Aufgaben:Exercise 3.5: Circuit with R, L and C

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Two-port network with  $R$,  $L$ and  $C$

We consider a two-port network with the resistance  $R = 100 \ \rm \Omega$  in the longitudinal branch, while in the transverse branch an inductance  $L$  and a capacitance  $C$  are connected in series. The pole–zero diagram is drawn below.

Note the normalization of the complex frequency  $p = {\rm j} \cdot 2 \pi f$  to the value  $1/T$  with  $T = 1 \ \rm µ s$.  As a consequence, for example the pole at  $-1$  in reality is at  $-10^6 \cdot \ \rm 1/s$ .

To compute time functions, the residue theorem can be applied:

• For  $N$  simple poles, the output  $y(t)$  is composed of  $N$  natural oscillations (residuals) .
• For a simple pole at  $p_{{\rm x}i}$,  the following holds for the residual:

$${\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} \hspace{-0.7cm}\{Y_{\rm L}(p)\cdot {\rm e}^{p t}\}= Y_{\rm L}(p)\cdot (p - p_{{\rm x}i})\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} \hspace{0.05cm} .$$

However, this approach only works if the number  $Z$  of zeros is less than  $N$. In this exercise, for example if the step response  $\sigma(t)$  is computed.  In this case,  $Z = 2$  and  $N = 3$ apply since the step function at the input must additionally be taken into account by  $X_{\rm L}(p) = 1/p$ .

• This approach does not work for the computation of the impulse response  $h(t)$  due to  $Z = N =2$ .
• Here, the fact that the integral over the impulse response  $h(t)$  results in the step response  $\sigma(t)$  can be considered.

Please note:

• The exercise belongs to the chapter  Inverse Laplace Transform.

Questions

1

What is the function of the two-port network considered here?  Is it

	a low-pass filter,
	a high-pass filter,
	a band-pass filter,
	a band-stop filter?

2

Compute  $L$  and  $C$  for the given pole–zero configuration.  Consider the normalization value  $1/T$  and the resistance  $R = 100 \ \rm \Omega$.

$L \hspace{0.24cm} = \ $ $\ \rm µ H$

$C \hspace{0.2cm} = \ $ $\ \rm nF$

3

Compute the output signal  $y(t) = \sigma(t)$ if a step function  $x(t) = \gamma(t)$ is applied to the input.  Enter the following signal values:

$y(t = 0) \ = \ $

$y(t = 0.5 \ \rm µ s) \ = \ $

$y(t = 2.0 \ \rm µ s) \ = \ $

$y(t = 5.0 \ \rmµ s) \ = \ $

4

Compute the impulse response  $h(t)$, in particular for times  $t = 0$  and  $t = 1 \ \rm µ s$.  Which of the following statements are true?

	$h(t)$  includes a Dirac function at  $t = 0$.
	The continuous part of  $h(t)$  is negative in the whole range.
	The continuous part of  $h(t)$  has a maximum.

Solution

Solution

(1)  Suggested solution 4 is correct:

• At extremely low frequencies  $(f \rightarrow 0)$,  the capacitance  $C$  has an infinitely large resistance and at very high frequencies  $(f \rightarrow \infty)$  the inductance  $L$.
• In both cases,  $Y(f) = X(f)$   ⇒   $H(f) = 1$ holds.
• In contrast, the LC series connection acts as a short circuit at the resonance frequency  $f_0$  and  $H(f = f_0) = 0$ holds.
• The followiong follows from the block diagram alone:   It is a band-stop filter.

(2)  The following  $p$–transfer function  $($without the normalization factor  $1/T)$ is obtained from pole–zero diagram:

$$H_{\rm L}(p)= \frac {(p - {\rm j} \cdot 2)(p + {\rm j} \cdot 2)} {(p +1)(p +4 )}= \frac {p^2 +4} {p^2 + 5 \cdot p +4} \hspace{0.05cm} .$$

• The capacitance for the circuit is obtained considering the voltage divider properties with the reactance  $p \cdot L$  of the inductance and the reactance  $1/(p \cdot C)$  of the capacitance:

$$H_{\rm L}(p)= \frac { p\cdot L +1/(pC) } {R + p \cdot L +1/(pC) }= \frac { p^2 +1/(pC) } {p^2 + p \cdot {R}/{L} +1/(pC) }\hspace{0.05cm} .$$

• Taking into account the normalization factor  $1/T= 10^6 \cdot \rm 1/s$ and by comparison, the following is found:

$${R}/{L} \hspace{0.25cm} = \hspace{0.2cm} 5 \cdot 10^{6 }\, {\rm 1/s} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}L= \frac{100\, {\rm \Omega}}{5 \cdot 10^6 \, {\rm 1/s}}\hspace{0.15cm}\underline{= 20\,{\rm µ H} \hspace{0.05cm}} ,$$

$${1}/({LC}) \hspace{0.25cm} = \hspace{0.2cm}4 \cdot 10^{12 }\, {\rm 1/s^2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}C= \frac{1}{4 \cdot 10^{12 }\, {\rm 1/s^2}\cdot 2 \cdot 10^{-5 }\, {\rm \Omega \cdot s} } \hspace{0.15cm}\underline{= 12.5\,{\rm nF}} \hspace{0.05cm} .$$

