electrical_engineering_1:circuits_under_different_frequencies [MEXLE Wiki]

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====== 7 Networks at variable frequency ======

Further content can be found at this [[https://www.electronics-tutorials.ws/accircuits/series-circuit.html|Tutorial]] or that [[https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-natural-and-forced-response/a/ee-rlc-natural-response-intuition|Tutorial]]

==== Introduction ====

In the previous chapters, it was explained what the "influence of a sinusoidal current flow" of capacitors and inductors looks like. To describe this, the impedance was introduced. This can be understood as a complex resistance for sinusoidal excitation.

It applies to the capacitor:

\begin{align*} 
\underline{U}_C = \frac{1}{{\rm j}\omega \cdot C} \cdot \underline{I}_C \quad \rightarrow \quad 
\underline{Z}_C = \frac{1}{{\rm j}\omega \cdot C} 
\end{align*}

and for the inductance

\begin{align*} 
\underline{U}_L = {\rm j}\omega \cdot L \cdot \underline{I}_L \quad \rightarrow \quad 
\underline{Z}_L = {\rm j}\omega \cdot L 
\end{align*}

Complex impedances can be dealt with in much the same way as ohmic resistances in Electrical Engineering 1 (see: [[:electrical_engineering_1:simple_circuits|simple DC Circuits]], [[electrical_engineering_1:non-ideal_sources_and_two_terminal_networks|linear Sources and two-terminal network]], [[:electrical_engineering_1:network_analysis|Analysis of DC Networks]]). In these transformations, the fraction $ j\omega \cdot$ is preserved. Circuits with impedances such as inductors and capacitors will show a frequency dependence accordingly.

<callout>

=== Targets ===

After this lesson, you should:

  - know that …
  - know that … is formed.
  - be able to … can …

</callout>

===== 7.1 From Two-Terminal Network to Four-Terminal Network =====

<WRAP> <imgcaption imageNo01 | Two-Terminal Network to Four-Terminal Network> </imgcaption> \\ {{drawio>ZweipolundVierpol.svg}} \\ </WRAP>

Until now, components such as resistors, capacitors, and inductors have been understood as two-terminal. This is also obvious since there are only two connections. In the following however circuits are considered, which behave similarly to a voltage divider: On one side a voltage $U_\rm I$ is applied, and on the other side $U_\rm O$ is formed with it. This results in 4 terminals. The circuit can and will be considered as a four-terminal network in the following. However, the input and output values will be complex.

For a four-terminal network, the relation of "what goes out" (e.g. $\underline{U}_\rm O$ or $\underline{U}_2$) to "what goes in" (e.g. voltage $\underline{U}_\rm I$ or $\underline{U}_1$) is important. Thus, the output and input variables ($\underline{U}_\rm O$) and ($\underline{U}_\rm I$) give the quotient:

\begin{align*} 
\underline{A} & = {{\underline{U}_{\rm O}^\phantom{O}}\over{\underline{U}_{\rm I}^\phantom{O}}} \\ 
              & \text{with} \; \underline{U}_{\rm O} = U_{\rm O} \cdot {\rm e}^{{\rm j} \varphi_{u\rm O}} \\ 
              & \text{and}  \; \underline{U}_{\rm I} = U_{\rm I} \cdot {\rm e}^{{\rm j} \varphi_{u\rm I}} \\ 
              \\ 
\underline{A} & = \frac {\underline{U}_{\rm O}^\phantom{O}}{\underline{U}_{\rm I}^\phantom{O}} 
                = \frac {U_{\rm O}            \cdot {\rm e}^{{\rm j}  \varphi_{u\rm O}}}{U_{\rm I}\cdot {\rm e}^{{\rm j} \varphi_{u\rm I}}} \\ 
              & = \frac {U_{\rm O}}{U_{\rm I}}\cdot {\rm e}^{{\rm j} (\varphi_{u\rm O}-\varphi_{u\rm I})} \\ 
\end{align*} 

\begin{align*} 
\boxed{\underline{A} = \frac {\underline{U}_{\rm O}^\phantom{O}}{\underline{U}_{\rm I}^\phantom{O}} = \frac {U_\rm O}{U_\rm I}\cdot {\rm e}^{{\rm j} \Delta\varphi_{u}}} 
\end{align*}

