Chapter 4 Second Order Circuit (11th & 12th week) SEE 1023 Circuit Theory Chapter 4 Second Order Circuit (11th & 12th week) Prepared by : jaafar shafie

Second Order Circuit In first order circuit, the RC and RL circuits are represented in first order differential equation. This is due to the existence of only one storage element at any particular circuit. In this chapter, two storage elements will exist in a particular circuit. Thus, this circuit are characterized by second order differential equation. A circuit with second order differential equation is called SECOND ORDER CIRCUIT.

Second Order Circuit Types of second order circuit that may exist:- Series RLC circuit, Parallel RLC circuit, RLL circuit, RCC circuit. V R L C I R L C V R2 L2 L1 R1 C2 R2 C1 I

Second Order Circuit As usual, the circuit will be analyzed in two parts:- Source–free Circuit (natural response) Energy is initially stored in the element – thus no effect of current or voltage sources. Circuit with source (forced response) Current or voltage sources is directly connected to the first order circuits. Before the circuits are being analyzed, one should find the initial and steady state value of the capacitor voltage and inductor current and it’s derivative; i.e.:- v(0), i(0), dv(0)/dt, di(0)/dt, v(), i().

Second Order Circuit Consider the circuit shown below where the switch has been closed for a long time. Find:- v(0+), i(0+), dv(0+)/dt, di(0+)/dt, v(), i(). 24V 4Ω 0.4F t=0s 1H + v(t) — i(t)

Second Order Circuit When t=0+s=0–s, the inductor is shorted and the capacitor is opened. The equivalent circuit is shown below, When the switch is opened, the equivalent circuit is thus, 24V 4Ω + v(0+) - i(0+) v(0+)=(4/8)24=12V i(0+)= 24/8 =3A i(0+) 24V 4Ω 1H 0.4F i(0+)= iC (0+) =3A and it is know that

Second Order Circuit Applying KVL to the circuit when switch is opened, thus, i(0+) 24V 4Ω 1H 0.4F

Second Order Circuit The steady state value, i() 24V 4Ω + v() -

Natural Response - Series RLC Circuit Natural response is obtained with Series RLC circuit without source. Energy is initially stored in the L and C, where the initial voltage at capacitor is VO and initial current at inductor is IO. Applying KVL to the loop, IO R L C + VO -

Natural Response - Series RLC Circuit To eliminate the integral, differentiate with respect to t and rearrange the terms such that Now, we have second order differential equation. It is known that in first-order differential equation, the current is where A and s are constants. Substitute Aest to the second order equation.

Natural Response - Series RLC Circuit Thus, we may write as or And we should find the value of A, thus Aest must not equal to zero. The only part that should equal to zero is known as CHARACTERISTIC EQUATION

Natural Response - Series RLC Circuit Solve the characteristic equation, one might find the two roots, which are and

Natural Response - Series RLC Circuit On the other hand, we may represent the two roots as and where The two solutions for s (i.e. s1 and s2) shows that there are two values of the current, which are The total response of the current would be

Natural Response - Series RLC Circuit Thus, we may found three type of response which are 1st Type :  > ω0 , the response is called OVERDAMPED 2. 2nd Type :  = ω0 , the response is called CRITICALLY DAMPED 3. 3rd Type :  < ω0 , the response is called UNDERDAMPED

Natural Response - Series RLC Circuit 1st Type :  > ω0 , OVERDAMPED In this type,  > ω0, or C > 4L/R2. It is found that both roots, (i.e. s1 and s2) are negative and real. Thus, the current response is which decays to zero when t increased. A1 and A2 are determined from the initial inductor current and the rate of change of current. Solve for A1 and A2

Natural Response - Series RLC Circuit For example A series RLC circuit has R=20Ω, L=1mH and C=100F. If i(0+)=1A and vC(0+)=18V, find the current response. *It is clear that C > 4L/R2, and the response is the 1st type which is ‘overdamped’. Step1: Find the value of s1 and s2

Natural Response - Series RLC Circuit Step2: Find the value of A1 and A2 from the initial values. apply KVL to the loop thus,

