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| learn:courses:real-analog-chapter-10:start [2017/05/04 20:25] – [Example10.8: RC Circuit Steady-state Sinusoidal Response] 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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| $$i(t)=0.98 \cos(2t - 78.7^{\circ})$$ | $$i(t)=0.98 \cos(2t - 78.7^{\circ})$$ | ||
| - | 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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