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| learn:courses:real-analog-chapter-10:start [2017/01/13 19:40] – [Example 10.2: RC Circuit Sinusoidal Steady-state Response via Time-domain Analysis] 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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| + | <-- | ||
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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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| Substituting (10.35) into equation (10.33) and taking the appropriate derivative results in: | Substituting (10.35) into equation (10.33) and taking the appropriate derivative results in: | ||
| - | $$j \omega \underline{Y}e^{j \omega t} + \frac{!}{RC} \underline{Y}e^{j \omega t} = \frac{V_P}{RC}e^{j \omega t} (Eq. 10.37)$$ | + | $$j \omega \underline{Y}e^{j \omega t} + \frac{1}{RC} \underline{Y}e^{j \omega t} = \frac{V_P}{RC}e^{j \omega t} (Eq. 10.37)$$ |
| We can divide equation (10.37) by $e^{j \omega t}$ to obtain: | We can divide equation (10.37) by $e^{j \omega t}$ to obtain: | ||
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| Equation (10.38) can be solved for $\underline{Y}$: | Equation (10.38) can be solved for $\underline{Y}$: | ||
| - | $$\left\( j \omega + \frac{1}{RC} \right\) \underline{Y} = \frac{V_P}{RC} \Rightarrow \underline{Y} = \frac{\frac{V_P}{RC}}{j \omega + \frac{1}{RC}} | + | $$\left( j \omega + \frac{1}{RC} \right) \underline{Y} = \frac{V_P}{RC} \Rightarrow \underline{Y} = \frac{\frac{V_P}{RC}}{j \omega + \frac{1}{RC}} |
| So that: | So that: | ||
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| Since our physical input is the real part of the conceptual input, and __since all circuit parameters are real valued__, our physical output is the real part of equation (10.42) and the forced response is: | Since our physical input is the real part of the conceptual input, and __since all circuit parameters are real valued__, our physical output is the real part of equation (10.42) and the forced response is: | ||
| - | $$y_f(t)=\frac{V_P}{\sqrt{1+(\omega RC)^2}} \cos \left\[ \omega t - \tan^{-1} (\omega RC) \right\] (Eq. 10.43)$$ | + | $$y_f(t)=\frac{V_P}{\sqrt{1+(\omega RC)^2}} \cos \left[ \omega t - \tan^{-1} (\omega RC) \right] |
| Which agrees with our result from the time-domain analysis of example 10.2. | Which agrees with our result from the time-domain analysis of example 10.2. | ||
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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}$$ | + | And the phasor $\underline{Y}$ can be written as $\underline{Y} = \frac{5}{\sqrt{2}}e^{-j45^{\circ}}=\frac{5}{\sqrt{2}} \angle - 45^{\circ}$ |
| We can create a phasor diagram of the input phasor $\underline{U}$ and the output phasor $\underline{Y}$ | We can create a phasor diagram of the input phasor $\underline{U}$ and the output phasor $\underline{Y}$ | ||
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| $$i_L(t)= I_{\underline{L}}e^{j \omega t} (Eq. 10.46)$$ | $$i_L(t)= I_{\underline{L}}e^{j \omega t} (Eq. 10.46)$$ | ||
| - | Where the phasor $I_{underline{L}}=|I_L| \angle \theta$. | + | Where the phasor $I_{\underline{L}}=|I_L| \angle \theta$. |
| Substituting our assumed input and output in phasor form into equation (10.44) results in: | Substituting our assumed input and output in phasor form into equation (10.44) results in: | ||
| - | $$Lj \omega I_{underline{L}}e^{j \omega t} + RI_{\underline{L}}e^{j \omega t} = \underline{U}e^{j \omega t} (Eq. 10.47)$$ | + | $$Lj \omega I_{\underline{L}}e^{j \omega t} + RI_{\underline{L}}e^{j \omega t} = \underline{U}e^{j \omega t} (Eq. 10.47)$$ |
| As in Example 10.4, we divide through by $e^{j \omega t}$ to obtain the frequency domain governing equation: | As in Example 10.4, we divide through by $e^{j \omega t}$ to obtain the frequency domain governing equation: | ||
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| $$I_{\underline{L}}= \frac{\underline{U}}{R+j \omega L} = \frac{V_P \angle 30^{\circ}}{R+ j \omega L} (Eq. 10.49)$$ | $$I_{\underline{L}}= \frac{\underline{U}}{R+j \omega L} = \frac{V_P \angle 30^{\circ}}{R+ j \omega L} (Eq. 10.49)$$ | ||
| - | So that the phasor $I_{underline{L}}$ has magnitude and phase: | + | So that the phasor $I_{\underline{L}}$ has magnitude and phase: |
| $$|I_L|= \frac{V_P}{\sqrt{R^2+(\omega L)^2}} \\ | $$|I_L|= \frac{V_P}{\sqrt{R^2+(\omega L)^2}} \\ | ||
| - | \theta = 30^{\circ} -\tan^{-1} \left\( \frac{\omega L}{R} \right\) \\ | + | \theta = 30^{\circ} -\tan^{-1} \left( \frac{\omega L}{R} \right) \\ |
| (Eq. 10.50)$$ | (Eq. 10.50)$$ | ||
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| The exponential form of the inductor current is therefore: | The exponential form of the inductor current is therefore: | ||
| - | $$i_L(t)=\frac{V_P}{\sqrt{R^2+(\omega L)^2}} e^{j \left\[ \omega t + 30^{\circ}-\tan^{-1} \left\( \frac{\omega L}{R} \right\) \right\]} (Eq. 10.51)$$ | + | $$i_L(t)=\frac{V_P}{\sqrt{R^2+(\omega L)^2}} e^{j \left[ \omega t + 30^{\circ}-\tan^{-1} \left( \frac{\omega L}{R} \right) \right]} (Eq. 10.51)$$ |
| And the actual physical inductor current is: | And the actual physical inductor current is: | ||
| - | $$i_L(t)=\frac{V_P}{\sqrt{R^2+(\omega L)^2}} \cos \left\[ \omega t + 30^{\circ} -\tan^{-1} \left\( \frac{\omega L}{R} \right\) \right\] (Eq. 10.52)$$ | + | $$i_L(t)=\frac{V_P}{\sqrt{R^2+(\omega L)^2}} \cos \left[ \omega t + 30^{\circ} -\tan^{-1} \left( \frac{\omega L}{R} \right) \right] |
| ---- | ---- | ||
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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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| Dividing through by $e^{j \omega t}$ and noting that $j^2=-1$, results in: | Dividing through by $e^{j \omega t}$ and noting that $j^2=-1$, results in: | ||
| - | $$\left\[ \frac{1}{LC} - \omega^2 + j \frac{R}{L} \omega \right\] \underline{V} = \frac{2}{LC}$$ | + | $$\left[ \frac{1}{LC} - \omega^2 + j \frac{R}{L} \omega \right] \underline{V} = \frac{2}{LC}$$ |
| So that: | So that: | ||
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| The magnitude and phase of $\underline{V}$ are: | The magnitude and phase of $\underline{V}$ are: | ||
| - | $$|V|=\frac{\frac{2}{LC}}{\sqrt{ \left\( \frac{1}{LC} - \omega^2 \right\) ^2 + \left\( \frac{R}{L} \omega \right\) ^2}}$$ | + | $$|V|=\frac{\frac{2}{LC}}{\sqrt{ \left( \frac{1}{LC} - \omega^2 \right) ^2 + \left( \frac{R}{L} \omega \right) ^2}}$$ |
| - | $$\angle \underline{V} = -\tan^{-1} \left\( \frac{\frac{R \omega}{L}}{\frac{1}{LC}-\omega^2} \right\)$$ | + | $$\angle \underline{V} = -\tan^{-1} \left( \frac{\frac{R \omega}{L}}{\frac{1}{LC}-\omega^2} \right)$$ |
| And the capacitor voltage is: | And the capacitor voltage is: | ||
| - | $$v(t)= \frac{\frac{2}{LC}}{\sqrt{ \left\( \frac{1}{LC} - \omega^2 \right\) + \left\( \frac{R}{L} \omega \right\) ^2}} \cos \left\{ \omega t - \tan^{-1} \left\( \frac{\frac{R \omega}{L}}{\frac{1}{LC} - \omega^2} \right\) \right\} (Eq. 10.58)$$ | + | $$v(t)= \frac{\frac{2}{LC}}{\sqrt{ \left( \frac{1}{LC} - \omega^2 \right) + \left( \frac{R}{L} \omega \right) ^2}} \cos \left\{ \omega t - \tan^{-1} \left( \frac{\frac{R \omega}{L}}{\frac{1}{LC} - \omega^2} \right) \right\} (Eq. 10.58)$$ |
| The complex arithmetic in this case becomes a bit tedious, but the complexity of the frequency-domain approach is nowhere near that of the time-domain solution of the second-order differential equation. | The complex arithmetic in this case becomes a bit tedious, but the complexity of the frequency-domain approach is nowhere near that of the time-domain solution of the second-order differential equation. | ||
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| ---- | ---- | ||
