Revision: August 21, E Main Suite D Pullman, WA (509) Voice and Fax

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1 .7.: Frequency doma system characterization Revision: August, 00 E Ma Suite D Pullman, WA 9963 ( Voice and Fax Overview In previous chapters, we wrote the differential equation governg the relationship between a circuit s put and put (the putput equation and used this differential equation to determe the response of a circuit to some put. We also characterized the timedoma behavior of the system by examg 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 ga, while the response of a second order circuit is characterized by its natural frequency, dampg ratio and DC ga. It is important to recognize that these characterizations were dependent of specific put parameters; they depended upon the type of response (e.g. a step function or a natural response, but were dependent of detailed formation such as the amplitude of the step put or the actual values of the itial conditions. We will now use the steady state susoidal response to characterize a circuit s behavior. As the case of our timedoma characterization, this characterization will allow the system s behavior to be defed terms of its response to susoidal puts, but the characterization will be dependent of details such as the put susoid s amplitude or phase angle. (The put susoid s frequency will, however, still be of prime importance. Before begng this module, you should be able to: Calculate impedances for resistors, capacitors, and ductors (chapter.7.3 Analyze frequency doma circuits to determe phasor representations of circuit put and put signals (chapter.7.4 Draw phasor diagrams of circuit put and put signals (chapter.7.3 After completg this module, you should be able to: Defe the frequency response of a system Defe the magnitude response and phase response of a system Determe the magnitude and phase responses of a circuit This module requires: N/A As we have seen, for the case which a susoidal put is applied to a lear system, the system s forced response consists of a susoid with the same frequency as the put susoid, but general havg a different amplitude and phase from the put susoid. Figure shows the general behavior, block diagram form. Changes the amplitude and phase angle between the put and put signals are often used to characterize the circuit s putput relationship at the put frequency, ω. In this chapter, we will demonstrate how this characterization is performed for puts with discrete frequencies (as the case of circuits with one or several puts, each with a sgle frequency Doc: XXXYYY page of 6

2 component. Later chapters will extend these concepts to the case which frequency is considered to be a contuous variable. Figure. Susoidal steadystate putput relation for a lear time variant system. In previous chapters (.7.0 through.7.4 we have simplified the analysis of a system s steady state susoidal response significantly. We first represented the susoidal signals as complex exponentials order to facilitate our analysis. We subsequently used phasors to represent our complex exponential signals, as shown Figure ; this allowed us to represent and analyze the circuit s steady state susoidal response directly the frequency doma. U A θ Y B φ Figure. Phasor representation of susoidal puts and puts. In the frequency doma analyses performed to date, we have generally determed the system s response to a specific put signal with a given frequency, amplitude, and phase angle. We now wish to characterize the system response to an put signal with a given frequency, but an arbitrary amplitude and phase angle. As dicated previously chapter.7.0, we will see that the putput relationship governg the system reduces to a relationship between the put and put signal amplitudes and the put and put signal phases. The circuit can thus be represented phasor form as shown Figure 3. The system s effect on a susoidal put consists of an amplitude ga between the put and put signals ( A B Figure 3 and a phase difference between the put and put signals ( φ θ Figure 3. U A θ B ( φ θ A Y B φ Figure 3. Frequency doma representation of circuit putput relationship Rather than perform a rigorous demonstration of this property at this time, we will simply provide a simple example to illustrate the basic concept. page of 6

3 Example : A susoidal voltage, v (t, is applied to the circuit to the left below. Determe the frequencydoma relationship between the phasor representg v (t and the phasor representg the put voltage v (t. V C V Sce the frequency is unspecified, we leave frequency as an dependent variable, ω, our analysis. In the frequency doma, therefore, the circuit can be represented as shown to the right above. The frequency doma circuit is a simple voltage divider, so the relation between put and put is: V C R + C V + V RC The factor is a complex number, for given values of ω, R, and C. It constitutes a + RC multiplicative factor which, when applied to the put, results the put. This multiplicative factor is often used to characterize the system s response at some frequency, ω. We will call this multiplicative factor the frequency response function, and denote it as H(. For a particular frequency, H( is a complex number, with some amplitude, H(, and phase angle, H(. For our example, the magnitude and phase of our frequency response function are: H( H( tan + Accordg 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 dividual numbers, and the phase angle of the result is the sum of the dividual phase angles. Thus, if the put voltage is represented phasor form as V V V and the put voltage is V V φ, it is easy to obta the put voltage from the put voltage and the frequency response function: V V H( V V + H( page 3 of 6

