Frequency response, amplitude response, phase response

In this chapter, we will deal with the frequency response, amplitude response and phase response of two-port networks. These are used to describe the frequency-dependent behaviour of electrical networks in the form of two-port networks, as shown in Figure 1. Using the example of simple passive filter circuits, we will introduce the terms frequency response, amplitude response and phase response and explain their meaning and application.

1 Definition of frequency response, amplitude response and phase response

The frequency response \(\underline {F}(\mathrm {j}\omega )\), also known as the complex amplitude response, describes the ratio of the output signal to the input signal of a linear time-invariant system (LTI system) under sinusoidal excitation. It offers a way to investigate the frequency-dependent behaviour of two-port networks, is a special case of the Laplace transfer function from system theory, and is defined as follows:

\begin {equation} \underline {F}(\mathrm {j}\omega ) = \frac {\underline {U}_2}{\underline {U}_1}\label {eq:def:F} \end {equation}

Due to the linearity of LZI systems, the frequency and sine waveform are retained at the output. The amplitude and phase may change from the input to the output depending on the frequency. This is particularly evident in the notation in polar coordinates.

\begin {equation} \underline {F}(\mathrm {j}\omega ) = A(\omega )\cdot \mathrm e^{\mathrm j\varphi (\omega )}\label {eq:def:Fpolar} \end {equation} The amplitude response \(A(\omega )\) describes the relative amplitude change between output and input voltage and corresponds to the magnitude of the frequency response. The phase response \(\varphi (\omega )\), on the other hand, describes the absolute phase change between output and input voltage and corresponds to the phase of the frequency response. Using \(\underline {F}(\mathrm {j}\omega )\), we can define both as follows:

\begin {align} \label {eq:def:A} A(\omega ) & = |\underline {F}(\mathrm {j}\omega )|& & = \sqrt {\left (\Re \{\underline {F}(\mathrm {j}\omega )\}\right )^2 + \left (\Im \{\underline {F}(\mathrm {j}\omega )\}\right )^2}& & = \frac {|\underline {U}_2|}{|\underline {U}_1|}\\ \label {eq:def:phi} \varphi (\omega ) & = \angle \underline {F}(\mathrm {j}\omega )& & = \arctan \left (\frac {\Im \{\underline {F}(\mathrm {j}\omega )\}}{\Re \{\underline {F}(\mathrm {j}\omega )\}}\right )& & = \angle \underline {U}_2 - \angle \underline {U}_1 \end {align}

2 Frequency response using the example of a simple low-pass filter

The two-port network in Figure 2 represents a simple low-pass filter (low-pass for short). The name low-pass is derived from the behaviour that low-frequency signals pass through almost unchanged, while high-frequency signals are strongly attenuated (filtered). The functionality and behaviour of the low-pass filter can be clearly illustrated by means of a frequency response analysis.

The low-pass filter shown consists of a series connection of \(R\) and \(C\) with input voltage \(\underline {U}_1\) across both elements and output voltage \(\underline {U}_2\) across \(C\). Since the individual elements \(R\) and \(C\) are linear, time-invariant (LTI) components, the low-pass filter composed of them is also an LTI system. This means that when the circuit is excited with a sinusoidal input voltage \(u_1(t)\) (left), the output voltage \(u_2(t)\) (right) is also sinusoidal and has the same frequency as the input voltage. \begin {equation} \begin {aligned} u_1(t)& = \hat {U}_1 \cdot \sin (\omega t)\\\xrightarrow {LZI}\ u_2(t)& = \hat {U}_2 \cdot \sin (\omega t + \varphi ) \end {aligned} \end {equation}

The LZI property is a prerequisite for determining the frequency response and applying complex AC calculations. An additional prerequisite is the assumption of sinusoidal excitation in the steady state, which we use as a basis for our analysis. Instead of the input and output voltages \(u_1(t)\) and \(u_2(t)\) in the time domain, we consider the corresponding complex input and output voltages \(\underline {U}_1\) and \(\underline {U}_2\) in the frequency domain, as shown in the circuit diagram.

