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learn:courses:real-analog-chapter-11:start [2017/01/18 21:40] – [First Order High-pass Filters] Martha | learn:courses:real-analog-chapter-11:start [2023/02/08 20:37] (current) – external edit 127.0.0.1 | ||
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====== Real Analog: Chapter 11 ====== | ====== Real Analog: Chapter 11 ====== | ||
- | ====== 11. Introduction and Chapter | + | [[{}/ |
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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 [[/ |
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- //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 frequency response for this circuit is $H(j \omega) = \frac{1}{j \omega + 2}$. Therefore, the cuttof | + | The frequency response for this circuit is $H(j \omega) = \frac{1}{j \omega + 2}$. Therefore, the cutoff |
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