Some Problems of Complex Signal Representation

Abstract

The time domain signal is based on the decomposition of the unit step signal, the complex signal is represented by the Heaviside Function, and the problem of the definition of the original jump time in the new function is proposed, based on the analysis and comparison of simple signal and complex signal in time domain and frequency domain, the problems needing attention in using $\varepsilon (t)$ to express signal are put forward. It is concluded that no definition or special definition of the “0” moment in the original unit step signal does not affect the composition of the composite function.

1 The Introduction

Complex signals can be easily expressed by linear combination of step signals and delay signals. In addition [1, 2], the step function is used to represent the action interval of the signal, so that the piecewise defined function can be expressed into a unified form by the step function, and the function is cut or the piecewise defined function is unified into the function defined on the whole number line, which often makes the function representation simple and easy, and simplifies the operation, and reduces the error. The study of some characteristics of complex signals becomes convenient and easy. Using the characteristic of linear time-invariant system [3], the spectrum of complex signal can be studied and discussed through the spectrum of unit step signal and the characteristics of frequency domain, so as to reduce the calculation difficulty of complex signal spectrum.

2 Complex Functions Are Represented by Unit Step Functions

Generally, in the definition of the unit step function ε(t) [4], the time of “0” is undefined or defined as “0.5” according to requirements, i.e. $\varepsilon (t) = \left\{ {\begin{array}{*{20}l} 1 \hfill & {t > 0} \hfill \\ {0.5} \hfill & {t = 0} \hfill \\ 0 \hfill & {t < 0} \hfill \\ \end{array} } \right.$ or $\varepsilon (t) = \left\{ {\begin{array}{*{20}l} 1 \hfill & {{\text{t}} > {0}} \hfill \\ {\text{no definition}} \hfill & {{\text{t}} = {0}} \hfill \\ 0 \hfill & {{\text{t}} < {0}} \hfill \\ \end{array} } \right.$ [5, 6], Thus, when complex functions are represented by linear combinations of unit step functions [7], undefined points occur within the defined interval [8]. As shown in Fig. 1, 2 and 3, is $f(t)$ equal to the sum of $f_{1} (t)$ and $f_{2} (t)$? Since the unit step function is undefined at time “0”, should the value at time “0” be added to the sum of $f_{1} (t)$ and $f_{2} (t)$ to equal f(t)? Can you express the Fourier transform of $f(t)$ using the Fourier transform of $f_{1} (t)$ and the linear properties of the Fourier transform?

Fig. 1

.

Fig. 2

.

Fig. 3

.

2.1 $f_{1} (t)$, $f_{2} (t)$ for $f(t)$

Using unit step signals ε(t) to describe, $f_{1} (t)$, $f_{2} (t)$, and $f_{1} (t)$, $f_{2} (t)$ for $f(t)$

$$ f_{1} (t) = ( - t + 1)[\varepsilon (t) - \varepsilon (t - 1)],\;\;f_{2} (t) = (t + 1)[\varepsilon (t + 1) - \varepsilon (t)]. $$

According to the definition 1 of ε(t), the functions in the above two equations are not defined at the time of “0”, “1” and “-1”. Is $f(t)$ properly represented by the sum of $f_{1} (t)$ and $f_{2} (t)$? The waveform shows that $f_{1} (t)$ and $f_{2} (t)$ are not defined at the “0” moment, but the value of $f(t)$ is “1”. Does this mean that the value of “0” moment is missing? The following is a demonstration of the relationship between the frequency domain and the time domain.

2.2 $F_{1} (\omega )$ for $F(\omega )$

That's the sum of the Fourier transform of $f_{1} (t)$ and the Fourier transform of $f_{2} (t)$, compared to the Fourier transform of $F(\omega )$. Since $f_{1} (t) = ( - t + 1)[\varepsilon (t) - \varepsilon (t - 1)]{ ,}$ using the linear, time-shift, and frequency-domain differential properties of common Fourier transform, we can ge:

$$ F_{1} (\omega ) = \int_{ - \infty }^{\infty } {( - t + 1)e^{ - j\omega t} dt} = \int_{0}^{1} {( - t + 1)e^{ - j\omega t} dt} $$

$$ F_{2} (\omega ) = \int_{ - 1}^{0} {(t + 1)e^{ - j\omega t} dt} $$

$$ F_{1} (\omega ) + F_{2} (\omega ) = \int_{ - 1}^{0} {(t + 1)e^{ - j\omega t} dt} + \int_{0}^{1} {( - t + 1)e^{ - j\omega t} dt} $$

