Aufgaben:Exercise 1.4Z: Representation of Oscillations

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Two representations of a harmonic oscillation

Here, we consider a harmonic oscillation  $z(t)$, which is shown in the graph together with the corresponding analytical signal  $z_+(t)$ . These signals can be described mathematically as follows:

$$z(t) = A_{\rm T} \cdot \cos(2 \pi f_{\rm T} t + \phi_{\rm T})= A_{\rm T} \cdot \cos(2 \pi f_{\rm T}( t - \tau)) \hspace{0.05cm},$$

$$ z_+(t) = A_{\rm 0} \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t}$$

The two amplitude parameters  $A_{\rm T} $  and  $A_0$  are each dimensionless, the phase value  $ϕ_{\rm T} $  is supposed to lie between  $\text{±π}$ , and the duration τ $τ$  is non-negative.

Subtask   (4)  refers to the equivalent lowpass signal  $z_{\rm TP}(t)$, which is related to  $z_+(t)$  as follows:

$$z_{\rm TP}(t) = z_+(t) \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t}.$$

Further note that  $ϕ_{\rm T}$  ppears in the above equation with a positive sign.  See "Notes on Nomenclature" below for reasons for the differential usage of  $φ_{\rm T}$  and  $ϕ_{\rm T} = – φ_{\rm T}$.

Notes on Nomenclature:

• In this tutorial, as is common in other literature, the phase enters the equations with a negative sign when describing harmonic oscillation, Fourier series, and Fourier integrals, whereas in the context of modulation methods, the phase is always given a plus sign.
• To distinguish these two variants, we use  $\phi_{\rm T}$ and $\varphi_{\rm T} = - \phi_{\rm T}$.  Both symbols denote the lowercase Greek "phi", with the notation  $\phi$  used predominantly in Anglo-American contexts, and $\varphi$ in German.
• The phase values  $\varphi_{\rm T} = 90^\circ$ and $\phi_{\rm T} = -90^\circ$  are thus equivalent and both represent the sine function:

$$\cos(2 \pi f_0 t - 90^{\circ}) = \cos(2 \pi f_0 t - \varphi_{\rm T}) = \cos(2 \pi f_0 t + \phi_{\rm T}) = \sin(2 \pi f_0 t ).$$

Further hints:

• This exercise belongs to the chapter  General Model of Modulation.
• Particular reference is made to the page  Describing the physical signal using the equivalent lowpass signal.
• Further information on this topic can be found in the following chapters of the book "Signal Representation":

(1)   Harmonic Oscillation,

(2)  Analytical Signal and its Spectral Function  and

(3)  Equivalent Low-Pass Signal and its Spectral Function.

• In our tutorial $\rm LNTwww$, the plot of the analytical signal  $s_+(t)$  in the complex plane is sometimes referred to as the "pointer diagram", while the "locus curve" gives the time course of the equivalent lowpass signal  $s_{\rm TP}(t)$ . We refer you to the corresponding interactive Applets

(1)  Physical & Analytic Signal,

(2)  Physical Signal & Equivalent Lowpass Signal.

Questions

1

Calculate the signal parameters  $A_{\rm T}$,  $f_{\rm T}$  and  $ω_{\rm T}$.

$A_{\rm T} \ = \ $

$f_{\rm T} \ = \ $ $\ \text{Hz}$

$\omega_{\rm T} \ = \ $ $\ \text{1/s}$

2

Determine the phase  $\phi_{\rm T}$  $($zwischen $±180^\circ)$ and the duration  $τ$.

$\phi_{\rm T} \ = \ $ $\ \text{Grad}$

$τ \ = \ $ $\ \text{ms}$

3

At what time  $t_1 > 0$  does the analytical signal  $z_+(t)$  first become imaginary?

$t_1 \ = \ $ $\ \text{ms}$

4

What is the equivalent low-pass signal  $z_{\rm TP}(t)$?  Enter the value at  $t = 1 \text{ ms}$ ms to check.

${\rm Re}\big[z_{\rm TP}(t = 1\ \rm ms)\big] \ = \ $

${\rm Im}\big[z_{\rm TP}(t = 1\ \rm ms)\big] \ = \ $

5

Which of these statements are valid for all harmonic oscillations?

