RF Laboratory Manual - Passive Filter Design.

1 RF Laboratory Manual - Passive Filter Matzner, S. Levy, H. Moalem and D. Group InsertionLossMethod .. Elliptic Filter Solution .. BandPassFilterTransformation .. RequiredEquipment .. Chebyshev3dBequalrippleLPFD esign .. Measurement .. Measurement .. Elliptic LPF design .. Measurement .. Final 27 CONTENTS34 CONTENTSPRELAB EXERCISE1. Using Matlab, draw a graph of PLR as a function of the normalized fre-quency of a Chebyshev LPF,3dBequal ripple,N=1,3,5,7(see Figure 3).2. Compare Butterworth and Chebyshev LPF with1dBequal ripple,N=3,fc=1 GHz, Zin=Zout=50 :1. Calculate the elements Calculate the transfer function of Draw the graph of the transfer function of eachfilter (only magnitude)up to Calculate the stopband attenuation (dB/octave) of BACKGROUND THEORYA Filter is a two port network used to control the frequency response at acertain point in a system by providing transmission within the passband of thefilter and attenuation in the stopband of thefilter.

2 The basicfilter types arelow-pass, high-pass, bandpass and band-reject (notch) Group DelayThe group delay is the derivative of the transmission phase with respect to theangular frequency and it is a measure of the distortion in the signal introducedby phase differences for different frequencies. It is defined as: g( )= d d Where is the transmission phase in radians and is the angular frequency inradians per second. From this definition, we can conclude that a linear phase(in respect to frequency) is represented by a constant group delay. The groupdelay is the slope of the graph ofS21(phase) as a function of Passive FiltersA passivefilter is one which can be made of inductors and capacitors.

3 In theButterworth and Chebyshev cases, the total number of capacitors and induc-tors is equal to the highest power of frequency in the frequency polynomial,and gives us the order of thefilter. In the Ellipticfilter case, the number ofcapacitors indicates the order of thefilter. The lumped elements values havebeen computed and tabulated for eachfilter type for the normalized frequency c=1radsecand source and load impedances ofZin=ZLoad=1 . Insertion Loss MethodIdealfilter would have no insertion loss and a linear phase response in thepassband, an infinite attenuation in the stopband and matched at the input andBACKGROUND THEORY7output. It is impossible to practically built such afilter, therefore compromisesmust be made.

4 design by insertion loss method, allows a high degree of controlover thefilter width, stopband slope and phase characteristic. Depend on theapplication, the necessary trade offdesign can be response is defined by the Power Loss Ratio (PLR) method:PLR( )=Power available from sourcePower delivered to load= | ( )|2( )Where ( )is the reflection coefficient looking into the input of thefilternetwork. Pay attention that the PLR can be described as1/|S12|2,under theassumption that thefilter input and output are matched. If| ( )|2is an evenfunction of ,it can be expressed as:| ( )|2=M( 2)M( 2)+N( 2)( )WhereMandNare real polynomials of the equation( ) in equation ( ) yields:PLR=1+M( 2)N( 2)( ) Butterworth Filter TheoryAnother name of the Butterworth Filter is maximallyflat magnitude has a maximallyflat (has no ripples) Filter response.

5 Butterworthfilter transfer function contains only poles. The Butterworthfilter has a morelinear phase response in the passband than the Chebyshev and PLR of the low passfilter is specified by:PLR=1+k2 c 2N( )WhereNis the order of thefilter. c- The cutofffrequency of frequency = c, which is at the edge of the passband, the PLR is equalto1+ ,this point is the 3dBpoint .Figure 1 shows the PLRof a Butterworth LPF as a function of the normalized frequency for 1 - Butterworth PLR. Blue line - Order N=1. Pink line - Order N= line - Order N= can be seen from Figure 1 that for higher orders, the attenuation out-side the passband is higher, which means that thefilter frequency response attenuation outside of the passband is increasing monotonically withfrequency for > c.

6 The rate of the increasing of the insertion loss out sidethe passband is20N Filter THEORY9-70-60-50-40-30-20-1000246810 Normalized frequencyINsertion(dB)Figure 2 - Butterworth low passfilter, theoretical response for N=1 (upperblue line), N=2 (midle pink line), N=3 (lower red line).Table 1 contains the elements values for Butterworth LPF for : Elements values for Butterworth Chebyshev Filter TheoryChebyshevfilters have a narrower transition region between the passband andthe stopband and more passband ripple (type I) or stopband ripple (type II)than the poles can be derived by moving the poles of the normalizedButterworth low-pass transfer function to the right, by multiplying the realparts of the poles positions by a constantKrand the imaginary parts by aconstantKj,where bothK0s are smaller than 1.