<callout icon="fa fa-exclamation" color="red" title="Reminder:">

  * The complex-valued quotient ${\underline{U}_{\rm O}}/{\underline{U}_{\rm I}}$ is called the **transfer function**.
  * The frequency-dependent magnitude of the quotient $A(\omega)={U_{\rm O}}/{U_{\rm I}}$ is called **amplitude response**  and the angular difference $\Delta\varphi_{u}(\omega)$ is called **phase response**.

</callout>

The frequency behavior of the amplitude response and the frequency response is not only important in electrical engineering and electronics but will also play a central role in control engineering.

~~PAGEBREAK~~ ~~CLEARFIX~~

===== 7.2 RL Series Circuit =====

<WRAP> <imgcaption imageNo02 | RL-series> </imgcaption> \\ {{drawio>RLReihenschaltung.svg}} \\ <WRAP>

First, a series connection of a resistor $R$ and an inductor $L$ shall be considered (see <imgref imageNo02 >). This structure is also called RL-element. \\ 
Here, $\underline{U}_{\rm I}= \underline{X_\rm I} \cdot \underline{I}_{\rm I}$ with $\underline{X}_{\rm I} = R + {\rm j}\omega \cdot L$ and corresponding for $\underline{U}_{\rm O}$: 
\begin{align*} 
\underline{A} = \frac {\underline{U}_{\rm O}^\phantom{O}}{\underline{U}_{\rm I}^\phantom{O}} 
              = \frac {\omega L}{\sqrt{R^2 + (\omega L)^2}}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan \frac{\omega L}{R} \right)} 
\end{align*}

This results in the following for

  * the amplitude response: $A = \frac {\omega L}{\sqrt{R^2 + (\omega L)^2}}$ and
  * the phase response: $\Delta\varphi_{u} = \arctan \frac{R}{\omega L} = \frac{\pi}{2} - \arctan \frac{\omega L}{R}$

The main focus should first be on the amplitude response. Its frequency response can be derived from the equation in various ways.

  - Extreme frequency consideration of this RL circuit (in the equation and the system)
  - Plotting amplitude and frequency response
  - Determination of prominent frequencies

These three points are now to be gone through.

==== 7.2.1 RL High Pass ====

For the first step, we investigate the limit consideration: We look at what happens when the frequency $\omega$ runs to the definition range limits, i.e. $\omega \rightarrow 0$ and $\omega \rightarrow \infty$:

  * For $\omega \rightarrow 0$, $A = \frac {\omega L}{\sqrt{R^2 + (\omega L)^2}} \rightarrow 0$ as the numerator approaches zero and the denominator remains greater than zero.
  * For $\omega \rightarrow \infty$, $A \rightarrow 1$, because in the root in the denominator $(\omega L)^2$ becomes larger and larger in the ratio $R^2$ to . So the root tends to $\omega L$ and thus to the numerator.

It can thus be seen that:

  * at small frequencies there is no voltage $U_2$ at the output.
  * at high frequencies $A = \frac {U_{\rm O}}{U_{\rm I}} = \rightarrow 1$, so the voltage at the output is equal to the voltage at the input.

