1. Fourier analysis

Was given this proof to do for a tutorial:

Prove that the non-zero frequency spectral lines of the periodic unipolar square-wave pulse train, figure 1 where V = 1 volt, are indeed at the same amplitude as the amplitude of the Fourier Transform of a single pulse from within the pulse train.

2. anyone??

3. Originally Posted by Yaris
Was given this proof to do for a tutorial:

Prove that the non-zero frequency spectral lines of the periodic unipolar square-wave pulse train, figure 1 where V = 1 volt, are indeed at the same amplitude as the amplitude of the Fourier Transform of a single pulse from within the pulse train.

Depending how you define things, the n-th harmonic of the square wave
(I will assume unit amplitude for the square wave
multiplying by a positive constant has no effect on what follows - other
than increasing every thing by a constant factor that is )
has complex amplitude :

$
A_n=\int_0^2 \chi_{[0,1]}(t) \exp(2 \pi\ \bold{i}\ n f_0 t) dt
$
,

where $\chi_{[0,1]}$ is the characteristic function of the interval
[0,1].

So:

$
A_n=\int_0^1 \exp(2 \pi\ \bold{i}\ n f_0 t) dt
$
,

and this corresponds to a nominal frequency $nf_0$, where $f_0$ is 1/2 Hz.

Now the Ft of a single pusle (starting at t=0) is:

$
\mathcal{F}\chi_{[0,1]}(f)=\int_{-infty}^{infty} \chi_{[0,1]}(t) \exp(2 \pi\ \bold{i}\ f t) dt
$
.

So

$
\mathcal{F}\chi_{[0,1]}(f)=\int_0^1 \exp(2 \pi\ \bold{i}\ f t) dt
$
.

Hence:

$
\mathcal{F}\chi_{[0,1]}(n f_0)=A_n
$

Now this is for the single pulse starting from t=0, if it starts elsewhere we
have a translation which corresponds to multiplying the Ft by a complex
number of unit modulus so the moduli of the amplitudes are still equal.

RonL

Note I say depending on how you define things because the definition of
what is the forward and backward transform varies from author to author,
as does where one places the normalising constants and so does the
convention of angular frequency of normal frequency.