Characteristic Input Functions

Unit Step Function

„Heavyside” unit step function is one of the most important and most frequently used input function at the system analysis. Figure 8. presents both the curve and mathematical description of the function.

Figure 8.

$$ 1(t)=\left\{\begin{matrix} 1, $ for $ 0\leq t\\ 0, $ for $ t<0 \end{matrix}\right.$

Figure 9. shows the curve and the mathematical description of the shifted unit step function:

Figure 9.

$$ 1(t-\tau)=\left\{\begin{matrix} 1, $ for $ \tau<t\\ 0, $ for $ t<\tau \end{matrix}\right.$

Impulse function (Dirac-delta)

Let’s consider Figure 10. for the introduction of ideal impulse function.

Figure 10.

Basic property of the $$f_{0}(t)$ function:

$$ f_{0}(t)= \begin{cases} 1, & \text{ for } 0 \leq t <1 \\ 0, & \text{ otherwise } \end{cases}$

As well as $$f_{0}(t)$ should satisfy the following condition: $$\int_{-\infty}^{\infty}f_{0}(t)dt = 1$ too.

Define $$f_{1}(t)$ function in the following way (see Figure 10.)

$$ f_{1}(t)= \begin{cases} \frac{1}{\tau_{1}}, & \text{ for } 0 \leq t <\tau_{1}, \quad \text{and} \quad \tau_{1}<1\\ 0, & \text{ otherwise } \end{cases}$

Consequently is possible to see $$\int_{-\infty}^{\infty}f_{1}(t)dt = 1$ .

Define similar way $$f_{2}(t)$ and then $$f_{n}(t)$ .

$$ \begin{matrix} f_{2}(t)= \begin{cases} \frac{1}{\tau_{2}}, & \text{ for } 0 \leq t <\tau_{2}<\tau_{1}, \quad \text{and} \quad \tau_{1}<1\\ 0, & \text{ otherwise } \end{cases}\Rightarrow \int_{-\infty}^{\infty}f_{2}(t)dt = 1\\ \vdots\\ f_{n}(t)= \begin{cases} \frac{1}{\tau_{n}}, & \text{ for } 0 \leq t <\tau_{n}<\dots<\tau_{2}<\tau_{1}, \quad \text{and} \quad \tau_{1}<1\\ 0, & \text{ otherwise} \end{cases}\Rightarrow \int_{-\infty}^{\infty}f_{n}(t)dt = 1 \end{matrix} \end{cases}$

Determine the limit as follow:

$$ \delta(t)=\lim_{\tau \to 0} f(t)= \lim_{\tau \to 0} \begin{cases} \frac{1}{\tau}, & \text{ for } 0 \leq t <\frac{1}{\tau}\\ 0, & \text{ otherwise } \end{cases},\quad \text{and} \quad \int_{-\infty}^{\infty}\delta(t)dt=1$

The obtained $$\delta(t)$ function is called Dirac-delta, Dirac impulse, unit impulse function. Main feature of Dirac-delta is, everywhere in the $$(-\infty;\infty)$ interval is zero except in $t=0 point, where value of the function i.e. the “amplitude” is infinite, while the improper integral value i.e. “the intensity” is 1.
Let’s define the shifted Dirac-delta function as follow (see Figure 11.).

Figure 11.

The shifted Dirac-delta can be expressed by using the above mentioned definitions and interpretations as follow:

$$ \delta(t-t_{0})= \begin{cases} \lim_{\tau \to 0} \frac{1}{\tau}, & \text{ where } t_{0} \leq t <\tau+t_{0}\\ 0, & \text{ otherwise } \end{cases},\quad \text{and} \quad \int_{-\infty}^{\infty}\delta(t-t_{0})dt=1$

The shifted $$\delta(t-t_0)$ function is presented on the Figure 12. :

Figure 12.

Let’s attempt to find relation between the “Heaviside” unit step function and Dirac-delta unit impulse function. Let’s consider the set of continuously differentiable functions in the $$-\infty < t < \infty$ interval at first:

$$ f(t) = \int_{a}^{t}\frac{df(\tau)}{d\tau}d\tau, \text{ where } f(a) = 0$

According to the classical mathematical analysis Heavyside unit step function cannot be continuously
differentiable in the $$-\infty < t < \infty$ interval, i.e.

$$ \frac{d1(t)}{dt}= \begin{cases} 0, & \text{ for } t < 0\\ \text{not interpreted ,} & \text{ for } t=0\\ 0, & \text{ for } t>0 \end{cases}$

Let’s consider the Figure 13. in order to make relation between the Dirac delta unit impulse function and Heavyside unit step function.