Result: \\ 
The RL element shown here therefore only allows large frequencies to pass (= pass through) and small ones are filtered out. \\ 
The circuit corresponds to a **high pass**. \\ \\

This can also be derived from understanding the components: 
  * At small frequencies, the current in the coil and thus the magnetic field changes only slowly. So only a negligibly small reverse voltage is induced. The coil acts like a short circuit at low frequencies. 
  * At higher frequencies, the current generated by $U_I$ through the coil changes faster, the induced voltage $U_{\rm i} = - {\rm d}I / {\rm d}t$ becomes large. \\ As a result, the coil inhibits the current flow and a voltage drops across the coil. 
  * If the frequency becomes very high, only a negligible current flows through the coil - and hence through the resistor. The voltage drop at $R$ thus approaches zero and the output voltage $U_\rm O$ tends towards $U_\rm I$.

The transfer function can also be decomposed into amplitude response and frequency response. \\ 
Often these plots are not given in with linear axis but:
  * the amplitude response with a double logarithmic coordinate system and
  * the phase response single logarithmic coordinate system. 

By this, the course from low to high frequencies is easier to see. The following simulation in <imgref imageNo5> shows the amplitude response and frequency response in the lower left corner.

<WRAP centeralign> 
<imgcaption imageNo5 | RL high pass filter> 
</imgcaption>\\
{{url>https://www.falstad.com/afilter/circuitjs.html?cct=$+1+0.000005+5+50+5+50%0A%25+4+1630997.1347384118%0Ar+64+80+224+80+0+35%0AO+224+80+336+80+0%0Ag+224+208+224+240+0%0A170+64+80+32+80+3+20+1000+5+0.1%0Al+224+80+224+208+0+0.001+0%0Ao+3+16+0+34+5+0.00009765625+0+-1+in%0Ao+1+16+0+34+2.5+0.00009765625+1+-1+out%0A 500,500 noborder}}
</WRAP>


For further consideration, the equation of the transfer function $\underline{A} = \dfrac {\underline{U}_{\rm O}^\phantom{O}}{\underline{U}_{\rm I}^\phantom{O}}$ is to be rewritten so that it becomes independent of component values $R$ and $L$.\\ 
This allows for a generalized representation. This representation is called **normalization**:

<WRAP centeralign>
\begin{align*} 
\large{\underline{A}  = \frac {\underline{U}_{\rm O}^\phantom{O}}{\underline{U}_{\rm I}^\phantom{O}} 
                      = \frac {\omega L}    {\sqrt{R^2 +    (\omega L)^2}}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan \frac{\omega L}{R} \right)}}
 \quad  \quad \vphantom{\HUGE{I \\ I}} \large{\xrightarrow{\text{normalization}}} \vphantom{\HUGE{I \\ I}} \quad \quad \quad 
\large{\underline{A}_{norm} 
                      = \frac {\omega L / R}{\sqrt{1  + (\omega L / R)^2}}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan \frac{\omega L}{R} \right)} } 
\large{               = \frac {x}           {\sqrt{1  + x^2             }}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan x \right)} }
\end{align*} 
</WRAP>


This equation behaves quite the same as the one considered so far.

~~PAGEBREAK~~ ~~CLEARFIX~~

<imgref imageNo03 > shows the two plots. On the x-axis, $x = \omega L / R$ has been plotted as the normalization variable. This represents a weighted frequency.

<WRAP> <imgcaption imageNo03 | Amplitude and phase response of the RL high-pass filter> </imgcaption> {{drawio>AmplitudenPhasengangRLHochpass.svg}} </WRAP>

Here, too, the behavior determined in the limit value observation can be seen: 
  * at small frequencies $\omega$ (corresponds to small $x$), the amplitude response tends toward zero. 
  * At high frequencies, the ratio $U_{\rm O} / U_{\rm I} = 1 $ is established.

Interesting in the phase response is the point $x = 1$.

  * Further to the left of this point (i.e. at smaller frequencies) a tenfold increase of the frequency $\omega$ produces a tenfold increase of $U_{\rm O} / U_{\rm I}$.
  * Further to the right of this point (i.e. at higher frequencies) $U_{\rm O} / U_{\rm I} = 1$ remains.

So this point marks a limit. Far to the left, the ohmic resistance is significantly greater than the amount of impedance of the coil: $R \gg \omega L$. far to the right is just the opposite.

