R1 + C +

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Vi _ R1 = 100 C = 47 F R2 = 500

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69 Compute the frequency at which the phase shift introduced by the circuit of Example 63 is equal to 10 610 Compute the frequency at which the output of the circuit of Example 63 is attenuated by 10 percent (ie, VL = 09VS ) 611 Compute the frequency at which the output of the circuit of Example 66 is attenuated by 10 percent (ie, VL = 09VS )

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Frequency Response and System Concepts

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612 Compute the frequency at which the phase shift introduced by the circuit of Example 66 is equal to 20 613 Compute the frequencies 1 and 2 for the band-pass lter of Example 67 (with R = 1 k ) for equating the magnitude of the band-pass lter frequency response to 1/ 2 (this will result in a quadratic equation in , which can be solved for the two frequencies)

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COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM

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The transient analysis methods illustrated in the preceding chapter for rst- and second-order circuits can become rather cumbersome when applied to higher-order circuits Moreover, solving the differential equations directly does not reveal the strong connection that exists between the transient response and the frequency response of a circuit The aim of this section is to introduce an alternate solution method based on the notions of complex frequency and of the Laplace transform The concepts presented in this section will demonstrate that the frequency response of linear circuits is but a special case of the general transient response of the circuit, when analyzed by means of Laplace methods In addition, the use of the Laplace transform method allows the introduction of systems concepts, such as poles, zeros, and transfer functions, that cannot be otherwise recognized Complex Frequency In 4, we considered circuits with sinusoidal excitations such as v(t) = A cos( t + ) which we also wrote in the equivalent phasor form V(j ) = Aej = A The two expressions just given are related by v(t) = Re(Vej t ) (642) (641) (640)

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As was shown in 4, phasor notation is extremely useful in solving AC steady-state circuits, in which the voltages and currents are steady-state sinusoids of the form of equation 640 We now consider a different class of waveforms, useful in the transient analysis of circuits, namely, damped sinusoids The most general form of a damped sinusoid is v(t) = Ae t cos( t + ) (643)

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As one can see, a damped sinusoid is a sinusoid multiplied by a real exponential, e t The constant is real and is usually zero or negative in most practical circuits Figures 631(a) and (b) depict the case of a damped sinusoid with negative and with positive , respectively Note that the case = 0 corresponds exactly to a sinusoidal waveform The de nition of phasor voltages and currents given in 4 can easily be extended to account for the case of damped sinusoidal waveforms by de ning a new variable, s, called the complex frequency: s = + j (644)

Part I

Circuits

1 05 v (t) 0 05 1 0 08 06 t (s) Damped sinusoid (exponential decay) 02 04 1 v (t)

6 4 2 0 2 4 6 8 0 08 06 t (s) Damped sinusoid (exponential growth) 02 04 1

Figure 631(a) Damped sinusoid: negative

Figure 631(b) Damped sinusoid: positive

You may wish to compare this expression with the term j in equation 565 Note that the special case = 0 corresponds to s = j , that is, the familiar steadystate sinusoidal (phasor) case We shall now refer to the complex variable V(s) as the complex frequency domain representation of v(t) It should be observed that from the viewpoint of circuit analysis, the use of the Laplace transform is analogous to phasor analysis; that is, substituting the variable s wherever j was used is the only step required to describe a circuit using the new notation