Figure 13.

$$ 1(t,\tau)$ function can be described in the following way:

$$ 1(t,\tau)=\begin{cases} 0, & \text{ for } t<0\\ \frac{t}{\tau}, & \text{ for } 0\leqslant t <\tau \quad \text{ and }\quad 0<\tau<1\\ 1, & \text{ for } t\geqslant \tau \end{cases}$

The relation between the Heavyside function and the $$1(t,\tau)$ is as follow:

$$ 1(t)=\lim_{\tau \to 0 }1(t,\tau)$

Consequently the derivative of $$1(t,\tau)$ function by variable is.

$$ \frac{\partial1(t,\tau)}{\partial t}=\begin{cases} 0, & \text{ for } t<0\\ \frac{1}{\tau}, & \text{ for } 0\leqslant t <\tau\\ 0, & \text{ for } t\geqslant \tau \end{cases} = \begin{cases} \frac{1}{\tau},& \text{ for } 0\leqslant t <\tau\\ 0, & \text{ otherwise } \end{cases} \quad \text {and} \quad 0<\tau<1$

Since variable $$\tau$ has changed in the (0,1) interval, consequently $$ \frac{\partial 1(t,\tau)}{\partial t}$ derived function resulted same
impulse function with unit area (i.e. intensity) as was discussed above at the explanation of Dirac-delta.
Apply the following denoting:

$$ \delta(t,\tau)=\frac{\partial1(t,\tau)}{\partial t}= \begin{cases} 0,& \text{ for } t<0\\ \frac{1}{\tau},&\text{ for } 0\leqslant t <\tau\\ 0, & \text{ for } t\geqslant \tau \end{cases}$

$$1(t,\tau)$ and $$ \delta(t,\tau)$ have the following relation too.

$$ 1(t,\tau)=\int_{-\infty}^{t}\delta(t,\tau)dt$

Consequently, determine the following limit:

$$ \lim_{\tau \to 0}1(t,\tau)=\lim_{\tau \to 0}\int_{-\infty}^{t}\delta(t,\tau)dt\quad\Rightarrow \quad 1(t)=\int_{-\infty}^{t}\delta(t)dt$

It is possible to say, the derivative of Heavyside unit step function is defined at $t=0 point, as follows:
$$ \frac{\partial 1(t)}{\partial t} = \delta(t)$
The operation is called weak derivative and it is denoted by $$ \frac{\delta}{\delta t}$ .

Next will be discussed several consequences of Dirac-delta and the weak derivative.
Let $f(t) continuously differentiable function in the $$ (-\infty;\infty)$ interval. The value of $f(t) at $t=0 point can be determined by Dirac-delta as follow

$$ f(0)=\int_{-\infty}^{\infty}f(t)\delta(t)dt$

Value of the same $f(t) function at any $t_0 point within the $$-\infty<t_0<\infty$ interval can be determined by the shifted Dirac delta:

$$ f(t_0)=\int_{-\infty}^{\infty}f(t)\delta(t-t_0)dt$

Let’s define the analysis of that function set having finite “jump” discontinuities within the domain. (see Figure 14.)

Figure 14.

$$ f(t)=\begin{cases} f_1(t),\quad \text{for}\quad t>t_0 \quad \text{and}\quad f_1(t)\quad \text{ is continuously differentiable in the } (t_0;\infty) \text{ interval}\\ f_2(t) ,\quad \text{for}\quad t \leq t_0 \quad \text{and}\quad f_2(t)\quad \text{ is continuously differentiable in the } (-\infty;t_0] \text{ interval} \end{cases}$

The finite discontinuity of $f(t) function at $t=t_0 point should be satisfied the following condition
$$ \left |\lim_{t \to t_0+0} f(t)- \lim_{t \to t_0-0} f(t) \right |<\infty$ Consequently, the derivative of $f(t) function can be described by the weak derivative in the $$-\infty<t<\infty$ interval as follow:

$$ \frac{\delta f(t)}{\delta t}=\begin{cases} \frac{df_1(t)}{dt}, \quad \text{ for } \quad t_0<t \\ \\ \left [\lim_{t \to +t_0} f(t)- \lim_{t \to -t_0} f(t) \right ]\delta(t-t_0), \quad \text{ for } \quad t_0=t \\ \\ \frac{df_2(t)}{dt}, \quad \text{ for } \quad t<t_0 \end{cases}$

The weak derivative function can be presented as follow (Figure 15.).