The point $x=1$ just marks the cut-off frequency. \\ It holds

<WRAP group>
<WRAP quarter column > </WRAP>
<WRAP quarter column > 
\begin{align*} 
\vphantom{\HUGE{I }} \\
\underline{A}_{\rm norm} = \frac{x}{\sqrt{1 + x^2}}    \cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - arctan x \right)} 
                         = \frac{U_{\rm O}}{U_{\rm I}} \cdot {\rm e}^{{\rm j}\varphi} 
\end{align*}
</WRAP>
<WRAP quarter column > 
\begin{align*} 
\left\{\begin{array}{l}
x \ll 1 & \widehat{=}& \omega L \ll R \, : \quad & \frac{U_{\rm O}}{U_{\rm I}}=x                  &, \varphi = \frac{\pi}{2}  \, \widehat{=} \, 90° \\
x \gg 1 & \widehat{=}& \omega L \gg R \, : \quad & \frac{U_{\rm O}}{U_{\rm I}}=1                  &, \varphi = 0              \; \widehat{=} \, 0° \\
x   = 1 & \widehat{=}& \omega L =   R  , : \quad & \frac{U_{\rm O}}{U_{\rm I}}=\frac{1}{\sqrt{2}} &, \varphi = \frac{\pi}{4}  \, \widehat{=} \, 45°
\end{array} \right.
\end{align*}
</WRAP>
<WRAP quarter column > </WRAP>
</WRAP>



<callout icon="fa fa-exclamation" color="red" title="Reminder:">

  * The **cut-off frequency** $f_\rm c$ for high-pass and low-pass filters is the frequency at which the ohmic resistance just equals the value of the impedance.
  * The cut-off frequency separates a range in which the filter allows signals through from one in which they are suppressed (=blocked).
  * At the cut-off frequency, the phase $\varphi = 45°$ and the amplitude $A = \frac{1}{\sqrt{2}}$.
  * In German the cut-off Frequency is called //Grenzfrequenz// $f_{\rm Gr}$

These statements apply to single-stage passive filters, i.e. one RL or one RC element. Multistage filters are considered in circuit engineering. </callout>

The cut-off frequency, in this case, is given by:

\begin{align*} 
R               &= \omega L \\
\omega _{\rm c} &= \frac{R}{L} \\
2 \pi f_{\rm c} &= \frac{R}{L} \quad \rightarrow \quad \boxed{f_{\rm c} = \frac{R}{2 \pi \cdot L}} \end{align*}

==== 7.2.2 RL Low Pass ====

<WRAP> <imgcaption imageNo04 | Circuit, pointer diagram, and amplitude and phase response of RL low-pass filter> </imgcaption> {{drawio>AmplitudenPhasengangRLTiefpass.svg}} </WRAP>

So far, only one variant of the RL element has been considered, namely the one where the output voltage $\underline{U}_{\rm O}$ is tapped at the inductance. \\
 Here we will briefly discuss what happens when the two components are swapped.

In this case, the normalized transfer function is given by:

\begin{align*} 
\underline{A}_{\rm norm} = \frac {1}{\sqrt{1 + (\omega L / R)^2}}\cdot {\rm e}^{-{\rm j} \; \arctan \frac{\omega L}{R} } 
\end{align*}

The cut-off frequency is again given by $f_{\rm c} = \frac{R}{2 \pi \cdot L}$.

