Difference between revisions of "Aufgaben:Exercise 3.5: Circuit with R, L and C"
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| − | [[File:P_ID1776__LZI_A_3_5.png|right|frame|Two-port network with $R$, | + | [[File:P_ID1776__LZI_A_3_5.png|right|frame|Two-port network with $R$, $L$, $C$]] |
| − | 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. | + | 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 $-1$ is at $-10^6 \cdot \ \rm 1/s$ in reality. | 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 $-1$ is at $-10^6 \cdot \ \rm 1/s$ in reality. | ||
The residue theorem can be applied to compute time functions: | The residue theorem can be applied to compute time functions: | ||
| − | *For $N$ simple poles, the output $y(t)$ is composed of $N$ natural oscillations ( | + | *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: | *For a simple pole at $p_{{\rm x}i}$, the following holds for the residual: | ||
:$${\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} | :$${\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} | ||
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\hspace{0.05cm} .$$ | \hspace{0.05cm} .$$ | ||
| − | :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 $N = 3$ hold since the step function at the input must additionally be taken into account by $X_{\rm L}(p) = 1/p$ . | + | :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 $N = 3$ hold 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 $Z = N =2$ . | *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. | + | *Here, the fact that the integral over the impulse response $h(t)$ results in the step response $\sigma(t)$ can be considered. |
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| − | + | Please note: | |
| − | |||
| − | |||
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*The exercise belongs to the chapter [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]]. | *The exercise belongs to the chapter [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]]. | ||
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{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? | {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? | ||
|type="[]"} | |type="[]"} | ||
| − | + $h(t)$ includes a Dirac function at $t = 0$. | + | + $h(t)$ includes a Dirac delta function at $t = 0$. |
- The continuous part of $h(t)$ is negative in the whole range. | - The continuous part of $h(t)$ is negative in the whole range. | ||
+ The continuous part of $h(t)$ has a maximum. | + The continuous part of $h(t)$ has a maximum. | ||
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===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' <u>Suggested solution 4</u> is correct: | + | '''(1)''' <u>Suggested solution 4</u> is correct: |
*At extremely low frequencies $(f \rightarrow 0)$, the capacitance $C$ has infinite resistance and at very high frequencies $(f \rightarrow \infty)$ the inductance $L$. | *At extremely low frequencies $(f \rightarrow 0)$, the capacitance $C$ has infinite resistance and at very high frequencies $(f \rightarrow \infty)$ the inductance $L$. | ||
| − | *In both cases, $Y(f) = X(f)$ ⇒ $H(f) = 1$ holds. | + | *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. | + | *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 <u>band-stop filter</u>. | *The followiong follows from the block diagram alone: It is a <u>band-stop filter</u>. | ||
| − | '''(2)''' The following $p$–transfer function $($without the normalization factor $1/T)$ is obtained from pole–zero diagram: | + | '''(2)''' The following $p$–transfer function $($without the normalization factor $1/T)$ is obtained from the pole–zero diagram: |
:$$H_{\rm L}(p)= \frac {(p - {\rm j} \cdot 2)(p + {\rm j} \cdot | :$$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} | 2)} {(p +1)(p +4 )}= \frac {p^2 +4} {p^2 + 5 \cdot p +4} | ||
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{R + p \cdot L +1/(pC) }= \frac { p^2 +1/(pC) } | {R + p \cdot L +1/(pC) }= \frac { p^2 +1/(pC) } | ||
{p^2 + p \cdot {R}/{L} +1/(pC) }\hspace{0.05cm} .$$ | {p^2 + p \cdot {R}/{L} +1/(pC) }\hspace{0.05cm} .$$ | ||
| − | *Taking into account the normalization factor $1/T= 10^6 \cdot \rm 1/s$ and by comparison, the following is found: | + | *Taking into account the normalization factor $1/T= 10^6 \cdot \rm 1/s$ and by comparison, the following is found: |
:$${R}/{L} \hspace{0.25cm} = \hspace{0.2cm} 5 \cdot 10^{6 }\, {\rm 1/s} | :$${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 | \hspace{0.3cm}\Rightarrow \hspace{0.3cm}L= \frac{100\, {\rm \Omega}}{5 \cdot 10^6 \, {\rm | ||