Figure 15.

$$ \frac{\delta f(t)}{\delta t}=\begin{cases} \frac{df_1(t)}{dt},\quad \text{for} \quad t_0<t \\ \\ I\delta(t-t_0),\quad \text{for} \quad t_0=t, \\ \\ \frac{df_2(t)}{dt},\quad \text{for} \quad t<t_0 \end{cases}\quad \text{where}\quad I=\left [\lim_{t \to +t_0} f(t)- \lim_{t \to -t_0} f(t) \right ]$

Periodic Function

Stationary state of any linear systems is investigated very often by periodic input signal. $f(t) function is periodic by , if $f(t)=f(t+nT) , where $n=1,2,...,k,... Natural numbers (see Figure 16.)

Figure 16.

Periodicity of the function is $$ T=t_2-t_1=t_3-t_2=...=t_n-t_{n-1}$

According to the Fourier theory any $f(t) periodic function can be approximated by the following series:

$$\tilde{f}(t)=A_{0}+\sum_{k=1}^{n}\left(A_{k}\text{cos}\left(k\frac{2\pi}{T}t\right)+B_{k}\text{sin}\left(k\frac{2\pi}{T}t\right)\right)$

Maximal error of the approximation is: $$ max\left | f(t)-\tilde{f}(t) \right |$ . Let’s introduce $$\omega=\frac{2\pi}{T}$ generalized cycle frequency
(See the detailed at Appendix: Deriving of Fourier Theory, Fourier Series):

Consequently, $$\tilde{f}(t)=A_{0}+\sum_{k=1}^{n}\left(A_{k}\text{cos}(k\omega t)+B_{k}\text{sin}(k\omega t)\right)$

Next problem is arisen: is it possible to expand $f(t) by infinite trigonometric series
If $f(t) consistently convergent in the $$(-\infty;\infty)$ interval, then $f(t) function can be expanded by Fourier series as follow

$$f(t)=A_{0}+\sum_{k=1}^{\infty}\left(A_{k}\text{cos}(k\omega t)+B_{k}\text{sin}(k\omega t)\right)$ , where $$A_{0}=\frac{1}{T}\int_{0}^{T}f(t)dt; A_{k}=\frac{2}{T}\int_{0}^{T}f(t)\text{cos}\left(k\omega t\right)dt; B_{k}=\frac{2}{T}\int_{0}^{T}f(t)\text{sin}(k\omega t)dt$ .

$$ \left.\begin{matrix} \text{cos}(k\omega t)=\frac{e^{jk\omega t}+e^{-jk\omega t}}{2}\\ \\ j\text{sin}(k\omega t)=\frac{e^{jk\omega t}-e^{-jk\omega t}}{2} \end{matrix}\right\}$ by means of these Euler formulas $f(t) can be written as follow
(See the detailed at Appendix: Description of Fourier Series by complex expression) :

$$ f(t)=A_{0}+\sum_{k=1}^{\infty}c_{k}e^{jk\omega t} + \sum_{k=-1}^{-\infty}c_{k}e^{jk\omega t} = \sum_{k=-\infty}^{\infty}c_{k}e^{j\omega_{k} t};$ ahol $$ c_{k}=\frac{1}{T} \int_{0}^{T}f(t)e^{-jk\omega t}dt$

The next examples illustrate, how are presented the above described characteristic and periodic input functions in the imaging. The illustration are the response functions of 2D emission imaging systems (they are created by round shape detector gamma cameras)

Figure 17. Input function of 2D imaging system f(x,y)=A0*1(x,y) so called 2D Heavyside function. The figure shows the response on the A0*1(x,y) input function in case of round shape detector (i.e. the response of homogeny input).

Figure 18. Response of point source. The point spread function (PSF) is obtained from an extra small size (comparing to the detector size) and high concentration activity. It can be considered like the response of 2D Dirac-delta /i.e. f(x,y)=A0*delta(x,y).

Figure 19. The obtained response is originated by periodically located constant activity line source in ‘Y’ /i.e. vertical/ direction.

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