kameren
Guest
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 Posted:
Wed Dec 14, 2005 5:16 pm Post subject:
Matlab problem(pam signal) |
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sin signal frequency 10^4/(2*pi) Hz
sampling rate 8 kHz
Applied to a sample-and-hold circuit to produce a flat-topped PAM signal
with pulse duration 500us.
how do i compute the waveform of the PAM signal?
how do i compute and denote the magnitude spectrum of the PAM signal?
how do i compute the envelope?
Thus far this is what i have. i'm still adjusting the length part. hop
someone can help. thanks
clf;
fs=8000;
Ts=1/fs;
td=1.25e-5;
f0=10000/(2*pi);
t=[0:td:100*td];
y=sin(2*pi*f0*t);
axis([0 0.1 -1 1]);
plot(t,y,'r-');
hold on;
t2=[0:Ts:100*td];
y2=sin(2*pi*f0*t2);
plot(t2,y2,'b*');
y3=zeros(1,length(t));
x=(length(t)/length(t2));
for n=1:length(t2)
for m=1:(length(t)/length(t2))
y3((n-1)*floor(x)+m)=y2(n);
end
end
plot (t,y3(1:length(t)),'m' |
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Fred Marshall
Guest
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| Posted:
Thu Dec 15, 2005 9:16 am Post subject:
Re: Matlab problem(pam signal) |
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"kameren" <t.tharmarasan@gmail.com> wrote in message
news:v-OdnSnDP_Y0oz3eRVn-sg@giganews.com...
| Quote: | sin signal frequency 10^4/(2*pi) Hz
sampling rate 8 kHz
Applied to a sample-and-hold circuit to produce a flat-topped PAM signal
with pulse duration 500us.
how do i compute the waveform of the PAM signal?
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It looks like you did. The only thing that might be a bit misleading is
that the sampling frequency appears to be very close to being a multiple of
the signal frequency: 5.0266. You might want to to two things:
1) make it very much not such a multiple like relating them by a multiple of
pi.
2) make it exactly a multiple just to see what happens (understanding it is
a very special case).
| Quote: | how do i compute and denote the magnitude spectrum of the PAM signal?
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abs(fft(y))
| Quote: | how do i compute the envelope?
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Here is one, rather simple way:
r(i) = abs(y(i)); interim rectified version of the input.
s(i) =s (i-1) + k*[r(i)-s(i-1)] where k<1
s(i) = k*r(i) + (1-k)*s(i-1); the output
k is a coefficient that works to lowpass filter the result to remove the
sampling frequency.
You will recognize this as an IIR filter.
If r(i) goes to zero and stays there, then the output will decay
exponentially to zero.
If r() is a step, then the output will rise exponentially to the value of
the step.
If k is large, the exponential rise/fall will be fast.
If k is small, the exponential rise/fall will be slow.
Generally, k is set high enough to respond as rapidly to the peak values as
needed and low enough to respond slowly to the sampling frequency.
If k is too high then the sampling frequency will be evident.
Now, if you meant the "complex envelope" that's a different matter!
Fred |
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kameren
Guest
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| Posted:
Wed Dec 21, 2005 5:17 pm Post subject:
Re: Matlab problem(pam signal) |
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| Quote: | sin signal frequency 10^4/(2*pi) Hz
sampling rate 8 kHz
Applied to a sample-and-hold circuit to produce a flat-topped PAM signal
with pulse duration 500us.
how do i compute the waveform of the PAM signal?
how do i compute and denote the magnitude spectrum of the PAM signal?
how do i compute the envelope?(to see if the envelope is 20khz)
k completed the part i needed but i'm confused with the envolope.
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my final form of matlab code:
clear all
clf;
fs=8000;
Ts=1/fs;
td=1.25e-5;
f0=10000/(2*pi);
t=[0:td:100*td];
y=sin(2*pi*f0*t);
SUBPLOT(3,1,1), plot(t,y,'b'); %input signal(message)
xlabel('Time(s)');
ylabel('Amplitude')
hold;
t2=[0:Ts:100*td];
y2=sin(2*pi*f0*t2);
plot(t2,y2,'g*'); %sampled at 8kHz
y3=zeros(1,length(t));
m=1;
for n=1:101;
if t(n)== t2(m);
m=m+1;
y3(n)=y2(m-1);
else
y3(n)=y2(m-1);
end
end
plot (t,y3,'m'); %sample and hold
z=ones(1,length(t2));
%-----------------------------Getting Pulse signal at sampling frequenc
8khz------------------------------------
c=1;
for n=1:101;
if t(n)==t2(c);
z2(n)=z(c);
c=c+1;
else
z2(n)=0;
end
end
%plot (t,z2)
st=z2.*y; % resultant of input signal and sampled pulse signal
%plot(t,st)
%------------------------------Procedure to get better graphics of a Fla
top Pam Signal-------------------------
for k=1:101;
if st(k)~=0&k<101
st2(k-1)=st(k);
st2(k)=st(k);
st2(k+1)=st(k);
else
st2(100)=st(101);
st2(101)=st(101);
end
end
SUBPLOT(3,1,2),
plot (t,st2) % Pam signal
xlabel('Time(s)');
ylabel('Amplitude');
Sf= fft (st2,512); % in frequency domain
mag= abs(Sf); % magnitude of Sf
f = (0:511)*fs/length(Sf);
SUBPLOT(3,1,3),
plot(f,mag);
%-------------------------------------disregard-----------------------------------------------
%plot (t,y6)
%y9=(y6.*y);
%plot (t,y9)
%Y=fft (y3,101); %S(f)
%mag= abs (Y); %magnitude of S(f)
%f = (0:100)*fs/length(Y);
%plot(f,mag) %magnitude spectrum
%y4=(square(2*pi*8000*t));
%y5=ones(1,length(t));
%y6=(y5+y4)/2; %pulse sampling at 8000hz
%plot(t,y,'b');
%powerdB=20*log10(mag);
%plot (t,abs(hilbert(y)))
%stem(t2,y2)
%plot (abs(fft(y)))
%plot(linspace(-5*1e9,5*1e9,512),fftshift(abs(fft(Pulse,512)))) |
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