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| learn:courses:real-analog-chapter-10:start [2017/01/17 18:29] – 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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| + | <-- | ||
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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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| $$\angle H(j2) = -\tan^{-1} (\omega RC) = -\tan^{-1} (2 \cdot 2 \Omega \cdot 0.25F) = -\tan^{-1} (1) = -45^{\circ}$$ | $$\angle H(j2) = -\tan^{-1} (\omega RC) = -\tan^{-1} (2 \cdot 2 \Omega \cdot 0.25F) = -\tan^{-1} (1) = -45^{\circ}$$ | ||
| - | The output amplitude is then the product of $|V_{in}|$ and $|H(j2)|$ and the output phase is the sum of $\angle V_{in}$ and $\angle H(j2)$ | + | The output amplitude is then the product of $|V_{in}|$ and $|H(j2)|$ and the output phase is the sum of $\angle V_{in}$ and $\angle H(j2)$, so that: |
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| + | $$|V_{out}| = |V_{in}| \cdot |H(j2)| = 3 \cdot \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$$ | ||
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| + | $$\angle V_{out} = \angle V_{in} + \angle H(j2) = 20^{\circ} + (-45^{\circ}) = -25^{\circ}$$ | ||
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| + | And the time-domain output voltage is: | ||
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| + | $$v_{out}(t) = \frac{3}{\sqrt{3}} \cos(2t-25^{\circ})$$ | ||
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| + | When $v_{in} = 7 \cos(4t-60^{\circ})$, | ||
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| + | $$|H(j4)|= \frac{1}{\sqrt{1+(\omega RC)^2}}=\frac{1}{\sqrt{5}}$$ | ||
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| + | And: | ||
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| + | $$\angle H(j4) = -\tan^{-1}(\omega RC) = -63.4^{\circ}$$ | ||
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| + | The output amplitude is then the product of $|V_{in}|$ and $|H(j4)|$ and the output phase is the sum of $\angle V_{in}$ and $\angle H(j4)$ so that the time-domain output voltage in this case is: | ||
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| + | $$v_{out}(t) = \frac{7}{\sqrt{5}} \cos(2t-123.4^{\circ})$$ | ||
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| + | ---- | ||
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| + | From the above examples we can see that, once the frequency response function is calculated for a circuit __as a function of frequency__, | ||
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| + | We conclude this section with one additional example, to illustrate the use of the frequency response function and superposition to determine a circuit’s response to multiple inputs of different frequencies. | ||
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| + | ---- | ||
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| + | ==== Example 10.20 ==== | ||
| + | Use the results of examples 10.18 and 10.19 above to determine the response $v_{out}(t)$ of the circuit shown below if the input voltage is $v_{in}(t) = 3 \cos(2t+20^{\circ}) + 7 \cos(4t-60^{\circ})$. Plot the input and output waveforms. | ||
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| + | Recall, from section 10.5, that superposition is the __only__ valid approach for performing frequency domain analysis of circuits with inputs at multiple frequencies. Also recall that each frequency can be analyzed separately in the frequency domain, but that the superposition process (the summation of the individual contributions) must be done in the time domain. For this problem, we have contributions at two different frequencies: | ||
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| + | $$v_1(t) = \frac{3}{\sqrt{2}} \cos(2t-25^{\circ})$$ | ||
| + | |||
| + | And: | ||
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| + | $$v_2(t) = \frac{7}{\sqrt{5}} \cos(2t-123.4^{\circ})$$ | ||
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| + | The overall response is then: | ||
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| + | $$v_{out}(t) = v_1(t) + v_2(t) = \frac{3}{\sqrt{2}} \cos(2t-25^{\circ}) + \frac{7}{\sqrt{5}} \cos(2t-123.4^{\circ})$$ | ||
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| + | A plot the input and output waveforms is shown below: | ||
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| + | {{ : | ||
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| + | ---- | ||
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| + | ==== Section Summary ==== | ||
| + | * The frequency response function or //frequency response// describes a circuit’s input-output relationship directly in the frequency domain, __as a function of frequency__. | ||
| + | * The frequency response is a complex function of frequency $H(j \omega)$ (that is, it is a complex number which depends upon the frequency). This complex function is generally expressed as a magnitude and phase, $|H(j \omega)|$ and $\angle H(j \omega)$, respectively. $|H(j \omega)|$ is called the //magnitude response// of the circuit, and $\angle H(j \omega)$ is called the //phase response// of the circuit. The overall idea is illustrated in the block diagram below: | ||
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| + | {{ : | ||
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| + | * The magnitude response of the circuit is the ratio of the output amplitude to the input amplitude. This is also called the gain of the system. Thus, in the figure above, the output amplitude $B = |H(j \omega)| \cdot A$. Note that the magnitude response or gain of the system is a function of frequency, so that inputs of different frequencies will have different gains. | ||
| + | * The phase response of the circuit is the difference between the output phase angle and the input phase angle. Thus, in the figure above, the output phase $\phi = \angle H(j \omega) + \theta$. Like the gain, the phase response is a function of frequency – inputs at different frequencies will, in general, have different phase shifts. | ||
| + | * Use of the frequency response to perform circuit analyses can be particularly helpful when the input signal contains a number of sinusoidal components at different frequencies. In this case, the response of the circuit to each individual component can be determined in the frequency domain using the frequency response and the resulting contributions summed in the time domain to obtain the overall response. | ||
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| + | ---- | ||
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| + | ==== Exercises ==== | ||
| + | **1.** Determine the voltage across the capacitor in the circuit below if $u(t) = 4 \cos(t+30^{\circ}) + 2cos(2t-45^{\circ})$. (Hint: this may be easier if you find the response to the input as a function of frequency, evaluate the response for each of the above frequency components, and superimpose the results.) | ||
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| + | **2.** Determine the voltage across the resistors in the circuit below if $u_1(t) = 4 \cos(2t)$ and if $u_2(t) = \cos(4t)$. (Hint:this may be easier if you find the response to the input as a function of frequency, evaluate the response for each of the above frequency components, and superimpose the results.) | ||
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| + | [[{}/ | ||
| + | [[{}/ | ||