4.1: Analog Signals and Their Frequency Analysis

Learning Objectives

After completing this section, you will be able to:

1. Define analog (continuous-time) signals.
2. Conduct analog frequency analysis using the Laplace transform.

This section discusses the analog signals in the physical domain and their frequency responses. Continuous-time signals (or analog signal) often represent the physical variables or parameters in the physical domain. For example, in a vehicle mechatronic system, the vehicle speed, acceleration, and cabin temperature are among the physical parameters. These variables can be converted into analog signals using speed, acceleration, and temperature sensors. On the other hand, analog control signals for electric motors or other actuators are generated by the microcontroller in the information domain. These are then converted into analog signals using D/A converters. In certain applications, such as pulse-width modulated signal (Section 4.6), digital signals can be amplified and used directly to drive a DC motor. Physically, the analog signal in a mechatronic system is often represented by a voltage signal \(u(t)\) as a function of time or a current signal \(i\left(t\right)\). Mathematically, both voltage \(u(t)\) and current \(i(t)\) signals can be represented by piecewise continuous functions of time \(t\). The definition of a piecewise continuous function is shown below.

Definition 4.1 [18]

A function \(f(t)\) is called piecewise continuous (or sectionally continuous) on an interval \(α<t<β\) if there are finitely many points (\(t_i)\) \(α=t_0<t_1<\cdots <t_n=β\ \)such that

1. \(f(t)\) is continuous on each subinterval \(t_0<t<t_1\), \(t_1<t<t_2\), ···,\(t_{n-1}<t<t_n\).
2. On each subinterval \(\left(t_{k-1}, t_{k 1}\right)(k=1,2, \ldots, n)\) both limits \(\lim _{t \rightarrow t_{k-1}^{+}} f(t)\) and \(\lim _{t \rightarrow t_k^{-}} f(t)\) exist for \(k= 1,2, \ldots, n-1\), i.e., these limits are finite.

Remark 4.1

Note that function \(f(t)\) does not need to be defined at the points \(t_k,\ \ (k=1,\ 2,\ \dots ,\ n-1)\). In addition, if function \(f\left(t\right)\) is defined on an interval \(α\le t<β\), then \({\mathop{\mathrm{lim}}_{t\to α^+} f(t)\ }\) exists and \(f\left(α\right)\) is finite. For \(f\left(t\right)\) defined on an interval \(α<t\le β\), \({\mathop{\mathrm{lim}}_{t\to β^-} f(t)\ }\) exists and \(f\left(β\right)\) is finite.

Table 4.1: Typical excitation functions and their Laplace transforms

Description	\(f\left(t\right),\ \ t\ge 0\)	\(F\left(s\right)\triangleq L\left[f\left(t\right)\right]\)
Unit impulse	\(δ\left(t\right)\)	\(1\)
Unit Step	\(u\left(t\right)\)	\(\dfrac{1}{s}\)
Unit ramp	\(t\)	\(\dfrac{1}{s^2}\)
Sinusoid	\({ \sin ω_0t\ }\)	\(\dfrac{ω_0}{s^2+ω^2_0}\)

Table 4.1 considers continuous-time (analog) signals that are used to study system characteristics such as excitation functions. You can see the table summarizes their piecewise-continuous properties at time 0 and frequency responses. Let’s start with the unit impulse input \(δ\left(t\right)\). The unit impulse function \(δ(t)\) is defined in equation \ref{4.1} and its Laplace transform is 1, as indicated in Table 4.1.

\[δ\left(t\right)=\left \{ \begin{array}{cc} \infty , & t=0 \\ 0, & t\neq 0 \end{array} \right. \]

and

\[\int^{\infty }_0{δ\left(t\right)dt}=1 \label{4.1}\]

The time domain unit impulse function was discussed in Chapter 3. Here, we are looking into its frequency response. Note that the Laplace transform of the unit impulse \(δ\left(t\right)\) is 1 (a real number) and, as a result, its magnitude will be 1 and phase is 0. Figure 4.3 (generated using the MATLAB command bode) shows this finding. Note that the unit impulse function \(δ\left(t\right)\) is an idealized continuous-time signal. It cannot be realized physically due to its infinity function value at time zero. Especially by Definition 4.1 and Remark 4.1, the unit impulse function \(δ\left(t\right)\) is not piecewise continuous; at its boundary of time equal to zero, the associated function value is unbounded.

Now let’s consider two continuous-time (analog) excitation signals: the unit step and unit ramp functions defined in Table 4.1. Both functions are defined over \(t\in \left[0,\ \infty \right)\) and continuous over \(t\in (0,\ \infty )\). Because the limit \({\mathop{\mathrm{lim}}_{t\to 0^+} f(t)\ }\) exists for both functions, they are both piecewise continuous based on Definition 4.1 and Remark 4.1. The unit step function \(u(t)\) is defined as below and its Laplace transform is \(1/s\) as indicated in Table 4.1.

\[u\left(t\right)=\left\{ \begin{array}{cc} 0, & t<0 \\ 1, & t\ge 0 \end{array} \right.\ \label{4.2}\]

The unit ramp function \(r(t\)\()\) is defined as below and its Laplace transform is \(1/s^2\) as indicated in Table 4.1.

\[r\left(t\right)=\left\{ \begin{array}{cc} 0, & t<0 \\ t, & t\ge 0 \end{array} \right. \label{4.3}\]

Time domain unit step and ramp functions were discussed in Chapter 3. Here, we are looking mainly into their frequency responses. The Laplace transform of the unit step \(u\left(t\right)\) is \(1/s\) and unit ramp \(1/s^2\). As a result, the magnitudes of unit step and ramp decrease at rates of -20 dB and -40 dB, respectively, with their phases remain at \({-90}^{{}^\circ }\) and \({-180}^{{}^\circ }\), respectively. Those facts can be found in Figure 4.4 (generated using the MATLAB command bode). Both unit step and ramp functions are piecewise continuous.

