square wave can be found in mahemaical ables. The frequencies and heir ampliudes are given below. The number of frequencies ha we will consider is: 3 n,.. The fundamenal frequency f ha we will use is khz and he corresponding radian frequency is: The fundamenal frequency and is odd harmonics are: For he square wave, he ampliudes of he harmonics are: The square wave is hen: Ω fund a n π π Ω n ( n + Ω fund n + n = ( a exp i Ω n ( n The imaginary par is used because we wan o selec he sine (no he cosine and is harmonics. The ime scale of he problem is he inverse of he frequency Ω fund. This ime scale is used o define a range of imes for ploing:.,.. Ω fund Ω fund Plo of he sum represening he square wave F ( sin( Ω

2 9/7/7 Miscellaneous exercises Filered square wave The plo of he square wave was made using poins on he ime axis. If fewer poins had been used, he deails in he square wave (a he "corners" would no have been ploed. Try i: Wha happens if he number of harmonics ( in he sum is halved? Doubled? II. Square wave ino an RC low pass filer: V in V ou Our RC filer is shown in he diagram a he righ. The values of he circui componens are R = kω and C =.5 μf. R ohms C.5 6 Farads The corresponding impedances are: Z R Z ω ( ic ω The ime scale of ineres is he RC ime consan: The 3 db poin of his filer is: Thus our filer easily passes he fundamenal: RC = 5 5 ω 3db RC Ω fund = ω 3db = 4 For a volage divider wih impedances Z (ω and Z (ω, he volage division raio (gain Gim is: Gim Z ( + Z Z ω The "im" indicaes ha we have used complex impedance o ge he complex circui response. n = ( n exp( i Ω n A range of frequencies for ploing is: Noe ha he response funcion Gim(ω goes inside of he summaion. ω plo.,.. RC RC The absolue value of he response funcion of he filer is: Gim( ω plo ω plo

3 9/7/7 Miscellaneous exercises Filered square wave 3 The low-pass filered square wave F ( Noe ha he filer has rounded he some (bu no all of he "corners" on he square wave. Thus he low pass filer doesn' jus "smooh" he curve. The effec of he filer is more complex. Try i: How is he filered wave changed if R is doubled? Halved? III. Square wave ino an RC high pass filer This is really easy o do a his poin because all ha is necessary is o replace Z wih Z and vice versa. Gim ( Z ω Z + Z Noe ha he numeraor is Z and no Z. n = ( n exp( i Ω n The absolue value of he response funcion of he filer is: Gim( ω plo ω plo

4 9/7/7 Miscellaneous exercises Filered square wave 4 The high-pass filered square wave F ( Try i: Why is he ampliude. for his wave and. for he square wave? IV. RLC bandpass filer The circui a righ is an RLC filer which has a response ha is peaked a he resonan frequency. For his circui, we can apply he same ype of analysis ha is used for he RC circui. V in The circui elemen values are: V ou R 3 L 3 C. 6 KΩ milli Henries. micro Farads The LC par of he circui has a resonan frequency: ω CL ω = 5 This frequency is much higher han he fundamenal frequency of our square wave. We will define a range of frequencies for ploing ha go from. o imes ω. ω. ω,. ω.. ω

5 9/7/7 Miscellaneous exercises Filered square wave 5 The RLC circui can be reaed as a volage divider wih complex impedances. The firs impedance is simply: And he second impedance is he parallel combinaion of L and C: The gain ha we defined above mus be redefined wih he new impedances: Z R ( ic ω Z ω Gim Z ω + il ω Z ( + Z The absolue value of he complex response funcion has a peak a he resonan frequency of he LC par of he circui: The wave afer passage hrough he filer is: n = ( n exp( i Ω n Gim ω The RLC filered square wave. F ( The plo shows ha he ransiions ("edges" of he square wave cause he LC par of he circui o oscillae a he resonan frequency. Try i: Are he oscillaions in he plo more or less heavily damped if he value of R is doubled? Try i: Are he number of fas oscillaions in one period of he sine wave equal o he number ha you expec from comparing ω wih Ω fund?