Hi,
I will present a little initial analysis which will be helpful for
you to
work this out. Like I have already said the problem is a little
unwieldy and will take time to complete in one sitting.
The wave form is periodic ( the problem mentions repetitive rate of
4us ), hence the Fourier series expansion for a standard periodic
function holds good here.
Note: 'us' here refers to Microseconds
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 rectangular signal in the
form
given above.
Lets see how this can be done.
Let the signal be along the x-axis in such a way that it extends from
-0.1 us to + 0.1 us
(I hope you can visualise this)
That done, now let me define the function for you:
The definition will look like this,
f(x) = 15 when x lies between -0.1 and +0.1
f(x) = 0 for values of x at 1< |x| <4
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 rectangular 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 evaluating 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
To find the amplitude of the 90th harmonic, put n=90 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.
Hence you will have your answer.
_____________________________________________________________________________
Now I will try and begin a little bit of the solution along the steps
I have outlines earlier.
Let us solve for the fundamental amplitude first,
From the equations that I have shown above,
A0 = 1/4(Integral)(-4 to 4)15dx
Now between -0.1 and +0.1 the function has a finite non zero value.
The
values at the rest of 'x' is 0.
Hence,
A0 = 1/4(Integral)(-0.1 to+0.1)15dx
=> (Integral)(-0.1 to +0.1)(15/4)dx
=> 2(Integral)(0 to 0.1)(15/4)dx
=> 2*(1.5/4) Amp.
=> 0.75 Amp.
Hence the amplitude of the Fundamental is 1.75 amp. the frequency of
course will be 1/4 us^-1
or 0.25 MHz
Now let us see the case of the 90th harmonic.
A90 + jB90 = 1/4(Integral)(-.01 to +0.1)15[cos(90*pi*x/4) +
jSin(90*pi*x/4)]dx
=> 1/4(Integral)(-0.1 to +0.1)15[cos(22.5*pi*x) + jSin(22.5*pi*x)]dx
=> (Integral)(-0.1 to +0.1)(15/4)exp(22.5*j*pi*x)dx
-------equation 2
=> (Integral)(-0.1 to +0.1)3.75exp(22.5*pi*j*x)dx
=> Put (22.5*pi*j*x) = t
Then dt = 22.5*pi*j*dx
Since x-> -0.1 t-> -2.25*pi*j and since x -> +0.1 t -> +2.25*pi*j
(Integral )(-2.25*pi*j to +2.25*pi*j)3.75exp(t)(1/22.5*pi*j)dt
Moving on this way,
You will arrive at the A90 + jB90 format of the Amplitude of the 90th
harmonic.
I would suggest that you follow the steps I have given above and use
the formalas shown. The evaluation of independent steps are pretty
simple as it involves basic integral calculus and complex numbers.
Thank you for using this service.
Have a nice day.
Regards,
Shiv Reddy |
