## Compression property of fourier transformation and bandwidth of a signal.

When we record a speech signal $x(t)$ for 10 minutes, the intensity of the speech signal is distributed more or less uniformly throughout the 10-minute interval. But, if we take the FT of $x(t)$, the result will not be uniform. What we will observe is that the signal is highly concentrated around 0Hz, then drop quickly in intensity towards higher frequencies. This is a result of the Riemann-Lebesgue Lemma. The lemma says that

\begin{eqnarray}

\lim_{\omega \rightarrow \pm \infty} \int_{-\infty}^{\infty} x(t) \sin(\omega t) dt = 0\\

\lim_{\omega \rightarrow \pm \infty} \int_{-\infty}^{\infty} x(t) \cos(\omega t) dt = 0

\end{eqnarray}

If we multiply the first equation with $-i$, add it to second equation, and invoke the euler’s law $\cos(\omega t)-i\sin(\omega t)=\cos(-\omega t)+i\sin(-\omega t) = e^{-i \omega t}$ we arrive at

\begin{eqnarray}

\lim_{\omega \rightarrow \pm \infty} \int_{-\infty}^{\infty} x(t) e^{-i \omega t} dt = \lim_{\omega \rightarrow \pm \infty }X(\omega) =0

\end{eqnarray}

which proves the claim.

This compression property guarantees that most real life signals have a “bandwidth”, ie, their fourier transforms are zero outside of an interval $[-BW,BW]$. This property is very useful in data compression. For example, what is known as .jpeg format is nothing but discrete cosine transformation, which is a close relative of fourier transformation. See https://www.youtube.com/watch?v=Q2aEzeMDHMA , https://www.youtube.com/watch?v=0me3guauqOU and many other related videos on youtube.

**FILTERING**

Filtering means elimination of some frequencies from a given signal. There are four different ways of filtering:

- Lowpass
- High-pass
- Bandpass
- Band-reject

We will only explain lowpass filtering and bandpass filtering.

**Lowpass filtering**

Lowpass filtering corresponds to the reduction elimination of high-frequency information contained in a signal $x(t)$. All lowpass filters has a quantity known as Bandwith (BW). The filter reduces or eliminates the frequencies above its bandwidth..

In this lecture, we will only be interested with sharp filters, ie, filters which pass all the frequencies below the BW, unattenuated, and zeroes out all the frequencies above the BW. Naturally, the low pas filter will be of the form

\begin{eqnarray}

\chi_{[-BW,BW]} (\omega)

\end{eqnarray}

in the frequency domain. Lowpass-filtering a given signal $x(t)$ involves three steps:

- Take the fourier transform of the signal, $x(t) \rightarrow X(\omega)$ to go from time domain to frequency domain.
- Multiply $X(\omega)$ with the lowpass filter to find the filtered signal $F(\omega) = \chi_{[-BW,BW]} (\omega) X(\omega)$:
- Return the filtered signal back to time domain via an inverse fourier transformation: $f(t) = IFT[ F(\omega) ] $.

Or, in the last chapter, we have derived

\begin{eqnarray}

FT[ \frac{\sin(t K)}{\pi t} ] = \chi_{(-K,K)}(\omega)

\end{eqnarray}

via duality. We also know that multiplication in time is equivalent to convolution in frequency. Therefore the same lowpass signal can be obtained solely in time domain via convolution, without going into frequency domain:

\begin{eqnarray}

f(t) = x(t) * \frac{\sin(t BW)}{\pi t}

\end{eqnarray}

Here $\frac{\sin(t BW)}{\pi t}$ is the time domain representation of the lowpass filter.

**Bandpass filtering**

Bandpass filtering is the elimination of all frequencies in a signal $x(t)$ except the ones occuring in the frequency interval $[-H,-L] \cup [L,H]$. Naturally, in the frequency domain, a bandpass filter will look like as

\begin{eqnarray}

\chi_{[-H,-L]} (\omega)+\chi_{[L,H]} (\omega)

\end{eqnarray}

As in lowpass filtering, bandpass-filtering a given signal $x(t)$ involves three steps:

- Take the fourier transform of the signal, $x(t) \rightarrow X(\omega)$ to go from time domain to frequency domain.
- Multiply $X(\omega)$ with the bandpass filter to find the filtered signal $F(\omega) = (\chi_{[-H,-L]} (\omega)+\chi_{[L,H]}(\omega) )X(\omega)$:
- Return the filtered signal back to time domain via an inverse fourier transformation: $f(t) = IFT[ F(\omega) ] $.