	The spectrum  $Z(f)$  consists of two Dirac functions at  $±f_{\rm T}$.
	The spectrum  $Z_+(f)$  has one Dirac function at  $–f_{\rm T}$.
	The spectrum  $Z_{\rm TP}(f)$  contains a Dirac function at  $f = 0$.
	The analytical signal  $z_+(t)$  is always complex.
	The equivalent lowpass signal  $z_{\rm TP}(t)$  is always complex.

Solution

Solution

(1)  In the graphical representation of the time function   $z(t)$ , one can identify

• the (normalized) amplitude  $A_{\rm T}\hspace{0.15cm}\underline{ = 2}$  and the period  $T_0=2$  milliseconds.
• Therefore, the signal frequency is   $f_{\rm T} = 1/T_0\hspace{0.15cm}\underline{ = 500}$  Hz and the angular frequency is  $ω_{\rm T}= 2πf_{\rm T} \hspace{0.15cm}\underline{ = 3141.5}$  1/s.

(2)  The analytical signal is generally:

$$z_+(t) = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}(\omega_{\rm T}\cdot \hspace{0.05cm}t + \phi_{\rm T})} = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \phi_{\rm T}} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.05cm}t }\hspace{0.05cm}.$$

• At the same time the relationship:

$$A_0 = z_+(t = 0) = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \phi_{\rm T}} \hspace{0.05cm}.$$

• The complex amplitude  $A_0$  can be read from the upper plot.

$$A_0 = - \sqrt{2} - {\rm j} \cdot \sqrt{2} = A_{\rm 0} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} 0.75 \pi} \hspace{0.05cm}.$$

• A comparison of both equations leads to the result:

$$ \phi_{\rm T} = - 0.75 \pi \hspace{0.15cm}\underline {= - 135^{\circ}} \hspace{0.05cm}.$$

• Thereby, the following relationship exists with the duration  $τ$:

$$\phi_{\rm T} = - 2 \pi \cdot f_{\rm T} \cdot \tau \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \tau = \frac{-\phi_{\rm T}}{2 \pi \cdot f_{\rm T}} = \frac{0.75 \pi}{2 \pi \cdot 0.5\,{\rm kHz}} \hspace{0.15cm}\underline {= 0.75 \,{\rm ms}} \hspace{0.05cm}.$$

(3)  Das analytische Signal legt in der Zeit  $T_0$  genau eine Umdrehung zurück.

• Ausgehend von  $A_0$  erreicht man somit nach  $t_1 = T_0/8\hspace{0.15cm}\underline{ = 0.25}$  ms zum ersten Mal, dass das analytische Signal imaginär ist:

$$z_+(t_1) = - 2 {\rm j}.$$

• Wegen der Beziehung  $z(t) = {\rm Re}[z_+(t)]$  tritt zu diesem Zeitpunkt  $t_1$  auch der erste Nulldurchgang des Signals  $z(t)$  auf.

(4)  Mit dem Ergebnis der Teilaufgabe  (2)  erhält man:

$$ z_{\rm TP}(t) = z_+(t) \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm T}\cdot \hspace{0.05cm}t} = A_0 = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \phi_{\rm T}} = {\rm const.}$$

• Somit gilt für alle Zeiten  $t$  und damit auch für  $t = 1$ ms:

$${\rm Re}[z_{\rm TP}(t)] = - \sqrt{2} \hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm},$$

$$ {\rm Im}[z_{\rm TP}(t)] = - \sqrt{2}\hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm}.$$

(5)  Richtig sind die Aussagen 1, 3 und 4:

• Die einzige Diracfunktion von  $Z_+(f)$  liegt bei  $f = f_{\rm T}$  und nicht bei  $–f_{\rm T}$.
• Das analytische Signal einer harmonischen Schwingung ist immer komplex.
• Das äquivalente TP–Signal einer harmonischen Schwingung ist meistens komplex.  Ausnahme:

$$z(t) = ±A_{\rm T} · \cos(ω_{\rm T} · t) \ \Rightarrow \ z_{\rm TP}(t) = ±A_{\rm T}.$$