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3-Lesson_Notes_Lecture20_ee205 - Lecture 20 In this lecture...

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Lecture 20 In this lecture the following material is covered: We explore the complex frequency ( s ) plane Natural response of a circuit is obtained from pole location Examples are worked to find the complete response (forced and natural) through complex frequency domain analysis Demonstrate how the location of poles of the poles can affect the natural response Finally, a test on the material of this chapter will be conducted . The Complex Frequency (s) Plane The complex frequency plane or the s-plane is sketched in Fig.22-1. Recollect that, s j σ ω = + s-plane Figure 22-1 j ω σ The horizontal (x) axis of the plane is real part σ and the vertical (y) axis being the imaginary part ω . Also, observe that both σ and ω are frequencies; one being real part and the other imaginary part of complex frequency. From the several of examples solved you may have noticed that response of a circuit depends both on σ as well as ω . The first term
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controls how quickly the transients (or oscillations) will die down or grow. The second term controls the frequency of the oscillations in the response . If we denote a pole by a symbol ‘x’ and a zero by a symbol ‘ Ο ’, then depicting the poles and zeroes of a function in the s-plane is referred to as pole-zero plot . Consider for example the following function, 2 2 50( 3)( 4 8) ( ) ( 4)( 2 2) s s s H s s s s s + + + = + + + (22-1) The poles are at s =0, -1, -1+j1, -1-j1; the zeroes are at s =-3, -2+j2, -2- j2, respectively. The pole-zero plot for the function is shown in Fig. 22-2. s-plane -1 -2 -3 -4 0 -1 -2 1 2 X o X X X o o Figure 22-2 j ω σ 1` Natural Response from Transfer (or, Network) Function In earlier chapters we learnt how to find the natural response of a circuit. We write the circuit equations in terms of differential equations and solve for the response by setting the input to zero. To refresh are our memory, let us consider an example.
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