Applets:Period Duration of Periodic Signals

Open Applet in a new tab       Version with Exercises and Solutions in German

Applet Descripition

This applet draws the course and calculates the period duration  $T_0$  of the periodic function

$$x(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right).$$

Please note:

• The phases  $\varphi_i$  must be entered here in radians.  Conversion from the input value:  

$$\varphi_i \text{[in radians]} =\varphi_i \text{[in degrees]}/360 \cdot 2\pi.$$

• The maximum value  $x_{\rm max}$  and a signal value  $x(t_*)$  at a given time  $t_*$ are also output.

Theoretical background

A periodic signal  $x(t)$  is present exactly when it is not constant and if for all arbitrary values of  $t$  and all integer values of  $i$  with an appropriate  $T_{0}$  applies:  

$$x(t+i\cdot T_{0}) = x(t).$$

• $T_0$  is called the  period duration   and  $f_0 = 1/T_0$  the  basic frequency.

• For a harmonic oscillation  $x_1(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)$  applies  $f_0 = f_1$  and  $T_0 = 1/f_1$,  independent of the phase  $\varphi_1$  and the amplitude  $A_1 \ne 0$.

$\text{Calculation Rule: }$  If the periodic signal  $x(t)$  consists of two parts  $x_1(t)$  and  $x_2(t)$  like in this applet, then applies for the basic frequency and the period duration with  $A_1 \ne 0$,  $f_1 \ne 0$,  $A_2 \ne 0$,  $f_2 \ne 0$:

$$f_0 = {\rm gcd}(f_1, \ f_2) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}T_0 = 1/f_0.$$

Here  $\rm gcd$  denotes the greatest common divisor.

$\text{Examples:}$   In the following  $f_0'$,  $f_1'$  and $f_2'$  denote signal frequencies normalized to $1\ \rm kHz$:

(a)   $f_1' = 1.0$,   $f_2' = 3.0$   ⇒   $f_0' = {\rm gcd}(1.0, \ 3.0) = 1.0$   ⇒   $T_0 = 1.0\ \rm ms$;

(b)   $f_1' = 1.0$,   $f_2' = 3.5$   ⇒   $f_0' = {\rm gcd}(1.0, \ 3.5)= 0.5$   ⇒   $T_0 = 2.0\ \rm ms$;

(c)   $f_1' = 1.0$,   $f_2' = 2.5$   ⇒   $f_0' = {\rm gcd}(1.0, \ 2.5) = 0.5$   ⇒   $T_0 = 2.0\ \rm ms$;

(d)   $f_1' = 0.9$,   $f_2' = 2.5$   ⇒   $f_0' = {\rm gcd}(0.9, \ 2.5) = 0.1$   ⇒   $T_0 = 10.0 \ \rm ms$;

(e)   $f_2' = \sqrt{2} \cdot f_1' $   ⇒   $f_0' = {\rm gcd}(f_1', \ f_2') \to 0$   ⇒   $T_0 \to \infty$  ⇒   the signal  $x(t)$  is not periodic.

$\text{Note:}$  The period duration could also be determined as  least common multiple  $\rm (lcm)$  according to  $T_0 = {\rm lcm}(T_1, \ T_2)$:

(c)   $T_1 = 1.0\ \rm ms$,   $T_2 = 0.4\ \rm kHz$   ⇒   $T_0 = {\rm lcm}(1.0, \ 0.4) \ \rm ms = 2.0\ \rm ms$

With all other parameter values, however, there would be numerical problems, for example

(a)   $T_1 = 1.0\ \rm ms$  and  $T_2 = 0.333\text{...} \ \rm ms$  have no  "least common multiple"  due to the limited representation of real numbers.

Exercises

• First select the number  (1, 2, ... )  of the exercise.
  
• An exercise description is displayed.  Parameter values are adjusted.
  
• Solution after pressing "Show solution".
  
• The number  0  corresponds to a  "Reset":  Same setting as at the program start.
  
• $A_1'$  and  $A_2'$  denote the signal amplitudes normalized to  $1\ \rm V$.
  
• $f_0'$,  $f_1'$  and  $f_2'$  are the frequencies normalized to  $1\ \rm kHz$.

(1)   Consider  $A_1' = 1.0, \ A_2' = 0.5, \ f_1' = 2.0, \ f_2' = 2.5, \ \varphi_1 = 0^\circ \ \varphi_2 = 90^\circ$.  How large is the period  $T_0$?

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$The period is  $T_0 = 2.0 \ \rm ms$   due to   $\rm{gcd}(2.0, 2.5) = 0.5$.

(2)   Vary  $\varphi_1$  and  $\varphi_2$  in the whole possible range  $\pm 180^\circ$.  How does this affect the period  $T_0$?

