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| ====== Real Analog: Chapter 11 ====== | ====== Real Analog: Chapter 11 ====== | ||
| - | ====== 11. Introduction and Chapter | + | [[{}/ |
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| + | <-- | ||
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| + | ===== 11. Introduction and Chapter Objectives | ||
| In section 10.6, we saw that a system’s frequency response provided a steady-state input-output relationship for a system, as a function of frequency. We could apply this frequency response to the phasor representation of the input signal in order to determine the system’s steady-state sinusoidal response – we simply evaluated the frequency response at the appropriate frequencies to determine the effect of the system on the input sinusoids. This approach had the potential for simplifying our analysis considerably, | In section 10.6, we saw that a system’s frequency response provided a steady-state input-output relationship for a system, as a function of frequency. We could apply this frequency response to the phasor representation of the input signal in order to determine the system’s steady-state sinusoidal response – we simply evaluated the frequency response at the appropriate frequencies to determine the effect of the system on the input sinusoids. This approach had the potential for simplifying our analysis considerably, | ||
| - | In [[https:// | + | In [[/ |
| - // | - // | ||
| - //Designing systems to provide a desired frequency response//. Audio components in stereo systems are generally designed to produce a desired frequency response. A graphic equalizer, for example, can be used to boost (or amplify) some frequency ranges and attenuate other frequency ranges. When adjusting the settings on an equalizer, you are essentially directly adjusting the system’s frequency response to provide a desired system response. | - //Designing systems to provide a desired frequency response//. Audio components in stereo systems are generally designed to produce a desired frequency response. A graphic equalizer, for example, can be used to boost (or amplify) some frequency ranges and attenuate other frequency ranges. When adjusting the settings on an equalizer, you are essentially directly adjusting the system’s frequency response to provide a desired system response. | ||
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| In Figure 11.2, the output is determined by multiplying the phasor representation of the input by the system’s frequency response. It is important to keep in mind that the arguments of this multiplication are complex functions of frequency – both the input and the frequency response at any frequency are complex numbers, so the output at any frequency is also a complex number. We typically use polar form to represent these complex numbers, so the __amplitude of the output signal is the product of the amplitude of the input signal and the magnitude response of the system__ and the __phase of the output signal is the sum of the phase of the input and the phase response of the system__. Mathematically, | In Figure 11.2, the output is determined by multiplying the phasor representation of the input by the system’s frequency response. It is important to keep in mind that the arguments of this multiplication are complex functions of frequency – both the input and the frequency response at any frequency are complex numbers, so the output at any frequency is also a complex number. We typically use polar form to represent these complex numbers, so the __amplitude of the output signal is the product of the amplitude of the input signal and the magnitude response of the system__ and the __phase of the output signal is the sum of the phase of the input and the phase response of the system__. Mathematically, | ||
| - | $$|\underline{Y}(j \omega)|=|\underline{U}(j \omega)| \cdot |H (j \omega)| | + | $$|\underline{Y}(j \omega)|=|\underline{U}(j \omega)| \cdot |H (j \omega)| |
| And | And | ||
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| $$\underline{U}_{j \omega} = \begin{cases} | $$\underline{U}_{j \omega} = \begin{cases} | ||
| - | j \omega, & 0 < \leq 1 \\ | + | j \omega, & 0 < \omega |
| j(2-\omega), | j(2-\omega), | ||
| 0, & \text{otherwise} | 0, & \text{otherwise} | ||
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| And: | And: | ||
| - | $$\angle H(j \omega) = =\tan^{-1}(\frac{\omega}{2})$$ | + | $$\angle H(j \omega) = -\tan^{-1}(\frac{\omega}{2})$$ |
| Plotting these functions results in the graphical frequency response shown below: | Plotting these functions results in the graphical frequency response shown below: | ||
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| ==== Checking the Frequency Response ==== | ==== Checking the Frequency Response ==== | ||
| A circuit’s __amplitude response__ is relatively easy to determine at low and high frequencies. For very low ($\omega \rightarrow 0$) and very high ($\omega \rightarrow \infty$) frequencies, | A circuit’s __amplitude response__ is relatively easy to determine at low and high frequencies. For very low ($\omega \rightarrow 0$) and very high ($\omega \rightarrow \infty$) frequencies, | ||
