Well, again
For checking your understanding please do the following exercises:
Up to now, we had analyzed DC signals (chapters 1. - 4.) and abrupt voltage changes for (dis)charging capacitors (chapter 5.). In households, we use alternating voltage (AC) instead of a constant voltage (DC). This is due to at least three main facts
This does not mean that DC power lines are useless or only full of disadvantages:
Besides the applications in power systems AC values are also important in communication engineering. Acoustic and visual signals like sound and images can often be considered as wavelike AC signals. Additionally, also for signal transfer like Bluetooth, RFID, and antenna design AC signals are important.
To understand these systems a bit more, we will start this chapter with a first introduction to AC systems.
Voltages and currents in the following chapters will be time-dependent values. As already used in chapter 5. for the time-dependent values lowercase letters will be written.
By these time-dependent values, any temporal form of the voltage/current curves is possible (see Abbildung 1).
In the following, we will investigate mainly pure AC signals.
There are some important characteristic values when investigating AC signals (Abbildung 2). For the signal itself, these are:
Additionally, there are also characteristic values related to time:
Mathematically, the AC voltages and currents can be written as:
$$u(t)=\hat{U}\cdot \sin(\omega t + \varphi_U)$$
$$i(t)=\hat{I}\cdot \sin(\omega t + \varphi_I)$$
Between the AC voltages and currents, there is also another important characteristic: The phase difference $\Delta \varphi$ is given by $\Delta \varphi = \varphi_U - \varphi_I$. The phase difference shows how far the momentary value of the current is ahead of the momentary value of the voltage.
The initial phase $\varphi_0$ has a direction/sign which has to be considered. In the case a) in the picture the zero-crossing of the sinusoidal signal is before $t=0$ or $\omega t =0$. Therefore, the initial phase $\varphi_0$ is positive.
Similarly also for the phase difference $\Delta \varphi$ the direction has to be taken into account. In the following image, the zero-crossing of the voltage curve is before the zero-crossing of the current. This leads to a positive phase difference $\Delta \varphi$.
To analyze AC signals more, often different types of averages are taken into account. The most important values are:
The arithmetic mean is given by the (equally weighted) averaging of the signed measuring points.
For finite values the arithmetic mean is given by:
$$\overline{X}={{1}\over{n}}\cdot \sum_{i=1}^n x_i$$
For functions, it is given by: $$\boxed{\overline{X}={{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} x(t) {\rm d}t}$$
For pure AC signals, the arithmetic mean is $\overline{X}=0$, since the unsigned value of the integral between the upper half-wave and $0$ is equal to the unsigned value of the integral between the lower half-wave and $0$.
Since the arithmetic mean of pure AC signals with $\overline{X}=0$ does not really give an insight into the signal, different other (weighted) averages can be used.
One of them is the rectified value. For this, the signal is first rectified (visually: negative values are folded up onto the x-axis) and then averaged.
For finite values, the rectified value is given by:
$$\overline{|X|}={{1}\over{n}}\cdot \sum_{i=1}^n |x_i|$$
For functions, it is given by: $$\boxed{\overline{|X|}={{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} |x(t)| {\rm d}t}$$
\begin{align*} \overline{|X|} &= {{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} |\hat{X}\cdot \sin(\omega t + \varphi) | {\rm d}t \\ \end{align*}
Without limiting the generality, we use $\varphi=0$ and $t_0 = 0$ \begin{align*} \overline{|X|} &= {{1}\over{T}}\cdot \int_{t=0 }^{T } |\hat{X}\cdot \sin(\omega t ) | {\rm d}t \\ \end{align*}
Since $sin(\omega t)\geq0$ for $t\in [0,\pi]$, the integral can be changed and the absolute value bars can be excluded like the following
\begin{align*}
\overline{|X|} &= {{1}\over{T}}\cdot 2 \cdot \int_{t=0}^{T/2} \hat{X}\cdot \sin( {{2\pi}\over{T}} t ) {\rm d}t \\
&= 2 \cdot {{1}\over{T}}\cdot [-\hat{X}\cdot {{T}\over{2\pi}}\cdot \cos( {{2\pi}\over{T}} t )]_{t=0}^{T/2} \\
&= 2 \cdot {{1}\over{T}}\cdot {{T}\over{2\pi}}\cdot \hat{X}\cdot [-\cos( {{2\pi}\over{T}} t )]_{t=0}^{T/2} \\
&= {{1}\over{\pi}}\cdot \hat{X} \cdot [1+1] \\
\boxed{\overline{|X|}
= {{2}\over{\pi}}\cdot \hat{X} \approx 0.6366 \cdot \hat{X}}\\
\end{align*}
Often it is important to be able to compare AC signals to DC signals by having equivalent values. But what does equivalent mean?
