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| learn:courses:real-analog-chapter-10:start [2017/05/04 19:09] – old revision restored (2017/05/04 18:48) Martha | learn:courses:real-analog-chapter-10:start [2022/10/04 11:38] (current) – external edit 127.0.0.1 | ||
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| ====== Real Analog: Chapter 10 ====== | ====== Real Analog: Chapter 10 ====== | ||
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| $$\frac{d^2v(t)}{dt^2} + \frac{R}{L} \frac{dv(t)}{dt} + \frac{1}{LC} v(t) = \frac{1}{LC} v_S(t) (Eq. 10.53)$$ | $$\frac{d^2v(t)}{dt^2} + \frac{R}{L} \frac{dv(t)}{dt} + \frac{1}{LC} v(t) = \frac{1}{LC} v_S(t) (Eq. 10.53)$$ | ||
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| By the definition of impedance, we can determine the current through the capacitor to be: | By the definition of impedance, we can determine the current through the capacitor to be: | ||
| - | $$\underline{I} = \frac{\underline{Y}}{\underline{Z}_C} = \frac{\underline{Y}}{\frac{1}{j \omega C}} = j \omega C \underline{Y}%% | + | $$\underline{I} = \frac{\underline{Y}}{\underline{Z}_C} = \frac{\underline{Y}}{\frac{1}{j \omega C}} = j \omega C \underline{Y}$$ |
| The voltage across the resistor can now, by the definition of impedance, be written as $\underline{V}_R = R \cdot \underline{I} = R (j \omega C \underline{Y})$. We now apply KVL for phasors to the circuit to the right above, which leads to: | The voltage across the resistor can now, by the definition of impedance, be written as $\underline{V}_R = R \cdot \underline{I} = R (j \omega C \underline{Y})$. We now apply KVL for phasors to the circuit to the right above, which leads to: | ||
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| Solving for $\underline{Y}$ in this equation provides $\underline{Y} = \frac{V_P \angle 0^{\circ}}{1+ j \omega RC}$ | Solving for $\underline{Y}$ in this equation provides $\underline{Y} = \frac{V_P \angle 0^{\circ}}{1+ j \omega RC}$ | ||
| - | By the rules of complex arithmetic, we can determine the magnitude and phase angle of &\underline{Y}$ to be: | + | By the rules of complex arithmetic, we can determine the magnitude and phase angle of $\underline{Y}$ to be: |
| $$|\underline{Y}| = \frac{V_P}{\sqrt{1+(\omega RC)^2}}$$ | $$|\underline{Y}| = \frac{V_P}{\sqrt{1+(\omega RC)^2}}$$ | ||
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| $$i(t)=0.98 \cos(2t - 78.7^{\circ})$$ | $$i(t)=0.98 \cos(2t - 78.7^{\circ})$$ | ||
| - | Which is consistent | + | Which is consistent |
| === Important Result: === | === Important Result: === | ||
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| - | When $v_{in}(t) = 3 \cos(2t+20^{\circ})$, | + | When $v_{in}(t) = 3 \cos(2t+20^{\circ})$, |
| $$|H(j2)|= \frac{1}{\sqrt{1+(\omega RC)^2}} = \frac{1}{\sqrt{1+(2 \cdot 2\Omega \cdot 0.25F)^2}} = \frac{1}{\sqrt{1+1^2}}=\frac{1}{\sqrt{2}}$$ | $$|H(j2)|= \frac{1}{\sqrt{1+(\omega RC)^2}} = \frac{1}{\sqrt{1+(2 \cdot 2\Omega \cdot 0.25F)^2}} = \frac{1}{\sqrt{1+1^2}}=\frac{1}{\sqrt{2}}$$ | ||
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| $$|V_{out}| = |V_{in}| \cdot |H(j2)| = 3 \cdot \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$$ | $$|V_{out}| = |V_{in}| \cdot |H(j2)| = 3 \cdot \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$$ | ||
| - | $$\angle V_{out} = \angle V_{in} + \angle H(j2) = 20^{\circ} + 9-45^{\circ}) = -25^{\circ}$$ | + | $$\angle V_{out} = \angle V_{in} + \angle H(j2) = 20^{\circ} + (-45^{\circ}) = -25^{\circ}$$ |
| And the time-domain output voltage is: | And the time-domain output voltage is: | ||
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| $$v_{out}(t) = \frac{3}{\sqrt{3}} \cos(2t-25^{\circ})$$ | $$v_{out}(t) = \frac{3}{\sqrt{3}} \cos(2t-25^{\circ})$$ | ||
| - | When $v_{in} = 7 \cos(4t-60^{\circ}$, | + | When $v_{in} = 7 \cos(4t-60^{\circ})$, $\omega = 4$ rad/sec, $|V_{in}|=7$ and $\angle V_{in} = -60^{\circ}$. For this value of $\omega$, and the given values of R and C, the magnitude and phase of the frequency response function are: |
| $$|H(j4)|= \frac{1}{\sqrt{1+(\omega RC)^2}}=\frac{1}{\sqrt{5}}$$ | $$|H(j4)|= \frac{1}{\sqrt{1+(\omega RC)^2}}=\frac{1}{\sqrt{5}}$$ | ||
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