| + | ==== Section Summary ==== | ||
| + | * The steady-state response of a linear time invariant system to a sinusoidal input is a sinusoid with the same frequency as the input sinusoid. Only the amplitude and phase angle of the output sinusoid can be different from the input sinusoid, so the solution is entirely characterized by the magnitude and phase angle of the output sinusoid. | ||
| + | * The steady-state response of a system to a sinusoidal input can be determined by assuming a form of the solution, substituting the input signal and the output signal into the governing differential equation and solving for the amplitude and phase angle of the output sinusoid. | ||
| + | * The solution approach is simplified if the sinusoidal signals are represented as complex exponentials. The approach is further simplified if these complex exponentials are represented in phasor form – the phasor is a complex number which provides the amplitude and phase angle of the complex exponential. | ||
| + | * The above solution approaches convert the governing differential equation into an algebraic equation. If complex exponentials in phasor form are used to represent the signals of interest, the governing algebraic equation can have complex coefficients. | ||
| + | * The relationships between the steady state sinusoidal inputs and outputs are described by a relationship between the amplitudes (generally a __ratio__ between the output amplitude and the input amplitude) and the phase angles (generally a __difference__ between the output and input phase angles). | ||
| + | * These relationships are often displayed graphically in a phasor diagram. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | **1.** The differential equation governing a circuit is: | ||
| + | |||
| + | $$2\frac{dy(t)}{dt}+ 6y(t)=u(t)$$ | ||
| + | |||
| + | Where $u(t)$ is the input and $y(t)$ is the output. Determine the steady-state response of the circuit to an input $u(t)=2 \cos(3t)$. | ||
| + | |||
| + | **2.** For the circuit shown below, $u(t)$ is the input and $y(t)$ is the output. | ||
| + | - Write the differential equation relating $u(t)$ and $y(t)$. | ||
| + | - Determine $y(t)$, $t \rightarrow \infty$, if $u(t)=3 \cos(2t)$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 10.4: Phasor Representations of Circuit Elements ===== | ||
| + | In section 10.3, we determined the sinusoidal steady-state response of an electrical circuit by transforming the circuit’s governing differential equation into the //frequency domain// or //phasor domain//. This transformation converted the differential equation into an algebraic equation. This conversion significantly simplified the subsequent circuit analysis, at the relatively minor expense of performing some complex arithmetic. In this module, we will further simplify this analysis by transforming the __circuit itself__ directly into the frequency domain and __writing the governing algebraic equations directly__. | ||
| + | |||
| + | This approach eliminates the necessity of ever writing the differential equation governing the circuit (as long as we are only interested in the circuit’s sinusoidal steady-state response). This approach also allows us to apply analysis techniques previously used only for purely resistive circuits to circuits containing energy storage elements. | ||
| + | |||
| + | ==== Phasor Domain Voltage-current Relationships ==== | ||
| + | In section 10.2, we introduced phasors as a method for representing sinusoidal signals. Phasors provide the magnitude and phase of the sinusoid. For example, the signal $v(t)=V_P \cos(\omega t + \theta)$ has amplitude $V_P$ and phase angle $\theta$. This information can be represented in phasor form as: | ||
| + | |||
| + | $$\underline{V}=V_Pe^{j \theta}$$ | ||
| + | |||
| + | In which complex exponentials are used to represent the phase. Equivalently, | ||
| + | |||
| + | $$\underline{V} = V_P \angle \theta$$ | ||
| + | |||
| + | Note that the phasor does not provide the frequency of the signal, $\omega$. To include frequency information, | ||
| + | |||
| + | $$v(t)= \underline{V}e^{j \omega t}$$ | ||
| + | |||
| + | In section 10.3, we used phasor representations to determine the steady-state sinusoidal response of electrical circuits by representing the signals of interest as complex exponentials in phasor form. When signals in the governing differential equation are represented in this form, the differential equation becomes an algebraic equation, resulting in a significant mathematical simplification. In section 10.3, it was also noted that the mathematics could be simplified further by representing the circuit itself directly in the phasor domain. In this section, we present the phasor form of voltage-current relations for our basic circuit elements: resistors, inductors, and capacitors. The voltage-current relations for these elements are presented individually in the following sub- sections. | ||
| + | |||
| + | ==== Resistors ==== | ||
| + | The voltage-current relationship for resistors is provided by Ohm’s Law: | ||
| + | |||
| + | $$v(t) = R \cdot i(t) (Eq. 10.59)$$ | ||
| + | |||
| + | If the voltage and current are represented in phasor form as: | ||
| + | |||
| + | $$v(t)= \underline{V}e^{j \omega t} (Eq. 10.60)$$ | ||
| + | |||
| + | And: | ||
| + | |||
| + | $$i(t)= \underline(I)e^{j \omega t} (Eq. 10.61)$$ | ||
| + | |||
| + | Equation (10.59) can be written: | ||
| + | |||
| + | $$\underline{V}e^{j \omega t} = R \cdot \underline{I}e^{j \omega t} (Eq. 10.62)$$ | ||
| + | |||
| + | Cancelling the $e^{j \omega t}$ term from both sides results in: | ||
| + | |||
| + | $$\underline{V}= R \cdot \underline{I} | ||
| + | |||
| + | The voltage-current relationship for resistors (Ohm’s Law) is thus identical in the time and frequency domains. Schematically, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Equation (10.63) shows that, in the frequency domain, the voltage and current in a resistor are related by a __purely real, constant__ multiplicative factor. Thus, the sinusoidal voltage and current for a resistor are simply scaled versions of one another – there is no phase difference in the voltage and current for a resistor. This is shown graphically in Figure 10.11. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | A representative phasor diagram of the resistor’s voltage and current will appear as shown in Figure 10.12 – the phasors representing voltage and current will always be in the same direction, though their lengths will typically be different. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ==== Inductors ==== | ||
| + | The voltage-current relationship for inductors is: | ||
| + | |||
| + | $$v(t)= L \cdot \frac{di(t)}{dt} | ||
| + | |||
| + | As with the resistive case presented above, we assume that the voltage and current are represented in phasor form as $v(t)=\underline{V}e^{j \omega t}$ and $i(t)=\underline{I}e^{j \omega t}$, respectively. Substituting these expressions into equation (10.64) results in: | ||
| + | |||
| + | $$\underline{V}e^{j \omega t} = L \cdot \frac{d}{dt} \left[ \underline{I}e^{j \omega t} \right] = L (j \omega) \underline{I}e^{j \omega t} (Eq. 10.65)$$ | ||
| + | |||
| + | Dividing equation (10.65) by $e^{j \omega t}$ and re-arranging terms slightly results in the phasor domain or frequency domain representation of the inductor’s voltage-current relationship: | ||
| + | |||
| + | $$\underline{V} = j \omega L \cdot \underline{I} | ||
| + | |||