Most importantly, these „equivalent values“ are used to compare the output power of a system. One of these equivalent values is the supply voltage value of $230~V$ (or in some countries $110V$).
How do we come to these values?
We want to find the voltage $U_{DC}$ and $I_{DC}$ of a DC source, that the output power $P_{DC}$ on a resistor $R$ is similar to the output power $P_{AC}$ of an AC source with the instantaneous values $u(t)$ and $i(t)$. For this, we have to consider the instantaneous power $p(t)$ for a distinct time $t$ and integrate this over one period $T$.
\begin{align*} P_{\rm DC} &= P_{\rm AC} \\ U_{DC} \cdot I_{\rm DC} &= {{1}\over{T}} \int_{0}^{T} u(t) \cdot i(t) {\rm d}t \\ R \cdot I_{\rm DC}^2 &= {{1}\over{T}} \int_{0}^{T} R \cdot i^2(t) {\rm d}t \\ I_{\rm DC}^2 &= {{1}\over{T}} \int_{0}^{T} i^2(t) {\rm d}t \\ \rightarrow I_{\rm DC} &= \sqrt{{{1}\over{T}} \int_{0}^{T} i^2(t) {\rm d}t} \end{align*}
A similar approach can be used on instantaneous voltage $u(t)$. Generally, the RMS value of $X$ is given by
\begin{align*}
\boxed{X_{\rm RMS} = \sqrt{{{1}\over{T}} \int_{0}^{T} x^2(t) {\rm d}t}}
\end{align*}
What is the meaning of RMS? Simple:
By this abbreviation, one can also not forget in which order the formula has to be written… Often the RMS value is also called effective value (in German: Effektivwert).
\begin{align*} X &= \sqrt{{{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} x^2(t) {\rm d}t} \\ &= \sqrt{{{1}\over{T}}\cdot \int_{t=0} ^{T} \hat{X}^2\cdot \sin^2( \omega t) {\rm d}t} \\ &= \sqrt{{{1}\over{T}}\cdot \int_{t=0} ^{T} \hat{X}^2\cdot {{1}\over{2}}\cdot (1- \cos(2\cdot \omega t)) {\rm d}t} \\ &= \sqrt{{{1}\over{T}}\cdot \hat{X}^2\cdot {{1}\over{2}}\cdot [t + {{1}\over{2\omega }}\cdot \sin(2\cdot \omega t)]_{0}^{T}} \\ &= \sqrt{{{1}\over{T}}\cdot \hat{X}^2\cdot {{1}\over{2}}\cdot (T - 0 + 0 - 0)} \\ &= \sqrt{{{1}\over{2}}\cdot \hat{X}^2} \\ \boxed{X = {{1}\over{\sqrt{2}}}\cdot \hat{X} \approx 0.707\cdot \hat{X}}\\ \end{align*}
The following simulation shows the different values for averaging a rectangular, a sinusoidal, and a triangular waveform.
Be aware that one has to wait for a full period to see the resulting values on the right outputs of the average generating blocks.
In the chapters 2. Simple Circuits and 3 Non-ideal Sources and Two-terminal Networks we already have seen, that it is possible to reduce complex circuitries down to equivalent resistors (and ideal sources). This we will try to adopt for AC components, too.
