Exercise 3.2Z: Sinc-Squared Spectrum with Diracs

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$\rm sinc^2$– spectrum with Diracs

The sketched spectrum  ${X(f)}$  of a time signal  ${x(t)}$  is composed of

• a continuous component  $X_1(f)$,

• plus three discrete spectral lines   ⇒   "Dirac functions".

The continuous component with  $f_0 = 200\, \text{kHz}$  and  $X_0 = 10^{–5} \text{ V/Hz}$ is as follows:

$$X_1( f ) = X_0 \cdot {\mathop{\rm sinc}\nolimits} ^2 ( {{f}/{f_0}} ),\quad {\rm where is}\quad {\mathop{\rm sinc}\nolimits} (x) = {\sin (\pi x)}/(\pi x).$$

• The spectral line at  $f = 0$  has the weight  $–\hspace{-0.08cm}1\,\text{V}$.
• In addition, there are two lines at frequencies  $\pm f_0$,  both with weight  $0.5\,\text{V}$.

Hints:

• This exercise belongs to the chapter  Fourier Transform and its Inverse.
• Further information on this topic can be found in the (German language) learning video  Kontinuierliche und diskrete Spektren   ⇒   "Continuous and discrete spectra".

• It can be assumed as known:  A triangular pulse  $y(t)$  with amplitude  ${A}$,  the absolute duration  $2T$  and symmetrical about  $t = 0$  $($i.e.:  the signal values are  $\ne 0 $  only between  $–T$  and  $+T$ )  has the following spectral function:

$$Y( f ) = A \cdot T \cdot {\rm sinc}^2 ( f T ).$$

Question

1

What are the values of the parameters  ${A}$  (maximum) and  ${T}$  (one-sided duration) of the triangular signal component  $x_1(t)$?

$A\ = \ $  $\text{V}$

$T\ = \ $  $\text{$µ$s}$

2

What is the DC component  ${B}$  of the signal?

$B\ = \ $  $\text{V}$

3

What is the amplitude  $C$  of the periodic component of  $x(t)$?

$C\ = \ $  $\text{V}$

4

What are the maximum and minimum values of the signal  $x(t)$?

$x_\text{max}\ = \ $  $\text{V}$

$x_\text{min}\hspace{0.2cm} = \ $  $\text{V}$

Solution

Solution

Area of the triangular pulse

(1)  The one-sided duration of the symmetrical triangular pulse is  $T = 1/f_0\hspace{0.15 cm}\underline{ = 5 \,{\rm µ s}}$.

• The spectral value  $X_0 = X_1(f = 0)$  indicates the pulse area of  $x_1(t)$  an.
• This is equal to  ${A} \cdot {T}$.  From this follows:

$$A = \frac{X_0 }{T} = \frac{ 10^{-5}\rm V/Hz }{5 \cdot 10^{-6}{\rm s}}\hspace{0.15 cm}\underline{= 2\;{\rm V}}.$$

(2)  The DC component is given by the Dirac weight at  $f = 0$ . One obtains  ${B} \hspace{0.15 cm}\underline{= -1 \,\text{V}}$.

(3)  The two spectral lines at  $\pm f_0$  together give a cosine signal with amplitude  ${C} \hspace{0.15 cm}\underline{= 1 \text{V}}$.

(4)  The maximum value occurs at time  ${t} = 0$    (here the triangular pulse and cosine signal are maximum):

$$x_{\text{max}} = A + B + C \hspace{0.15 cm}\underline{= +2 \text{V}}.$$

• The minimum values of  ${x(t)}$  result when the triangular pulse has decayed and the cosine function delivers the value  $–\hspace{-0.08 cm}1 \,\text{V}$ :

$$x_\text{min} = {B} - {C}\hspace{0.15 cm}\underline{ = -2\, \text{V}}.$$