$$ = - \int_{0}^{1} {te^{ - j\omega t} dt} + \int_{0}^{1} {e^{ - j\omega t} dt} + \int_{ - 1}^{0} {te^{ - j\omega t} dt} + \int_{ - 1}^{0} {e^{ - j\omega t} dt} $$

$$ = - \int_{0}^{ - 1} {te^{j\omega t} dt} + \int_{ - 1}^{1} {e^{ - j\omega t} dt} + \int_{ - 1}^{0} {te^{ - j\omega t} dt} $$

$$ = \int_{ - 1}^{1} {e^{ - j\omega t} dt} + \int_{ - 1}^{0} {t(e^{ - j\omega t} + e^{j\omega t} )dt} $$

$$ = \int_{ - 1}^{1} {e^{ - j\omega t} dt} + 2\int_{ - 1}^{0} {t\cos \omega tdt} $$

Take Two integrals separately

$$ \int_{ - 1}^{1} {e^{ - j\omega t} dt} = - \frac{1}{j\omega }e^{ - j\omega t} \left| {_{ - 1}^{1} } \right. = - \frac{1}{j\omega }(e^{ - j\omega } - e^{j\omega } ) = \frac{1}{j\omega }2j\sin \omega = 2\frac{\sin \omega }{\omega } $$

(2.2–1)

$$ 2\int_{ - 1}^{0} {t\cos \omega tdt} = \frac{2}{\omega }\int_{ - 1}^{0} {td\sin \omega t} = \frac{2}{\omega }(t\sin \omega t\left| {_{ - 1}^{0} } \right. - \int_{ - 1}^{0} {\sin \omega t} dt) $$

$$ = \frac{2}{\omega }( - \sin \omega + \frac{1}{\omega } - \frac{1}{\omega }\cos \omega ) $$

$$ = - \frac{2}{\omega }\sin \omega + \frac{2}{{\omega^{2} }} - \frac{2}{{\omega^{2} }}\cos \omega $$

(2.2–2)

Add (2.2–1) and (2.2–2):

$$ \frac{2}{{\omega^{2} }} - \frac{2}{{\omega^{2} }}\cos \omega = \frac{4}{{\omega^{2} }}\sin^{2} \frac{\omega }{2} = s_{a}^{2} (\frac{\omega }{2}) $$

(2.2–3)

From the Fourier transform of the commonly used signal, we can see that the Fourier transform $F(\omega ) = s_{a}^{2} (\frac{\omega }{2})$ of the signal $f(t)$ in Fig. 3 is the same as formula (2.2–3), And by the one-to-one correspondence between the Fourier transform and the primitive function, we get $f(t) = f_{1} (t) + f_{2} (t)$.

2.3 The Temporal Interpretation of $f(t) = f_{1} (t) + f_{2} (t)$ Holds

From the time domain, $f(t) = f_{1} (t) + f_{2} (t)$

$$ f_{2} (t) = (t + 1)[\varepsilon (t + 1) - \varepsilon (t)],f_{1} (t) = ( - t + 1)[\varepsilon (t) - \varepsilon (t - 1)]{,} $$

$$ f(t) = f_{1} (t) + f_{2} (t) = ( - t + 1)[\varepsilon (t) - \varepsilon (t - 1)] + (t + 1)[\varepsilon (t + 1) - \varepsilon (t)] $$

$$ = ( - t + 1)\varepsilon (t) - ( - t + 1)\varepsilon (t - 1) + (t + 1)\varepsilon (t + 1) - (t + 1)\varepsilon (t) $$

$$ = - 2t\varepsilon (t) - ( - t + 1)\varepsilon (t - 1) + (t + 1)\varepsilon (t + 1) $$

The function graph is shown below 2.3–1(a).

Fig. 4

.

As you can see from the figure, the value of $f(t)$ at t = 0 can be determined by $(t + 1)\varepsilon (t + 1)$, so Fig. 4(a) is the sum of three straight lines to Fig. 4(b). The fact that $f_{1} (t)$ and $f_{2} (t)$ are undefined at the “0” moment and that the value of $f(t)$ is “1” does not mean that the value of the “0” moment is missing and that it does not require $f_{1} (t)$ and $f_{2} (t)$ to add the value of “0” to get $f(t)$. When the functions defined by the step signal form a combined function, some overlapping undefined points can be naturally compensated in the process of function combination.

3 The Conclusion

Similar to the above, many functions defined by $\varepsilon (t)$ when the combination of some overlap undefined points in the process of function combination can be made up naturally, without adding. As the Common Gate Function $G_{\tau } (t)$ (Fig. 5).

Fig. 5.

Function $G_{\tau } (t)$

What is reasonable and right to deal with an undefined “0” moment? In this paper, two examples of Combined functions are given, and the problems needing attention in using $\varepsilon (t)$ to express signals are put forward.