# Fourier Transform

• October 12th 2009, 10:08 PM
Forum user
Fourier Transform
[IMG]file:///Users/admin/Library/Caches/TemporaryItems/moz-screenshot.jpg[/IMG]Can anyone please help me with this question. The hint for this question is :

1a)Note that the signal is periodic. It is better to work out the spectrum of just one pulse and then use results from tutorial and laboratory experiment S1. Note also that convolving two triangular pulses gives a triangular one.

b) Do it graphically. The second waveform is also equal to the differences of two triangular pulses.

Please help me with this. I am stuck with this for very long. Thank You Very Much!!! :)
• October 12th 2009, 10:25 PM
CaptainBlack
Quote:

Originally Posted by Forum user
[IMG]file:///Users/admin/Library/Caches/TemporaryItems/moz-screenshot.jpg[/IMG]Can anyone please help me with this question. The hint for this question is :

1a)Note that the signal is periodic. It is better to work out the spectrum of just one pulse and then use results from tutorial and laboratory experiment S1. Note also that convolving two triangular pulses gives a triangular one.

b) Do it graphically. The second waveform is also equal to the differences of two triangular pulses.

Please help me with this. I am stuck with this for very long. Thank You Very Much!!! :)

Well lets write the functional form for a single pulse:

$
f(t)=\begin{cases} 2+t,&0\le t<2\\
2-t,&-2<=t<0\\
0,&\text{otherwise}\end{cases}
$

Then the Fourier transform of this is:

$
\mathcal{F}f(\omega) = \int_0^2 (2+t)e^{-i\omega t}\;dt+\int_{-2}^0 (2-t)e^{-i\omega t}\;dt
$

(you may have to adjust this slightly to whatever form of definition you are using for the FT)

You should know how to go from the FT of a single pulse to that of a periodic pulse train.

CB
• October 12th 2009, 10:31 PM
Forum user
Thank u very much!!! I will go n try it out. :)
• October 12th 2009, 10:36 PM
CaptainBlack
Quote:

Originally Posted by Forum user
Note also that convolving two triangular pulses gives a triangular one.

The convolution of two rectangular pulses is a triangular pulse, the convolution of two triangular pulses is not triangular.

CB
• October 12th 2009, 10:47 PM
Forum user
ok! noted. Thank u!
i typed wrongly. So sorry! :)
• October 13th 2009, 04:22 AM
Forum user
erm may I know why is it 2+T and 2-T?
• October 13th 2009, 06:27 AM
CaptainBlack
Quote:

Originally Posted by Forum user
erm may I know why is it 2+T and 2-T?

at t=0 the signal has value 2, and for positive t the slope is +1, and negative it is -1.

Alternativly, between 0 and 2 we have a linear function that goes from 2 to 4, and between -2 and 0 we have a linear function that goes from 4 to 2.

CB
• October 13th 2009, 07:09 AM
Forum user
oic... thank u very much! :)