First-order filters

1 EE 2301st- order filters 1 First-order filtersThe general form for the transfer function of a first order filter is:T(s)=Go s+Zos+PoThere will always be a single pole at s = Po. The pole must be real (there is only one, so no complex conjugates are not possible) and it must be negative (for stability). There will always be a zero, which can be at s = 0, as s (zero at infinity), or somewhere else, s = Zo. (Note the zero can have a positive value.) There may be a gain factor, Go, which might be 1 or smaller (for a passive circuit with a voltage divider) or have a magnitude greater than 1 for an active circuit .

2 The two most important cases are the zero at infinity, which is a low-pass filter and the zero at zero, which is the high-pass (s)=a1s+a0b1s+boHowever, we will typically recast this into a standard form:EE 2301st- order filters 2 Low-passIn the case were a1 = 0, we have a low-pass (s)=Go Pos+PoT(s)=aob1s+boIn standard form, we write it as: j xpole at P0s-planezero as s The reason for this form will become clear as we proceed. We will ignore the gain initially and focus on sinusoidal behavior by letting s = j .Pos+Pos=j =PoPo+j Re-expressing the complex value in magnitude and phase form:PoPo+j =[PoP2o+ 2]exp(j LP) LP= arctan( Po)EE 2301st- order filters 3By looking at the magnitude expression, we can see the low-pass behavior.

3 For low frequencies ( << Po): . At high frequencies ( >> Po): . At low frequencies, the magnitude is 1 (the output is equal to the input) and at high frequencies, the magnitude goes down inversely with frequency, consistent with the notion of a low-pass response. We can also examine the phase at the extremes. For low frequencies ( 0): . At high frequencies ( + ): .PoP2o+ 2 PoP2o=1 PoP2o+ 2 Po 2=Po LP= arctan( Po) 0 LP= arctan( Po) 90 EE 2301st- order filters 4M=PoP2o+ 2We can use the functions to make the magnitude and phase as frequency response plots. LP= arctan( Po) (dB)-50-40-30-20-10010101001000100001000 00phase ( )-90-75-60-45-30-15010100100010000100000 angular frequency (rad/s)angular frequency (rad/s)angular frequency (rad/s)magnitude - linear scalemagnitude - Bode plotphaseEE 2301st- order filters 5 Use the standard definition for cut-off frequency, which is the frequency at which the magnitude is down by from the value in the pass-band.

4 For our low-pass function, the pass band is at low frequencies, and the magnitude there is 1. (Again, we are hiding Go by assuming that it is unity. If Go 1, then everything is scaled by Go.)2 Cut-off frequencyM=12=PoP2o+ 2cWith a bit of algebra, we find that . The cut-off frequency is defined by the pole. Tricky! Thus, in all of our equations, we could substitute c for Po. c=PoWe can also calculate the phase at the cut-off frequency. LP= arctan( cPo)= 45 The cut-off frequency points are indicated in the plots on the previous 2301st- order filters 6 /3= arctan F 7/3(M )=*R FM + F7/3(V)=*R FV+ F7/3(V)=*R +V F7/3(M )=*R +M F /3= arctan F Again, we could just as easily use real frequency rather than angular frequency.

5 As an exercise: re-express all of the above formulas using f instead of .To emphasize the importance of the corner frequency in the low-pass function, we can express all the previous results using c in place of Po. On the left are the functions in "standard" form. On the right, the functions are expressed in a slightly different form that is sometimes easier to use. 7/3(M ) = *R F + F<latexit sha1_base64="ZcBktQQdyEg5 AHKJu2kZSXSy48k=">AAAC53icbZFBaxNBFMdn16o12prq0ctgEFIsIRsE PSgUFOqhhwhNU8imy+zk7 WbM7Mw681 YSlv0M3sSrH8uLn8 XZJC1J6oOFP7///83svBfnUljsdv94/r29+w8e7j 9qPH5ycPi0efTs0urCcBhwLbW5ipkFKRQMUKCEq9 wAy2 IJw3j2sfaH38 FYodUFLnIYZyxVIhGcoUNRcx7 GIqWhiyC9iMrzfhVKSLD9 NdQZpCw0Ip3icR1aZT7 QjYazSNNbK0wM4+WqjUa8 KkP7zeANuO69prfeda+qomar2+kui94 VwVq0yLr60ZE3 DCeaFxko5 JJZOwq6OY5 LZlBwCVUjLCzkjM9 YCiMnFcvAjsvliCr6ypEJTbRxn0K6pJsdJcusXWS xS2

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