| + | In the frequency domain, therefore, the inductor’s phasor voltage is proportional to its phasor current. The constant of proportionality is, unlike the case of the resistor, an __imaginary number__ and is a __function of the frequency__, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The factor of j in the voltage-current relationship of equation (10.66) introduces a $90^{\circ}$ phase shift between inductor voltage and current. Since $j=e^{j90^{\circ}}$, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ==== Capacitors ==== | ||
| + | The voltage-current relationship for capacitors is: | ||
| + | |||
| + | $$i(t)=C \cdot \frac{dv(t)}{dt} | ||
| + | |||
| + | As with the previous cases, we assume that the voltage and current are represented in phasor form as $v(t)=\underline{V}e^{j \omega t}$ and $i(t)=\underline{I}e^{j \omega t}$, respectively. Substituting these expressions into equation (10.67) results in: | ||
| + | |||
| + | $$\underline{I}e^{j \omega t} = C \cdot \frac{d}{dt} \left[ \underline{V}e^{j \omega t} \right] = C(j \omega) \underline{V} e^{j \omega t} (Eq. 10.68)$$ | ||
| + | |||
| + | Dividing the above by $e^{j \omega t}$ results in the phasor domain or frequency domain representation of the capacitor' | ||
| + | |||
| + | $$\underline{I}=j \omega C \cdot \underline{V} | ||
| + | |||
| + | To be consistent with our voltage-current relationship for resistors and capacitors, we write the voltage in terms of the current. Thus, | ||
| + | |||
| + | $$\underline{V} = \frac{1}{j \omega C} \cdot \underline{I} | ||
| + | |||
| + | In the frequency domain, therefore, the capacitor’s phasor voltage is proportional to its phasor current. The constant of proportionality is an __imaginary number__ and is a __function of the frequency__, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The factor of $\frac{1}{j}$ in the voltage-current relationship of equation (10.70) introduces a $90^{\circ}$ phase shift between inductor voltage and current. Since $\frac{1}{j}=e^{-j90^{\circ}} = 1 \angle - 90^{\circ}$, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ==== Impedance and Admittance ==== | ||
| + | The frequency domain voltage-current characteristics presented in the previous subsections indicate that the voltage difference across a circuit element can be written in terms of a multiplicative factor (which can be a complex number) times the current through the element. In order to generalize and formalize this concept, we define // | ||
| + | |||
| + | $$\underline{Z} = \frac{\underline{V}}{\underline{I}} | ||
| + | |||
| + | Therefore, if the phasor voltage and current for a circuit element are given by: | ||
| + | |||
| + | $$\underline{V}=V_Pe^{j \theta}$$ | ||
| + | |||
| + | And: | ||
| + | |||
| + | $$\underline{I}=I_Pe^{j \phi}$$ | ||
| + | |||
| + | Then the impedance is: | ||
| + | |||
| + | $$\underline{Z} = \frac{\underline{V}}{\underline{I}} = \frac{V_P}{I_P}e^{j(\theta_Z)} | ||
| + | |||
| + | Or alternatively, | ||
| + | |||
| + | $$\underline{Z} \frac{V_P}{I_P} \angle \theta_Z | ||
| + | |||
| + | 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: | ||
| + | |||
| + | $$|\underline{Z}|=\frac{V_P}{I_P}=\frac{|\underline{V}|}{|\underline{I}|} | ||
| + | |||
| + | And the angle of the impedance is the difference between the voltage phase angle and the current phase angle: | ||
| + | |||
| + | $$\theta_Z= \angle \underline{Z} = \angle \underline{V} - \angle \underline{I} = \theta - \phi (Eq. 10.75)$$ | ||
| + | |||
| + | The impedance can also be represented in rectangular coordinates as: | ||
| + | |||
| + | $$\underline{Z} = R + jX (Eq. 10.76)$$ | ||
| + | |||
| + | Where $R$ is the real part of the impedance (called the // | ||
| + | |||
| + | $$|\underline{Z}|= \sqrt{R^2+X^2} \\ | ||
| + | |||
| + | \angle \underline{Z} = \tan^{-1} \left( \frac{X}{R} \right)$$ | ||
| + | |||
| + | And: | ||
| + | |||
| + | $$R=Re \left\{ \underline{Z} \right\} = | \underline{Z} | \cos \theta_Z$$ | ||
| + | |||
| + | $$X=Im \left\{ \underline{Z} \right\} = | \underline{Z} | \sin \theta_Z$$ | ||
| + | |||
| + | Impedance is an extremely useful concept, in that it can be used to represent the voltage-current relations for any two-terminal electrical circuit element in __the frequency domain__, as indicated in Figure 10.19. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The // | ||
| + | |||
| + | $$\underline{Y}=\frac{1}{\underline{Z}} | ||
| + | |||
| + | Admittance is also a complex number, and is written in rectangular coordinates as: | ||
| + | |||
| + | $$\underline{Y}=G+jB | ||
| + | |||
| + | Where $G$ is called the // | ||
| + | |||
| + | Impedances and admittances for the three electrical circuit elements presented previously in this section are provided in Table 10.1 below. These results are readily obtained from the previously presented phasor domain voltage-current relationships and the definitions of impedance and admittance. __The relations provided in Table 10.1 should be committed to memory__. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.7: Provide the Phasor-domain Representation of the Circuit Below ==== | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The input amplitude is 20 volts, and the input phase is $30^{\circ}$, | ||
| + | |||
| + | The frequency of the input voltage is $\omega = 10 rad/sec$. Thus, the impedances of the passive circuit elements are as follows: | ||
| + | |||
| + | * Resistor: $\underline{Z} = R = 2 \Omega$ | ||
| + | * Inductor: $\underline{Z} = j \omega L = j(10rad / sec)(0.1H) = j 1 \Omega$ | ||
| + | * Capacitor: $\underline{Z} = \frac{1}{j \omega C} = \frac{1}{j(10rad / sec) (\frac{1}{30}F)} = \frac{3}{j} \Omega = -j 3 \Omega$ | ||
| + | |||
| + | The phasor-domain circuit is shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * Voltage-current relation for our passive circuit elements in the frequency domain are: | ||
| + | * Resistor: $\underline{V}=R\underline{I}$ | ||
| + | * Inductor: $\underline{V}=j \omega L \cdot \underline{I}$ | ||
| + | * Capacitor: $\underline{V} = \frac{1}{j \omega C} \cdot \underline{I}$ | ||
| + | * The impedance of a circuit element is the ratio of the phasor voltage to the phasor current in that element: | ||
| + | * Resistor: $\underline{Z}=R$ | ||
| + | * Inductor: $\underline{Z}=j \omega L$ | ||
| + | * Capacitor: $\underline{Z} = \frac{1}{j \omega C}$ | ||
| + | * Impedance is, in general, a complex number. Units of impedance are ohms ($\Omega$). The real part of impedance is the // | ||
| + | * Admittance is the inverse of impedance. | ||
| + | * Admittance is, in general, a complex number. The real part of admittance is // | ||
| + | * Impedance and admittance are, in general, functions of frequency. | ||
| + | * Impedance and admittance are not phasors. They are complex numbers – there is no sinusoidal time domain function corresponding to impedance or admittance. (Phasors, by definition, are a way to describe a time- domain sinusoidal function.) | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | **1.** | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | **2.** Sketch a diagram of the input and output phasors for exercise 1 above. | ||
| + | |||
| + | **3.** Determine the impedance of the circuit elements shown below if $V_{in}(t) = 2 \cos(4t)$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | **4.** Determine the impedance of the circuit elements in exercise 3 if $V_{in}(t) = 3 \cos(8t)$. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 10.5: Direct Frequency Domain Circuit Analysis ===== | ||
| + | In section 10.3, we determined the steady-state response of electrical circuits to sinusoidal signals using __phasor representations of the signals__ involved, and __time-domain representations of the circuit element voltage-current relations__. Applying KVL and KCL in this manner resulted in governing equations in which the time dependence had been removed, which converted the governing equations from differential equations to algebraic equations. Unknowns in the resulting algebraic equations were the phasor representations of the signals. These equations could then be solved to determine the desired signals in phasor form; these results could then be used to determine the time-domain representations of the signals. | ||
