Convolution in a Continous Linear Time Invariant System

AI Thread Summary
Convolution in continuous linear time-invariant systems can be achieved by representing analog signals as sums of scaled and shifted Dirac delta functions, despite the practical limitations of unit impulses in analog systems. While digital samples approximate analog signals, they preserve features of frequencies below half the sampling frequency, necessitating anti-aliasing filters for accurate representation. The Dirac delta function, which is not a true function but a limit of functions, allows continuous waveforms to be expressed mathematically for convolution integrals. In contrast, discrete signals can be represented as sums of scaled and shifted unit impulse sequences, but this does not apply to analog signals due to their continuous nature. Understanding these concepts is crucial for accurate system response analysis in signal processing.
N.Saravanan
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Dear Experts,
For convolution to work any input signal we should be able to represent the input signal in terms of appropriately scaled and shifted unit impulses. This one holds good for discrete time system in which the input signal can be represented as sum of scaled shifted unit impulses. But is it possible to represent an analog input signal as sum of scaled and shifted unit impulse. If so how? Why I ask is unlike in discrete system for which the unit impulse has no width, the practical unit impulse in analog system has negligible width. The unit impulse signal raises to value 1 from 0 in a very short time interval and falls back to zero again. Sum of scaled and shifted unit impulses repeat this action at a faster rate. So if we represent an analog input signal by scaled and shifted unit impulses the representation is actually a signal which touches the zero axis at intermediate intervals. But the original input analog signal need not touch the zero axis. So won't the signal approximation in continuous time produce a distorted input signal. So if we convolve this distorted zero touching input signal will we get the actual response of a system to any input? Kindly please explain the concept.

Thank You,
N.Saravanan.
 
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N.Saravanan said:
But is it possible to represent an analog input signal as sum of scaled and shifted unit impulse. If so how? Why I ask is unlike in discrete system for which the unit impulse has no width, the practical unit impulse in analog system has negligible width.
Yes that's true. What you can say is that digital samples approximate an analog signal.

How approximate? Think of the Fourier series representation of the analog signal. Digital samples preserve most features of the frequencies below 1/2 the sampling frequency, and a poor job of samples above that frequency. Indeed, you may need an anti-aliasing low-pass filter to remove those frequencies before sampling.
Beyond that, the accuracy of the approximation depends on the nature of the analog signal.
 
A continuous waveform may be expressed as a sum of infinitely-many scaled and shifted Dirac delta functions, not unit impulses. A delta function is mis-named--it's not a true function but rather the limit of a series of functions such that its height is infinite height, its width zero and its area unity. You can read about how it works and how it relates to convolution integrals from Wikipedia or any of dozens of online explanations.
 
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Hi Saravanan

The dirac pulse:

$$

\delta(t) = \begin{cases} \infty & \text{ $ t = 0 $} \\ 0 & \text{Otherwise} \end{cases}
$$
And its discrete time equivalent (equivalent in the sense that it plays much of the same role as the pulse) the unit impulse sequence:

$$
\delta[ n ] = \begin{cases}1 & \text{n = 0} \\ 0 & \text{Otherwise} \end{cases}
$$

Below are the following theorems:

1. All CT and DT functions can be expressed as a sum of scaled and shifted dirac pulses and unit impulse sequences:
$$
f(t) = \displaystyle \int_{-\infty}^{\infty} f(\tau) \cdot \delta(t - \tau) \,\,\,\, \text{d}\tau
$$
and
$$
x[ n ] = \displaystyle \sum_{k \to -\infty}^{k \to \infty} x[ k ] \cdot \delta[ n - k]
$$
If its a continuous time signal, you express is as a superposition of scaled and shifted dirac pulses, if its a discrete time signal, you express it as a sum of scaled and shifted unit impulse sequences.

I think its as simple as that, as to why it works, you can refer to the two properties below:

1. The unit area property of the dirac pulse and the unit sample sequence.
$$
\displaystyle \sum_{n \to -\infty}^{n \to \infty} \delta[ n ] = \displaystyle \int_{-\infty}^{\infty} \delta(t) \,\,\,\,\text{dt} = 1
$$
2- The sifting property of the dirac pulse and the unit sample sequence:
$$
f(t_{0}) = \displaystyle \int_{-\infty}^{\infty} f(t) \cdot \delta(t - t_{0} ) \,\,\, \text{dt}
$$
and
$$
x[ n_{0} ] = \displaystyle \sum_{k \to -\infty}^{k \to \infty} x[ k ] \cdot \delta[ k - n_{0} ]
$$

Is it possible to express an analogue signal as a sum of scaled and shifted discrete unit impulse sequences? I don't believe this is possible, because remember, all discrete signals are sampled signals of the originals, hence, shifting a discrete time signal is simply multiplying the sample period with some integer. So its not possible, because you can only shift between integer multiples of your sampling time!

But, it is perfectly possible to express a DT signal as a sum of scaled and shifted unit impulses.
 
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