| - | * __Capacitors at low and high frequencies__: | + | * __Capacitors at low and high frequencies__: |
| - | * __Inductors at low and high frequencies__: | + | * __Inductors at low and high frequencies__: |
| Please note that the above statements are relative only to the amplitude response. | Please note that the above statements are relative only to the amplitude response. | ||
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| {{ : | {{ : | ||
| - | As $\omega \rightarrow 0$, the circuit becomes a voltage divider, and $v_{out} = \frac{3k \Omega}{2k \Omega + 3k \Omega} \cdot v_{in}$, so that the circuit;s gain is $\frac{3}{5}$. As $\omega \rightarrow \infty$, $v_{out}$ is measured across a short circuit, and the gain is zero. | + | As $\omega \rightarrow 0$, the circuit becomes a voltage divider, and $v_{out} = \frac{3k \Omega}{2k \Omega + 3k \Omega} \cdot v_{in}$, so that the circuit's gain is $\frac{3}{5}$. As $\omega \rightarrow \infty$, $v_{out}$ is measured across a short circuit, and the gain is zero. |
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| The maximum magnitude response is approximately //K// when $\omega \rightarrow \infty$ and the magnitude response is zero at $\omega = 0$. Thus, the filter is passing high frequencies and stopping low frequencies. __The differential equation describes a high-pass filter__. | The maximum magnitude response is approximately //K// when $\omega \rightarrow \infty$ and the magnitude response is zero at $\omega = 0$. Thus, the filter is passing high frequencies and stopping low frequencies. __The differential equation describes a high-pass filter__. | ||
| - | The magnitude response of the filter is shown in Figure 11.8. As with the non-ideal low-pass filter, the frequency response is a smooth curve, rather than the discontinuous function shown in Figure 11.6. Again, there is no single frequency that obviously separates the passband from the stopband, so we must choose a relatively arbitrary point to define the boundary between the passband and the stopband. Consistent with our choice of cutoff frequency for the low-pass filter, the cutoff frequency for a high-pass filter is defined as the frequency at which the magnitude response is $\frac{1}{\sqrt{2}}# times the magnitude response at $\omega \rightarrow \infty$. For the magnitude response given by equation (1.10), the cutoff frequency is $\omega = \omega_c$. This point is indicated on Figure 11.8. | + | The magnitude response of the filter is shown in Figure 11.8. As with the non-ideal low-pass filter, the frequency response is a smooth curve, rather than the discontinuous function shown in Figure 11.6. Again, there is no single frequency that obviously separates the passband from the stopband, so we must choose a relatively arbitrary point to define the boundary between the passband and the stopband. Consistent with our choice of cutoff frequency for the low-pass filter, the cutoff frequency for a high-pass filter is defined as the frequency at which the magnitude response is $\frac{1}{\sqrt{2}}$ times the magnitude response at $\omega \rightarrow \infty$. For the magnitude response given by equation (1.10), the cutoff frequency is $\omega = \omega_c$. This point is indicated on Figure 11.8. |
| {{ : | {{ : | ||
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| * The cutoff frequency is also called the corner frequency, the 3dB frequency, or the half-power point. | * The cutoff frequency is also called the corner frequency, the 3dB frequency, or the half-power point. | ||
| * The cutoff frequency for __both__ low-pass and high-pass filters is defined as the frequency at which the magnitude is $\frac{1}{\sqrt{2}}$ times the __maximum__ value of the magnitude response. | * The cutoff frequency for __both__ low-pass and high-pass filters is defined as the frequency at which the magnitude is $\frac{1}{\sqrt{2}}$ times the __maximum__ value of the magnitude response. | ||
| - | * It can be seen from examples in section 11.2 that the phase response of a first order low-pass filter is $0^{\circ}$ at $\omaga = 0$ and decreases to $-90^{\circ}$ as $\omega \rightarrow \infty$. The phase response is $-45^{\circ}$ at the cutoff frequency. | + | * It can be seen from examples in section 11.2 that the phase response of a first order low-pass filter is $0^{\circ}$ at $\omega = 0$ and decreases to $-90^{\circ}$ as $\omega \rightarrow \infty$. The phase response is $-45^{\circ}$ at the cutoff frequency. |
| * It can be seen from examples in section 11.2 that the phase response of a first order high-pass filter is $90^{\circ}$ at $\omega = 0$ and decreases to $0^{\circ}$ as $\omega \rightarrow \infty$. The phase response is $45^{\circ}$ at the cutoff frequency. | * It can be seen from examples in section 11.2 that the phase response of a first order high-pass filter is $90^{\circ}$ at $\omega = 0$ and decreases to $0^{\circ}$ as $\omega \rightarrow \infty$. The phase response is $45^{\circ}$ at the cutoff frequency. | ||
