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# Modifying transmit signal

by on March 11, 2007

In DSP-Proakis, I found the problem 2.56 interesting as it provides insight into delta modulation. The task is to show that,

$x(n) = \sum_{k=-\infty}^{\infty}[x(k)-x(k-1)]u(n-k)$

where
$x(n)$ is any discrete time signal and
$u(n)$ is the unit step function.

My take:
On a close look, we can see that the above equation is the convolution of $u(n)$ with the difference of $x(n)$ and $x(n-1)$, i.e. to show that
$x(n) = [x(n)-x(n-1)] \ast u(n)$

where $\ast$ is the convolution operator.

The term $[x(n)-x(n-1)]$ can be written as $[x(n) - x(n-1)] = x(n) \ast [1\ \ -1]$.
With this, the equation reduces to
$x(n) = x(n) \ast [1\ \ -1] \ast u(n)$

As the convolution of $[1 \ \ -1]$ and $u(n)$ is the unit impulse response;$[1 \ \ -1] \ast u(n) = \delta(n)$, hence the proof.

Quite interesting to see that, even if the transmitter sends only the difference between the signal $x(n)$ and the delayed version $x(n-1)$, the receiver can successfully recover the original waveform by successively accumulating the received samples.

Isn’t this is a stripped down version of delta modulation (without the quantizer block)?

Also note that, as the frequency response of $[1 \ \ -1]$ has nulls in DC and significantly attenuates the near DC frequencies, the transmitter can do some sort of spectral shaping if so desired.

However for transmissions with significant low frequency components, the suppression of DC and near DC frequencies are not desirable and out came Delta-Sigma modulation.

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