# 1.3 Key discrete-time test signals

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In our study of discrete-time signals and signal processing, there are five very important signals that we will use to both illustrate signal processing concepts, and also to probe or test signal processing systems: the delta function , the unit step function , the unit pulse function , the real exponential function , sinusoidal functions , and complex exponential functions . This module will consider the first four; sinusoids and complex exponentials are particularly important, so a separate model will cover them. Each of these signals will be introduced as infinite-length signals, but they all have straightforward finite-length equivalents.

## The discrete-time delta function

The delta function is probably the simplest nontrivial signal. It is represented mathematically with (no surprise) the Greek letter delta: $\delta[n]$. It takes the value 0 for all time points, except at the time point 0 where it peaks up to the value 1:$\delta[n]=\begin{cases}1&n=0 \\ 0&\textrm{otherwise}\end{cases}$ In a variety of important settings, we will often see the delta function shifted by a particular time value. The delta function $\delta[n-m]$ is 0, except for a peak of 1 at time $m$: One of the reasons the shifted delta function is so useful is that we can use it to select, or sample, a value of another signal at some defined time value. Suppose we have some signal $x[n]$, and we would like to isolate that signal's value at time $m$. What we can do is multiply that signal by a shifted delta signal. We can say $y[n]=x[n]\delta[n-m]$, but since that $y[n]$ will be zero for all $n$ except at $n=m$, it is equivalent to express it as $y[n]=x[m]\delta[n-m]$, where now $x[m]$ is no longer a function, but a constant. The following figure shows how this operation isolates a particular time sample of $x[n]$:

## The unit step function

The unit step function can be thought of like turning on a switch. Usually identified as $u[n]$, it is $0$ for all $n \lt 0$, and then at $n=0$ it "switches on" and is $1$ for all $n \geq 0$: $u[n]=\begin{cases}1&n \lt 0\\ 1&n\geq 0\end{cases}$: As with the delta function, it will also be useful for us to shift the step function: And, as you might have guessed, we can use a shifted step function in a similar way to the delta function by multiplying it with another signal. Whereas the delta function selected a single value of a certain signal (zeroing out the rest), the step function isolates a portion of a signal after a given time. Below, a step function is used to zero out all the values of $x[n]$ for $n\lt 5$, keeping the rest: Supposing a signal $x[n]$ were not causal, setting $m$ to zero and performing the operation $x[n]u[n]$ would zero out all values of $x[n]$ before $n=0$, thereby making the result causal.

## The unit pulse function

The unit pulse $p[n]$ is very similar to the unit step function in how it "switches on" from 0 to 1, but then it also "switches off" at a later time. We will say it "switches on" at time $N_1$, and "off" at time $N_2$: $p[n] = \begin{cases}0&n\lt N_1 \\ 1&N_1 \leq n \leq N_2 \\ 0&n\gt N_2\\ \end{cases}$ Of course, rather than use the above piece-wise notation, it is also possible to express the pulse as the difference of two step functions: $p[n] = u[n-N_1]- u[n-(N_2+1)]$.

## The real exponential function

Finally, we have the real exponential function, which takes a real number $a$ (that we are going to assume is positive) and raises it to the power of $n,$ where $n$ is the time index: $r[n] = a^n$, $a\in R$, $a\geq 0$. So at $n=0$, $r[n]=a^0$, at $n=1$ it equals $a$, is $a^2$ at $n=2$, and so on. As the name suggests, the signal will exponentially increase or decrease, depending on the value of $a$.

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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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