| * For both low-pass and high-pass filters, the cutoff frequency is the inverse of the time constant for the circuit, so that $\omega_c = \frac{1}{\tau}$. | * For both low-pass and high-pass filters, the cutoff frequency is the inverse of the time constant for the circuit, so that $\omega_c = \frac{1}{\tau}$. | ||
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| ==== Example 11.12: First Order Low-pass Filter ==== | ==== Example 11.12: First Order Low-pass Filter ==== | ||
| The circuit below is the circuit from example 11.6. The input is $v_{in}(t)$ and the output is $v_{out}(t)$. | The circuit below is the circuit from example 11.6. The input is $v_{in}(t)$ and the output is $v_{out}(t)$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | In example 11.6, the frequency response was determined to be: | ||
| + | |||
| + | $$H(j \omega) = \frac{2}{2+ j \omega}$$ | ||
| + | |||
| + | The maximum value of the magnitude response is one at a frequency of zero radians/ | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 11.13: First Order High-pass Filter ==== | ||
| + | The circuit below is the circuit from example 11.7. The input is $v_s(t)$ and the output is $v(t)$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The frequency response of the circuit was previously determined to be: | ||
| + | |||
| + | $$H(j \omega) = \frac{\underline{V}}{\underline{V}_S} = \frac{j2 \omega}{1+ j2 \omega}$$ | ||
| + | |||
| + | The maximum value of the magnitude response is one as $\omega \rightarrow \infty$ and goes to zero at a frequency of zero radians/ | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * Filters are frequency-selective systems. These systems provide a desired relationship between the input signal spectrum and the output signal spectrum. The filter does this by //passing// certain frequencies from the input to the output and // | ||
| + | * The range of frequencies which are passed is called the // | ||
| + | * The range of frequencies which are stopped is called the // | ||
| + | * Filters are broadly categorized as follows: | ||
| + | * //Low-pass filters// pass low frequencies and stop high frequencies | ||
| + | * //High-pass filters// pass high frequencies and stop low frequencies | ||
| + | * //Band-pass filters// pass a range of frequencies between two ranges of stopped frequencies | ||
| + | * // | ||
| + | * In this chapter, we were concerned only with first-order low-pass and high-pass filters. These filters are primarily characterized by the following parameters: | ||
| + | * Cutoff frequency: the cutoff is defined as the frequency at which the magnitude is $\frac{1}{\sqrt{2}}$ times the __maximum__ value of the magnitude response. | ||
| + | * //DC gain//: the DC gain is the ratio of the output amplitude to the input amplitude for a __constant__ input (a cosine function with zero frequency). Low pass filters have a relatively high DC gain and a correspondingly large response to a constant input. High pass filters have a low DC gain; they have little or no response to constant inputs. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | **1.** What is the cutoff frequency of the circuit below? (You may want to use your results from exercise 1 of section 11.1.). | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | **2.** Use the circuit behavior at high and low frequencies and your time constant calculated in exercise 1 above to sketch the frequency response of the circuit of exercise 1. Label your frequency response to include DC gain and cutoff frequency. Is the circuit a high-pass or low-pass filter? | ||
| + | |||
| + | **3.** Calculate the time constant and the cutoff frequency for the circuit below, if u(t) is the input and $y(t)$ is the output. Verify that the cutoff frequency is the inverse of the time constant. Use the circuit behavior at high and low frequencies to determine whether the circuit is a high-pass or low-pass filter. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | **4.** Calculate the frequency response of the circuit of exercise 3. Compare your frequency response to your results of exercise 3. Resolve any differences between the two. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 11.4: Introduction to Bode Plots ===== | ||
| + | Plotting a systems’ frequency response on a linear scale, as done in sections 11.3 and 11.4, has a number of drawbacks, especially for higher-order systems((Higher order systems are often modeled as a series of lower-order systems (or in the technical parlance, // | ||
| + | * Use of logarithms converts the operation of multiplication and division to addition and subtraction. This can simplify the creation of frequency response plots for higher order systems. | ||
| + | * Frequencies and amplitudes of interest commonly span many orders of magnitude. Logarithmic scales improve the presentation of this type of data. | ||