According to equation 1, the frequency response \(\underline {F}(\mathrm {j}\omega )\) corresponds to the complex voltage ratio of output to input voltage \(\underline {U}_2/\underline {U}_1\). According to the complex voltage divider rule, the voltage ratio corresponds to the ratio of the impedance at the output (capacitance impedance) and the impedance at the input (sum of the capacitance and resistance impedances), or \(\frac {\underline {Z}_C}{R + \underline {Z}_C}\) for short.

Equation 4 shows the determined frequency response of the low-pass filter. For simplification, the denominator and numerator have been rationalised, the denominator sorted and the term (optionally) transformed into Cartesian coordinates. (Compare annotated side calculation ??):

\begin {equation} \begin {aligned} \underline {F}(\mathrm {j}\omega ) & = \frac {\underline {U}_2}{\underline {U}_1} = \frac {\frac {1}{\mathrm {j}\omega C}}{R + \frac {1}{\mathrm {j}\omega C}} = \frac {1}{\mathrm {j}\omega CR + 1} \\ & = \frac {1}{1 + \mathrm {j}\omega CR} = \frac {1 - \mathrm {j}\omega CR}{1 + (\omega CR)^2} \end {aligned}\label {eq:tiefpass:frequenzgang} \end {equation}

We have thus determined \(\underline {F}(\mathrm {j}\omega )\) as a function of \(\omega \) and the component values \(C\) and \(R\). Using absolute values, we can determine the amplitude response according to (??) and the phase response by calculating the phase according to (??).

The form \(\frac {1}{1+\mathrm {j}\omega CR}\) is suitable for calculating the amount.

Since the magnitude in the numerator is directly readable as \(1\), only the magnitude in the denominator needs to be determined, as shown in (??). Alternatively, the Cartesian form from (4) can also be used in (??).

For the phase shift \(\varphi (\omega )\), the Cartesian notation in the form \(\frac {1-\mathrm {j}\omega CR}{1+(\omega CR)^2}\) is suitable. In this notation, the real and imaginary parts can be easily read. As a purely real quantity, the denominator has no influence on the phase as shown in (??).

\begin {align} A(\omega ) &= |\underline {F}(\mathrm {j}\omega )| = \left |\frac {1}{1+\mathrm {j}\omega CR}\right | &&= \frac {\left |\qquad \!1\!\qquad \right |}{\left |1+\mathrm {j}\omega CR\right |} &= \frac {1}{\sqrt {1 + (\omega CR)^2}} \label {eq:tiefpass:ampli}\\ \varphi (\omega ) &= \angle \underline {F}(\mathrm {j}\omega ) = \angle \left (\frac {1 - \mathrm {j}\omega CR}{\cancel {1+(\omega CR)^2}}\right ) &&= \arctan \left ( \frac { \frac {-\omega CR}{\cancel {1+(\omega CR)^2}} }{ \frac {1}{\cancel {1+(\omega CR)^2}} } \right ) &= \arctan \left (-\omega CR\right ) \label {eq:tiefpass:phase} \end {align}

Alternatively, the phase response can also be determined from the non-Cartesian form of the frequency response. The phase response is calculated from the phase angle of the numerator subtracted from the phase angle of the denominator:

\begin {align} \varphi (\omega ) &= \angle \left (\frac {1}{1+\mathrm {j}\omega CR}\right ) = \underbrace {\arctan \left (\frac {0}{\cancel {1}}\right )}_{\varphi _{Zaehler}}-\underbrace {\arctan \left (\frac {\omega CR}{\cancel {1}}\right )}_{\varphi _{Nenner}} &= \arctan \left (-\omega CR\right ) \label {eq:tiefpass:phase:alternative} \end {align}

This approach is particularly useful if the numerator and/or phase can be read directly from the counter. This saves one calculation step when converting to Cartesian coordinates.