| + | |||
| + | In section 10.4, we replaced the time-domain voltage-current relations for passive electrical circuit elements with impedances, which provide voltage-current relations for the circuit elements __directly in the frequency domain__. At the end of section 10.4, we used these impedances to schematically represent a circuit directly in the frequency domain. | ||
| + | |||
| + | In this section, we will use this frequency-domain circuit representation to perform circuit analysis directly in the frequency domain using phasor representations of the signals and impedance representations of the circuit elements. This will allow us to write the algebraic equations governing the phasor representation of the circuit __directly__, | ||
| + | |||
| + | Performing the circuit analysis directly in the frequency domain using impedances to represent the circuit elements can result in a significant simplification of the analysis. In addition, many circuit analysis techniques which were previously applied to resistive circuits (e.g. circuit reduction, nodal analysis, mesh analysis, superposition, | ||
| + | - Provide examples of applying these analysis methods to frequency-domain circuits, and | ||
| + | - Note any generalizations relative to using phasors in these analysis methods | ||
| + | |||
| + | Throughout this section, the reader should firmly keep in mind that we are dealing __only with the steady-state responses of circuits to sinusoidal forcing functions__. It is sometimes easy to lose track of this fact, since the sinusoidal nature of the signal is often not explicitly stated, but any time we deal with impedances and phasors, we are working with sinusoidal signals. | ||
| + | |||
| + | ==== Kirchhoff’s Voltage Law ==== | ||
| + | Kirchhoff’s Voltage Law states that the sum of the voltage differences around any closed loop is zero. Therefore, if $v_1(t), | ||
| + | |||
| + | $$\sum_{k=1}^{N} v_k(t)=0 | ||
| + | |||
| + | Substituting the phasor representation of the voltages results in: | ||
| + | |||
| + | $$\sum_{k=1}^{N} \underline{V}_k e^{j \omega t}=0 (Eq. 10.80)$$ | ||
| + | |||
| + | Dividing equation (10.80) by $e^{j \omega t}$ results in: | ||
| + | |||
| + | $$\sum_{k=1}^{N} \underline{V}_k = 0 (Eq. 10.81)$$ | ||
| + | |||
| + | So that KVL states that the sum of the phasor voltages around any closed loop is zero. | ||
| + | |||
| + | |||
| + | ==== Kirchhoff' | ||
| + | Kirchhoff' | ||
| + | |||
| + | $$\sum_{k=1}^{N} i_k(t) = 0 (Eq. 10.82)$$ | ||
| + | |||
| + | Substituting the phasor representation of the currents results in: | ||
| + | |||
| + | $$\sum_{k=1}^{N} \underline{I}_k e^{j \omega t} = 0 (Eq. 10.83)$$ | ||
| + | |||
| + | Dividing equation (10.83) by $e^{j \omega t}$ results in: | ||
| + | |||
| + | $$\sum_{k=1}^{N} \underline{I}_k = 0 (Eq. 10.84)$$ | ||
| + | |||
| + | So that KCL states that the sum of the phasor currents entering (or leaving) a node is zero. | ||
| + | |||
| + | === Important Result: === | ||
| + | KVL and KCL apply directly in the frequency domain. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example10.8: | ||
| + | In this example, we will revisit example 10.3. In that example, we determined the capacitor voltage in the circuit to the left below, using phasor analysis techniques applied to the circuit’s time-domain governing equation. In this example, we will represent the circuit itself directly in the frequency domain, using impedance representations of the circuit element. The frequency-domain representation of the circuit is shown to the right below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | By the definition of impedance, we can determine the current through the capacitor to be: | ||
| + | |||
| + | $$\underline{I} = \frac{\underline{Y}}{\underline{Z}_C} = \frac{\underline{Y}}{\frac{1}{j \omega C}} = j \omega C \underline{Y}$$ | ||
| + | |||
| + | 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: | ||
| + | |||
| + | $$V_P \angle 0^{\circ} = R(j \omega C \underline{Y}) + \underline{Y}$$ | ||
| + | |||
| + | 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: | ||
| + | |||
| + | $$|\underline{Y}| = \frac{V_P}{\sqrt{1+(\omega RC)^2}}$$ | ||
| + | |||
| + | $$\angle \underline{Y} = -\tan^{-1} (\omega RC)$$ | ||
| + | |||
| + | And the time-domain solution for $y(t)$ is thus: | ||
| + | |||
| + | $$y(t) = \frac{V_P}{\sqrt{1+(\omega RC)^2}} \cos \left[ \omega t - \tan^{-1} (\omega RC) \right]$$ | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Parallel and Series Impedances & Circuit Reduction ==== | ||
| + | Consider the case of N impedances connected in series, as shown in Figure 10.20. Since the elements are in series, and since we have seen that KCL applies to phasors, the phasor current $\underline{I}$ flows through each of the impedances. Applying KVL for phasors around the single loop, and incorporating the definition of impedance, we obtain: | ||
| + | |||
| + | $$\underline{V} = \underline{I} (\underline{Z}_1 + \underline{Z}_2 + \dots + \underline{Z}_N) = 0 (Eq. 10.85)$$ | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | If we define $\underline{Z}_{eq}$ as the equivalent impedance of the series combination, | ||
| + | |||
| + | $$\underline{Z}_{eq} = \underline{Z}_1 + \underline{Z}_2 + \dots + \underline{Z}_N | ||
| + | |||
| + | So that __impedances in series sum directly__. Thus, impedances in series can be combined in the same way as resistances in series. | ||
| + | |||
| + | By extension of the above result, we can develop a voltage divider formula for phasors. Without derivation, we state that the phasor voltage across the k< | ||
| + | |||
| + | $$\underline{V}_k = \underline{V} \frac{\underline{Z}_k}{\underline{Z}_1 + \underline{Z}_2 + \dots + \underline{Z}_N} | ||
| + | |||
| + | So that our __voltage division relationships for resistors in series apply directly in the frequency domain for impedances in series__. | ||
| + | |||
| + | We now consider the case of N impedances connected in parallel, as shown in Figure 10.21. Since the elements are in parallel, and KVL applied to each loop shows that all circuit elements share the same phasor voltage difference $\underline{V}$. Applying KCL for phasors at the upper node, and incorporating the definition of admittance, we obtain: | ||
| + | |||
| + | $$\underline{I}=\underline{V} (\underline{Y}_1 + \underline{Y}_2 + \dots + \underline{Y}_N) = 0 (Eq. 10.88)$$ | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | If we define $\underline{Y}_{eq}$ as the equivalent impedance of the series combination, | ||
| + | |||
| + | $$\underline{I} = \underline{V} \cdot \underline{Y}_{eq} | ||
| + | |||
| + | Where: | ||
| + | |||
| + | $$\underline{Y}_{eq} = \underline{Y}_1 + \underline{Y}_2 + \dots + \underline{Y}_N | ||
| + | |||
| + | So that __admittances in series sum directly__. Converting our admittances to impedances indicates that the equivalent impedance of a parallel combination of N impedances as shown in Figure 10.21 is: | ||
| + | |||
| + | $$\underline{Z}_{eq} = \frac{1}{\frac{1}{\underline{Z}_1} + \frac{1}{\underline{Z}_2} + \dots + \frac{1}{\underline{Z}_N}} | ||
| + | |||
| + | Thus, impedances in parallel can be combined in the same way as resistances in parallel. | ||
| + | |||
| + | By extension of the above result, we can develop a current divider formula for phasors. Without derivation, we state that the phasor current across the k< | ||