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$The period  $T_0 = 2.0 \ \rm ms$  remains the same for all  $\varphi_1$  and  $\varphi_2$.

(3)   Select the default setting   ⇒   "Recall Parameters".  Vary  $A_1'$  in the entire possible range  $0 \le A_1' \le 1$.

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$The period  $T_0 = 2.0 \ \rm ms$  remains the same with the exception of  $A_1' =0$.  In the latter case:  $T_0 = 0.4 \ \rm ms$.

(4)   Choose the default setting   ⇒   "Recall Parameters"  and vary  $f_2'$.  Does this affect  $T_0$?  Which value is the result for  $f_2' = 0.2$?

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$The period jumps back and forth.  For  $f_2' = 0.2$  the result is  $T_0 = 5.0 \ \rm ms$  because of $\ \rm{gcd} (2.0,0.2)=0.2$.

(5)   Consider  $A_1' = 1.0, \ A_2' = 0.5, \ f_1' = 0.2, \ f_2' = 2.5, \ \varphi_1 = 0^\circ \ \varphi_2 = 90^\circ$.  How large is the period  $T_0$?  Save this setting with  "Store Parameters".

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$The period is  $T_0 = 10.0 \ \rm ms$    due to  $\rm{gcd}(0.2, 2.5) = 0.1$.

(6)   Select the last setting   ⇒  "Recall Parameters"  and change  $f_2' = 0.6$.  Save this setting with  "Store Parameters".

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$The period is  $T_0 = 5.0 \ \rm ms$  due to  $\rm{gcd}(0.2,0.6) = 0.2$.

(7)   How large is the maximum signal value  $x_{\rm max}$  with the same settings?`

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$ $x_{\rm max} =x(t_* + i \cdot T_0) = 1.38 \ {\rm V} < A_1 + A_2$  with  $t_* = 0.3 \ \rm ms$  and  $T_0 = 5.0 \ \rm ms$.

(8)   What changes with  $\varphi_2 = 0^\circ$   ⇒   Sum of two cosine waves?

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$ $t_* = 0$,  $T_0 = 5.0 \ \rm ms$  ⇒   $x_{\rm max} =x(t_* + i \cdot T_0) = 1.5 \ {\rm V}=A_1 + A_2$.

(9)   Now consider  $\varphi_1 = \varphi_2 = 90^\circ$   ⇒   Sum of two sine waves?

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$The maximum signal value is now  $x_{\rm{max}} = 1.07 \ \rm V < A_1 + A_2$.  This value results from  $T_0 = 5.0 \ \rm ms$  and  $t_* = 0.6 \ \rm ms$  or  $t_* = 1.9 \ \rm ms$.

Applet Manual

Bildschirmabzug der englischen Version

    (A)     Parameter input for harmonic oscillation 1

    (B)     Parameter input for harmonic oscillation 2 and time  $t_*$.

    (C)     Numerical output of the main result  $T_0$; graphical illustration by red line

    (D)     Save parameter sets

    (E)     Retrieve parameter sets

    (F)     Output of  $x_{\rm max}$  and the signal values  $x(t_*) = x(t_* + T_0)= x(t_* + 2T_0)$

    (G)     Graphic field for displaying the signals

                  The signal values  $x(t_*) = x(t_* + T_0)= x(t_* + 2T_0)$  are marked by green dots

                  At the bottom of the graphic field you will find the following buttons:

                  (1)     Zoom funktions:    „$+$” (Zoom In),    „$-$” (Zoom Out),    $\rm o$ (Reset)

                  (2)     Move with    "←"    (Section to the left, ordinate to the right),    „$\uparrow$”    „$\downarrow$”,     „$\rightarrow$”

    (H)     Task selection according to the task number

In all applets top right:    Changeable graphical interface design   ⇒   Theme:

• Dark:   black background  (recommended by the authors).
• Bright:   white background  (recommended for beamers and printouts)
• Deuteranopia:   for users with pronounced green–visual impairment
• Protanopia:   for users with pronounced red–visual impairment

About the Authors

This interactive calculation tool was designed and implemented at the  Institute for Communications Engineering  at the  Technical University of Munich.

• The first German version was created in 2004 by  Ji Li  as part of her diploma thesis with “FlashMX – Actionscript” (Supervisor: Günter Söder).
• In 2017 the program was redesigned by David Jobst (Bachelor thesis LB, Supervisor:  Tasnád Kernetzky ) via „HTML5”.
• The English version was done in 2020 by  Carolin Mirschina.  Translation with www.DeepL.com/Translator (free version).

Once again: Open Applet in new Tab

Open Applet in a new tab       Version with Exercises and Solutions in German