how to determine fundamental freq and plot this equation

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..

Answer 1

[Wayne King]

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;

[bym]

+1 vote for Wayne's patience

Answer 2

[Wayne King]

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);

[Wayne King]

Then just

y = sum(X,2);

Answer 3

[moonman]

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?

Answer 4

[moonman]

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*

[Wayne King]

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]

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

Answer 5

[moonman]

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?

[Wayne King]

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]

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]

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 :)

Answer 6

[moonman]

King i m waiting for u....

[Wayne King]

you're right, look at the plot with the axis tick labels set.

Answer 7

[moonman]

so the period is 1/3. am i right

What is fundamental frequency

[moonman]

The defination of fundamental frequncy says that it is such frequency where the frequencies are all integer multiple of common frequency

[Wayne King]

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]

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]

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]

I meant yes to your first question, not the question on the Nyquist rate ,the Nyquist rate is 60 samples/second.

Answer 8

[moonman]

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