| + | |||
| + | $$\underline{I}_k = \underline{I} \frac{\frac{1}{\underline{Z}_k}}{\frac{1}{\underline{Z}_1} + \frac{1}{\underline{Z}_2} + \dots + \frac{1}{\underline{Z}_N}} | ||
| + | |||
| + | So that our __current division relationships for resistors in parallel apply directly in the frequency domain for impedances in parallel__. | ||
| + | |||
| + | === Important Result: === | ||
| + | All circuit reduction techniques for resistances apply directly to the frequency domain for impedances. Likewise, voltage and current divider relationships apply to phasor circuits in the frequency domain exactly as they apply to resistive circuits in the time domain. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.9 ==== | ||
| + | Use circuit reduction techniques to determine the current phasor $\underline{I}$ leaving the source in the circuit below. (Note: the circuit below is the frequency domain circuit we obtained in example 10.7.) | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Since __impedances in series add directly__, the inductor and resistor can be combined into a single equivalent impedance of $(2+j1)\Omega$, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Using the reduced circuit to the right above and the definition of impedance, we can see that: | ||
| + | |||
| + | $$\underline{I} = \frac{20 \angle 30^{\circ}}{2.37 \angle -18^{\circ} \Omega} = \frac{2}{2.37} \angle \left[ 30^{\circ} - (-18^{\circ}) \right]A$$ | ||
| + | |||
| + | So that: | ||
| + | |||
| + | $$\underline{I} = 8.44 \angle 48^{\circ}A$$ | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.10 ==== | ||
| + | Use circuit reduction techniques to determine the current, $i(t)$ through the inductor in the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | With $\omega=2$ rad/sec, the frequency domain representation of the circuit is as shown in the figure to the left below; in that figure, we have also defined the current phasor $\underline{I}_S$ leaving the source. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | We now employ circuit reduction techniques to determine the phasor $\underline{I}$. To do this, we first determine the circuit impedance seen by the source; this impedance allows us to determine the source current $\underline{I}_S$. The current $\underline{I}$ can be determined from a current divider relation and $\underline{I}_S$. | ||
| + | |||
| + | The impedances of the series combination of the capacitor and the $4\Omega$ resistor is readily obtained by adding their individual impedances, as shown in the figure to the left below. This equivalent impedance is then in parallel with the inductor’s impedance; the equivalent impedance of this parallel combination is as shown in the circuit to the right below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The source current is then, by the definition of impedance, $\underline{I}_S = \frac{5 \angle 0^{\circ}}{(4-j4) \Omega + 2 \Omega} = 0.69 \angle - 33.7^{\circ}$. | ||
| + | |||
| + | The circuit to the left above, along with our voltage divider formula, provides: | ||
| + | |||
| + | $$\underline{I} = \frac{(4-j4)\Omega}{(4-j4)\Omega + j4\Omega} \cdot \underline{I}_S = (1-j1) \Omega \cdot 0.69 \angle - 33.7^{\circ} = 0.98 \angle - 78.7^{\circ}$$ | ||
| + | |||
| + | And the current $i(t)=0.98 \cos(2t -78.7^{\circ})$. | ||
| + | |||
| + | |||
| + | ==== Nodal and Mesh Analysis ==== | ||
| + | Nodal analysis and mesh analysis techniques have been previously applied to resistive circuits in the time domain. In nodal analysis, we applied KCL at independent nodes and used Ohm’s Law to write the resulting equations in terms of the node voltages. In mesh analysis, we applied KVL and used Ohm’s Law to write the resulting equations in terms of the mesh currents. | ||
| + | |||
| + | In the frequency domain, as we have seen in previous sub-sections, | ||
| + | * The circuit excitations and responses are represented by phasors | ||
| + | * Phasor representations of node voltages and mesh currents are used | ||
| + | * Impedances are used in the place of resistances | ||
| + | |||
| + | Application of nodal and mesh analysis to frequency-domain circuit analysis is illustrated in the following examples. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.11 ==== | ||
| + | Use nodal analysis to determine the current $i(t)$ in the circuit of example 10.10. | ||
| + | |||
| + | The desired frequency-domain circuit was previously determined in Example 10.10. Nodal analysis of the frequency-domain circuit proceeds exactly as was done in the case of resistive circuits. The reference voltage, $V_R = 0$, and our single node voltage, $V_A$, for this circuit are defined on the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Applying KCL in phasor form at node A provides: | ||
| + | |||
| + | $$\frac{5 \angle 0^{\circ} - \underline{V}_A}{2 \Omega} - \frac{\underline{V}_A}{(4-j4) \Omega} - \frac{\underline{V}_A}{j4 \Omega} = 0$$ | ||
| + | |||
| + | Solving for $\underline{V}_A$ gives $\underline{V}_A = 3.92 \angle 11.31^{\circ}V$. By the definition of impedance, the desired current phasor $\underline{I}= \frac{\underline{V}_A}{j4 \Omega} = \frac{3.92 \angle 11.31^{\circ}}{4 \angle 90^{\circ}} = 0.98 \angle - 78.7^{\circ}$ so that $i(t)=0.98 \cos(2t - 78.7^{\circ})$, | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.12 ==== | ||
| + | Use mesh analysis to determine the current i(t) in the circuit of examples 10.10 and 10.11. | ||
| + | |||
| + | The desired frequency-domain circuit was previously determined in Example 10.10. Mesh analysis of the frequency-domain circuit proceeds exactly as for resistive circuits. The figure below shows our choice of mesh loops; the series resistor-capacitor combination has been combined into a single equivalent resistance in the figure below, for clarity. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | KVL around the mesh loop $\underline{I}_1$ provides: | ||
| + | |||
| + | $$5 \angle 0^{\circ} - 2 \cdot \underline{I}_1 - (4-j4)(\underline{I}_1-\underline{I}_2) = 0$$ | ||
| + | |||
| + | KVL around the mesh loop $\underline{I}_2$ provides: | ||
| + | |||
| + | $$(4-j4)(\underline{I}_2-\underline{I}_1) + j4 \cdot \underline{I}_2 = 0$$ | ||
| + | |||
| + | The second equation above can be simplified to provide: $\underline{I}_2 = (1-j)\underline{I}_1$. Using this result to eliminate $\underline{I}_1$ in the mesh equation for loop $\underline{I}_1$ and simplifying provides: | ||
| + | |||
| + | $$5 \angle 0^{\circ} = \left[ \frac{(6-j4)}{1-j}+ (j4-4) \right] \underline{I}_2$$ | ||
| + | |||
| + | So that $\underline{I}_2 = 0.98 \angle - 78.7^{\circ}$. The mesh current $\underline{I}_2$ is simply the desired current $\underline{I}$, | ||
| + | |||
| + | $$i(t)=0.98 \cos(2t - 78.7^{\circ})$$ | ||
| + | |||
| + | Which is consistent without results from examples 10.10 and 10.11. | ||
| + | |||
| + | === Important Result: === | ||
| + | Nodal and mesh analysis methods apply to phasor circuits exactly as they apply to resistive circuits in the time domain. Impedances simply replace resistances, | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Superposition ==== | ||
| + | The extension of superposition to the frequency domain is an extremely important topic. Several common analysis techniques you will encounter later in this course and in future courses (frequency response, Fourier Series, and Fourier Transforms, for example) will depend heavily upon the superposition of sinusoidal signals. In this sub-section, | ||
| + | |||
