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What is the relationship between the pole zero diagram and bode plot? Case example: a 2nd order RLC low pass filter: - $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{\frac{1}{j\omega C}}{R+j\omega L +\frac{1}...
#3: Post edited
- What is the relationship between the pole zero diagram and bode plot?
- Case example: a 2nd order RLC low pass filter: -
- ![Image alt text](https://electrical.codidact.com/uploads/uXVdDX5xEQTBgCncyP2m5tYU)
- $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{\frac{1}{j\omega C}}{R+j\omega L +\frac{1}{j\omega C}} = \dfrac{1}{j\omega RC +j^2\omega^2 LC +1} = \dfrac{\frac{1}{LC}}{\frac{1}{LC}-\omega^2 +j\omega\frac{RC}{LC}}
- $$
Or- $$\dfrac{V_{OUT}}{V_{IN}} =\dfrac{\frac{1}{LC}}{s^2 + s\frac{RC}{LC} + \frac{1}{LC}}
$$
- What is the relationship between the pole zero diagram and bode plot?
- Case example: a 2nd order RLC low pass filter: -
- ![Image alt text](https://electrical.codidact.com/uploads/uXVdDX5xEQTBgCncyP2m5tYU)
- $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{\frac{1}{j\omega C}}{R+j\omega L +\frac{1}{j\omega C}} = \dfrac{1}{j\omega RC +j^2\omega^2 LC +1} = \dfrac{\frac{1}{LC}}{\frac{1}{LC}-\omega^2 +j\omega\frac{RC}{LC}}
- $$
- Or, in "s" terms: -
- $$\dfrac{V_{OUT}}{V_{IN}} =\dfrac{\frac{1}{LC}}{s^2 + s\frac{RC}{LC} + \frac{1}{LC}}
- $$
- Therefore we know the poles are defined when the denominator equals zero and: -
- $$s = \dfrac{-R}{2L} ± j\sqrt{\dfrac{1}{LC} - \dfrac{R^2}{4L^2}}$$
- But how does that affect the bode plot?
#2: Post edited
What is the relationship between the pole zero diagram and bode plot - use the case of a 2nd order RLC low pass filter as an example: -![Image alt text](https://electrical.codidact.com/uploads/uXVdDX5xEQTBgCncyP2m5tYU)
- What is the relationship between the pole zero diagram and bode plot?
- Case example: a 2nd order RLC low pass filter: -
- ![Image alt text](https://electrical.codidact.com/uploads/uXVdDX5xEQTBgCncyP2m5tYU)
- $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{\frac{1}{j\omega C}}{R+j\omega L +\frac{1}{j\omega C}} = \dfrac{1}{j\omega RC +j^2\omega^2 LC +1} = \dfrac{\frac{1}{LC}}{\frac{1}{LC}-\omega^2 +j\omega\frac{RC}{LC}}
- $$
- Or
- $$\dfrac{V_{OUT}}{V_{IN}} =\dfrac{\frac{1}{LC}}{s^2 + s\frac{RC}{LC} + \frac{1}{LC}}
- $$