# Discrete time aperiodic signals

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This module describes discrete time aperiodic signals.

## Introduction

This module describes the type of signals acted on by the Discrete Time Fourier Transform.

## Relevant spaces

The Discrete Time Fourier Transform maps arbitrary discrete time signals in ${l}^{2}(\mathbb{Z})$ to finite-length, discrete-frequency signals in ${L}^{2}(\left[0 , 2\pi \right))$ .

## Periodic and aperiodic signals

When a function repeats itself exactly after some given period, or cycle, we say it's periodic . A periodic function can be mathematically defined as:

$f(n)=f(n+mN)\forall m\colon m\in \mathbb{Z}$
where $N> 0$ represents the fundamental period of the signal, which is the smallest positive value of N for the signal to repeat. Because of this, you may also see a signal referred to as an N-periodic signal.Any function that satisfies this equation is said to be periodic with period N. Periodic signals in discrete time repeats themselves in each cycle. However, only integers are allowed as time variable in discrete time. We denote signals in such case as f[n], n = ..., -2, -1, 0, 1, 2, ... Here's an example of a discrete-time periodic signal with period N:

We can think of periodic functions (with period $N$ ) two different ways:

1. as functions on all of $\mathbb{R}$
2. or, we can cut out all of the redundancy, and think of them as functions on an interval $\left[0 , N\right]()$ (or, more generally, $\left[a , a+N\right]()$ ). If we know the signal is N-periodic then all the information of the signal is captured by the above interval.

An aperiodic DT function, however, $f(n)$ does not repeat for any $N\in \mathbb{R}$ ; i.e. there exists no $N$ such that this equation holds. This broader class of signals can only be acted upon by the DTFT.

Suppose we have such an aperiodic function $f(n)$ . We can construct a periodic extension of $f(n)$ called ${f}_{No}(n)$ , where $f(n)$ is repeated every ${N}_{0}$ seconds. If we take the limit as ${N}_{0}\to$ , we obtain a precise model of an aperiodic signal for which all rules that govern periodic signals can be applied, including Fourier Analysis (with an important modification). For more detail on this distinction, see the module on the Discete Time Fourier Transform .

## Conclusion

A discrete periodic signal is completely defined by its values in one period, such as the interval [0,N].Any aperiodic signal can be defined as an infinite sum of periodic functions, a useful definition that makes it possible to use Fourier Analysis on it by assuming all frequencies are present in the signal.

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20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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Sherica
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Tamia
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Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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J, combine like terms 7x-4y
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Samantha
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Asali
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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Cied
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Porter
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Cesar
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Stotaw
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Azam
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Azam
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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