Hi,
I wish I can reproduce the signal by drawing a little diagram to make
this solution a little easier. Since you have explicitly asked that I
only help you start with the calculation for the problem, I will
present a little initial analysis which will be helpful for you to
work this out.
The wave form is periodic ( the problem mentions repetitive rate of
40us ), hence the Fourier series expansion for a standard periodic
function holds good here.
Step 1:
The Fourier series for the given signal would be,
If Fourier series is defined for a period from c to c+2*pi.
f(x) = A0/2 + (summation)(n=1 to infinity)[An*Cos(nx) +
Bn*Sin(nx)]-(equation 1)
Here the values of A0, An and Bn will give you the answers for:
A0 --> Answer for Amplitude of the fundamental.
An + jBn--> Answer for the amplitude of the nth harmonic.
A0 = (1/2*pi)(Integral)(c to c+2*pi)f(x)dx
An = (1/2*pi)(Integral)(c to c+2*pi)f(x)Cos(nx)dx
Bn = (1/2*pi)(Integral)(c to c+2*pi)f(x)Sin(nx)dx
Note 1: c here is the base which can be given any value depending on
the period of the signal.
Note 2: 'j' used above refers to the complex operator, sqrt(-1).
Step 2:
Now comes the problem of expressing your triangular signal in the form
given above.
Lets see how this can be done.
Assume that the triangular wave is positioned in such a way that its
peak coincides with the y-axis. This means that the triangular pulse
is symmetrical about the y-axis. The coordinates will then be, -1 us
and 1 us.
(I hope you can visualise this)
That done, now let us define the function for you:
The definition will look like this,
f(x) = 20(1 - |x| ) when x lies between -1 and 1
f(x) = 0 for values of x at 1< |x| <20.
I havent examined the function properly, if you find that it is not
feasible please let me clarify.
Step 3:
Now moving on with the rest of the solution.
The values of A0, An and Bn are given thus:
A0 = (1/2*pi)(Integral)(0 to 2*pi)f(x)dx
An = (1/2*pi)(Integral)(0 to 2*pi)f(x)Cos(nx)dx
Bn = (1/2*pi)(Integral)(0 to 2*pi)f(x)Sin(nx)dx
Note that I have replaced the base c with 0(zero) which is commonly
done in problems of this nature.
Step 4:
At this stage you will have to make a small transformation.
let z=pi*x/q
Now if x-> -q to +q then z-> -pi to +pi
To understand the need for this transformation, you will have to go
back to your function. It is defined as a triangular waveform, hence
you will have to perform the transformation (which anyway is linear)
so that you have the same period but it becomes easier to express it
in the time scale adopted.
Your function f(x) now becomes f(q*z/pi)
Note 1: The important point to remember here is that 'c' from step 1
has been given a value of pi, so that the limits extend from -pi to
+pi.
Step 5:
Now redefine the series in equation 1 above for this new value
replacing x by q*z/pi.
f(q*z/pi) is periodic and the coefficients will give you the answers
for the amplitude.
Step 6:
Proceed by evaluation the Fourier series from the above information,
A0 = (1/q)(Integral)(-q to +q)f(x)dx
An = (1/q)(Integral)(-q to -q)Cos(nx)f(x)dx
Bn = (1/q)(Integral)(-q to -q)Sin(nx)f(x)dx
I will stop here so that you can evaluate the rest of the result on
your own after this initial 'start'.
To find the amplitude of the 60th harmonic, put n=60 in the equations.
Final Step:
(A little note on Fourier Response)
The Fourier expansion series it self will give you the Fourier
Response of the signal. The signal being periodic will consist of a
fundamental and many harmonics. the essential point to remember though
is the fact that the harmonic frequencies are all multiples of the
fundamental harmonic. In the case of a non-periodic function, the
frequencies of the harmonic will lie all over the spectrum. Hence, the
coefficients taken in the form give below will give the Fourier
Response.
Ao -> fundamental
sqrt[An(^2) + Bn(^2)] -> harmonic amplitude.
I am sure you can work out the problem with this information. I am
glad i got to answer this question because this is one field I am
particluarly interested in.
If you are unsure about any part of this answer or will like more
information please ask for a clarification. I will be happy to provide
you with the same.
Thank you,
Regards,
Shiv Reddy |
Clarification of Answer by
shivreddy-ga
on
21 Nov 2002 11:58 PST
Hi,
I am glad you have completed the solution to this problem. I have
worked on the final results along similar lines that I have already
presented in the answer above. I must admit though that the
calculations that follows are pretty hurriedly done.
Let us solve for the fundamental amplitude first,
From the equations that I have shown above,
A0 = 1/20(Integral)(-20 to 20)20(1-|x|)dx
Now between -1 and 1 the function has a finite non zero value. The
values at the rest of 'x' is 0.
Hence,
A0 = 1/20(Integral)(-1 to 1)20(1-|x|)dx
=> (Integral)(-1 to 1)(1-|x|)dx
=> 2(Integral)(0 to 1)(1-x)dx
=> 2 ( 1 - 1/2 )
=> 1
Hence the amplitude of the Fundamental is 1 amp.
Now let us see the case of the 6oth harmonic.
A60 + jB60 = 1/20(Integral)20(-1 to 1)(1 - |x|)[cos(60*pi*x/20) +
jSin(60*pi*x/20)]dx
=> 1/20(Integral)(-1 to 1)20(1 - |x|)[cos(3*pi*x) + jSin(3*pi*x)]dx
=> (Integral)(-1 to 1)(1 - |x|)exp(3*j*pi*x)dx -------equation 1
=> (Integral)(-1 to 1)exp(3*pi*j*x)dx - (Integral)(-1 to
1)|x|exp(3*pi*j*x)dx
=> [First integral becomes zero since Sin(3*pi*anything)=0]
Put (3*pi*j*x) = t
Then dt = 3*pi*j*dx
Since x-> -1 t-> -3*pi*j and since x -> 1 t -> 3*pi*j
Moving on this way,
=> Second Integral = -1/[9*pi(squared)](Integral)(-3*pi*j to
+3*pi*j)texp(t)dt
------- equation 2
This integral can be evaluated the same way as this one:
(Differential of w.r.t. x)xexp(x) = xexp(x) + exp(x)
Note: exp(x) is the exponential x.
=> (x + 1)exp(x)
hence from the above,
(Integral)xexp(x)dx = (x - 1)exp(x)
This result can be used to solve the integral in equation 2.
You will get the final line as,
1/[9*pi(squared)][(t-1)exp(t){from -3*pi*j to 0} + (t-1)exp(t){from
0to -3*pi*j)]
Which gives 2/(9*pi*pi)
That is, the amplitude of the 60th harmonic is 2/[9*pi(squared)]
I hope your results match with this. Please post your results here and
the way you have arrived at them. I must inform you however that these
set of clarifications go beyond the scope of the question. You could
post an analysis check as a separate question and I will be happy to
work on it for you.
Thank You,
Regards,
Shiv Reddy
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