We want to analyze how the relationship between the current through a component and the voltage drop on this component behaves when an AC current is applied.
We start with Ohm's law, which states, that the instantaneous voltage $u(t)$ is proportional to the instantaneous current $i(t)$ by the factor $R$. $$u(t) = R \cdot i(t)$$
Then we insert the functions representing the instantaneous signals: $x(t)= \sqrt{2}{X}\cdot \sin(\omega t + \varphi_x)$: $$\sqrt{2}{U}\cdot \sin(\omega t + \varphi_u) = R \cdot \sqrt{2}{I}\cdot \sin(\omega t + \varphi_i)$$
Since we know, that $u(t)$ must be proportional to $i(t)$ we conclude that for a resistor $\varphi_u=\varphi_i$!
\begin{align*} R &= {{\sqrt{2}{U}\cdot \sin(\omega t + \varphi_i)}\over{\sqrt{2}{I}\cdot \sin(\omega t + \varphi_i) }} \\ &= {{U}\over{I}} \end{align*}
This was not too hard and quite obvious. But, what about the other types of passive two-terminal networks - namely the capacitance and inductance?
For the capacitance we have the basic formula: $$C={{Q}\over{U}}$$ This formula is also true for the instantaneous values: $$C={{q(t)}\over{u(t)}}$$ Additionally, we know, that the instantaneous current is defined by $i(t)={{{\rm d}q(t)}\over{{\rm d}t}}$.
By this we can set up the formula: \begin{align*} i(t) &= {{{\rm d}q(t)}\over{{\rm d}t}} \\ &= {{\rm d}\over{{\rm d}t}}\left( C \cdot u(t) \right) \end{align*}
Now, we insert the functions representing the instantaneous signals and calculate the derivative: \begin{align*} \sqrt{2}{I}\cdot \sin(\omega t + \varphi_i) &= {{\rm d}\over{{\rm d}t}}\left( C \cdot \sqrt{2}{U}\cdot \sin(\omega t + \varphi_u) \right) \\ &= C \cdot \sqrt{2}{U}\cdot \omega \cdot \cos(\omega t + \varphi_u) \\ \\ {I}\cdot \sin(\omega t + \varphi_i) &= C \cdot {U}\cdot \omega \cdot \sin(\omega t + \varphi_u + {{1}\over{2}}\pi) \tag{6.3.1} \end{align*}
Equating coefficients in $(6.3.1)$ leads to: \begin{align*} I &= C \cdot U \cdot \omega \\ {{U}\over{I}} &= {{1}\over{\omega \cdot C}} \end{align*} and: \begin{align*} \omega t + \varphi_i &= \omega t + \varphi_u + {{1}\over{2}}\pi \\ \varphi_i &= \varphi_u + {{1}\over{2}}\pi \\ \varphi_u -\varphi_i &= - {{1}\over{2}}\pi \end{align*}
The phase shift of $- {{1}\over{2}}\pi$ can also be seen in Abbildung 6 and Abbildung 5.
Only for a pure resistor as a two-terminal network, the impedance $Z_R$ is equal to the value of the resistance: $Z_R=R$.
For the pure capacitive as a two-terminal network, the impedance $Z_C$ is $Z_C={{1}\over{\omega \cdot C}}$.
The inductance will here be introduced shortly - the detailed introduction is part of electrical engineering 2.
For the capacitance $C$ we had the situation, that it reacts to a voltage change ${{\rm d}\over{{\rm d}t}}u(t)$ with a counteracting current:
$$i(t)= C \cdot {{\rm d}\over{{\rm d}t}}u(t)$$
This is due to the fact, that the capacity stores charge carriers $q$.
It appears that „the capacitance does not like voltage changes and reacts with a compensating current“.
When the voltage on a capacity drops, the capacity supplies a current - when the voltage rises the capacity drains a current.
