You don't seem to have anywhere in your expressions.
i've used the formula
An=(2/T) int f(t) cos (N2pit/2)dt
This is the same as romsek as I seem to under stand that one better. I have come out with
For int(3t+3)cos(ntpi/2) between -1 & 0
I come out with -12(cos(pi*n/2)-1)/(pi^2*n^2)
Please tell me this is correct. I have been round the houses with this one.
that what I got in the end i think.
An = ((16πnsin(3πn/2)+ (12cos(3πn/2))-(16cos(tn/2)-8)) / π^2 * n^2
Bn= 12sin(3πn/2) - 16πncos(3πn/2) – 4sin(πn/2) / π^2 * n^2
The only difference between my answers and yours i can see is the An you got 4 and i got 8. Slightly different from you beg formula one i used is 2/T which is 2/4, you have put 1/4, not sure which one is correct.
Next step i guess is to sketch in the frequency domain to show that the wave is out between 600 & 800hz?
regards
Ah thanks that clears it up, I think i actually am starting to understand this. Apart from now i need the next bit.
Find the amplitude at the fundamental frequency and at the next six higher harmonics.
Make a plot of this waveform in the frequency domain, showing amplitude against frequency in Hertz.
Can you give me an idea of how to attempt this bit?
regards
the sheet I posted has all this.
Having evaluated the general formula for the sine and cosine series in terms of $n$ simply plug in $n$.
The fundamental frequency is $n=0$
the final table has $\omega_n$ in KHz, the a's and b's, and then the magnitude
I guess you should divide the $\omega$'s by $2\pi$ to get the frequency. Sorry about that.
your period is 4ms
so the fundamental frequency is $f_0 = 1/T = 0.25 KHz$
the left column of the table is $2 \pi n f_0$ where $f_0$ is in KHz.
I should have just made this $n f_0$. I apologize.
the second column is $a_n$, the third $b_n$, the 4th $\sqrt{a_n^2 + b_n^2}$ which is the magnitude of the frequency response at $n f_0$
Thanks for that. Last bit of the question is,
Does the above waveform comply with this restriction ?
If not , can you suggest a way of circumventing this problem ?
I'm guessing as when n=0 then frequency will be 0, i assume this means that amp will be zero. But how do i accuratly plot this to confirm out of tollerance?
I know there is a problem and you have to increase the time range by 2.
I'm guessing i need to just increase all the ranges i.e -2<t<0, 0<t<2, 2<t<6? and do the same as you have to get the results.
regards
so from what you are saying looking at you results
freq Amp
0.25
0.5 2.55453
0.7 1.75197
1 1.27324
So between 500 and 1000hz the value for the amp is between 2.55453 & 1.27324 therefore doesnt comply with the restriction?.
regards