%Run?from?editor?Debug（F5）
%This?M?file?constructs?a?two?pole?two?zero?bandpass?filter?using?the?Z?Transform?
%transfer?function?and?analyzes?the?characteristics?of?the?filter?such?as?frequency?response?and
%phase?using?the?Matlab?filter?function?with?input?vector?x
%to?get?output?y=filter（bax）.Frequency?domain?plots?are?also?provided?to
%analyze?the?bandpass?filter.
%The?z?transform?is?just?basically?a?transformation?from?a?linear?continuous?system?to
%a?sampled?system.
%The?center?frequency?and?3dB?bandwidth（fBW）?of?the?BPF?filter?can?be?set?to?
%<100Hz?to?>1GHz?by?appropriate?setting?of?the?sampling?frequency（fs）?and
%center?frequency（fcent）?and?by?use?of?the?following?formula（BW=（1-a）/sqrt（a））?where
%BW=2*pi*fcent/fs?and?（a）?is?the?imaginary?pole?magnitudes.?To?get?0dB?of?attenuation
%at?the?center?frequency（fcent）the?poles?must?be?at?an?angle?of?2*pi*fcent/fs=pi/2.
%This?corresponds?to?+/-j?on?the?unit?circle?in?the?z?domain.?The?symbolic?math
%processor?in?Matlab?is?used?with?the?“solve“?function?to?determine?the
%value?as?shown?typed?in?the?command?window?“solve（‘2*（1-a）/sqrt（a）=BW‘）“.
%The?purpose?of?constructing?this?M?file?was?instigated?by?the?following?facts.
%I?have?some?knowledge?of?Laplace?transforms?and?the?s?plane.?Someone?mentioned?
%using?z?transforms?and?I?said?“What?is?a?z?transform?that?sounds?complicated?“.
%So?I?am?trying?to?understand?what?it‘s?all?about.?Anyway?that‘s?why.
%I?have?added?references?at?the?end?of?the?program?that?I?have?found
%helpful?and?hopefully?playing?with?the?program?will?give?you?a?better
%understanding?of?the?theory.?Always?remember?to?set?your?axis?settings?on
%plot?routines.?The?following?describes?the?BPF.?
%Bandpass?filter?2?poles?2?zeros
%H（z）=（z+1）*（z-1）/（z+jp）*（z-jp）=（z^2-1）/（z^2-（p^2）
%zeros?at?+/-1
%pole?at?+j/-j
%
%????????????????????????????????pi/2
%??????????????????????Z?plane???!+j（imag?axis）
%????????????????????????????????x
%?????????????????????fs/2???????!
%2?poles2?zeros???（-1）0---------!---------（+1）real?axis
%??????????????????????pi????????!
%????????????????????????????????x
%????????????????????????????????!-j
%
%
%==================================================
%Design?example
%==================================================
clear
fcent=6500e6;%center?frequency?of?bandpass?filter
fBW=1000e6;%3dB?bandwidth
fs=26e9;%sampling?frequency（set?~4?times?fcent）
%f/fs=.5set?f=.5*fs
f=[0:1e6:13e9];%keep?interval?steps?high?for?lower?
%computer?processing?time
Fn=fs/2;%nyquist?frequency-used?in?fft?plots
w=2*pi*f/fs;%set?so?w=pi?or?3.14
z=exp（w*j）;%z?definition
BW=2*pi*fBW/fs;
%solve（‘2*（1-a）/sqrt（a）=BW‘）%-solving?for?a?gives?a=.8862
%One?could?also?use?the?symbolic?math?processor?and?construct?a?graph?to
%solve?for?（a）.
%type?in?command?window?the?following:
%syms?BW?a
%BW=（2*（1-a）/sqrt（a））
%ezplot（BW）axis（[0?1?0?1]）ylabel（‘BW‘）grid?on
a=.8862;
p=（j^2*a^2）;%multiply?（j*a）*（-j*a）
gain=.105;%set?gain?for?un

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件???????1093??2006-03-10?16:10??Nyquist_filter.m

?????文件???????6268??2006-03-10?16:23??BPF?via?Z.m

-----------?---------??----------?-----??----

?????????????????7579????????????????????3