Differences
This shows you the differences between two versions of the page.
| Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
| learn:courses:real-analog-chapter-11:start [2017/05/05 13:51] – [Checking the Frequency Response] Martha | learn:courses:real-analog-chapter-11:start [2023/02/08 12:37] (current) – external edit 127.0.0.1 | ||
|---|---|---|---|
| Line 1: | Line 1: | ||
| ====== Real Analog: Chapter 11 ====== | ====== Real Analog: Chapter 11 ====== | ||
| - | --> | + | [[{}/ |
| - | * {{ : | + | --> |
| + | | ||
| + | | ||
| + | * {{ : | ||
| + | * [[http:// | ||
| + | * [[http:// | ||
| + | * Chapter 11 Videos: | ||
| + | * [[http:// | ||
| + | * [[http:// | ||
| + | * [[http:// | ||
| + | * {{ : | ||
| * {{ : | * {{ : | ||
| - | * {{ : | + | * {{ : |
| * {{ : | * {{ : | ||
| - | * {{ : | + | * {{ : |
| * {{ : | * {{ : | ||
| - | * {{ : | + | * {{ : |
| * {{ : | * {{ : | ||
| - | * {{ : | + | * {{ : |
| * {{ : | * {{ : | ||
| - | * {{ : | + | * {{ : |
| * {{ : | * {{ : | ||
| - | * {{ : | + | * {{ : |
| * {{ : | * {{ : | ||
| - | * {{ : | + | * {{ : |
| * {{ : | * {{ : | ||
| - | * {{ : | + | * {{ : |
| * {{ : | * {{ : | ||
| - | * {{ : | + | * {{ : |
| * {{ : | * {{ : | ||
| * {{ : | * {{ : | ||
| Line 26: | Line 35: | ||
| <-- | <-- | ||
| - | ====== 11. Introduction and Chapter Objectives | + | ===== 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. | ||
| Line 364: | Line 373: | ||
| {{ : | {{ : | ||
| - | 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. |
| Line 682: | Line 691: | ||
| {{ : | {{ : | ||
| - | 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 |
| {{ : | {{ : | ||
| Line 710: | Line 719: | ||
| {{ : | {{ : | ||
| + | |||
| + | [[{}/ | ||
| + | [[{}/ | ||