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:

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

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

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  $τ$   is non-negative.

Subtask  (4)  refers to the equivalent low-pass 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  $ϕ_{\rm T}$  appears 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 oscillations,  Fourier series,  and Fourier integrals.  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 low-pass signal.
• You will find further information on these topics in these chapters of the book „Signal Representation”:

(1)   Harmonic Oscillation,

(2)  Analytical Signal and its Spectral Function, 

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

• In our tutorial $\rm LNTwww$, the plot of the analytical signal $s_+(t)$  n 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 and analytic signal,

(2)  Physical signal and equivalent low-pass 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  $\phi_{\rm T}$  $($between  $±180^\circ)$  and the duration  $τ$.

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

$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  $t = 1 \text{ ms}$  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 delta functions at  $±f_{\rm T}$.
	The spectrum  $Z_+(f)$  has one delta Dirac function at  $–f_{\rm T}$.
	The spectrum  $Z_{\rm TP}(f)$  contains a Dirac delta function at  $f = 0$.
	The analytical signal  $z_+(t)$  is always complex.
	The equivalent low-pass 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)  The analytical signal covers exactly one revolution in the time  $T_0$ .

• Thus,  starting from  $A_0$  after   $t_1 = T_0/8\hspace{0.15cm}\underline{ = 0.25}$  ms,  we reach the first moment that the analytical signal is imaginary:

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

• Because of the relation  $z(t) = {\rm Re}[z_+(t)]$,  the first zero crossing of the signal  $z(t)$  also occurs at time  $t_1$.

(4)  Using the result from subtask   (2),  we obtain:

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

• Thus,  for all times  $t$  and hence also  $t = 1$ 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)  Statements 1, 3 and 4  are correct:

• The only Dirac delta function of  $Z_+(f)$  is at  $f = f_{\rm T}$  and not at  $–f_{\rm T}$.
• The analytical signal of a harmonic oscillation is always complex.
• The equivalent low-pass signal of a harmonic oscillation is usually complex.  The exception:

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