(3)  The step function at the input is accounted for by  $X_{\rm L}(p) = 1/p$ . This results in

$$Y_{\rm L}(p)= \frac {p^2 +4} {p \cdot (p +1)\cdot(p +4 )} \hspace{0.05cm} ,$$

whereof the time function  $y(t)$  can be determined by applying the residue theorem:

$$y_1(t) \hspace{0.25cm} = \hspace{0.2cm} \frac {p^2 +4} { (p +1)\cdot(p +4 )} \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}0}= 1 \hspace{0.05cm} ,$$

$$ y_2(t) \hspace{0.25cm} = \hspace{0.2cm} \frac {p^2 +4} { p\cdot(p +4 )} \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-1}= - {5}/{3}\cdot {\rm e}^{ \hspace{0.05cm}-t} \hspace{0.05cm} ,$$

$$ y_3(t) \hspace{0.25cm} = \hspace{0.2cm} \frac {p^2 +4} { p\cdot(p +1 )} \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-4}= {5}/{3}\cdot {\rm e}^{ \hspace{0.05cm}-4t}$$

$$\Rightarrow \hspace{0.3cm}y(t)= y_1(t)+y_2(t)+y_3(t)= 1- {5}/{3}\cdot {\rm e}^{ \hspace{0.05cm}-t/T}+\ {5}/{3}\cdot {\rm e}^{ \hspace{0.05cm}-4t/T} \hspace{0.05cm} .$$

Step response of the RLC resonant circuit

Here, it is taken into account that the constant  $10^6 \cdot \rm 1/s$  which is not considered in this calculation is compensated for by the time normalization to  $T = 1 \ \rm µ s$ .

The signal values which are looked for are:

$$y(t = 0) \hspace{0.05cm}\underline{= 1.000}\hspace{0.05cm}, \hspace{0.15cm}y(t = 0.5\,{\rm µ s}) \hspace{0.05cm}\underline{= 0.215}\hspace{0.05cm}, $$

$$y(t = 2\,{\rm µ s}) \hspace{0.05cm}\underline{= 0.775}\hspace{0.05cm}, \hspace{0.15cm}y(t = 5\,{\rm µ s}) \hspace{0.05cm}\underline{= 0.989}\hspace{0.05cm}. $$

The graph shows the signal curve.  The searched-for numerical values are inscribed again.

The following can be seen from this representation:

• Since extremely high frequencies are not affected by the system (band-stop filter), the jump from  $0$  to  $1$  with infinite edge steepness can also be seen in the output signal  $y(t)$ .
• The limit of  $y(t)$  for  $t → \infty$  consequently also yields the value $1$ because of  $H(f = 0) = 1$ .
• There is a drop in the signal curve due to the LC resonance frequency at  $f_0 = 1/\pi$  (in  $\rm MHz)$ .
• The signal minimum of  $\approx 0.215$  is at approximately  $t = 0.5 \ \rm µ s$.

Impulse response of the RLC low-pass filter

(4)  Suggested solutions 1 and 3 are correct:

• The impulse response  $h(t)$  is obtained from the step response  $\sigma(t)=y(t)$  by differentiation:

$$h(t)= \frac{{\rm d}\hspace{0.1cm}y(t)}{{\rm d}t}= \delta (t) + \frac {5}{3T}\cdot {\rm e}^{ \hspace{0.05cm}-t/T}- \frac {20}{3T}\cdot {\rm e}^{ \hspace{0.05cm}-4t/T} \hspace{0.05cm} .$$

• Der erste Lösungsvorschlag ist somit richtig, da die Differentiation einer Sprungfunktion die Diracfunktion liefert.
• Für den kontinuierlichen Anteil von  $h(t)$  erhält man folgende Zahlenwerte:

$$T \cdot h(t = 0 )\hspace{0.25cm} = \hspace{0.2cm} {5}/{3}- {20}/{3}= -5 \hspace{0.05cm} ,$$

$$ T \cdot h(t = T )\hspace{0.25cm} = \hspace{0.2cm} {5}/{3}\cdot {\rm e}^{ \hspace{0.05cm}-1}- {20}/{3}\cdot {\rm e}^{ \hspace{0.05cm}-4}= {5}/{3}\cdot 0.368- {20}/{3}\cdot 0.018\approx 0.491 \hspace{0.05cm} .$$

• Da  $h(t)$  im Grenzfall für  $t → \infty$  gegen Null strebt, ist der dritte Lösungsvorschlag ebenfalls richtig im Gegensatz zum zweiten.
• Der Verlauf von  $h(t)$  ist in der nebenstehenden Grafik dargestellt.