The following digression explains in more detail how to consider the numerator and denominator of frequency responses separately:

Digression: Calculating complex numbers in (Cartesian) Cartesian and polar coordinates

The frequency response usually takes the form of a complex fractional rational fraction. This means that both the denominator and the numerator can be represented as complex polynomials:

\begin {equation} \label {eq:def:F:komplexgebrochenrationalerbruch} \underline {F}(\mathrm {j}\omega ) = \frac {\underline {F}_Z(\mathrm {j}\omega )}{\underline {F}_N(\mathrm {j}\omega )} = \frac {a_n \omega ^n + \mathrm{d}ots + a_1 \omega + a_0}{b_m \omega ^m + \mathrm{d}ots + b_1 \omega + b_0} \qquad \text {mit} \qquad a_i, b_j \in \mathbb {C} \qquad n,m,\omega \in \mathbb {R} \end {equation}

If \(\underline {F}(\mathrm {j}\omega )\) is now available as a complex fraction with numerator and denominator in Cartesian coordinates, we can determine \(A(\omega )\) and \(\varphi (\omega )\) by separately calculating the magnitude and phase of the numerator and denominator. For this purpose, the formulas from (??) and (??) are applied for the numerator (index \(Z\)) and denominator (index \(N\)). Since the amplitude and phase response, as shown in (2), represent the polar coordinates (amplitude and phase) of the frequency response, the following calculation scheme results:

\begin {align} \label {eq:def:F:bruch} \underline {F}(\mathrm {j}\omega ) & = \frac {\underline {F}_Z(\mathrm {j}\omega )}{\underline {F}_N(\mathrm {j}\omega )}& & = \frac {|\underline {F}_Z| \cdot \mathrm {e}^{\mathrm {j}\angle {\underline {F}_Z}}}{|\underline {F}_N| \cdot \mathrm {e}^{\mathrm {j}\angle {\underline {F}_N}}}& & = \frac {A_Z}{A_N} \cdot \mathrm {e}^{\mathrm {j} (\varphi _Z - \varphi _N)} \vphantom {\Bigg |}\\ \label {eq:def:A:bruch} A(\omega ) & = |\underline {F}(\mathrm {j}\omega )|& & = \frac {|\underline {F}_Z|}{|\underline {F}_N|}& & = \frac {A_Z}{A_N}& \vphantom {\Bigg |}\\ \label {eq:def:phi:bruch} \varphi (\omega ) & = \angle \underline {F}(\mathrm {j}\omega )& & = \angle {\underline {F}_Z} - \angle {\underline {F}_N}& & = \varphi _Z - \varphi _N \end {align}

In general, polar coordinates (with magnitude and argument) are useful for multiplying and dividing complex numbers. Multiplication/division of complex numbers leads to multiplication/division of the magnitudes and addition/subtraction of the arguments. For addition and subtraction, the Cartesian form with real and imaginary parts is better suited for calculations, as these can be added/subtracted directly.

3 Boundary behaviour using the example of a simple low-pass filter

This chapter examines the boundary behaviour of the first-order low-pass filter from the previous chapter (see Figure 2) for very high and very low frequencies. To this end, the boundary values of the amplitude response \(A(\omega )\) and the phase response \(\varphi (\omega )\) are determined, and the results are described, interpreted and presented graphically. Limit values:

For the amplitude response from equation ??, the following limit values result for \(f \rightarrow 0\) and \(f \rightarrow \infty \):

\begin {align} &&A(\omega ) &=\frac {1}{\sqrt {1 + (\omega CR)^2}} & & & & &&\nonumber \\ && & & \lim _{\omega \rightarrow 0}A(\omega ) &=\frac {1}{1} & &=1 &&\label {eq:tiefpass:ampli:lim}\\ && & & \lim _{\omega \rightarrow \infty }A(\omega )&=\frac {1}{\sqrt {\infty }} & &=0 &&\nonumber \end {align}

The phase shift from equation ?? yields the following limit values: \begin {align} &&\varphi (\omega ) &=\arctan \left (-\omega CR\right )& & & & &&\nonumber \\ && & & \lim _{\omega \rightarrow 0}\varphi (\omega ) &=\arctan (0) & &=0^\circ &&\label {eq:tiefpass:phase:lim}\\ && & & \lim _{\omega \rightarrow \infty }\varphi (\omega ) &=\arctan (-\infty ) & &=-90^\circ &&\nonumber \end {align}

Equation ?? shows that the low-pass filter hardly attenuates at very low frequencies (\(A(\omega ) \rightarrow 1\)), but attenuates strongly at very high frequencies (\(A(\omega ) \rightarrow 0\)).

