Confusion in CT Fourier Transform Proof

Question

I am confused trying to understand the Proof of Fourier Transform from Oppenheim book Signals and Systems. I am pasting the equations directly from the book:

$$\widetilde{x}(t)=\sum_{k=-\infty}^{+\infty}a_ke^{jk\omega_0t} \tag{4.3} $$ $$a_k=\frac{1}{T}\int_{-T/2}^{T/2}\widetilde{x}(t)e^{-jk\omega_0t}dt \tag{4.4} $$

where $\omega_0=2\pi/T$. Since $\widetilde{x}(t)=x(t)$ for $|t| < T/2$, and also, since $x(t)=0$ outside this interval, eq. $(4.4)$ can be rewritten as

$$a_k=\frac{1}{T}\int_{-T/2}^{T/2}x(t)e^{-jk\omega_0t}dt=\frac{1}{T}\int_{-\infty}^{+\infty}x(t)e^{-jk\omega_0t}dt$$

1. My question is In the last equation on the L.H.S, there is no $\widetilde{x}(t)$ but it was mentioned in the eq. $(4.4)$
2. And then we have: $$Ta_k= \left.\frac{2 \sin(\omega T_1)}{\omega}\right\rvert_{\omega = k\omega_0}$$ That is, with $\omega$ thought of as a continuous variable, the function ($2\sin\omega T_1/ \omega$) represents envelope of $Ta_k$. Here I want to know what is meant by envelope?I want to add more details from the book exactly " Specifically we have plot $T_0a_k$ rather than $a_k$ and we have also modified the horizontal spacing in each plot. here is the plot " c) I want to know that how $T_0a_k$ will be drawn in continuous?

Answer 1

@robertbristow-johnson can you please explain exactly the answer?

oh dear. it's a lot of work and it should be in some textbook.

okay, using your O&S notation:

$$ \tilde{x}(t) = \sum\limits_{k=-\infty}^{+\infty} a_k \ e^{j k \omega_0 t} $$

where

$$ \tilde{x}(t) = \tilde{x}(t+T) \quad \forall t $$

$$ \omega_0 \triangleq \frac{2 \pi}{T} $$

and

$$ a_k = \frac{1}{T} \int\limits_{t_0}^{t_0 + T} \tilde{x}(t) \ e^{-j k \omega_0 t} \ dt $$

and where $t_0$ can be any real value: $-\infty < t_0 < +\infty$

we choose $t_0$ to be $t_0 = -\frac{T}{2}$ so that

$$ a_k = \frac{1}{T} \int\limits_{-\frac{T}{2}}^{+\frac{T}{2}} \tilde{x}(t) \ e^{-j k \omega_0 t} \ dt $$

That's Fourier Series.

A Riemann sum is a sum of areas of rectangles that, for decently well-behaved functions, converges to the area of an integral in the limit that the rectangles get thinner and thinner:

$$ \int\limits_{t_1}^{t_2} f(t) \ dt = \lim_{N \to +\infty} \sum\limits_{n=0}^{N-1} f\left(t_1 + n\frac{t_2 - t_1}{N} \right) \ \frac{t_2 - t_1}{N} $$

for $t_1 < t_2$.

this (Riemann integration) works for most ordinary functions that are decently well-behaved. for nastier functions, another method of defining integration may have to be used, but we're not dealing with that here.

so do you understand Fourier series and Riemann integration?

now, define

$$ \tilde{x}(t) = \begin{cases} x(t), & \text{if } |t| < \frac{T}{2} \\ x(t - mT), & \text{if } |t| > \frac{T}{2} \text{ and where } m \triangleq \left\lfloor \frac{t}{T} + \frac{1}{2} \right\rfloor \end{cases} $$

$\lfloor u \rfloor$ is the $\operatorname{floor}(u)$ function that is equal to the largest integer no larger the argument $u$. (it's always rounding down.) $\left\lfloor u + \frac{1}{2} \right\rfloor$ is always rounding to nearest.

