Well, again
For checking your understanding please do the following exercises:
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*}
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.
First, a series connection of a resistor $R$ and an inductor $L$ shall be considered (see figure 2). 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 main focus should first be on the amplitude response. Its frequency response can be derived from the equation in various ways.
These three points are now to be gone through.
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 $\omega \rightarrow \infty$:
It can thus be seen that:
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:
The transfer function can also be decomposed into amplitude response and frequency response.
Often these plots are not given in with linear axis but:
By this, the course from low to high frequencies is easier to see. The following simulation in figure 3 shows the amplitude response and frequency response in the lower left corner.
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:
\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*}
This equation behaves quite the same as the one considered so far.
figure 4 shows the two plots. On the x-axis, $x = \omega L / R$ has been plotted as the normalization variable. This represents a weighted frequency.
Here, too, the behavior determined in the limit value observation can be seen:
Interesting in the phase response is the point $x = 1$.
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
\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*}
\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*}
These statements apply to single-stage passive filters, i.e. one RL or one RC element. Multistage filters are considered in circuit engineering.
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*}
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 $f_{\rm c} = \frac{R}{2 \pi \cdot L}$.
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*}
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 $f_{\rm c} =\frac{1}{2 \pi\cdot RC}$
As seen before, the circuits with complex impedances can be interpreted as four-terminal networks. There, we will again look at “output versus input”, i.e: $A_V = {{U_O}\over{U_I}} \rightarrow \underline{A}_V = {{\underline{U}_O}\over{\underline{U}_I}} $.
In this chapter, we look at a combination where all three components resistor $R$, capacitor $C$, and inductance $L$ are used.
If a resistor $R$, a capacitor $C$, and an inductance $L$ are connected in series, the result is a series resonant circuit. In this case, it is not clearly defined, what the output voltage is. Consequently, it must be considered how the voltages behave across all the individual components in the following. The total voltage (= input voltage $U_I$) results to:
\begin{align*} \underline{U}_I = \underline{U}_R + \underline{U}_L + \underline{U}_C \end{align*}
Since the current in the circuit must be constant, the total impedance can be determined here in a simple way:
\begin{align*} \underline{U}_I &= R \cdot \underline{I} + {\rm j} \omega L \cdot \underline{I} + \frac {1}{j\omega C } \cdot \underline{I} \\ \underline{U}_I &= \left( R + {\rm j} \omega L - {\rm j} \cdot \frac {1}{ \omega C } \right) \cdot \underline{I} \\ \underline{Z}_{\rm eq} &= R + {\rm j} \omega L - {\rm j} \cdot \frac {1}{ \omega C } \end{align*}
By this, the magnitude of the (input) voltage $U_I$, the (input or total) impedance $Z$, and the phase result to:
\begin{align*} U_I &= \sqrt{U_R^2 + (U_Z )^2} = \sqrt{U_R^2 + (U_L - U_C)^2} \end{align*}
\begin{align*} Z &= \sqrt{R^2 + X^2} = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \end{align*}
\begin{align*} \varphi_u = \varphi_Z &= \arctan \frac{\omega L - \frac{1}{\omega C}}{R} \end{align*}
There are now 3 different situations to distinguish:
Again, there seems to be a singular frequency, namely when $U_L = U_C$ or $Z_L = Z_C$ holds:
\begin{align*} \frac{1}{\omega_0 C} & = \omega_0 L \\ \omega_0 & = \frac{1}{ \sqrt{LC}} \\ 2\pi f_0 & = \frac{1}{ \sqrt{LC}} \rightarrow \boxed{ f_0 = \frac{1}{2\pi \sqrt{LC}} } \end{align*}
The frequency $f_0$ is called resonance frequency.
| $\quad$ | $f \rightarrow 0$ | $\quad$ | $f = f_0$ | $\quad$ | $f \rightarrow \infty$ | |
|---|---|---|---|---|---|---|
| voltage $U_R$ at the resistor | $\boldsymbol{\small{0}}$ | $\boldsymbol{\LARGE{U}}$ since the impedances just cancel out | $ \boldsymbol{\small{0}}$ | |||
| voltage $U_L$ at the inductor | $\boldsymbol{\small{0}}$ because $\omega L$ becomes very small | $\boldsymbol{\omega_0 L \cdot I = \omega_0 L \cdot \frac{U}{R} = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}\cdot U}}$ | $\boldsymbol{\LARGE{U}}$ since $\omega L$ becomes very large |
|||
| voltage $U_C$ at the capacitor | $\boldsymbol{\LARGE{U}}$ because $\frac{1}{\omega C}$ becomes very large | $\boldsymbol{\frac{1}{\omega_0 C} \cdot I = \frac{1}{\omega_0 C} \cdot \frac{U}{R} = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}\cdot U}}$ | $\boldsymbol{\small{0}}$ because $\frac{1}{\omega C}$ becomes very small |
The calculation in the table shows that in the resonance case, the voltage across the capacitor or inductor deviates from the input voltage by a factor $\color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}}$. This quantity is called quality or Q-factor $Q_{\rm S}$:
\begin{align*} \boxed{ Q_{\rm S} = \left.\frac{U_C}{U} \right\vert_{\omega = \omega_0} = \left.\frac{U_L}{U} \right\vert_{\omega = \omega_0} = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}} } \end{align*}
The quality can be greater than, less than, or equal to 1. The quality $Q_{\rm S}$ does not have a unit and should not be confused with the charge $Q$.