4 Example : Use the frequency response function determed Example above to determe the response v (t of the circuit shown below to the followg put voltages: (a v (t 3cos(t+0 (b v (t 7cos(4t60 + v (t + 0.F v (t When v (t 3cos(t+0, ω rad/sec, V 3 and V 0. For this value of ω, and the given values of R and C, the magnitude and phase of the frequency response function are: H( j + H( j tan ( ω RC tan + ( Ω 0. F + ( Ω 0. F tan ( 4 The put amplitude is then the product of and H( j, so that: V V and H( j and the put phase is the sum of 3 V V H( j 3 V V + H( j 0 + ( 4 and the timedoma put voltage is: 3 v ( t cos( t When v (t 7cos(4t60, ω 4 rad/sec, V 7 and V 60. For this value of ω, and the given values of R and C, the magnitude and phase of the frequency response function are: H( j4 +, and H( j4 tan ( ω RC 63. The put amplitude is then the product of V V and H( j4 and the put phase is the sum of and H( j4 so that the timedoma put voltage this case is: 7 v ( t cos( t 3. page 4 of 6

5 From the above examples we can see that, once the frequency response function is calculated for a circuit as a function of frequency, we can determe the circuit s steadystate response to any put susoid directly from the frequency response function, with reanalyzg the circuit itself. We conclude this section with one additional example, to illustrate the use of the frequency response function and superposition to determe a circuit s response to multiple puts of different frequencies. Example 3: Use the results of examples and above to determe the response v (t of the circuit shown below if the put voltage is v (t 3cos(t+0 + 7cos(4t60. Plot the put and put waveforms. + v (t + 0.F v (t Recall, from chapter.7.4, that superposition is the only valid approach for performg frequency doma analysis of circuits with puts at multiple frequencies. Also recall that each frequency can be analyzed separately the frequency doma, but that the superposition process (the summation of the dividual contributions must be done the time doma. For this problem, we have contributions at two different frequencies: rad/sec and 4 rad/sec. Luckily, we have determed the dividual responses of the circuit to these two puts Example. Therefore, the time doma, the two contributions to our put will be: 3 7 v ( t cos( t and v ( t cos( t 3.. The overall response is then: 3 7 v ( t v ( t + v( t cos( t + cos( t 3. A plot the put and put waveforms is shown below: page of 6

6 Important pots: The frequency response function or frequency response describes a circuit s putput relationship directly the frequency doma, as a function of frequency. The frequency response is a complex function of frequency H( (that is, it is a complex number which depends upon the frequency. This complex function is generally expressed as a magnitude and phase, H( and H(., respectively. H( is called the magnitude response of the circuit, and H( is called the phase response of the circuit. The overall idea is illustrated the block diagram below: U A θ H( H( H( Y B φ The magnitude response of the circuit is the ratio of the put amplitude to the put amplitude. This is also called the ga of the system. Thus, the figure above, the put amplitude B H( A. Note that the magnitude response or ga of the system is a function of frequency, so that puts of different frequencies will have different gas. The phase response of the circuit is the difference between the put phase angle and the put phase angle. Thus, the figure above, the put phase φ H ( + θ. Like the ga, the phase response is a function of frequency puts at different frequencies will, general, have different phase shifts. Use of the frequency response to perform circuit analyses can be particularly helpful when the put signal contas a number of susoidal components at different frequencies. In this case, the response of the circuit to each dividual component can be determed the frequency doma usg the frequency response and the resultg contributions summed the time doma to obta the overall response. page 6 of 6