Equation ?? shows that the low-pass filter exhibits hardly any phase shift at very low frequencies (with hardly any attenuation) (\(\varphi (\omega ) \rightarrow 0^\circ \)), but exhibits a phase shift of up to \(-90^\circ \) at very high frequencies (strong attenuation).

Explanation of the limit behaviour based on the circuit diagram:

The boundary behaviour with regard to amplitude change and phase shift can be explained well using the circuit diagram 2.

In the limiting case of a DC voltage (\(f=0\)) at the input, the capacitance (\(X_C \rightarrow \infty \)) blocks the current. Its behaviour here corresponds to two open terminals, which means that the entire input voltage is applied to the output \(U_2\) across the capacitance \(C\). The relative amplitude change is therefore \(1\) (undamped) and the phase shift \(0^\circ \) (in phase).

At very high frequencies (\(f\rightarrow \infty \)), the capacitance behaves like a short circuit (\(X_C \rightarrow 0\)). This causes the output voltage to approach zero (strong attenuation, \(A\rightarrow 0\)) and the input voltage to be almost entirely across \(R\). The input voltage and current are therefore approximately in phase. The phase shift of the output voltage across the capacitance to the input voltage is therefore approximately \(-90^\circ \).

Visualisation:

To get an idea of the behaviour of amplitude and phase response, we can plot the functions graphically.

The amplitude response \(A(\omega )\) and the phase response \(\varphi (\omega )\) are shown in Figures 2a and 2b. Both graphs are shown with linear scales, whereby the scaling of both x-axes for \(\omega \left [\mathrm {Hz}\right ]\) are identical. A change in the factor \(CR\) in a linear representation merely leads to a compression or stretching of the graphs in the x-direction, which is why the values were not explicitly specified.

Chapter ?? introduces the representation of frequency response using a logarithmic scale. The definition of the cut-off frequency \(f_g\) (or the cut-off circular frequency \(\omega _g\)) for dividing the frequency range into a passband and a stopband is also explained there.

Digression on the derivation of the graph shape:

The graph shape of the amplitude response \(A(\omega )\) in linear representation can be explained by the fact that it approaches the horizontal straight line \(A=1\) for very low frequencies (\(f \rightarrow 0\)) and approaches the power function \(\frac {1}{\omega CR}\) for very high frequencies (\(f \rightarrow \infty \)). Figure 4 shows these approximations for \(A(x)=\frac {1}{\sqrt {1+x^2}}\) with \(x=\omega CR\).

The shape of the phase response can be easily derived from the shape of a (arc) tangent function. Figure 5 shows the tangent function and its inverse function, the arc tangent, as well as their respective reflections on the y-axis in a linear representation.

The inverse of the respective function can be obtained by reflecting the graph about the bisector (black dotted line) and swapping the x- and y-axes, provided that the respective function is continuously monotonically increasing or continuously monotonically decreasing. The tangent is therefore limited to the angle range \(-90^\circ \) to \(+90^\circ \) for the formation of the arc tangent with a value range from \(-\infty \) to \(+\infty \). Due to the non-representable pole points, the value range shown is limited to \(-6\) to \(+6\).

The phase response of the first-order low-pass filter corresponds to the plotted \(\arctan (-x)\) for \(x=\omega CR >0\). Since there are no negative frequencies in physics, \(\omega >0 \rightarrow x>0\) applies, which means that the value range of the phase response only covers half the value range of the arctangent, i.e. from \(-90^\circ \) to \(0^\circ \).

4 Frequency response using the example of a simple high-pass filter

A high-pass filter (high pass for short) works similarly to a low-pass filter. However, as the name suggests, high-pass filters allow high-frequency signals to pass through almost unattenuated (unfiltered), while low-frequency signals are strongly attenuated (filtered).

