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| learn:courses:real-analog-chapter-10:start [2017/01/17 19:31] – 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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| + | * Lecture Material: | ||
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| + | * Chapter 10 Videos: | ||
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| ====== 10. Introduction and Chapter Objectives ====== | ====== 10. Introduction and Chapter Objectives ====== | ||
| We will now study dynamic systems which are subjected to __sinusoidal__ forcing functions. Previously, in our analysis of dynamic systems, we determined both the //unforced response// (or // | We will now study dynamic systems which are subjected to __sinusoidal__ forcing functions. Previously, in our analysis of dynamic systems, we determined both the //unforced response// (or // | ||
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| - Represent the following sinusoids in phasor form: | - Represent the following sinusoids in phasor form: | ||
| - $3 \cos(5t-60^{\circ})$ | - $3 \cos(5t-60^{\circ})$ | ||
| - | - $-2 \cos(300t+45^{\circ}$ | + | - $-2 \cos(300t+45^{\circ})$ |
| - $\sin(6t)$ | - $\sin(6t)$ | ||
| - $7\cos(3t)$ | - $7\cos(3t)$ | ||
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| $$y(t)= \underline{Y} e^{j \omega t} (Eq. 10.35)$$ | $$y(t)= \underline{Y} e^{j \omega t} (Eq. 10.35)$$ | ||
| - | Where the phasor $\underline{Y} is a __complex number__ which can be expressed in either exponential or polar form: | + | Where the phasor $\underline{Y}$ is a __complex number__ which can be expressed in either exponential or polar form: |
| $$\underline{Y}=|Y|e^{j \theta_Y}=|Y| \angle \theta_Y | $$\underline{Y}=|Y|e^{j \theta_Y}=|Y| \angle \theta_Y | ||
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| \angle \theta_T = -\tan^{-1} (\omega RC) = -\tan^{-1} (1000 \cdot 1000 \cdot 1x10^{-6}) = - \frac{\pi}{4} = -45^{\circ}$$ | \angle \theta_T = -\tan^{-1} (\omega RC) = -\tan^{-1} (1000 \cdot 1000 \cdot 1x10^{-6}) = - \frac{\pi}{4} = -45^{\circ}$$ | ||
| - | And the phasor $\underline{Y}$ can be written as $\underline{Y} = \frac{5}{\sqrt{2}}e^{-j45^{\circ}}=\frac{5}{\sqrt{2}} \angle - 45^{\circ}$$ | + | And the phasor $\underline{Y}$ can be written as $\underline{Y} = \frac{5}{\sqrt{2}}e^{-j45^{\circ}}=\frac{5}{\sqrt{2}} \angle - 45^{\circ}$ |
| We can create a phasor diagram of the input phasor $\underline{U}$ and the output phasor $\underline{Y}$ | We can create a phasor diagram of the input phasor $\underline{U}$ and the output phasor $\underline{Y}$ | ||
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| - | In [[https:// | + | In [[/ |
| $$\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)$$ | ||
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| If the voltage and current are represented in phasor form as: | If the voltage and current are represented in phasor form as: | ||
| - | $$v(t)= \uunderline{V}e^{j \omega t} (Eq. 10.60)$$ | + | $$v(t)= \underline{V}e^{j \omega t} (Eq. 10.60)$$ |
| And: | And: | ||
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| Where $\theta_Z$ is the angle of $\underline{Z}$. The magnitude of the impedance is the ratio of the magnitude of the voltage to the magnitude of the current: | Where $\theta_Z$ is the angle of $\underline{Z}$. The magnitude of the impedance is the ratio of the magnitude of the voltage to the magnitude of the current: | ||
| - | $$|\underline{Z}|=\frace{V_P}{I_P}=\frac{|\underline{V}|}{|\underline{I}|} | + | $$|\underline{Z}|=\frac{V_P}{I_P}=\frac{|\underline{V}|}{|\underline{I}|} |
| And the angle of the impedance is the difference between the voltage phase angle and the current phase angle: | And the angle of the impedance is the difference between the voltage phase angle and the current phase angle: | ||
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| $$R=Re \left\{ \underline{Z} \right\} = | \underline{Z} | \cos \theta_Z$$ | $$R=Re \left\{ \underline{Z} \right\} = | \underline{Z} | \cos \theta_Z$$ | ||
| - | $$X=Im \left\{ \underline{Z} \right\{ = | \underline{Z} | \sin \theta_Z$$ | + | $$X=Im \left\{ \underline{Z} \right\} = | \underline{Z} | \sin \theta_Z$$ |
| Impedance is an extremely useful concept, in that it can be used to represent the voltage-current relations for any two-terminal electrical circuit element in __the frequency domain__, as indicated in Figure 10.19. | Impedance is an extremely useful concept, in that it can be used to represent the voltage-current relations for any two-terminal electrical circuit element in __the frequency domain__, as indicated in Figure 10.19. | ||
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| - | The // | + | The // |
| $$\underline{Y}=\frac{1}{\underline{Z}} | $$\underline{Y}=\frac{1}{\underline{Z}} | ||
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| ==== Section Summary ==== | ==== Section Summary ==== | ||
| * Voltage-current relation for our passive circuit elements in the frequency domain are: | * Voltage-current relation for our passive circuit elements in the frequency domain are: | ||
| - | * Resistor: $\underline{V}=R\inderline{I}$ | + | * Resistor: $\underline{V}=R\underline{I}$ |
| * Inductor: $\underline{V}=j \omega L \cdot \underline{I}$ | * Inductor: $\underline{V}=j \omega L \cdot \underline{I}$ | ||
| * Capacitor: $\underline{V} = \frac{1}{j \omega C} \cdot \underline{I}$ | * Capacitor: $\underline{V} = \frac{1}{j \omega C} \cdot \underline{I}$ | ||
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| By the definition of impedance, we can determine the current through the capacitor to be: | By the definition of impedance, we can determine the current through the capacitor to be: | ||
| - | $$\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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