| + | * Human senses are fundamentally logarithmic. The use of logarithmic scales is therefore more “natural”. (This is the reason for use of the Richter scale in measuring earthquake intensity, and the decibel scale in measuring sound levels. It is also the reason that increasing a musical tone by one octave corresponds to doubling its frequency.) | ||
| + | |||
| + | ==== Properties of Logarithms ==== | ||
| + | Since Bode plots employ logarithms extensively, | ||
| + | |||
| + | A plot of $log_{10}(x)$ vs. $x$ is shown in Figure 11.9 below. A few important features to note are: | ||
| + | |||
| + | * $log_{10}(x)$ is a real number only for positive values of x. | ||
| + | * $log_{10}(x)$ asymptotically approaches $-\infty$ as $x \rightarrow 0$. The slope of $log_{10}(x)$ becomes very large as $x \rightarrow 0$. | ||
| + | * The slope of $log_{10}(x)$ becomes small as $x \rightarrow \infty$. | ||
| + | * From the comments above relative to the slope of $log_{10}(x)$, | ||
| + | * $log_{10}(1) = 0$ | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The basic defining property of a base-10 logarithm is that $x=10^y$, then $y = log_{10} x$. This property leads to the following rules governing logarithmic operations: | ||
| + | |||
| + | **1.** Logarithms convert multiplication and division to addition and subtraction, | ||
| + | |||
| + | $$log_{10}(xy) = log_{10}x+log_{10}y$$ | ||
| + | |||
| + | And: | ||
| + | |||
| + | $$log_{10} \left( \frac{x}{y} \right) = log_{10}x-log_{10}y$$ | ||
| + | |||
| + | This property is especially useful for us, since determining the spectrum of an output signal results from the product of an input signal’s spectrum with the frequency response. Thus, the output spectrum on a logarithmic scale can be obtained from a simple addition. | ||
| + | |||
| + | **2.** Logarithms convert exponentiation to multiplication by the exponent, so that: | ||
| + | |||
| + | $$log_{10}(x^n)=nlog_{10}x$$ | ||
| + | |||
| + | |||
| + | ==== Decibel Scales ==== | ||
| + | Magnitude responses are often presented in terms of decibels (abbreviated dB). Decibels are a logarithmic scale. A magnitude response is presented in units of decibels according to the following conversion: | ||
| + | |||
| + | $$|H(j \omega)|_{dB} = 20log_{10} (|H(j \omega)|) | ||
| + | |||
| + | Strictly speaking, magnitudes in decibels are only appropriate if the amplitude response is unitless (e.g. the units of the input and output must be the same in order for the logarithm to be a mathematically appropriate operation). However, in practice, magnitude responses are often presented in decibels regardless of the relative units of the input and output – thus, magnitude responses are provided in decibels even if the input is voltage and the output is current or vice-versa. | ||
| + | |||
| + | === Brief Historical Note === | ||
| + | Decibel units are related to the unit “bel”, which are named after Alexander Graham Bell. Units of bels are, strictly speaking, applicable only to power. Power in bels is expressed as $log_{10}( \frac{P}{P}_{ref} )$, where $P_{ref}$ is a “reference” power. Bels are an inconveniently large unit, so these were converted to decibels, or tenths of a bel. Thus, power in decibels is $10log_{10}(\frac{P}{P}_{ref})$. Since the units of interest to electrical engineers are generally voltages or currents, which must be squared to obtain power, we obtain $20log_{10}(|H(j \omega)|)$. The significant aspect of the decibel unit for us is not, however, the multiplicative factor of “20”), but the fact that the unit is __logarithmic__. | ||
| + | |||
| + | We conclude this subsection with a table of common values for $|H(j \omega)|$ and their associated decibel values. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | |||
| + | ==== Bode Plots ==== | ||
| + | Bode plots are simply plots of the magnitude and phase response of a system using a particular set of axes. For Bode plots, | ||
| + | * Units of frequency are on a base-10 logarithm scale. | ||
| + | * Amplitudes (or magnitudes) are in decibels (dB) | ||
| + | * Phases are presented on a linear scale | ||
| + | |||
| + | === Notes === | ||
| + | * Since frequencies are on a logarithmic scale, frequencies separated by the same multiplicative factor are evenly separated on a logarithmic scale. Some of these multiplicative factors have special names. For example, frequencies separated by a factor of two are said to be separated by //octaves// on a logarithmic scale and frequencies separated by a factor of 10 are said to be separated by // | ||
| + | * Since decibels are intrinsically a logarithmic scale, magnitudes which are separated by the same multiplicative factor are evenly separated on a decibel scale. For example, magnitudes which are separated by a factor of 10 are separated by 20dB on a decibel scale. | ||
| + | |||
