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| learn:courses:real-analog-chapter-10:start [2017/01/13 22:42] – [Parallel and Series Impedances & Circuit Reduction] 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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| {{ : | {{ : | ||
| + | |||
| + | 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.) | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | |||
| + | |||
| + | [[{}/ | ||
| + | [[{}/ | ||