Circuits and Systems Theory

1. Circuits and Systems Theory Assist.Prof. Sibel ÇİMEN Electronics and Communication Engineering University of Kocaeli

2. Course Book : • Fundamentals of Electric Circuits, by Charles K. Alexander and Matthew N. O. Sadiku, McGraw Hill; 3rd edition (2007)

3. Reference Books: • Electric Circuits, by James W. Nilsson and Susan Riedel, Prentice Hall; 8th edition (2007) • Schaum'sOutline of Electric Circuits, by MahmoodNahvi and Joseph Edminister, McGraw-Hill; 4th edition (2002) • Introduction to Electric Circuits, by Richard C. Dorf and James A. Svoboda, Wiley, 7th edition (2006) • Schaum'sOutline of Basic Circuit Analysis, by John O'Malley and John O'Malley, McGraw-Hill; 2nd edition (1992)

4. Course Outline • Second Order DC Circuits (Fund. of Electric Circuits, CH 8) • Sinusoids and Phasors (Fund. of Electric Circuits, CH 9) • Sinusoidal Steady-State Analysis (Fund. of Electric Circuits, CH 10) • AC Power Analysis (Fund. of Electric Circuits, CH 11) • Frequency Response (Fund. of Electric Circuits, CH 14) • Laplace Transform (Fund. of Electric Circuits, CH 15

5. Second Order Circuits

6. 1.1. Introduction

7. 1.1. Introduction • Keep in mind; • Capacitor voltage always continuous… • Inductor current always continuous…

8. EXAMPLE 1.1. • The switch in Figure has been closed for a long time. İt is open at t=0. • Find; • )/dt, )/dt,

9. EXAMPLE 1.1. /Solution: • (a) If the switch is closed a long time before t = 0, it means that the circuithas reached dc steady state at t = 0. At dc steady state, the inductor actslike a short circuit, while the capacitor acts like an open circuit, so wehave the circuit in Fig. 8.(a) at t = 0−. Thus, As the inductor current and the capacitor voltage cannot change abruptly,

10. EXAMPLE 1.1. /Solution: • (b) At t = 0+, the switch is open; the equivalent circuit is as shown in Fig.8(b). The same current flows through both the inductor and capacitor. We now obtain byapplying KVL to the loop

11. EXAMPLE 1.1. /Solution: (c) Fort > 0, the circuit undergoes transience. But as t →∞, thecircuitreaches steady state again. The inductor acts like a short circuit and thecapacitor like an open circuit, so that the circuit becomes that shown inFig. 8(c), from which we have

12. EXAMPLE 1.2. • In figurecalculate; • , • )/dt, )/dt

13. EXAMPLE 1.2. /Solution: • (a) For t < 0, 3u(t) = 0. At t = 0−, since the circuit has reached steadystate, the inductor can be replaced by a short circuit, while the capacitoris replaced by an open circuit as shown in Fig. (a). From this figurewe obtain

14. EXAMPLE 1.2. /Solution: • For t > 0, 3u(t) = 3, so that the circuit is now equivalent to thatin Fig. (b). Since the inductor current and capacitor voltage cannotchange abruptly, Applying KCL at node ain Fig. (b) gives Applying KVL to the middle mesh in Fig.(b) yields

15. EXAMPLE 1.2. /Solution: But applying KVL to the right mesh in Fig. (b) gives

16. EXAMPLE 1.2. /Solution: • (c) As t →∞, the circuit reaches steady state. We have the equivalentcircuit in Fig.(a) except that the 3-A current source is now operative. • By current division principle,

17. 1.3. The Source-Free Series RLC Circuits (1.a) (1.b)

18. 1.3. The Source-Free Series RLC Circuits • Applying KVL aroundtheloop; (2) • To eliminate the integral, we differentiate with respect to t and rearrange terms:…. (3)

19. 1.3. The Source-Free Series RLC Circuits • with initial values equation (2)… (4) or Bobinin uçlarından akan akımın exp. Karakteristiği olduğunu biliyoruz… • equation (3)becomes… or i 0

20. 1.3. The Source-Free Series RLC Circuits Known as characteristic equation (5) İt’sroots; Where;

21. 1.3. The Source-Free Series RLC Circuits resonant frequency or undamped natural frequency (rad/s) neperfrequencyor damping factor (Np/s) İn terms of equation (5) gets… (6) The two values of s in Eq. (5) indicate that there are two possible solutions in Eq. (6); that is, Natural response of series RLC;

22. 1.3. The Source-Free Series RLC Circuits There are three types of solutions; Overdamped Case ( İn this situation; negative and real

23. 1.3. The Source-Free Series RLC Circuits Critically Damped Case (

24. 1.3. The Source-Free Series RLC Circuits Under Damped Case (

25. EXAMPLE 1.3. R=40 Ω, L=4H and C=1/4 F. Calculate the characteristic roots of the circuit. Is the natural response overdamped, underdamped, or critically damped? Solution: Sinceα > ω0, we conclude that the response is overdamped. This is alsoevident from the fact that the roots are real and negative.

26. EXAMPLE 1.4. Find i(t) in the circuit. Assume that the circuit has reached steady state at t=. Solution: For t < 0, the switch is closed. The capacitor acts like an open circuitwhile the inductor acts like a shunted circuit. The equivalent circuit isshown in Fig. (a). Thus, at t = 0,

27. EXAMPLE 1.4./ Solution • For t > 0, the switch is opened and the voltage source is disconnected.The equivalent circuit is shown in Fig. 8.11(b), which is asource-free series RLC circuit. Hence, the response is underdamped (α < ω); that is,

28. EXAMPLE 1.4./ Solution • We now obtain and using the initial conditions. At t = 0,