you can see that $\tilde{x}(t)$ satisfies periodicity: $ \tilde{x}(t) = \tilde{x}(t+T) \quad \forall t $. and you can also see that

$$ x(t) = \lim_{T\to +\infty} \tilde{x}(t) \quad \forall t $$

so we're taking a finite section of $x(t)$ and periodically extending that to get $\tilde{x}(t)$. but as the period gets longer and longer and goes to $\infty$ the periodic extension becomes the same as $x(t)$.

now if you plug in $a_k$ into the Fourier sum:

$$ \tilde{x}(t) = \sum\limits_{k=-\infty}^{+\infty} \frac{1}{T} \int\limits_{-\frac{T}{2}}^{+\frac{T}{2}} \tilde{x}(u) \ e^{-j k \omega_0 u} \ du \ e^{j k \omega_0 t} $$

and let $T \to +\infty$ ...

okay, i got back to this. (sorry for the delay.)

so we've been saying

$$ \begin{align} x(t) & = \lim_{T \to \infty} \tilde{x}(t) \\ & = \lim_{T \to \infty} \sum\limits_{k=-\infty}^{+\infty} \frac{1}{T} \int\limits_{-\frac{T}{2}}^{+\frac{T}{2}} \tilde{x}(u) \ e^{-j k \omega_0 u} \ du \ e^{j k \omega_0 t} \\ & = \lim_{T \to \infty} \sum\limits_{k=-\infty}^{+\infty} \frac{1}{T} \int\limits_{-\frac{T}{2}}^{+\frac{T}{2}} x(u) \ e^{-j k \omega_0 u} \ du \ e^{j k \omega_0 t} \\ \end{align} $$

(note the subtle difference between the last two lines. why can we do that?)

and we're saying

$$ \begin{align} \int\limits_{-\frac{T}{2}}^{+\frac{T}{2}} x(u) \ e^{-j k \omega_0 u} \ du & = \lim_{N \to +\infty} \sum\limits_{n=0}^{N-1} x\left(-\frac{T}{2} + n\frac{+\frac{T}{2} - (-\frac{T}{2})}{N} \right) \ e^{-j k \omega_0 \left(-\frac{T}{2} + n\frac{+\frac{T}{2} - (-\frac{T}{2})}{N} \right)} \ \frac{+\frac{T}{2} - (-\frac{T}{2})}{N} \\ & = \lim_{N \to +\infty} \sum\limits_{n=0}^{N-1} x\left(-\frac{T}{2} + n\frac{T}{N} \right) \ e^{-j k \omega_0 \left(-\frac{T}{2} + n\frac{T}{N} \right)} \ \frac{T}{N} \\ \end{align} $$

putting the two limits together:

$$ \begin{align} x(t) & = \lim_{T \to \infty} \sum\limits_{k=-\infty}^{+\infty} \frac{1}{T} \int\limits_{-\frac{T}{2}}^{+\frac{T}{2}} x(u) \ e^{-j k \omega_0 u} \ du \ e^{j k \omega_0 t} \\ & = \lim_{T \to \infty} \sum\limits_{k=-\infty}^{+\infty} \frac{1}{T} \lim_{N \to +\infty} \sum\limits_{n=0}^{N-1} x\left(-\frac{T}{2} + n\frac{T}{N} \right) \ e^{-j k \omega_0 \left(-\frac{T}{2} + n\frac{T}{N} \right)} \ \frac{T}{N} \ e^{j k \omega_0 t} \\ & = \lim_{T \to \infty} \lim_{N \to +\infty} \sum\limits_{k=-\infty}^{+\infty} \sum\limits_{n=0}^{N-1} x\left(-\frac{T}{2} + n\frac{T}{N} \right) \ e^{-j k \omega_0 \left(-\frac{T}{2} + n\frac{T}{N} \right)} \ \frac{T}{N} \ e^{j k \omega_0 t} \frac{1}{T} \\ & = \lim_{N \to +\infty} \sum\limits_{n=0}^{N-1} \lim_{T \to \infty} \sum\limits_{k=-\infty}^{+\infty} x\left(-\frac{T}{2} + n\frac{T}{N} \right) \ e^{-j k \omega_0 \left(-\frac{T}{2} + n\frac{T}{N} \right)} \ \frac{1}{N} \ e^{j k \omega_0 t} \\ \end{align} $$

(answer not done yet, still being developed.)