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

Course Book : Fundamentals of Electric Circuits, by Charles K. Alexander and Matthew N. O. Sadiku, McGraw Hill; 5rd edition (2007)

Reference Books: Electric Circuits, by James W. Nilsson and Susan Riedel, Prentice Hall; 8th edition (2007) Schaum's Outline of Electric Circuits, by Mahmood Nahvi 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's Outline of Basic Circuit Analysis, by John O'Malley and John O'Malley, McGraw-Hill; 2nd edition (1992)

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

Second Order Circuits

Overview The previous chapter introduced the concept of first order circuits. This chapter will expand on that with second order circuits: those that need a second order differential equation. RLC series and parallel circuits will be discussed in this context. The step response of these circuits will be covered as well. Finally the concept of duality will be discussed. Duality=çiftlik

Second Order Circuits The previous chapter considered circuits which only required first order differential equations to solve. However, when more than one “storage element”, i.e. capacitor or inductor is present, the equations require second order differential equations The analysis is similar to what was done with first order circuits This time, though we will only consider DC independent sources

1.1. Introduction

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

EXAMPLE 1.1. The switch in Figure has been closed for a long time. İt is open at t=0. Find; 𝑖(0 + ),𝑣( 0 + ) 𝑑𝑖(( 0 + )/dt, 𝑑𝑣(( 0 + )/dt, 𝑖(∞) , 𝑣(∞)

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

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 𝑣 𝐿 by applying KVL to the loop

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

EXAMPLE 1.2. In figure calculate; 𝑖 𝑙 ( 0 + ), 𝑣 𝑐 ( 0 + ), 𝑣 𝑅 ( 0 + ) 𝑑𝑖 𝐿 (( 0 + )/dt, 𝑑𝑣 𝑐 (( 0 + )/dt 𝑖 𝐿 ∞ , 𝑣 𝑐 ∞ , 𝑣 𝑅 (∞)

EXAMPLE 1.2. /Solution: (a) For t < 0, 3u(t) = 0. At t = 0−, since the circuit has reached steady state, the inductor can be replaced by a short circuit, while the capacitor is replaced by an open circuit as shown in Fig. (a). From this figure we obtain

EXAMPLE 1.2. /Solution: For t > 0, 3u(t) = 3, so that the circuit is now equivalent to that in Fig. (b). Since the inductor current and capacitor voltage cannot change abruptly, Applying KCL at node a in Fig. (b) gives Applying KVL to the middle mesh in Fig.(b) yields

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

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

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

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

1.3. The Source-Free Series RLC Circuits with initial values equation (2)… or (4) Based on the first order solutions, we can expect that the solution will be in exponential form. equation (3) becomes… or i

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

1.3. The Source-Free Series RLC Circuits 𝜔 0 : resonant frequency or undamped natural frequency (rad/s) ∝: neper frequency or damping factor (Np/s) İn terms of 𝜔 0 𝑎𝑛𝑑 ∝ 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;

1.3. The Source-Free Series RLC Circuits There are three types of solutions; Overdamped Case (∝> 𝝎 𝟎 ) İn this situation; Aşırı sönümlü Kritik sonumlu Az sonumlu 𝑠 1 𝑎𝑛𝑑 𝑠 2 negative and real From this, we should not expect to see an oscillation

1.3. The Source-Free Series RLC Circuits Critically Damped Case (∝= 𝝎 𝟎 ) Kritik Sönümlü Durum: Köklerin reel ve birbirine eşit olduğu durumdur. Bu durumda kökler ₁=₂= ve sönüm oranı =1 olur.

1.3. The Source-Free Series RLC Circuits Under Damped Case (∝< 𝝎 𝟎 ) ω0 is often called the undamped natural frequency ωd is called the damped natural frequency

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: Since α > ω0, we conclude that the response is overdamped. This is also evident from the fact that the roots are real and negative.

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

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 a source-free series RLC circuit. Hence, the response is underdamped (α < ω); that is,

EXAMPLE 1.4./ Solution We now obtain 𝐴 1 and 𝐴 2 using the initial conditions. At t = 0,