7 The poles would lie on an10 BACKGROUND THEORY ellipse of the unit circle. That means that like Butterworthfilters, Chebyshevfilters contain only poles. However, while the poles of the Butterworthfilterlie on a circle in the s-plane, those of the Chebyshevfilter lie on an Chebyshev phase response exhibits more linearity than the Elliptic oneand less linearity than the Butterworth one. The insertion loss of an N orderChebychev LPF is:PLR=1+k2T2N c ( )WhereTN(x)=xN N2 xN 2 1 x2 + N4 xN 4 1 x2 2 ( )TN(x)is the Nthorder Chebyshev polynom wherex= / c. Chebyshevpolynom result in a sharp sloop of thefilter response outside of the passbandand a ripple of 1+k2of the polynomTN(x)is oscillating between 1for|x| 1(the passband region).

8 The amplitude of the ripple is determinedbyk2. Like in the case of the Butterworthfilter, the response for thePLRisincreasing by at least 20 NdB/decade. Figure 3 shows the insertion loss of aChebychev 3dB ripple LPF for different orders:Figure 3 - Chebyshev Low Pass Filter response for 3 dB ripple, orders 1 to 4 shows the PLR of Chebychev and Butterworthfilters for N=3:CHEBYSHEV Filter THEORY11 Figure 4 - Butterworth and Chebyshev PLR for N= Chebyshev LPF ImplementationFor a Chebyshev LPF with a normalized cutofffrequency c=1and a nor-malized unity source impedance, we will derive the normalized elements valuesof the inductor, L, and capacitor, C. Figure 4 shows the structure of a secondorder (N=2) Chebyshev 4 - Low passfilter structure for N= THEORYThe PLR is:PLR=1+k2T2N( )=11 | ( )|2( )In order tofind ( ),wehavetocalculateZin(see Figure 4):Zin=R||1j C+j L=R1j CR+1j C+j L=j L+R(1 jR C)1+ 2R2C2 Where ( )=ZL Z0ZL+Z0( )By inserting equation ( ) in equation ( ), the PLR becomes:PLR=11 Zin 1 Zin+1 2=11 Zin 1 Zin+1 Z in 1Z in+1 =|Zin+1|22(Zin+Z in)( )Taking the real and imaginary parts of equation ( ), we get:Zin+Z in=2R1+ 2R2C2( )|Zin+1|2= R1+ 2R2C2+1 2+ L R2C1+ 2R2C2 2 Therefore.

9 PLR=1+14R (1 R)2+ 2 R2C2+L2 2 LCR2 +L2R2C2 4 ( )By recalling that the second order of a chebyshev polynom isT2(x)=2x2 1, we get:PLR=1+k2T22( )=1+k2 4 4 4 2+1 ( )By equating ( ) and ( ) we get:1+k2 4 4 4 2+1 =1+14R (1 R)2+ 2 R2C2+L2 2 LCR2 +L2R2C2 4 ( )Iftherippleisknown,onecansolvetheequati onat =0for R and get:R=2k2+1+2kp(k2+k)( )Equating the coefficients of 4and 2yield:CHEBYSHEV Filter THEORY134k2=14RL2C2R2( ) 4k2=14R C2R2+L2 2 LCR2 One can use the equations to obtain the values of the capacitor and inductorof the second order Chebyshev LPF. Thus constructing the famous table ofMatthaei, Young and Jones for Chebychev 3dB ripple in the passband. Thenormalized elements values are given in Table-2 : Elements values for 3dB equal ripple Chebychev Elliptic Filter TheoryAnother name for the Elliptic Filter is Cauer Filter .

10 ComparedwithBut-terworth and Chebyshevfilters, Ellipticfilters have the most rapid transition(narrow transition band). However, this does not come without a price. Ellip-ticfilters have a ripple in both the passband and stopband. This is the resultofapole-zeroconfiguration which consists of both poles and zeros. An El-lipticfilter is notorious for introducing large phase distortions, especially nearthe edge of the pass-band where the sharp amplitude characteristic implies astrongly non-linear phase PLR of the low passfilter is specified by:PLR=1+k2Zn2 c 2N( )WhereZn(x)is the Nthorder Elliptic an odd order,m=(N 1)/2andZn(x)is:Zn(x)=x(a22 x2)(a24 x2) (a2m x2)(1 a22x2)(1 a24x2) (1 a2mx2)For an even order,m=N/2andZn(x)is:Zn(x)=(a22 x2)(a24 x2) (a2m x2)(1 a22x2)(1 a24x2) (1 a2mx2)14 BACKGROUND THEORYThe zeros ofZnarea2,a4.