Natural Response - Series RLC Circuit Step3: It is found that, A1 = 0.9216, A2 = 78.36m Time -2ms 0ms 2ms 4ms 6ms 8ms 10ms 12ms 14ms 16ms 18ms I(L1) 0A 0.5A 1.0A

Natural Response - Series RLC Circuit 2nd Type :  = ω0 , CRITICALLY DAMPED In this type,  = ω0, or C = 4L/R2. It is found that both roots, (i.e. s1 and s2) are equal to – or –R/2L. Thus, the current response is It can be seen that solution for A3 could not be obtained with two initial condition [i(0) & di(0)/dt)]. Thus, there might be another method to find the response of critically damped.

Natural Response - Series RLC Circuit In this type,  = ω0 = R/2L = 1/LC. Thus, the second order differential equation become Let , thus In first order differential equation, it is found that , thus , or ,or

Natural Response - Series RLC Circuit By integrating the equation, one will found the response of critically damped response as Thus, the derivative of the current is, when t=0s,

Natural Response - Series RLC Circuit For example A series RLC circuit has R=20Ω, L=1mH and C=10F. If i(0+)=1A and vC(0+)=18V, find the current response. *It is clear that C = 4L/R2, and the response is the 2nd type which is ‘critically damped’. Step1: Find the value of s1 and s2

Natural Response - Series RLC Circuit Step2: Find the value of A1 and A2 from the initial values. apply KVL to the loop thus,

Natural Response - Series RLC Circuit Step3: It is found that, A1 = 8k, A2 = 1 Time -0.5ms 0ms 0.5ms 1.0ms 1.5ms 2.0ms I(L1) 0A 0.5A 1.0A

Natural Response - Series RLC Circuit 3rd Type :  < ω0 , UNDERDAMPED In this type,  < ω0, or C < 4L/R2. The roots can be written as where and The response can be further written as

Natural Response - Series RLC Circuit Rearranging the response such that, Applying Euler’s identities to the above equation, where Thus, Let B1=A1+A2 and B2=j(A1–A2), thus

Natural Response - Series RLC Circuit Differentiate the i(t), we have

Natural Response - Series RLC Circuit For example A series RLC circuit has R=20Ω, L=1mH and C=2F. If i(0+)=1A and vC(0+)=18V, find the current response. *It is clear that C < 4L/R2, and the response is the 3rd type which is ‘underdamped’. Step1: Find the value of s1 and s2

Natural Response - Series RLC Circuit Step2: Find the value of A1 and A2 from the initial values. apply KVL to the loop thus,

Natural Response - Series RLC Circuit Step3: It is found that, B1 = 1, B2 = 0.4, Time -0.4ms -0.2ms 0ms 0.2ms 0.4ms 0.6ms 0.8ms 1.0ms I(L1) -0.5A 0A 0.5A 1.0A

Natural Response - Series RLC Circuit In summary, the response for series RCL are 1st Type :  > ω0 , (OVERDAMPED) 2. 2nd Type :  = ω0 , (CRITICALLY DAMPED) 3rd Type :  < ω0 , (UNDERDAMPED)

Natural Response - Series RLC Circuit The comparison of the three responses are shown below Time 0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms I(L1) -0.5A 0A 0.5A 1.0A C=100F C=10F C=2F

Natural Response - Parallel RLC Circuit Natural response for Series RLC circuit is obtained without any connection to source. Applying KCL, thus we have Differentiate with t, IO R L C + VO - v

Natural Response - Parallel RLC Circuit Replace the first derivative as s and the second derivative as s2. Thus, the characteristic equation is obtained as follows, The roots are characterized by these equations

Natural Response - Parallel RLC Circuit On the other hand, we may represent the two roots as and where

Natural Response - Parallel RLC Circuit 1st Type :  > ω0 , OVERDAMPED In this type,  > ω0, or L > 4R2C. The roots, (i.e. s1 and s2) are negative and real. Thus, the voltage response is A1 and A2 are determined from the initial inductor current and the rate of change of current. Solve for A1 and A2

Natural Response - Parallel RLC Circuit For example A parallel RLC circuit has R=20Ω, L=10mH and C=1F. If iL(0+)=0.5A and v(0+)=10V, find the current response. Step1: Find the value of  and ω0 *It is clear that  > ωO , and the response is the 1st type which is ‘overdamped’.