For an inductance $L$ it is just the other way around: „the inductance does not like current changes and reacts with a compensating voltage drop“. Once the current changes the inductance will create a voltage drop that counteracts and continues the current: A current change ${{\rm d}\over{{\rm d}t}}i(t)$ leads to a voltage drop $u(t)$: $$u(t)= L \cdot {{\rm d}\over{{\rm d}t}}i(t)$$ The proportionality factor here is $L$, the value of the inductance, and it is measured in $[L] = 1~\rm H = 1~Henry$.
We can now again insert the functions representing the instantaneous signals and calculate the derivative: \begin{align*} \sqrt{2}{U}\cdot \sin(\omega t + \varphi_u) &= L \cdot {{\rm d}\over{{\rm d}t}}\left( \sqrt{2}{I}\cdot \sin(\omega t + \varphi_i) \right) \\ &= L \cdot \sqrt{2}{I}\cdot \omega \cdot \cos(\omega t + \varphi_i) \\ \\ {U}\cdot \sin(\omega t + \varphi_u) &= L \cdot {I}\cdot \omega \cdot \sin(\omega t + \varphi_i + {{1}\over{2}}\pi) \tag{6.3.2} \end{align*}
Equating coefficients in $(6.3.2)$ leads to: \begin{align*} U &= L \cdot {I}\cdot \omega \\ \boxed {Z_L = {{U}\over{I}} = \omega \cdot L} \end{align*} and: \begin{align*} \omega t + \varphi_u &= \omega t + \varphi_i + {{1}\over{2}}\pi \\ \varphi_u &= \varphi_i + {{1}\over{2}}\pi \\ \boxed{\varphi = \varphi_u -\varphi_i = + {{1}\over{2}}\pi } \end{align*}
The phase shift of $+ {{1}\over{2}}\pi$ can also be seen in Abbildung 8 and Abbildung 7.
One way to memorize the phase shift is given by the word CIVIL:
For the concept of AC two-terminal networks, we are also able to use the DC methods of network analysis to solve AC networks.
Calculate the rectified value of rectangular and triangular signals! Use similar symmetry simplifications as shown for AC signals. Compare it to the values shown in Abbildung 3.
Calculate the RMS value of rectangular and triangular signals! Use similar symmetry simplifications as shown for AC signals. Compare it to the values shown in Abbildung 3.
A coil has a impedance of $80~\Omega$ at a frequency of $500 ~\rm Hz$. At which frequencies the impedance will have the following values?
On the other hand, one can also use the rule of proportion here, and circumvent the calculation of inductance.
It is possible to calculate the reactance at other frequencies with the given reactance.
\begin{align*}
X_L&=2\pi fL\\
f &=\frac{X_L}{2\pi L}\\
&=\frac{X_L}{X_{L0}}f_0
\end{align*}
With the values given: \begin{equation*} f_1 = \frac{85 ~\Omega}{80~\Omega}\cdot500~{\rm Hz}\qquad f_2 = \frac{120~\Omega}{80~\Omega}\cdot500~{\rm Hz}\qquad f_3 = \frac{44 ~\Omega}{80~\Omega}\cdot500~{\rm Hz} \end{equation*}
A capacitor with $5 ~{\rm µF}$ is connected to a voltage source which generates $U_\sim = 200 ~{\rm V}$. At which frequencies the following currents can be measured?
A capacitor shall have a capacity of $4.7 ~{\rm µF} \pm 10~\%$. This capacitor shall be used with an AC voltage of $400~\rm V$ and $50~\rm Hz$. What is the possible current range which could be found on this component?
…
What's the point of complex numbers?
What's the point of complex numbers? (Alternative)
geometric interpretation of the complex multiplication
Why have the absolute values and the angles to be added for multiplication?
The video „Alternating Current AC Basics - Part 1“ of EEVblog explains the ideas behind averaged values (arithmetic mean, rectified value and RMS value)
An alternative derivation of the RMS value for a sinusidal function