how to determine fundamental freq and plot this equation
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for last one hour i am stcuk with this equation
x(t) = (summation sign k=-10 to k=10) k^2 * e^(j*6*k*pi*t)
How to sketch its spectrum and calculate its fundamental frequency and period
Plz help me. Give some guideline
Should i open this equation from -10 to +10 by hand and calculate all values..
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Akzeptierte Antwort
Wayne King
am 1 Okt. 2011
Yes, Think about exp(1j*2*pi*k*3*t) what value of T exists such that
exp(1j*2*pi*k*3*(t+T))= exp(1j*2*pi*k*3*t)
For that to happen:
exp(1j*2*pi*k*3*T)=1
The answer will depend on k and the smallest positive value of k gives you the fundamental frequency.
Confirm your math with:
Fs = 100;
t = (0:1/Fs:2-(1/Fs))';
X = zeros(200,21);
for k = -10:10
X(:,k+11) = k^2.*exp(1j*2*pi*k*3*t);
end
y = sum(X,2);
plot(t,real(y)); grid on;
Weitere Antworten (7)
Wayne King
am 1 Okt. 2011
Hi , One way. I'll just assume a sampling frequency of 100.
Fs = 100;
t = (0:1/Fs:1-(1/Fs))';
X = zeros(100,21);
for k = -10:10
X(:,k+11) = k^2.*exp(1j*2*pi*k*3*t);
end
y = sum(X,2);
1 Kommentar
moonman
am 1 Okt. 2011
2 Kommentare
Wayne King
am 1 Okt. 2011
what value of T makes exp(1j*2*pi*k*3*T)=1
think about exp(1j*theta) when is that equal to 1+j0 ? for what values of theta is that equal to 1+j0
moonman
am 1 Okt. 2011
3 Kommentare
Wayne King
am 1 Okt. 2011
Now look at the plot.
Fs = 100;
t = (0:1/Fs:2-(1/Fs))';
X = zeros(200,21);
for k = -10:10
X(:,k+11) = k^2.*exp(1j*2*pi*k*3*t);
end
y = sum(X,2);
plot(t,real(y)); grid on;
set(gca,'xtick',[1/3 2/3 3/3 4/3 5/3])
You see :)
moonman
am 1 Okt. 2011
5 Kommentare
Wayne King
am 1 Okt. 2011
yes. The highest frequency is exp(1j*2*pi*3*10), when k=10. The bandwidth is 30-(-30)=60, so the Nyquist rate is 60 samples/second.
Wayne King
am 1 Okt. 2011
I meant yes to your first question, not the question on the Nyquist rate ,the Nyquist rate is 60 samples/second.
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