Natural Response - Parallel RLC Circuit Find s1 and s2. The initial condition are

Natural Response - Parallel RLC Circuit Step3: It is found that, A1 = -18.09, A2 = 28.09 Time 0s 0.2ms 0.4ms 0.6ms 0.8ms 1.0ms 1.2ms 1.4ms 1.6ms 1.8ms 2.0ms V(R5:2) -10V -5V 0V 5V 10V

Natural Response - Parallel RLC Circuit 2nd Type :  = ω0 , CRITICALLY DAMPED In this type,  = ω0, or L = 4R2C. From the series RLC circuit, the response found in parallel RLC circuit is The derivative is represented as and at t=0s,

Natural Response - Parallel RLC Circuit For example A parallel RLC circuit has R=50Ω, L=10mH and C=1F. If iL(0+)=0.5A and v(0+)=10V, find the current response. Step1: Find the value of  and ω0 *It is clear that  = ωO , and the response is the 2nd type which is ‘critically damped’.

Natural Response - Parallel RLC Circuit Step2: Find the value of A1 and A2 from the initial values. apply KCL to the loop

Natural Response - Parallel RLC Circuit Step3: It is found that, A1 = -600k, A2 = 10 Time 0s 0.1ms 0.2ms 0.3ms 0.4ms 0.5ms 0.6ms 0.7ms 0.8ms 0.9ms 1.0ms V(R5:2) -20V -10V 0V 10V

Natural Response - Parallel RLC Circuit 3rd Type :  < ω0 , UNDERDAMPED In this type,  < ω0, or L < 4R2C. The roots can be written as where and The response can be further written as

Natural Response - Parallel RLC Circuit Applying the same method as in series, the response of the parallel is Let B1=A1+A2 and B2=j(A1–A2), thus

Natural Response - Parallel RLC Circuit Differentiate the v(t), we have

Natural Response - Parallel RLC Circuit For example A parallel RLC circuit has R=80Ω, L=10mH and C=1F. If iL(0+)=0.5A and v(0+)=10V, find the current response.

Natural Response - Parallel RLC Circuit In summary, the responses for parallel RCL are 1st Type :  > ω0 , (OVERDAMPED) 2. 2nd Type :  = ω0 , (CRITICALLY DAMPED) 3rd Type :  < ω0 , (UNDERDAMPED)

Step Response - Series RLC Circuit Step response is obtained in Series RLC circuit with source. Applying KVL to the loop, ,where Substitute i to the derivative terms, i R L C + v - VS 2nd order diff. equ for capacitor voltage

Step Response - Series RLC Circuit To obtain the total response, consider the KVL of the loop Differentiate the equation, we get It can be seen that the characteristic equation is the same with source-free series RLC. Thus, by looking at the roots of the characteristic equation, one may find the type of the response (either overdamped, critically damped or underdamped).

Step Response - Series RLC Circuit Furthermore, it can be concluded that the total response can be represented in terms of ‘transient’ and ‘steady-state’ value, where vt(t) is the ‘transient response’ and vss(t) is the ‘steady-state’ response. The vt(t) is the part which will determine the type of the response, and it reflects to the response for source-free series RLC circuit.

