Homework Three

Look carefully at the signs in your exponential for the Fourier transform ($e^{-j2\pi ft}$) and its inverse ($e^{j2\pi ft}$)... -Brandon
Edited errors described above--Nicholas.Christman 16:07, 3 December 2009 (UTC)

October 5th, 2009, class notes (as interpreted by Nick Christman)

---

Nick Christman

The topic covered in class on October 5th was about how to deal with signals that are not periodic.

Given the following Fourier series (Equation 1), what if the signal is not periodic?

$x(t)=x(t+T)=\sum _{n=-\infty }^{\infty }\alpha _{n}e^{\frac {j2\pi nt}{T}}$ where $\alpha _{n}={\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t')e^{-{\frac {j2\pi nt'}{T}}}\,dt'$

To investigate this potential disaster, let's look at what happens as the period increases (i.e. not periodic). Essentially, as $T\rightarrow \infty$ we can say the following:

${\frac {1}{T}}$	$\rightarrow$	$\,df$
${\frac {n}{T}}$	$\rightarrow$	$\,f$
$\sum _{n=-\infty }^{\infty }{\frac {1}{T}}$	$\rightarrow$	$\int _{-\infty }^{\infty }({\mbox{ }})\,df$

With this, we get the following (Equation 2):

$x(t)=\lim _{T\to \infty }\left[\sum _{n=-\infty }^{\infty }\left({\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t')e^{-{\frac {j2\pi nt'}{T}}}\,dt'\right)e^{\frac {j2\pi nt}{T}}\right]\equiv \int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }\,x(t')e^{-j2\pi ft'}\,dt'\right)e^{j2\pi tf}\,df$

Given the above equivalence, we say the following:

$\,X(f)=\int _{-\infty }^{\infty }\,x(t')e^{j2\pi ft'}\,dt'$

Therefore, we have obtained an equation to relate the Fourier analysis of a function in the time-domain to the frequency-domain:

$\,X(f)=\int _{-\infty }^{\infty }\,x(t)e^{j2\pi ft}\,dt$	$\equiv$	$\langle x(t)|e^{j2\pi tf}\rangle$	$\,x(t){\mbox{ projected onto }}e^{j2\pi tf}$
$\,x(t)=\int _{-\infty }^{\infty }\,X(f)e^{-j2\pi ft}\,df$	$\equiv$	$\langle X(f)|e^{-j2\pi tf}\rangle$	$\,X(f){\mbox{ projected onto }}e^{-j2\pi tf}$

From this we can see that $\,x(t)$ is the inverse Laplace transform of $\,X(f)$. Similarly, $\,X(f)$ is the Laplace transform of $\,x(t)$

The next thing we did was rearranged some limits within Equation 2 (given the above similarities) to give us the following:

$\int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }\,x(t')e^{-j2\pi ft'}\,dt'\right)e^{j2\pi tf}\,df$

$\equiv$

$\int _{-\infty }^{\infty }\,x(t')\left(\int _{-\infty }^{\infty }e^{j2\pi f(t-t')}\,df\right)\,dt'$

Notice that $e^{j2\pi f(t-t')}\equiv \delta (t-t')$

Similarly,

$\int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }\,X(f')e^{j2\pi f't}\,dt'\right)e^{-j2\pi tf}\,df$

$\equiv$

$\int _{-\infty }^{\infty }\,X(f')\left(\int _{-\infty }^{\infty }e^{j2\pi t(f'-f)}\,df\right)\,dt'$

Again, notice that $e^{j2\pi f(f'-f)}\equiv \delta (f'-f)=\delta (f-f')$

This is good news for both the time-domain and frequency domain, because these integrals will only be non-zero only when $\,t=\,t'{\mbox{ and }}\,f=\,f'$ respectively.

---

Back