Aufgaben:Exercise 4.3Z: Hilbert Transformator

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Hilbert transformator

The diagram describes a model of how, at least mentally,

• the analytical signal  $x_{+}(t)$  can be generated,
• from the real band-pass signal  $x(t)$.

The lower branch contains the so-called  "Hilbert transformer"  with the frequency response  $H_{\rm HT}(f)$. 

Its output signal  $y(t)$  is multiplied by the imaginary unit  $\rm j$  and added to the signal  $x(t)$ :

$$x_{\rm +}(t)= x(t) + {\rm j}\cdot y(t) .$$

As test signals are used, each with  $A = 1 \, \text{V}$  and  $f_0 = 10 \, \text{kHz}$:

$$x_1(t) = A \cdot {\cos} ( 2 \pi f_0 t ),$$

$$x_2(t) = A \cdot {\sin} ( 2 \pi f_0 t ),$$

$$x_3(t) = A \cdot {\cos} \big( 2 \pi f_0 (t - \tau) \big) \hspace{0.3cm}{\rm with}\hspace{0.3cm}\tau = 12.5 \hspace{0.1cm}{\rm µ s}.$$

Hints:

• This exercise belongs to the chapter  Analytical Signal and its Spectral Function.

• The following applies to the spectral function of the analytical signal:

$$X_{\rm +}(f)= \big[1 + {\rm sign}(f)\big] \cdot X(f).$$

Questions

1

Calculate the frequency response  $H_{\rm HT}(f)$  of the Hilbert transformer.  Which value is valid for the frequency  $f_0 = 10 \text{ kHz}$?

$\text{Re}[H_{\rm HT}(f = f_0)]\ = \ $

$\text{Im}[H_{\rm HT}(f = f_0)]\ = \ $

2

What is the Hilbert transform  $y_1(t)$  for the input signal  $x_1(t)$?  In particular, what value results at  $t = 0$?

$y_1(t = 0)\ = \ $  $\rm V$

3

What is the Hilbert transform  $y_2(t)$  for the input signal  $x_2(t)$?  Which value results in particular at  $t = 0$?

$y_2(t = 0)\ = \ $  $\rm V$

4

What is the Hilbert transform  $y_3(t)$  for the input signal  $x_3(t)$?  What value results for  $t=0$?  What is the phase delay  $\varphi_{\rm HT}$  of the Hilbert transformer?

$\varphi_{\rm HT}\ = \ $  $\text{Grad}$

$y_3(t = 0)\ = \ $  $\text{V}$

5

What is the analytical signal associated with  $x_3(t)$?  What are the values of the real and imaginary parts of this complex signal at time  $t = 0$?

$\text{Re}[x_{3+}(t = 0)]\ = \ $  $\text{V}$

$\text{Im}[x_{3+}(t = 0)]\ = \ $  $\text{V}$

Solution

Solution

(1)  For the spectral function at the model output holds:

$$X_{\rm +}(f)= \left(1 + {\rm j}\cdot H_{\rm HT}(f)\right) \cdot X(f).$$

• A comparison with the given relation

$$X_{\rm +}(f)= \left(1 + {\rm sign}(f)\right) \cdot X(f)$$

shows that  $H_{\rm HT}(f) = - {\rm j} \cdot \sign(f)$.

• Thus, the real part we are looking for is  ${\rm Re}[X_{\rm +}(f)]\hspace{0.15cm}\underline{=0}$  and the imaginary part is equal to  ${\rm Im}[X_{\rm +}(f)]\hspace{0.15cm}\underline{=-1}$.

(2)  From the spectral function

$$X_1(f) = {A}/{2}\cdot\delta (f + f_{0})+ {A}/{2}\cdot\delta (f - f_{0}).$$

becomes according to the Hilbert transformer:

$$Y_1(f) = {\rm j}\cdot {A}/{2}\cdot\delta (f + f_{0})-{\rm j}\cdot {A}/{2}\cdot\delta (f - f_{0}).$$

• Thus the signal at the output of the Hilbert transformer is:

$$y_1(t) = A \cdot {\sin} ( 2 \pi f_0 t ) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}y_1(t=0)\hspace{0.15 cm}\underline{ =0}.$$

(3)  Now the spectral functions at the input and output of the Hilbert transformer are:

$$X_2(f) = {\rm j}\cdot {A}/{2}\cdot\delta (f + f_{0})-{\rm j}\cdot {A}/{2}\cdot\delta (f - f_{0}),$$

$$Y_2(f) = -{A}/{2}\cdot\delta (f + f_{0})- {A}/{2}\cdot\delta (f - f_{0}).$$

• It follows that  $y_2(t) = - A \cdot \cos(2\pi f_0 t)$  and  $y_2(t = 0)\; \underline{= -\hspace{-0.08cm}1 \,\text{V}}$.

(4)  This input signal can also be represented as follows:

$$x_3(t) = A \cdot {\cos} ( 2 \pi f_0 t - 2 \pi \cdot {\rm 10 \hspace{0.05cm} kHz}\cdot {\rm 0.0125 \hspace{0.05cm} ms}) = A \cdot {\cos} ( 2 \pi f_0 t - \pi/4)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}y_3(t) = A \cdot {\cos} ( 2 \pi f_0 t - 3\pi/4).$$

• The signal phase is thus  $\varphi = \pi /4$.
• The Hilbert transformer delays this by  $\varphi_{\rm HT} \; \underline{= 90^\circ} \; (\pi /2)$.
• Therefore, the output signal  $y_3(t) = A \cdot \cos(2\pi f_0 t -3 \pi /4)$  and the signal value at time  $t = 0$  is  $A \cdot \cos(135^\circ) \; \underline{= -0.707 \,\text{V}}$.

(5)  The spectral function of the signal  $x_3(t)$  is:

$$X_3(f) = {A_0}/{2} \cdot {\rm e}^{{\rm j} \varphi}\cdot\delta (f + f_{\rm 0}) + {A_0}/{2} \cdot {\rm e}^{-{\rm j} \varphi}\cdot\delta (f - f_{\rm 0}) .$$

• For the analytical signal, the first component disappears and the component at  $+f_0$  is doubled:

$$X_{3+}(f) = {A_0} \cdot {\rm e}^{-{\rm j} \varphi}\cdot\delta (f - f_{\rm 0}) .$$

• By applying the  "Shifting Theorem" , the associated time function with  $\varphi = \pi /4$ is:

$$x_{3+}(t) = A_0 \cdot {\rm e}^{{\rm j}( 2 \pi f_{\rm 0} t \hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$

• Specifically, for time  $t = 0$:

$$x_{3+}(t = 0) = A_0 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \varphi} = A_0 \cdot{\cos} ( 45^\circ)-{\rm j}\cdot A_0 \cdot{\sin} ( 45^\circ)= \hspace{0.15 cm}\underline{{\rm 0.707 \hspace{0.05cm} V}-{\rm j}\cdot {\rm 0.707 \hspace{0.05cm} V}}.$$

Hint:  

• To get from  $x(t)$  to  $x_+(t)$,  just replace the cosine function with the complex exponential function.
• For example, the following applies to a harmonic oscillation:

$$x(t) = A \cdot {\cos} ( 2 \pi f_0 t -\hspace{0.05cm} \varphi) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{+}(t) = A \cdot {\rm e}^{{\rm j}( 2 \pi f_{\rm 0} t \hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$