Aufgaben:Exercise 5.2: Determination of the Frequency Response

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Measuring the frequency response  $H(f)$

We consider the illustrated measurement setup for the determination of the frequency response  $H(f)$  highlighted in blue.

• The input signal  $x(t)$  is white Gaussian noise with the noise power density  $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.
• Thus,  the auto-correlation function  $\rm (ACF)$  is:

$$\varphi _x ( \tau ) = {N_0 }/{2} \cdot \delta ( \tau ).$$

• The measured cross-correlation function  $\rm (CCF)$  between the signals  $x(t)$  and  $y(t)$  can be approximated as follows
  $($valid only for positive times  $t)$:

$$\varphi _{xy} \left( \tau \right) = K \cdot {\rm{e}}^{ - \tau /T_0 },\hspace{0.5cm}\text{with } \ K = 0.628 \cdot 10^{-12} \hspace{0.05cm} {\rm W}, \ T_0 = 1 \hspace{0.05cm} \rm ms.$$

• The ACF  $\varphi_y(\tau)$  of the output signal  $y(t)$ is also measured.

Notes:

• The exercise belongs to the chapter  Stochastic System Theory.
• Reference is also made to the chapter  Power-Spectral Density.
• Please also note the following Fourier transform  $($in  $\omega)$:

$$H( \omega ) = \frac{1}{{1 + {\rm{j}}\cdot \omega /\omega _0 }}\hspace{0.15cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\,\hspace{0.15cm}h(t) = \omega _0 \cdot {\rm{e}}^{ - \omega _0 t} \hspace{0.3cm}(t \ge 0).$$

For negative  $t$–values,  on the other hand,  $h(t) =0$  at all times.

Questions

1

Which of the following statements is true?  One can completely determine the frequency response  $H(f)$  by magnitude and phase if

	the functions  $\varphi_x(\tau)$  and  $\varphi_y(\tau)$  are known,
	the functions  $\varphi_x(\tau)$  and  $\varphi_{xy}(\tau)$  are known,
	the functions  $\varphi_{xy}(\tau)$  and  $\varphi_y(\tau)$  are known.

2

Calculate the impulse response  $h(t)$.  What is the value for  $t=T_0$?

$h(t = T_0) \ = \ $ $\ \cdot 10^{-3} \ \rm 1/s$

3

What is the frequency response  $H(f)$?  What value results for $f= 0$?

$H(f = 0) \ = \ $

4

Calculate the power-spectral density of the output signal  $y(t)$.  What value results for frequency  $f = 1/(2\pi T_0)$?

${\it \Phi}_y(f = 1/(2\pi T_0)) \ = \ $ $\ \cdot 10^{-12}\ \rm W/Hz$

Solution

Solution

(1)  Statements 2 and 3 are true.

• The following equations are valid:

$$\varphi _{xy} ( \tau ) = h( \tau ) * \varphi _x ( \tau )\quad \Rightarrow \quad H( f ) = \frac{{{\it \Phi} _{xy} ( f )}}{{{\it \Phi} _x ( f )}},$$

$$\varphi _y ( \tau) = \varphi _{xy} ( \tau) * h(- \tau)\quad \Rightarrow \quad H^{\star}( f ) = \frac{{{\it \Phi} _y ( f )}}{{{\it \Phi} _{xy} ( f )}}.$$

• In contrast,  the first statement is false:   The phase relations are lost in the ACF calculation.
• The associated spectral functions to  $\varphi_x(\tau)$  and  $\varphi_x(\tau)$  – namely  ${\it \Phi}_x(f)$  and  ${\it \Phi}_y(f)$  – are purely real,  so only the magnitude  $|H(f)|$  can be given.

(2)  For Dirac-shaped input ACF  $\varphi_x(\tau)$,  the impulse response  $h(t)$  is equal in shape to the CCF:

$$h(t) = \frac{{K \cdot {\rm{e}}^{ - t/T_0 } }}{N_0 /2} = 1.256 \cdot 10^{ - 2} \frac{1}{{\rm{s}}} \cdot {\rm{e}}^{ - t/T_0 } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}h(t = T_0)\hspace{0.15cm}\underline{ = 4.62 \cdot 10^{-3}\ \rm 1/s}.$$

(3)  The Fourier correspondence given is with  $T_0 = 1/\omega_0$  and the constant  $C= N_0/2 \cdot T_0/K$:

$$h(t) = \frac{C}{T_0 } \cdot {\rm{e}}^{ - t/T_0 }\hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\hspace{0.15cm} H( \omega ) = \frac{C}{{1 + {\rm{j}}\cdot \omega T_0 }}.$$

• The constant results in  $C = 0.08$.  With  $H(f) = 2 \pi \cdot H(\omega)$  it follows:

$$H(f) = \frac{0.5}{1 + {\rm{j\cdot 2\pi }}\cdot f \cdot T_0 } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} H(f= 0) \hspace{0.15cm}\underline{=0.5}.$$

(4)  For the output PSD,  in general or specifically here:

$${\it \Phi}_y (f) = {\it \Phi} _x (f) \cdot \left| {H(f)} \right|^2 = \frac{N_0 }{2} \cdot \frac{0.5^2 }{{\left( {1 + {\rm{j\cdot 2\pi }}\cdot f \cdot T_0 } \right)\left( {1 - {\rm{j\cdot 2\pi }}\cdot f \cdot T_0 } \right)}} = {N_0 }/{8} \cdot \frac{1}{1 + \left( {{\rm{2\pi }}\cdot f\cdot T_0 } \right)^2 }.$$

• At the given frequency  $f = 1/(2\pi T_0)$,    ${\it \Phi}_y (f)$  has dropped by half compared to its maximum at $f=0$ :

$${\it \Phi}_y (f = 1/(2 \pi T_0)) ={N_0 }/{16}\hspace{0.15cm} \underline{ = 6.25 \cdot 10^{ - 12} \;{\rm{W/Hz}}}.$$