| + | In all of our steady-state sinusoidal analyses, we have required that the circuit is __linear__. (The statement that the steady state response to a sinusoidal input is a sinusoid at the same frequency requires the system to be linear. Nonlinear systems do not necessarily have this characteristic.) Thus, all phasor circuits are linear and superposition must apply. Thus, if a phasor circuit has multiple inputs, we can calculate the response of the circuit to each input individually and sum the results to obtain the overall response. __It is important to realize, however, that the final step of summing the individual contributions to obtain the overall response can, in general, only be done in the time domain__. Since the phasor representation of the circuit response implicitly assumes a particular frequency, the phasor representations cannot be summed directly. The time domain circuit response, however, explicitly provides frequency information, | ||
| + | |||
| + | In fact, because the frequency-domain representation of the circuit depends upon the frequency of the input (in general, the impedances will be a function of frequency), the frequency domain representation of the circuit itself is, in general, different for different inputs. Thus, the __only__ way in which circuits with multiple inputs at different frequencies can be analyzed in the frequency domain is with superposition. | ||
| + | |||
| + | In the special case in which all inputs share a common frequency, the circuit response can be determined by any of our previous analysis techniques (circuit reduction, nodal analysis, mesh analysis, superposition, | ||
| + | |||
| + | Examples of the application of superposition to analysis of frequency-domain circuits are provided below. | ||
| + | |||
| + | === Important Result: === | ||
| + | In the case of multiple frequencies existing in the circuit, superposition is the only valid frequency-domain analysis approach. | ||
| + | |||
| + | Superposition applies directly in the frequency domain, insofar as contributions from individual sources can be determined by killing all other sources and analyzing the resulting circuit. In general, however, superimposing (summing) the contributions from the individual sources must be done in the time domain. | ||
| + | |||
| + | Superposition of responses to individual sources can be summed directly in the frequency domain (e.g. addition of the phasors representing the individual responses) is only appropriate if all sources have the same frequency. In this case (all source having the same frequency) any of our other modeling approaches are also valid. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.13 ==== | ||
| + | Determine the voltage $v(t)$ across the inductor in the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Since two different input frequencies are applied to the circuit, we __must__ use superposition to determine the response. The circuit to the left below will provide the phasor response $\underline{V}_1$ to the __current source__; the frequency is $\omega = 9$ rad/sec and the voltage source is killed. The circuit to the right below will provide the phasor response $\underline{V}_2$ to the __voltage source__; the frequency is $\omega = 3$ rad/sec and the current source is killed. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | To determine the voltage phasor resulting from the current source ($\underline{V}_1$ in the circuit to the left above), we note that the inductor and the $3\Omega$ resistor form a current divider. Thus, the current through the inductor resulting from the current source is $\underline{I}_1 = \frac{3 \Omega}{(3+j3)\Omega} \cdot 6 \angle 0^{\circ} = \frac{3 \angle 0^{\circ} \cdot \angle 0^{\circ}}{3 \sqrt{2} \angle 45^{\circ}} = \frac{6}{\sqrt{2}} \angle -45^{\circ}$. The voltage phasor $\underline{V}_1$ can then be determined by multiplying this current times the inductor' | ||
| + | |||
| + | $$\underline{V}_1 = j3 \Omega \cdot \frac{6}{\sqrt{2}} \angle - 45^{\circ} = 3 \angle 90^{\circ} \cdot \frac{6}{\sqrt{2}} \angle -45^{\circ}V$$ | ||
| + | |||
| + | And the time-domain voltage across the inductor due to the current source is: | ||
| + | |||
| + | $$v_1(t) = 9 \sqrt{2} \cos(9t + 45^{\circ})V$$ | ||
| + | |||
| + | To determine the voltage phasor resulting from the voltage source ($\underline{V}_2$ in the circuit to the right above), we note that the inductor and the $3 \Omega$ resistor now form a voltage divider. Thus, the voltage $\underline{V}_2$ can be readily determined by: | ||
| + | |||
| + | $$\underline{V}_2 = \frac{j1 \Omega}{(3+j1)\Omega} \cdot 4 \angle 30^{\circ} = \frac{1 \angle 90^{\circ} \cdot 4 \angle 30^{\circ}}{\sqrt{10} \angle 18.4^{\circ}} = \frac{4}{\sqrt{10}} \angle 101.6^{\circ}$$ | ||
| + | |||
| + | So that the time-domain voltage across the inductor due to the voltage source is: | ||
| + | |||
| + | $$v_2(t) = \frac{4}{\sqrt{10}} \cos(3t + 101.6^{\circ})V$$ | ||
| + | |||
| + | The overall voltage is then the sum of the contributions from the two sources, in the time domain, so: | ||
| + | |||
| + | $$v(t)=v_1(t)+v_2(t)$$ | ||
| + | |||
| + | And: | ||
| + | |||
| + | $$v(t)=9 \sqrt{2} \cos(9t + 45^{\circ}) + \frac{4}{\sqrt{10}} \cos(3t + 101.6^{\circ})V$$ | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.14 ==== | ||
| + | Determine the voltage $v(t)$ across the inductor in the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | This circuit is essentially the same as the circuit of Example 10.13, with the important difference that the frequency of the __voltage__ input has changed – the voltage source and current source both provide the same frequency input to the circuit, 9 rad/sec. We will first do this problem using superposition techniques. We will then use nodal analysis to solve the problem, to illustrate that multiple inputs at the same frequency do not require the use of superposition. | ||
| + | |||
| + | Individually killing each source in the circuit above results in the two circuits shown below. Note that the impedance of the inductor is now the same in both of these circuits. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The two circuits shown above will now be analyzed to determine the individual contributions to the inductor voltage; these results will then be summed to determine the overall inductor voltage. | ||
| + | |||
| + | The circuit to the left above has been analyzed in Example 6. Therefore, the voltage phasor $\underline{V}_1$ is the same as determined in Example 10.13: | ||
| + | |||
| + | $$\underline{V}_1 = 9 \sqrt{2} \angle 45^{\circ}V$$ | ||
| + | |||
| + | The voltage $\underline{V}_2$ in the circuit to the right above can be determined from application of the voltage divider formula for phasors: | ||
| + | |||
| + | $$\underline{V}_2 = \frac{j3\Omega}{(3+j3)\Omega} \cdot 4 \angle 30^{\circ} = \frac{3 \angle 90^{\circ} \cdot 4 \angle 30^{\circ}}{3 \sqrt{2} \angle 45^{\circ}} = 2 \sqrt{2} \angle 75^{\circ}$$ | ||
| + | |||
| + | Since both inputs have the same frequency, we can superimpose the phasor results directly (we could, of course, also determine the individual time domain responses and superimpose those responses if we chose): | ||
| + | |||
| + | $$\underline{V} = \underline{V}_1 + \underline{V}_2 = 9 \sqrt{2} \angle 45^{\circ} + 2 \sqrt{2} \angle 75^{\circ} = 15.24 \angle 50.3^{\circ}V$$ | ||
| + | |||
| + | So that the time domain inductor voltage is $v(t) = 15.24 \cos(9t + 50.3^{\circ})V$. Notice that the circuit response has only a single frequency component, since both inputs have the same frequency. | ||
| + | |||
| + | The superposition approach provided above is entirely valid. However, since both sources have the same input, we can choose any of our other analysis approaches to perform this problem. To emphasize this fact, we choose to do this problem using nodal analysis. | ||