Last, let’s consider the sinusoidal continuous-time (analog) excitation signal, \({sin \mathrm{(}ω_0t\mathrm{)}\ }\), defined in Table 4.1. The sinusoidal function is defined over \(t\in \left[0,\ \infty \right)\) and continuous over \(t\in \left[0,\ \infty \right)\). Recall that the unit sinusoidal function \({sin (ω_0t)\ }\) and its Laplace transform are given in Table 4.1, while \(ω_0=1\) is provided below for convenience.

\[L\left[\sin(t)\right]=\int_0^\infty e^{-st}\sin(t)\,dt=\dfrac{1}{s^2+1}\label{4.4}\]

The time domain unit sinusoidal function was also discussed in Chapter 3. Here, we are mainly looking into its frequency responses. Note that the Laplace transform of the unit sinusoidal function \({ \sin (t)\ }\) is shown in equation \ref{4.4}. As a result, the magnitude of unit sinusoidal frequency response is shown below, along with its phase.

\[\left|L\left[\sin(t)\right]\right|=\left|\dfrac{1}{1-\omega^2}\right|,\quad \angle\left(L\left[\sin(t)\right]\right)=-\tan^{-1}\left(\dfrac{1-\omega^2}{0}\right)\label{4.5}\]

It can be seen that the magnitude has a discontinuity at \(ω=1\) with infinity magnitude. See the top plot in Figure 4.5 for the case with natural frequency \(ω_0\) equal to one. In general, for a sinusoidal function, its frequency response magnitude always has discontinuity at its natural frequency \(ω_0\) with infinity magnitude. When the frequency \(ω\) is below its natural frequency \(ω_0\), the magnitude will be close to one. When the frequency \(ω\) is above its natural frequency \(ω_0\), the magnitude will decay at a rate of -40 dB (see the top plot below in Figure 4.5).

The phase of the frequency response can also be calculated using equation \ref{4.5}. When \(ω\) is below its natural frequency \(ω_0\), the phase is zero. When \(ω\) is above its natural frequency \(ω_0\), the phase is \(-{180}^{{}^\circ }\). As a result, the phase response also has a discontinuity at \(ω_0\); see the bottom plot of Figure 4.5. Both plots are generated using the MATLAB command bode.

Example 4.1

Consider a summation of sinusoidal signals with different frequencies and amplitudes in equation \ref{4.6}.

\[f\left(t\right)={ \sin (t)\ }+0.5{ \sin (4t)\ }+0.25{ \sin (3t)\ } \label{4.6}\]

The individual sinusoidal analog signals can be plotted in Figure 4.6. We can see three sinusoidal signals with frequencies of \(1\), \(4\), and \(3\) rad/s, respectively. The unit of the oscillation frequencies \(ω\) for sinusoidal function is in radian per second, not in Hertz (oscillation per second) represented by \(f\). They can be related as below.

\[ω=2πf \label{4.7}\]

The weighted summation function \(f(t)\) of \(\sin(t)\), \(0.5 \sin(4t)\), and \(0.25 \sin(3t)\) is plotted in Figure 4.7. It is still a periodic function with many frequency components, which can be seen from its frequency response (magnitude) in Figure 4.8.

To plot the frequency response of \(f(t)\), the Laplace transform is calculated as

\[L\left[f(t)\right]=\dfrac{1}{s^2+1}+0.5\dfrac{4}{s^2+16}+0.25\dfrac{3}{s^2+9} \label{4.8}\]

which can then be simplified into the following.

\[L\left[f(t)\right]=\dfrac{3.75s^4+57.75s^2+174}{s^6+26s^4+169s^2+144} \label{4.9}\]

The frequency response magnitude is plotted in Figure 4.8. We can see three peaks in the frequency response responding to the three sinusoidal signal frequencies at \(1\), \(3\), and \(4\) rad/s (see red labels on Figure 4.8). These are due to the denominator’s polynomial of \(s^6+26s^4+169s^2+144\) in \(L\left[f(t)\right]\). In addition, there are two zero magnitudes at \(2.027\) and \(3.360\) rad/s (see blue labels on Figure 4.8). These are due to the roots of the numerator’s polynomial of \(n\left(s\right)=3.75s^4+57.75s^2+174\) in \(L\left[f(t)\right]\). Normally, we expect that magnitude peaks will occur at the frequencies that the signal contains these frequency components. However, the zero magnitude frequencies cannot be directly correlated.

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