Step Response - Series RLC Circuit The transient responses are: While vss(t) is the final voltage value (i.e. when t=) across the capacitor. Normally, it will equal to Vs. overdamped critically damped underdamped

Step Response - Series RLC Circuit Thus, the complete responses for series RLC with source are: 1st Type :  > ω0 , (OVERDAMPED) 2. 2nd Type :  = ω0 , (CRITICALLY DAMPED) 3rd Type :  < ω0 , (UNDERDAMPED)

Step Response - Series RLC Circuit Example A series RLC circuit is shown below. If i(0+)=1A and vC(0+)=18V, find i(t) and v(t). Step 1: Find  and ω0. Determine type of response. It is found that  > ω0, type of response  overdamped. i(t) 20Ω 1mH 100F + v(t) - 20V

Step Response - Series RLC Circuit Step 3: The steady state voltage across capacitor is Step2: Find the value of A1 and A2 from the initial values. the derivative of the response at initial is thus,

Step Response - Series RLC Circuit It is found that the complete response is Next, find the voltage response if R is changed to 6.324Ω and then to 2Ω

Step Response - Series RLC Circuit Comparison of the three responses with different R value. 2Ω 6.324Ω 20Ω

Step Response - Parallel RLC Circuit Step response is obtained in Parallel RLC circuit with source. Applying KCL to the loop, ,where Substitute i to the derivative terms, i R L + v - C IS 2nd order diff. equ for capacitor voltage

Step Response - Parallel RLC Circuit It has the same characteristic equation to that of natural response of parallel/series RLC. Furthermore, it can also be concluded that the total response can be represented in terms of ‘transient’ and ‘steady-state’ value, where it(t) is the ‘transient response’ and iss(t) is the ‘steady-state’ response. The it(t) is the part which will determine the type of the response, and it reflects to the response for source-free parallel RLC circuit.

Step Response - Parallel RLC Circuit The transient responses are: While iss(t) is the final voltage value (i.e. when t=) across the capacitor. Normally, it will equal to Is. overdamped critically damped underdamped

Step Response - Parallel RLC Circuit Thus, the complete responses for parallel RLC with source are: 1st Type :  > ω0 , (OVERDAMPED) 2. 2nd Type :  = ω0 , (CRITICALLY DAMPED) 3rd Type :  < ω0 , (UNDERDAMPED)

General Second-Order Circuit Previously, only the second-order series and parallel circuits are considered. If the circuits were neither in series nor in parallel, what method should be used? In this topic, we would concentrate to find the response for a second-order circuit which is neither a series nor parallel circuit. The response might be in terms of voltage or current. Thus, generally the response are characterized as x(t). In general 2nd-order circuit, the most important part is to find the characteristic equation and find the two roots (s1 and s2). From the two roots, one should know the type of the response.

General Second-Order Circuit The type of the response are the same with series and parallel RLC circuit. Finally, find the initial and steady state value (x(0), dx(0)/dt and x()) to determine the constant value; i.e. A1 and A2. Generally, the response is the summation of the ‘transient’ and ‘steady-state’ value and can be expressed as where x is either in voltage or current.

General 2nd-Order Circuit – step by step Turn off the independent source to find the second order differential equation using KCL and/or KVL. Find the roots and determine the type of the response (i.e. O-D, C-D or U-D) The 2nd order differential equation would determine whether the response is voltage or current! Find the steady-state and initial value to determine A1 and A2.

General Second-Order Circuit - eg Consider a circuit shown below. Find the complete response of the voltage v(t)? Step 1: Turn off the independent source. Find the second order differential equation. i 4Ω 1H + v(t) - 0.5F 12V 2Ω t >0s i 4Ω 1H + v(t) - 0.5F 2Ω v using KCL and KVL to find the 2nd order differential equations,

General Second-Order Circuit - eg KCL at node ‘v’ KVL at mesh ‘i’ We just concern on ‘v’. Thus, substitute ‘i’ to the right hand side equation Thus, the characteristic equation is,

General Second-Order Circuit - eg Step 2: The two roots are s1 = -2, and s2 = -3 *It is obvious that the two roots are negative and real. Thus, the transient response is the 1st type (OVERDAMPED) Step 3: The standard response is The steady state voltage is The transient voltage is

General Second-Order Circuit - eg Step 4: The initial voltage and current is, At just after the switch is closed, the circuit is shown as iC i iR 4Ω 1H + v(t) - 0.5F 12V 2Ω

General Second-Order Circuit - eg The overall response is From the overall response, at initial It is found that A1 = 12 and A2 = -4. Thus,