| + | |||
| + | The frequency-domain circuit, with our definition of reference voltage and independent node, is shown in the figure below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | KCL at node A provides: | ||
| + | |||
| + | $$6 \angle 0^{\circ} = \frac{\underline{V}_A - 0}{j3\Omega} + \frac{\underline{V}_A - 4 \angle 30^{\circ}}{3\Omega}$$ | ||
| + | |||
| + | Solving the above equation for $\underline{V}_A$ provides $\underline{V}_A = 15.24 \angle 50.3V$ so that the inductor voltage as a function of time is: | ||
| + | |||
| + | $$v(t)=15.24 \cos(9t+50.3^{\circ})V$$ | ||
| + | |||
| + | Which is consistent with our result using superposition. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Thévenin’s & Norton’s Theorems, Source Transformations, | ||
| + | Application of Thévenin’s and Norton’s Theorems to frequency domain circuits is identical to their application to time domain resistive circuits. The only differences are: | ||
| + | * The open circuit voltage ($V_{OC}$) and short circuit current ($i_{SC}$) determined for resistive circuits is replaced by their phasor representations, | ||
| + | * The Thévenin resistance, $R_{TH}$, is replaced by a Thévenin impedance, $\underline{Z}_{TH}$. | ||
| + | |||
| + | Thus, the Thévenin and Norton equivalent circuits in the frequency domain are as shown in Figure 10.22. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Since Thévenin’s and Norton’s Theorems both apply in the frequency domain, the approaches we used for source transformations in the time domain for resistive circuits translate directly to the frequency domain, with impedances substituted for resistances and phasors used for voltage and current terms. | ||
| + | |||
| + | In order to determine the load necessary to draw the maximum power from a Thévenin equivalent circuit, we must re-derive the maximum power result obtained previously for resistive circuits, substituting impedances for admittances and using phasors for source terms. We will not derive the governing relationship, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.15 ==== | ||
| + | Determine the Thévenin equivalent circuit seen by the load in the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | In the circuit below, we have used the input frequency, $\omega = 2$ rad/sec, to convert the circuit to the frequency domain. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Removing the load and killing the source allows us to determine the Thévenin resistance of the circuit. The appropriate circuit is: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The parallel combination of two, $2\Omega$ resistors have an equivalent resistance of $1\Omega$. This impedance, in series with the $j1\Omega$ impedance, results in a Thévenin impedance $\underline{Z}_{TH} = (1 + j1)\Omega$. Replacing the source, but leaving the load terminals open-circuited, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Since there is no current through the inductor, due to the open-circuit condition, $\underline{V}_{OC}$ is determined from a simple resistive voltage divider formed by the two, $2\Omega$ resistors. Thus, the open-circuit voltage is: | ||
| + | |||
| + | $$\underline{V}_{OC} = \frac{2\Omega}{2\Omega + 2\Omega} \cdot 2 \angle 0^{\circ} = 1 \angle 0^{\circ}$$ | ||
| + | |||
| + | The resulting Thévenin equivalent circuit is shown below: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.16 ==== | ||
| + | Determine the Norton equivalent circuit of the circuit of example 10.15. | ||
| + | |||
| + | Since we determined the Thévenin equivalent circuit in Example 10.15, a source transformation can be used to determine the Norton equivalent circuit. Consistent with our previous source transformation rules, the short- circuit current, $\underline{I}_{SC}$, | ||
| + | |||
| + | $$\underline{I}_{SC} = \frac{\underline{V}_{OC}}{\underline{Z}_{TH}} = \frac{1 \angle 0^{\circ}}{(1+j1)\Omega} = \frac{1 \angle 0^{\circ}}{\sqrt{2} \angle 45^{\circ}} = \frac{1}{\sqrt{2}} \angle -45^{\circ}$$ | ||
| + | |||
| + | Since the impedance doesn’t change during a source transformation, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.17 ==== | ||
| + | Determine the load impedance for the circuit of Example 10.15 which will provide the maximum amount of power to be delivered to the load. Provide a physical realization (a circuit) which will provide this impedance. | ||
| + | |||
| + | The maximum power is delivered to the load when the load impedance is the complex conjugate of the Thévenin impedance. Thus, the load impedance for maximum power transfer is: | ||
| + | |||
| + | $$\underline{Z}_L = (1-j)\Omega$$ | ||
| + | |||
| + | And the loaded Thévenin circuit is: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | To implement this load, let us look at a parallel RC combination. With the frequency $\omega = 2$ rad/sec, the frequency domain load looks like: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Combining parallel impedances results in: | ||
| + | |||
| + | $$\underline{Z}_L = \frac{-j(\frac{R}{2C})}{R-\frac{j}{2C}}\Omega = \frac{\frac{R}{4C^2}-j\frac{R^2}{2C}}{R^2+\frac{1}{4C^2}}\Omega$$ | ||
| + | |||
| + | Setting $R = 2\Omega$ and $C = 0.25F$ makes $\underline{Z}_L = (1 - j )\Omega$, as desired, so the physical implementation of our load is as shown below: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * The following analysis methods apply in the frequency domain exactly as they do in the time domain for purely resistive circuits | ||
| + | * KVL and KCL | ||
| + | * Voltage and current dividers | ||
| + | * Circuit reduction techniques | ||
| + | * Nodal and mesh analysis | ||
| + | * Superposition, | ||
| + | * Thévenin’s and Norton’s theorems | ||
| + | |||
| + | One simply uses phasor representations for the voltages and/or currents in the circuit and impedances to represent the circuit element voltage-current relationships. The analysis techniques presented Chapters 1 through 4 are then applied exactly as they were for resistive circuits. | ||
| + | |||
| + | * One minor exception to the above statement is that, in order to draw maximum power from a circuit, the load impedance should be the complex conjugate of the impedance of the circuit’s Thévenin equivalent. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | **1.** Determine the impedance seen by the source for the circuit below if $u(t)=4 \cos(t+30^{\circ})$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | **2.** Determine the impedance seen by the source for the circuit below if $u(t) = 2 \cos(4000t)$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | **3.** For the circuit of exercise 1, determine the current delivered by the source. | ||
| + | |||
| + | **4.** For the circuit of exercise 2, determine the voltage across the $8\Omega$ resistors. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 10.6: Frequency Domain System Characterization ===== | ||
| + | In Chapters 7 and 8, we wrote the differential equation governing the relationship between a circuit’s input and output (the input-output equation) and used this differential equation to determine the response of a circuit to some input. We also characterized the time-domain behavior of the system by examining the circuit’s natural and step responses. We saw that the behavior of a first order circuit can be characterized by its time constant and DC gain, while the response of a second order circuit is characterized by its natural frequency, damping ratio and DC gain. It is important to recognize that these characterizations were independent of specific input parameters; they depended upon the type of response (e.g. a step function or a natural response), but were __independent of detailed information__ such as the amplitude of the step input or the actual values of the initial conditions. | ||