| + | One convenient aspect of the presentation of frequency responses in terms of Bode plots is the ability to generate a reasonable sketch of a frequency response very easily. In general, this approach consists of approximating the Bode plot of a system by its asymptotic behavior as a set of straight lines. This is called a “straight line approximation” of the Bode plot; the approach is illustrated for a typical low-pass filter in the following subsection. | ||
| + | |||
| + | ==== Bode Plots for First Order Low-pass Filters ==== | ||
| + | The frequency response of a general first order low-pass filter is provided in section 11.3 as: | ||
| + | |||
| + | $$H(j \omega) = \frac{K}{j \omega + \omega_c} | ||
| + | |||
| + | Thus, magnitude response of the circuit is: | ||
| + | |||
| + | $$|H(j \omega)| = \frac{K}{\sqrt{\omega^2 + \omega^2_c}} (Eq. 11.13)$$ | ||
| + | |||
| + | And the phase response of the circuit is: | ||
| + | |||
| + | $$\angle H(j \omega) = -\tan^{-1} \left( \frac{\omega}{\omega_c} \right) | ||
| + | |||
| + | To estimate the asymptotic behavior of the frequency response, we consider the behavior of equations (11.13) and (11.14) for the low frequency and high frequency cases. In general, we consider “low” frequencies to be frequencies which are __less than a factor of 10 below the cutoff frequency__ (i.e. $\omega < \frac{\omega_c}{10}$, | ||
| + | |||
| + | * __Low frequencies__: | ||
| + | * The magnitude response given by equation (2) is $|H(j \omega)| = \frac{K}{\sqrt{\omega^2 + \omega^2_c}}$. If $\omega << \omega$, the denominator is approximately $\sqrt{\omega^2_c}=\omega_c$ and the magnitude response $|H(j \omega)| \approx \frac{K}{\omega_c}$. If $\omega << \omega_c$, $\frac{\omega}{\omega_c} \approx 0$ and the phase response is approximately $\angle H(j \omega) \approx -\tan^{-1}(0) = 0^{\circ}$. | ||
| + | * __High frequencies__: | ||
| + | * If $\omega >> \omega_c$, the denominator of the amplitude response is $\sqrt{\omega^2 + \omega^2_c} \approx \sqrt{\omega^2} = \omega$. Therefore, for high frequencies, | ||
| + | |||
| + | === Summary: Low-pass Filter Straight-line Bode Plot Approximations === | ||
| + | The straight line approximation to the magnitude response is constant below the cutoff frequency, with a value (in decibels) of $20log_{10} \left( \frac{K}{\omega_c} \right)$. Above the cutoff frequency, the Bode plot straight-line approximation has a constant slope of -20 dB/decade. | ||
| + | |||
| + | The straight-line approximation to the phase response is zero degrees up to a frequency of $\frac{\omega_c}{10}$ and is $-90^{\circ}$ above a frequency of $10 \omega_c$. A straight line is used ot connect the $\frac{\omega_c}{10}$ and $10 \omega_c$ frequencies. | ||
| + | |||
| + | A straight-line approximation to the Bode plot for a typical low-pass circuit, with $K = \omega_c$ (so that the frequency response is $H(j \omega) = \frac{\omega_c}{j \omega + \omega_c}$ and the DC gain is 1, or 0dB) along with an exact curve is provided below in Figure 11.10. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | We conclude this section with a numerical example of the straight-line approximation to a Bode plot for a specific circuit. | ||
| + | |||
| + | ==== Example 11.14 ==== | ||
| + | Sketch a straight-line approximation to the Bode plot for the circuit below. The input is $v_{in}(t)$ and the output is $v_{out}(t)$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The frequency response for this circuit is $H(j \omega) = \frac{1}{j \omega + 2}$. Therefore, the cutoff frequency is $\omega_c=2$ rad/sec and the gain in decibels at low frequencies is $|H(j0)|_{dB} = 20log_{10} \left( \frac{1}{2} \right) \approx -6dB$. Thus, the straight-line magnitude response is -6dB below the cutoff frequency and decreases by 20dB/decade above the cutoff frequency. The straight-line phase response is $0^{\circ}$ below 0.2rad/sec, $-90^{\circ}$ above 20 rad/sec and a straight line between these frequencies. The associated plots are shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * Bode plots are a very useful format for plotting frequency responses. Bode plots provide magnitude responses and phase responses in the following format: | ||
| + | * Units of frequency are on a base-10 logarithm scale. | ||
| + | * Amplitudes (or magnitudes) are in decibels (dB) | ||
| + | * Phases are presented on a linear scale. | ||
| + | * Magnitude responses in decibels are calculated according to: | ||
| + | $$|H(j \omega)|_{dB} = 20log_{10}(|H(j \omega)|)$$ | ||
| + | * The use of logarithmic scales in Bode plots has a number of advantages. Logarithms convert multiplication and division into addition and subtraction, | ||
| + | * Bode plots also have the advantage of being approximated fairly well by straight-line approximations. This allows the engineer to sketch a fairly accurate frequency response plot with only a minimal number of calculations. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | **1.** Sketch a Bode plot (straight-line approximation) for the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | **2.** Sketch a Bode plot (straight-line approximation) for the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | [[{}/ | ||
| + | [[{}/ | ||