The reciprocal of the $Q$ is called attenuation $d_{\rm S}$. This is specified when using the circuit as a non-overshooting filter.
\begin{align*} \boxed{ d_{\rm S} = \frac{1}{Q_{\rm S}} = R \sqrt{\frac{C}{L}} } \end{align*}
Simulation in Falstad. Note: The simulation gives a highly simplified picture. The response of the microcontroller is shown reduced to a triangular signal, since the slope of the voltages cannot be represented. A real simulation requires a powerful SPICE program in which the conduction theory can be represented.
Further details can be found here (practice), here (layout), also Layout or Layout.
A $R$-$L$-$C$ series circuit uses a capacity of $C=100 ~\rm µF$. A voltage source with $U_I$ feeds the circuit at $f_1 = 50~\rm Hz$.
A given $R$-$L$-$C$ series circuit is fed with a frequency, $20~\%$ larger than the resonance frequency keeping the amplitude of the input voltage constant. In this situation, the circuit shows a current $30~\%$ lower than the maximum current value.
Calculate the Quality $Q = {{1}\over{R}}\sqrt{{{L}\over{C}}}$.
However, it is possible and there are multiple ways to solve it.
What we know
But first, add some more info, which is always true from resonant circuits at the resonant frequency:
From the task, the following is also known.
Solution 2: The fast path
We start with $Z = \sqrt{R^2 + (X_L + X_C)^2}$ for the cases: (1) at the resonant frequency $f_0$ and (2) at the given frequency $f = 1.2 \cdot f_0 $
\begin{align*} (1): && Z_0 &= R \\ (2): && Z &= \sqrt{R^2 + (X_L + X_C)^2} \\ \end{align*}
In formula $(2)$ the impedance $X_L$ and $X_C$ are:
With $X_{C0} = X_{L0}$ we get for $(1)$:
\begin{align*} Z &= \sqrt{R^2 + \left(1.2\cdot X_{L0} - {{1}\over{1.2}} X_{L0} \right)^2} \\ &= \sqrt{R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} \\ \end{align*}
Since we know that $Z = {{1}\over{0.7}} \cdot R$ and $Z_0 = R$, we can start by dividing $(2)$ by $(1)$:
\begin{align*} {{(2)}\over{(1)}} : && {{Z}\over{Z_0}} &= {{\sqrt{R^2 + (X_L + X_C)^2} }\over{R}} & &| \text{put in the info from before}\\ && {{1}\over{0.7}} &= {{\sqrt{R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} }\over{R}} & &| (...)^2 \\ && {{1}\over{0.7^2}} &= {{ {R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} }\over{R^2}} & &| \cdot R^2 \\ && {{1}\over{0.7^2}} \cdot R^2 &= {R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} & &| -R^2 \\ && \left({{1}\over{0.7^2}} -1\right) \cdot R^2 &= { X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} & &| : R^2 \quad | : \left(1.2 - {{1}\over{1.2}} \right)^2 \\ && {{X_{L0}^2}\over{R^2}} &={ { {{1}\over{0.7^2}} -1 } \over { \left(1.2 - {{1}\over{1.2}} \right)^2 } } & &| \sqrt{...} \\ && {{X_{L0}}\over{R}} &={ \sqrt{ {{1}\over{0.7^2}} -1 } \over { 1.2 - {{1}\over{1.2}} } } & &| \text{with } X_{L0} = \omega_0 \cdot L = {{1}\over{\sqrt{LC}}} \cdot L = \sqrt{ {L} \over {C} }\\ && {{1}\over{R}}\cdot \sqrt{ {L} \over {C} } &={ \sqrt{ {{1}\over{0.7^2}} -1 } \over { 1.2 - {{1}\over{1.2}} } } \end{align*}
\begin{align*} Q &= {{ \sqrt{{{1}\over{0.7^2}} - 1} }\over{ 1.2 - {{1}\over{1.2}} }} \\ &= 2.782... \\ \rightarrow Q &= 2.78 \end{align*}
Explanation (video): …