| + | |||
| + | We will now use the steady state sinusoidal response to characterize a circuit’s behavior. As in the case of our time-domain characterization, | ||
| + | |||
| + | When a sinusoidal input is applied to a linear system, the system’s __forced__ response consists of a sinusoid with the same frequency as the input sinusoid, but in general having a different amplitude and phase from the input sinusoid. Figure 10.24 shows the general behavior, in block diagram form. Changes in the amplitude and phase angle between the input and output signals are often used to characterize the circuit’s input-output relationship at the input frequency, $\omega$. In this chapter, we will demonstrate how this characterization is performed for inputs with discrete frequencies (as in the case of circuits with one or several inputs, each with a single frequency component). Later chapters will extend these concepts to the case in which frequency is considered to be a __continuous__ variable. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Previously in this chapter, we have (bit-by-bit) simplified the analysis of a system’s steady state sinusoidal response significantly. We first represented the sinusoidal signals as complex exponentials in order to facilitate our analysis. We subsequently used phasors to represent our complex exponential signals, as shown in Figure 10.25; this allowed us to represent and analyze the circuit’s steady state sinusoidal response directly in the frequency domain. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | In the frequency domain analyses performed to date, we have generally determined the system’s response to a __specific__ input signal with a given frequency, amplitude, and phase angle. We now wish to characterize the system response to an input signal with a given frequency, but an arbitrary amplitude and phase angle. As indicated previously in section 10.1, we will see that the input-output relationship governing the system reduces to a relationship between the output and input signal __amplitudes__ and the output and input signal __phases__. The circuit can thus be represented in phasor form as shown in Figure 10.26. The system’s effect on a sinusoidal input consists of an amplitude gain between the output and input signals ($\frac{B}{A}$ in Figure 10.26) and a __phase difference__ between the output and input signals ($\phi - \theta$ in Figure 10.26). | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Rather than perform a rigorous demonstration of this property at this time, we will simply provide some simple examples to illustrate the basic concept. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.18 ==== | ||
| + | A sinusoidal voltage, $v_{in}(t)$, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Since the frequency is unspecified, | ||
| + | |||
| + | $$\underline{V}_{out} = \frac{\frac{1}{j \omega C}}{R+ \frac{1}{j \omega C}} \cdot \underline{V}_{in} = \frac{1}{1+j \omega RC} \cdot \underline{V}_{in}$$ | ||
| + | |||
| + | The factor $\frac{1}{1+j \omega RC}$ is a complex number, for given values of $\omega$, R, and C. It constitutes a multiplicative factor which, when applied to the input, results in the output. This multiplicative factor is often used to __characterize__ the system’s response at some frequency, $\omega$. We will call this multiplicative factor the frequency response function, and denote it as $H(j \omega)$. For a particular frequency, $H(j \omega)$ is a complex number, with some amplitude, $|H(j \omega)|$ , and phase angle, $\angle H(j \omega)$. For our example, the magnitude and phase of our frequency response function are: | ||
| + | |||
| + | $$|H(j \omega)| = \frac{1}{\sqrt{1+(\omega RC)^2}}$$ | ||
| + | |||
| + | $$\angle H(j \omega) = - \tan^{-1}(\omega RC)$$ | ||
| + | |||
| + | According to the rules of multiplication of complex numbers, when two complex numbers are multiplied, the magnitude of the result is the product of the magnitudes of the individual numbers, and the phase angle of the result is the sum of the individual phase angles. Thus, if the input voltage is represented in phasor form as $\underline{V}_{in} = |V_{in}| \cdot |H(j \omega)|$ and the output voltage is $\underline{V}_{out} = |V_{out}| \angle \phi$, it is easy to obtain the output voltage from the input voltage and the frequency response function: | ||
| + | |||
| + | $$|V_{out}| = |V_{in}| \cdot |H(j \omega)|$$ | ||
| + | |||
| + | $$\angle V_{out} = \angle V_{in} + \angle H(j \omega)$$ | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 10.19 ==== | ||
| + | Use the frequency response function determined in Example 10.18 above to determine the response $v_{out}(t)$ of the circuit shown below to the following input voltages: | ||
| + | * $v_{in}(t)= 3 \cos(2t+20^{\circ})$ | ||
| + | * $v_{in}(t)= 7 \cos(4t-60^{\circ})$ | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | When $v_{in}(t) = 3 \cos(2t+20^{\circ})$, | ||
| + | |||
| + | $$|H(j2)|= \frac{1}{\sqrt{1+(\omega RC)^2}} = \frac{1}{\sqrt{1+(2 \cdot 2\Omega \cdot 0.25F)^2}} = \frac{1}{\sqrt{1+1^2}}=\frac{1}{\sqrt{2}}$$ | ||
| + | |||
| + | $$\angle H(j2) = -\tan^{-1} (\omega RC) = -\tan^{-1} (2 \cdot 2 \Omega \cdot 0.25F) = -\tan^{-1} (1) = -45^{\circ}$$ | ||
| + | |||
| + | 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 $\angle H(j2)$, so that: | ||
| + | |||
| + | $$|V_{out}| = |V_{in}| \cdot |H(j2)| = 3 \cdot \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$$ | ||
| + | |||
| + | $$\angle V_{out} = \angle V_{in} + \angle H(j2) = 20^{\circ} + (-45^{\circ}) = -25^{\circ}$$ | ||
| + | |||
| + | And the time-domain output voltage is: | ||
| + | |||
| + | $$v_{out}(t) = \frac{3}{\sqrt{3}} \cos(2t-25^{\circ})$$ | ||
| + | |||
| + | When $v_{in} = 7 \cos(4t-60^{\circ})$, | ||
| + | |||
| + | $$|H(j4)|= \frac{1}{\sqrt{1+(\omega RC)^2}}=\frac{1}{\sqrt{5}}$$ | ||
| + | |||
| + | And: | ||
| + | |||
| + | $$\angle H(j4) = -\tan^{-1}(\omega RC) = -63.4^{\circ}$$ | ||
| + | |||
| + | 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: | ||
| + | |||
| + | $$v_{out}(t) = \frac{7}{\sqrt{5}} \cos(2t-123.4^{\circ})$$ | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | From the above examples we can see that, once the frequency response function is calculated for a circuit __as a function of frequency__, | ||
| + | |||
| + | 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. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== 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. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | 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: | ||
| + | |||
| + | $$v_1(t) = \frac{3}{\sqrt{2}} \cos(2t-25^{\circ})$$ | ||
| + | |||
| + | And: | ||
| + | |||
| + | $$v_2(t) = \frac{7}{\sqrt{5}} \cos(2t-123.4^{\circ})$$ | ||
| + | |||
| + | The overall response is then: | ||
| + | |||
| + | $$v_{out}(t) = v_1(t) + v_2(t) = \frac{3}{\sqrt{2}} \cos(2t-25^{\circ}) + \frac{7}{\sqrt{5}} \cos(2t-123.4^{\circ})$$ | ||
| + | |||
| + | A plot the input and output waveforms is shown below: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== 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: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | * 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. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== 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.) | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | **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.) | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | |||
| + | |||
| + | [[{}/ | ||
| + | [[{}/ | ||