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s= 3
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1 1 2s + 5 = + s 2 + 5s + 6 s+2 s+3
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and using Table 61, we compute f (t) = e 2t + e 3t u(t)
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621 Find the inverse Laplace transform of each of the following functions:
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a F (s) = b F (s) = c F (s) = d F (s) = s2 1 + 5s + 6
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Transfer Functions, Poles, and Zeros It should be clear that the Laplace transform can be quite a convenient tool for analyzing the transient response of a circuit The Laplace variable, s, is an extension of the steady-state frequency response variable j already encountered in this chapter Thus, it is possible to describe the input-output behavior of a circuit using Laplace transform ideas in the same way in which we used frequency response ideas earlier Now, we can de ne voltages and currents in the complex frequency domain as V(s) and I(s), and denote impedances by the notation Z(s), where s replaces the familiar j We de ne an extension of the frequency response of a circuit, called the transfer function, as the ratio of any input variable to any output variable, ie: H1 (s) = Vo (s) Vi (s) or H2 (s) = Io (s) Vi (s) etc (653)
R1 + vC(t) L iO(t) C R2
As an example, consider the circuit of Figure 632 We can analyze it using a method analogous to phasor analysis by de ning impedances Z1 = R1 1 ZC = sC ZL = sL Z2 = R2 (654)
v i (t) + _
Then, we can use mesh analysis methods to determine that Io (s) = Vi (s) ZC (ZL + Z2 )ZC + (ZL + Z2 )Z1 + Z1 ZC (655)
Z1 ZL Z2
or, upon simplifying and substituting the relationships of equation 654, Io (s) 1 H2 (s) = = 2 + (R R C + L)s + R + R Vi (s) R1 LCs 1 2 1 2 (656)
+ Vi (s) + VC (s) ZC IO(s) _
If we were interested in the relationship between the input voltages and, say, the capacitor voltage, we could similarly calculate H1 (s) = VC (s) sL + R2 = 2 + (R R C + L)s + R + R Vi (s) R1 LCs 1 2 1 2 (657)
Figure 632 A circuit and its Laplace transform domain equivalent
6
Frequency Response and System Concepts
Note that a transfer function consists of a ratio of polynomials; this ratio can also be expressed in factored form, leading to the discovery of additional important properties of the circuit Let us, for the sake of simplicity, choose numerical values for the components of the circuit of Figure 632 For example, let R1 = 05 , C = 1 F, L = 05 H, and R2 = 2 Then we can substitute these values into 4 equation 657 to obtain H1 (s) = 05s + 2 s+4 =8 2 00625s 2 + 0375s + 25 s + 6s + 40 s+4 s 30000 + j 55678)(s 30000 j 55678) (658)
Equation 658 can be factored into products of rst-order terms as follows: H1 (s) = 8 (659)
where it is apparent that the response of the circuit has very special characteristics for three values of s: s = 4; s = +30000 + j 55678; and s = +30000 j 55678 In the rst case, at the complex frequency s = 4, the numerator of the transfer function becomes zero, and the response of the circuit is zero, regardless of how large the input voltage is We call this particular value of s a zero of the transfer function In the latter two cases, for s = +30000 j 55678, the response of the circuit becomes in nite, and we refer to these values of s as poles of the transfer function It is customary to represent the response of electric circuits in terms of poles and zeros, since knowledge of the location of these poles and zeros is equivalent to knowing the transfer function and provides complete information regarding the response of the circuit Further, if the poles and zeros of the transfer function of a circuit are plotted in the complex plane, it is possible to visualize the response of the circuit very effectively Figure 633 depicts the pole zero plot of the circuit of Figure 632; in plots of this type it is customary to denote zeros by a small circle, and poles by an
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