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'''(3)''' The step function at the input is accounted for by $X_{\rm L}(p) = 1/p$ . This results in | '''(3)''' The step function at the input is accounted for by $X_{\rm L}(p) = 1/p$ . This results in | ||
:$$Y_{\rm L}(p)= \frac {p^2 +4} {p \cdot (p +1)\cdot(p +4 )} | :$$Y_{\rm L}(p)= \frac {p^2 +4} {p \cdot (p +1)\cdot(p +4 )} | ||
| − | \hspace{0.05cm} ,$$ | + | \hspace{0.05cm}, $$ |
whereof the time function $y(t)$ can be determined by applying the residue theorem: | whereof the time function $y(t)$ can be determined by applying the residue theorem: | ||
:$$y_1(t) \hspace{0.25cm} = \hspace{0.2cm} | :$$y_1(t) \hspace{0.25cm} = \hspace{0.2cm} | ||
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[[File:P_ID1778__LZI_A_3_5_c.png|right|frame|Step response of the RLC resonant circuit]] | [[File:P_ID1778__LZI_A_3_5_c.png|right|frame|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 | + | 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 time normalization to $T = 1 \ \rm µ s$. |
The signal values which are looked for are: | The signal values which are looked for are: | ||
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The following can be seen from this representation: | 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 $y(t)$ | + | * 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 $H(f = 0) = 1$ | + | * 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 $\rm MHz)$ . | * 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 $t = 0.5 \ \rm µ s$. | *The signal minimum of $\approx 0.215$ is at approximately $t = 0.5 \ \rm µ s$. | ||
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[[File:P_ID1779__LZI_A_3_5_d.png|right|frame|Impulse response of the RLC low-pass filter]] | [[File:P_ID1779__LZI_A_3_5_d.png|right|frame|Impulse response of the RLC low-pass filter]] | ||
| − | '''(4)''' <u>Suggested solutions 1 and 3</u> are correct: | + | '''(4)''' <u>Suggested solutions 1 and 3</u> are correct: |
*The impulse response $h(t)$ is obtained from the step response $\sigma(t)=y(t)$ by differentiation: | *The impulse response $h(t)$ is obtained from the step response $\sigma(t)=y(t)$ by differentiation: | ||
:$$h(t)= \frac{{\rm d}\hspace{0.1cm}y(t)}{{\rm | :$$h(t)= \frac{{\rm d}\hspace{0.1cm}y(t)}{{\rm | ||
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\hspace{0.05cm}-4t/T} | \hspace{0.05cm}-4t/T} | ||
\hspace{0.05cm} .$$ | \hspace{0.05cm} .$$ | ||
| − | *The first suggested solution is thus correct since differentiation of a step function yields the Dirac function. | + | *The first suggested solution is thus correct since differentiation of a step function $\gamma(t)$ yields the Dirac delta function $\delta(t)$. |
*The following numerical values are obtained for the continuous part of $h(t)$ : | *The following numerical values are obtained for the continuous part of $h(t)$ : | ||
:$$T \cdot h(t = 0 )\hspace{0.25cm} = \hspace{0.2cm} {5}/{3}- {20}/{3}= -5 | :$$T \cdot h(t = 0 )\hspace{0.25cm} = \hspace{0.2cm} {5}/{3}- {20}/{3}= -5 | ||
Latest revision as of 11:12, 10 November 2021
Two-port network with $R$, $L$, $C$
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 $-1$ is at $-10^6 \cdot \ \rm 1/s$ in reality.
The residue theorem can be applied to compute time functions:
- 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 $N = 3$ hold 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
Solution
(1) Suggested solution 4 is correct:
- At extremely low frequencies $(f \rightarrow 0)$, the capacitance $C$ has infinite 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 the 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 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:
- $$h(t)= \frac{{\rm d}\hspace{0.1cm}y(t)}{{\rm d}t}= \delta (t) + \frac {5}{3T}\cdot {\rm e}^{ \hspace{0.05cm}-t/T}- \frac {20}{3T}\cdot {\rm e}^{ \hspace{0.05cm}-4t/T} \hspace{0.05cm} .$$
- The first suggested solution is thus correct since differentiation of a step function $\gamma(t)$ yields the Dirac delta function $\delta(t)$.
- The following numerical values are obtained for the continuous part of $h(t)$ :
- $$T \cdot h(t = 0 )\hspace{0.25cm} = \hspace{0.2cm} {5}/{3}- {20}/{3}= -5 \hspace{0.05cm} ,$$
- $$ T \cdot h(t = T )\hspace{0.25cm} = \hspace{0.2cm} {5}/{3}\cdot {\rm e}^{ \hspace{0.05cm}-1}- {20}/{3}\cdot {\rm e}^{ \hspace{0.05cm}-4}= {5}/{3}\cdot 0.368- {20}/{3}\cdot 0.018\approx 0.491 \hspace{0.05cm} .$$
- Since $h(t)$ tends to zero in the limiting case for $t → \infty$, the third proposed solution is also correct in contrast to the second one.
- The curve of $h(t)$ is shown in the adjacent diagram.
