how to determine fundamental freq and plot this equation

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moonman
moonman 2011-10-1
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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采纳的回答

Wayne King
Wayne King 2011-10-1
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;

更多回答（7 个）

Wayne King
Wayne King 2011-10-1
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);
moonman
moonman 2011-10-1
Can u tell me the fundamental frequency and period of this signal
Can i know fundamental freq and period of this signal without using matlab?
moonman
moonman 2011-10-1
I am badly stuck Can u explain me the solution
what u mean by
*For that to happen:
exp(1j*2*pi*k*3*T)=1*
  2 个评论
Wayne King
Wayne King 2011-10-1
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
moonman 2011-10-1
When the value of theta will be zero, at that time
exp(1j*2*pi*k*3*T)=1
and value of theta can be zero when k=0 ot T=0

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moonman
moonman 2011-10-1
When the value of theta will be zero, at that time exp(1j*2*pi*k*3*T)=1 and value of theta can be zero when k=0 ot T=0
ami right?
  3 个评论
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moonman
moonman 2011-10-1
I am lost and confused
The answer which is coming in my mind is that T should be 1/3
to satisfy the condition
Wayne King
Wayne King 2011-10-1
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 :)

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moonman
moonman 2011-10-1
King i m waiting for u....
moonman
moonman 2011-10-1
so the period is 1/3. am i right
What is fundamental frequency
  5 个评论
显示 3更早的评论
Wayne King
Wayne King 2011-10-1
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
Wayne King 2011-10-1
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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moonman
moonman 2011-10-2
Thanks Wayne King for taking me all along My problem is solved and u also taught me

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