how to determine fundamental freq and plot this equation

2011 10 1
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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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Wayne King
Wayne King 2011-10-1

1 个投票

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

0 个投票

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

0 个投票

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

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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*

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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

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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?

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Wayne King
Wayne King 2011-10-1
not just when theta=0 when theta = 2*k*pi, think about it. How can you tell 0 from 2*pi,4*pi, etc.
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

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King i m waiting for u....
moonman
moonman 2011-10-1

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so the period is 1/3. am i right
What is fundamental frequency

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moonman
moonman 2011-10-1
The defination of fundamental frequncy says that it is such frequency where the frequencies are all integer multiple of common frequency
Wayne King
Wayne King 2011-10-1
If the fundamental period is 1/3, the what is the fundamental frequency? It's just the reciprocal. If the whole waveform is repeating every 1/3 seconds, then there are 3 periods in a second, so the frequency is 3 cycles/second.
moonman
moonman 2011-10-1
The book says "determine the period of signal"
So it means we have to determine fundamental period, is it ok
And lastly and finally just for own curisity,
what will be the Nyquist Freq rate of this signal
Should it be 6 Hz???
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

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Thanks Wayne King for taking me all along My problem is solved and u also taught me

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