Doing all these without leaving time domain is also possible..

**MODULATION**

Consider a signal $x(t)$ which is centered at 0 Hz, and has a bandwith $B$. In other words, $X(\omega)$ lives in the interval $[-B,B]$. Is it possible to change the center frequency of this signal to $f$, $f>>B$? In other words, is it possible to transfer this signal to the frequency interval to $[-B-f, B-f] \cup [f-B,f+B]$ without loosing any information? The answer is yes, and this process is known as modulation.

Before discussing modulation, let us remember a property of convolution: Convolution with a dirac delta simply translates the signal.

\begin{eqnarray}

x(t) * \delta(t-t_0) = \int_{-\infty}^{\infty} x(t-s) \delta(s-t_0)ds = x(t-t_0)

\end{eqnarray}

modulation is simply achieved by multiplying the signal with $\cos(ft)$

\begin{eqnarray}

m(t) = x(t) \cos(ft)

\end{eqnarray}

where $m(t)$ denotes the modulated signal. To see this, remember the “modulation property” from the previous chapter, ie, multiplication in time domain is convolution in frequency domain. Hence

\begin{eqnarray}

FT[m(t)] &=& FT[x(t) \cos(ft)]\\

&=& \frac{1}{2\pi}FT[x(t)]*FT[ \cos(ft)]\\

&=& \frac{1}{2\pi}X(\omega)*\pi( \delta(\omega-f) + \delta(\omega+f) )\\

&=& \frac{1}{2}(X(\omega-f)+X(\omega+f))

\end{eqnarray}

Using this technique,

- we can put many speakers on the same telephone wire,
- we can listen to many different radio stations in the same city.

All we have to do id to assign a modulation signal $f_i$ to speaker $i$ (or radio station $i$) and guarantee that $f_i$ is at least $2B$ apart from its nearest neighbour $f_k$, where $B$ is the bandwidth. In case of the speech signal, $B$ is usually 40Khz. This results in a single time-domain signal on the wire or air

\begin{eqnarray}

m(t) &=& \sum_{i=1}^N m_i(t) = \sum_{i=1}^N \cos(f_i t) x_i(t)\\

M(\omega) &=& \frac{1}{2} \sum_{i=1}^N (X_i(\omega-f_i)+X_i(\omega+f_i))

\end{eqnarray}

where there are $N$ speakers (or radio stations), each generating a signal $x_i(t)$ to be transmitted.. Thanks to modulation, all these signals “live in the belly” of the composite signal $m(t)$ without mixing into each other.

For a second, just think how marvelous this process is. If I sum two numbers, I got the total and lost all information about the individual numbers. If I have 2 and 7 I sum them and get 2+7=9. But if I have 9, I cannot know whether I got it bu summing 2 and 7, 1 and 6, or 4 and 3. Modulation lets us get around this limitation in signal space. We can sum many individual signals and get a single signal, but out of this single signal we can get again the individual signals whenever we need..

**DEMODULATION**

The process of extracting the signal $x_i(t)$ from the signal $m(t)$ is called demodulation. The first step is to “filter” the composite signal $m(t)$ with a bandpass filter centered at $f_i$ and having bandwidth $B$ to extract $m_i(t)$:

\begin{eqnarray}

m(\omega) \left( \chi_{(-f_i-B,-f_i+B)}(\omega) + \chi_{(f_i-B,f_i+B)}(\omega) \right) = \frac{1}{2} (X_i(\omega-f_i)+X_i(\omega+f_i)) = m_i(t)

\end{eqnarray}

Basically each radio set does this: It receives the composite signal $m(t)$ at its antenna and filters it out to get the $m_i(t)$ of the desired station. Now the problem is to generate the signal $x_i(t)$ from $m_i(t)$.