Figure 6 shows an example of a first-order RC high-pass filter, which will be examined in more detail below. The high-pass filter consists of a simple series connection of resistor \(R\) and capacitor \(C\). The input voltage \(\underline {U}_1\) is applied across \(R\) and \(C\) in series, while the output voltage \(\underline {U}_2\) is tapped across \(R\). The setup is identical to the first-order RC low-pass filter, except that the output voltage is applied across \(R\) instead of \(C\).

The frequency response is derived in the same way as for the low-pass filter in section 2. Since the high-pass filter is an LZI system, the following applies: \begin {equation} \begin {aligned} u_1(t)& = \hat {U}_1 \cdot \sin (\omega t)\\ \xrightarrow {LZI} u_2(t)& = \hat {U}_2 \cdot \sin (\omega t + \varphi ) \end {aligned} \end {equation}

This means that for a sinusoidal input voltage \(u_1(t)\), the output voltage \(u_2(t)\) is also sinusoidal with the same frequency as the input voltage.

The frequency response can thus be determined according to equation 1 by setting the complex alternating voltages \(\underline {U}_2\) at the output and \(\underline {U}_1\) at the input in relation to each other. This gives us:

\begin {equation} \begin {aligned} \underline {F}(\mathrm {j}\omega ) & = \frac {\underline {U}_2}{\underline {U}_1} = \frac {R}{R + \frac {1}{\mathrm {j}\omega C}} = \frac {1}{1 + \frac {1}{\mathrm {j}\omega CR}} \\ & = \frac {1}{1 - \mathrm {j}\ \frac {1}{\omega CR}} = \frac {1 + \mathrm {j}\ \frac {1}{\omega CR}}{1 + \frac {1}{(\omega CR)^2}} \end {aligned}\label {eq:hochpass:frequenzgang} \end {equation} The following equations result for the amplitude response \(A(\omega )\) and the phase response \(\varphi (\omega )\):

\begin {align} A(\omega ) &= |\underline {F}(\mathrm {j}\omega )| = \left |\frac {1}{1-\mathrm {j}\frac {1}{\omega CR}}\right | &= \frac {1}{\sqrt {1 + \frac {1}{(\omega CR)^2}}} \label {eq:hochpass:ampli}\\ \varphi (\omega ) &= \angle \underline {F}(\mathrm {j}\omega ) = \angle \left (\frac {1 + \mathrm {j}\frac {1}{\omega CR}}{1 + \frac {1}{(\omega CR)^2}}\right ) &= \arctan \left (\frac {1}{\omega CR}\right )\label {eq:hochpass:phase} \end {align}

5 Boundary behaviour using the example of a simple high-pass filter

The boundary behaviour of the first-order high-pass filter from Chapter 4 is examined in this chapter for very high and very low frequencies. The limit values of the amplitude response \(A(\omega )\) and the phase response \(\varphi (\omega )\) are determined in the same way as for the low-pass filter described in Chapter 3.

Limit value:

The amplitude response from equation ?? yields the following limit values for \(f \rightarrow 0\) and \(f \rightarrow \infty \):

\begin {align} &&A(\omega ) &=\frac {1}{\sqrt {1 + \left (\frac {1}{\omega CR}\right )^2}}& & && &&\nonumber \\ && && \lim _{\omega \rightarrow \infty }A(\omega ) &=\frac {1}{\infty } &&=0 &&\\ && && \lim _{\omega \rightarrow 0}A(\omega ) &=\frac {1}{\sqrt {1}}&&=1 &&\nonumber \end {align}

For the phase response from equation ??, the following applies analogously: \begin {align} &&\varphi (\omega ) &=\arctan \left (\frac {1}{\omega CR}\right )& & && &&\nonumber \\ && && \lim _{\omega \rightarrow 0}\varphi (\omega ) &=\arctan (\infty ) &&=+90^\circ &&\\ && && \lim _{\omega \rightarrow \infty }\varphi (\omega ) &=\arctan (0) &&=0^\circ &&\nonumber \end {align}

This means that at very low frequencies, the high-pass filter attenuates strongly (\(A(\omega ) \rightarrow 0\)) with a phase shift of up to \(+90^\circ \) and at very high frequencies, it attenuates hardly at all (\(A(\omega ) \rightarrow 1\)) with a phase shift of up to \(0^\circ \).