<WRAP centeralign> 
<imgcaption imageNo6 | RL low pass filter> 
</imgcaption>\\
{{url>https://www.falstad.com/afilter/circuitjs.html?cct=$+1+0.000005+5+50+5+50%0A%25+4+1630997.1347384118%0Ar+224+208+224+80+0+35%0AO+224+80+336+80+0%0Ag+224+208+224+240+0%0A170+64+80+32+80+3+20+1000+5+0.1%0Al+64+80+224+80+0+0.001+0%0Ao+3+16+0+34+5+0.00009765625+0+-1+in%0Ao+1+16+0+34+2.5+0.00009765625+1+-1+out%0A 500,500 noborder}}
</WRAP>

~~PAGEBREAK~~ ~~CLEARFIX~~

===== 7.3 RC Series Circuit =====

==== 7.3.1 RC High Pass ====

<WRAP> <imgcaption imageNo05 | Circuit, pointer diagram, and amplitude and phase response of the RC high-pass filter> </imgcaption> {{drawio>AmplitudenPhasengangRCHochpass.svg}} </WRAP>

Now a voltage divider is to be constructed by a resistor $R$ and a capacity $C$. Quite similar to the previous chapters, the transfer function can also be determined here.

Here results as normalized transfer function:

\begin{align*} 
\underline{A}_{\rm norm} = \frac {\omega RC}{\sqrt{1 + (\omega RC)^2}}\cdot {\rm e}^{\frac{\pi}{2}-{\rm j} \; \arctan (\omega RC) } 
\end{align*}

In this case, the normalization variable $x = \omega RC$. Again, the cut-off frequency is determined by equating $R$ and the magnitude of the impedance of the capacitance:

\begin{align*} 
               R &= \frac{1}{\omega_{\rm c} C} \\
 \omega_{\rm c}  &= \frac{1}{RC} \\ 
2 \pi  f_{\rm c} &= \frac{1}{RC} \quad \rightarrow \quad 
\boxed{f_{\rm c}  = \frac{1}{2 \pi\cdot RC} } 
\end{align*}

<WRAP centeralign> 
<imgcaption imageNo7 | RC high pass filter> 
</imgcaption>\\
{{url>https://www.falstad.com/afilter/circuitjs.html?cct=$+1+0.000005+5+50+5+50%0A%25+4+1630997.1347384118%0Ac+64+80+224+80+0+0.000001+0%0Ar+224+80+224+208+0+35%0AO+224+80+336+80+0%0Ag+224+208+224+240+0%0A170+64+80+32+80+3+20+1000+5+0.1%0Ao+4+16+0+34+5+0.00009765625+0+-1+in%0Ao+2+16+0+34+2.5+0.00009765625+1+-1+out%0A 500,500 noborder}}
</WRAP>

~~PAGEBREAK~~ ~~CLEARFIX~~

==== 7.3.2 RC Low Pass ====

<WRAP> <imgcaption imageNo06 | Circuit, pointer diagram, and amplitude and phase response of RC low-pass filter> </imgcaption> {{drawio>AmplitudenPhasengangRCTiefpass.svg}} </WRAP>

Again, the voltage at the impedance is to be used as the output voltage. This results in a low-pass filter.

Here results as normalized transfer function:

\begin{align*} 
\underline{A}_{\rm norm} = \frac {1}{\sqrt{1 + (\omega RC)^2}}\cdot {\rm e}^{-{\rm j} \; \arctan (\omega RC) } 
\end{align*}

Also, the cut-off frequency is given by $f_{\rm c} =\frac{1}{2 \pi\cdot RC}$

~~PAGEBREAK~~ ~~CLEARFIX~~

<WRAP centeralign> 
<imgcaption imageNo8 | RC low pass filter> 
</imgcaption>\\
{{url>https://www.falstad.com/afilter/circuitjs.html?cct=$+1+0.000005+5+50+5+50%0A%25+4+1630997.1347384118%0Ac+224+208+224+80+0+0.000001+0%0Ar+64+80+224+80+0+35%0AO+224+80+336+80+0%0Ag+224+208+224+240+0%0A170+64+80+32+80+3+20+1000+5+0.1%0Ao+4+16+0+34+5+0.00009765625+0+-1+in%0Ao+2+16+0+34+2.5+0.00009765625+1+-1+out%0A 500,500 noborder}}
</WRAP>