This boundary behaviour can be explained well using the circuit diagram 6. At DC voltage (\(f=0\)), the capacitance blocks, causing the entire input voltage to drop across it.

At very low frequencies (\(f \to 0\)), the input voltage is approximately completely across the capacitance. The current leads the input voltage by up to \(+90^\circ \), resulting in a positive phase shift up to \(+90^\circ \) to the output voltage across the resistor.

At very high frequencies (\(f \rightarrow \infty \)), the capacitance behaves like a short circuit, causing the input voltage to be approximately completely across the resistor at the output. As a result, the output and input voltages are approximately equal in amplitude and phase.

Visualisation:

Figure 7 shows the amplitude response \(A(\omega )\) and the phase response \(\varphi (\omega )\) of the first-order highpass filter in linear representation with the same scaling and the same value range for the x-axes.

A logarithmic representation of the frequency response is introduced in Chapter ??.

6 Comparison of first-order passive filters with \(L\) and \(R\)

Low-pass and high-pass filters can also be implemented using an inductance \(L\) and a resistor \(R\). Table 1 compares the circuit diagrams for first-order RL filters and their RC counterparts. The formula for the frequency response is also listed for each circuit diagram.

\(\vphantom {\Big |}\)	Low-pass filter	\( \underline {F}(\mathrm {j}\omega )\)	High-pass filter	\( \underline {F}(\mathrm {j}\omega )\)	\(\omega _g\)
RL		\(\frac {1}{1+\mathrm {j} \omega \frac {L}{R}}\)		\(\frac {1}{1-\mathrm {j}\frac {R}{\omega L}}\)	\(\frac {R}{L}\)
RC		\(\frac {1}{1+\mathrm {j}\omega CR}\)		\(\frac {1}{1-\mathrm {j}\frac {1}{\omega CR}}\)	\(\frac {1}{CR}\)

A comparison of the circuit diagrams shows that the RL and RC variants differ only in the arrangement of \(L\) and \(R\) and \(R\) and \(C\), respectively. The frequency responses of the RL and RC variants differ only in the factor \(\frac {L}{R}\) and \(CR\) before \(\omega \). As will be explained in more detail in Chapter ??, this factor corresponds to the reciprocal of the cutoff frequency \(\omega _g\), which is included in the table for the sake of completeness.

7 Example filter circuit: Task

Filterschaltung: Aufgabe

Filter circuit: SolutionSolution a) Frequency response in general \begin {align*} \underline {F}(\mathrm {j}\omega ) &= \frac {R_1 || \underline {Z}_C}{R_2 + ( R_1 || \underline {Z}_C)} = \left .\frac {\frac {\frac {R_1}{\mathrm {j}\omega C}}{R_1 + \frac {1}{\mathrm {j}\omega C}}}{ R_2 + \frac {\frac {R_1}{\mathrm {j}\omega C}}{R_1 + \frac {1}{\mathrm {j}\omega C}}} \right | \cdot \frac {R_1 + \frac {1}{\mathrm {j}\omega C}}{R_1 + \frac {1}{\mathrm {j}\omega C}} \\[+2pt] &= \frac {\frac {R_1}{\mathrm {j}\omega C}}{R_1\cdot R_2 + \frac {R_2}{\mathrm {j}\omega C}+\frac {R_1}{\mathrm {j}\omega C}} = {hfid_polar}{\frac {R_1}{R_1+R_2+\mathrm {j}\omega CR_1R_2}}[2] &&\hat {=}{} \frac {a+jb}{c+jd} & a,b,c,d &\in \mathbb {R} \\[+6pt] &= {hfid_kart}{\frac {R_1(R_1+R_2)-\mathrm {j}\omega CR_1^2R_2}{(R_1+R_2)^2+(\omega CR_1R_2)^2}}[3] \text {(Cartesian)} &&\hat {=}{} \frac {a'+jb'}{c'} & a',b',c' &\in \mathbb {R} \end {align*}

\begin {align*} &\text {Amount:}& |A(\omega )| &= \frac {R_1}{\sqrt {(R_1+R_2)^2+(\omega CR_1R_2)^2}}& &\text {with}& |\underline {F}(\mathrm {j}\omega )| &= \frac {\sqrt {a^2+b^2}}{\sqrt {c^2+d^2}}\\ &\text {Phase:}& \varphi (\omega ) &= \arctan \left (\frac {-\omega C R_1^2R_2}{R_1(R_1+R_2)}\right )& &\text {with}& \varphi (\omega ) &= \arctan \left (\frac {b'}{a'}\right )\\[+2pt] && &= \arctan \left (\frac {-\omega C R_1R_2}{R_1+R_2}\right )\\[-4pt] &\text {Altern.:}& \varphi (\omega ) &= \arctan (0) - \arctan \left (\frac {\omega CR_1R_2}{R_1+R_2}\right )& &\text {with}& \varphi (\omega ) &= \overbrace {\varphi _{Z}}^{{\scriptscriptstyle {\arctan (\frac {b}{a})}\ \ }} - \overbrace {\varphi _{N}}^{{\ -\ \scriptscriptstyle {\arctan (\frac {d}{c})}}} \end {align*}

Solution b) Limit behaviour and sketch

\(A(\omega ) = \frac {R_1}{\sqrt {(R_1+R_2)^2+(\omega CR_1R_2)^2}}\)

\(\ \ \ \ \lim _{\omega \to 0} A(\omega ) = \frac {R_1}{R_1+R_2} = 0,9\)

\(\ \ \ \ \lim _{\omega \to \infty } A(\omega ) = \frac {R_1}{\infty } = 0\)

\(\varphi (\omega ) = \arctan (-\frac {\omega CR_1R_2}{R_1+R_2})\)

\(\lim _{\omega \to 0} \varphi (\omega ) = \arctan (0) = 0\)

\(\lim _{\omega \to \infty } \varphi (\omega ) = \arctan (-\infty ) = -90^\circ \)

Solution c) Determine specific operating point

Total: Frequency \(f_{1}\) where \(\hat {U}_2 = \frac {1}{2}\hat {U}_1\) and phase shift \(u_2\) relative to \(u_1\) at \(f_1\):\begin {align*} A(\omega _1) &= \frac {R_1}{\sqrt {(R_1+R_2)^2+(\omega CR_1R_2)^2}} \overset {!}{=} \frac {1}{2} & &\text {with}& A(\omega ) = \frac {U_2}{U_1} = \frac {\hat {U}_2}{\hat {U}_1} \overset {!}{=} \left .\frac {1}{2}\right |_{f=f_1} \\[+2pt] &\Rightarrow \frac {(R_1+R_2)^2 + (\omega _1 CR_1R_2)^2}{R_1^2} = 4 & &\Rightarrow & (R_1+R_2)^2 + (\omega _1 C R_1R_{\mathrm {C}})^2 = 4R_1^2 \\ &\Rightarrow (\omega _1 CR_1R_2)^2 = 4R_1^2 - (R_1+R_2)^2 & &\Rightarrow & \omega _1 = \pm \frac {\sqrt {4R_1^2 - (R_1^2 + 2R_1R_2 + R_2^2)}}{CR_1R_2} \\[+2pt] \omega _1 &= 1,3304\cdot 10^4\ \frac {1}{\mathrm {s}},\ f_1 = 2,117\ \mathrm {kHz} & &\text {with}& \omega _1 < 0 \text { unphysical},\ f=\frac {\omega }{2\pi } \\ \varphi (\omega _1) &= \arctan (-1,4967) = -56,25^\circ & &\text {with}& \varphi (\omega ) = \arctan \left (\frac {\Im \{\underline {F}\}}{\Re \{\underline {F}\}}\right ) \end {align*}