chapter 5 transient and steady state response(second-order circuit)
TRANSCRIPT
CHAPTER 5
Transient and Steady State
Response
(Second-Order Circuits)
Contents
Natural response of series RLC circuit
Natural response of parallel RLC circuit
Step response of series RLC circuit
Step response of parallel RLC circuit
What is second order?
• Circuits containing
two storage
elements.
• Second-order
circuit may have
two storage
elements of
different type or
the same type
Initial and final values
• Combination of R, L and C
• Find v(0), i(0), dv(0)/dt, di(0)/dt, i(∞) & v(∞)
• t(0-) the time just before switching event
• t(0+) the time just after switching event
• Assume the switching event take place at t=0
• Voltage polarity across capacitor
• Current direction across inductor
• Capacitor voltage always continuous v(0+) = v(0-)
• Inductor current always continuous i(0+)=i(0-)
Example
The switch in the figure shown has been closed for a long
time. It is open at t=0, Find:
(a) i(0+), v(0+)
(b) di(0+)/dt, dv(0+)/dt
(c) i(∞) , v(∞)
12 V
0.25 H4 Ω
0.1 F2 Ω
i
+V-
t=0
Exercise
The switch in the figure shown was open for a long time but
closed at t=0. Determine
(a) i(0+), v(0+)
(b) di(0+)/dt, dv(0+)/dt
(c) i(∞) , v(∞)
24 V
0.4 H
1/20 F2 Ω
i
+V-
t=0
10 Ω
The Source-Free Series RLC
• Applying KVL around the loop
𝑅𝑖 + 𝐿𝑑𝑖
𝑑𝑡+1
𝑐 −∞
𝑡
𝑖 𝑑𝑡 = 0
• Differentiate with respect to t
𝑑2𝑖
𝑑2+𝑅
𝐿
𝑑𝑖
𝑑𝑡+𝑖
𝐿𝐶= 0
• Finally,
𝑠2 +𝑅
𝐿𝑠 +
1
𝐿𝐶= 0
The Source-Free Series RLC
• Roots equation
𝑠1 = −𝑅
2𝐿+
𝑅
2𝐿
2
−1
𝐿𝐶
𝑠2 = −𝑅
2𝐿−
𝑅
2𝐿
2
−1
𝐿𝐶
or
𝑠1 = −𝛼 + 𝛼2 − 𝜔02
𝑠2 = −𝛼 − 𝛼2 − 𝜔02
where
𝛼 =𝑅
2𝐿, 𝜔0 =
1
𝐿𝐶
• 𝛼 (Np/s)
• 𝜔0 (rad/s)
The Source-Free Series RLC
Three type of solution
• If α > ω0 overdamped case
• If α = ω0 critically damped case
• If α < ω0 underdamped case
The Source-Free Series RLC
Overdamped case (α>ω0)
• Both roots S1 and S2 are negative and real
• The response is 𝑖 𝑡 = 𝐴1𝑒
𝑠1𝑡 + 𝐴2𝑒𝑠2𝑡
The Source-Free Series RLC
Critically damped case (α= ω0)
• Roots
𝑠1 = 𝑠2 = −𝛼 = −𝑅
2𝐿• The response is
𝑖 𝑡 = (𝐴2+𝐴1𝑡)𝑒−𝛼𝑡
The Source-Free Series RLC
Underdamped case(α<ω0)
• Roots
𝑠1 = −𝛼 + − 𝜔02 − 𝛼2 = −𝛼 +j𝜔𝑑
𝑠2 = −𝛼 − − 𝜔02 − 𝛼2 = −𝛼-j𝜔𝑑
where 𝜔𝑑 = 𝜔02 − 𝛼2
• The response is 𝑖 𝑡 = 𝑒−𝛼𝑡(𝐵1 cos𝜔𝑑𝑡 + 𝐵2 sin𝜔𝑑𝑡)
Example
Find i(t) for t > 0
+
v(t)
-
Exercise
Find i(t) in the circuit below. Assume that the
circuit has reached steady state at t=0-
Source Free Parallel RLC Circuits
• Initial inductor current and
initial voltage capacitor
𝑖 0 = 𝐼0 =1
𝐿 ∞
0
𝑣 𝑡 𝑑𝑡
𝑣 0 = 𝑉0• Applying KCL
𝑣
𝑅+1
𝐿 −∞
𝑡
𝑣𝑑𝑡 + 𝐶𝑑𝑣
𝑑𝑡= 0
Source Free Parallel RLC Circuits
• Derivatives with respect t and diving by C
𝑑2𝑣
𝑑𝑡2+1
𝑅𝐶
𝑑𝑣
𝑑𝑡+1
𝐿𝐶𝑣 = 0
or 𝑠2 +1
𝑅𝐶𝑠 +
1
𝐿𝐶
• Roots of the characteristics equation are
𝑠1,2 = −1
2𝑅𝐶±
1
2𝑅𝐶
2
−1
𝐿𝐶
Source Free Parallel RLC Circuits
or 𝑠1,2 = −𝛼 ± 𝛼2 −𝜔02
where 𝛼 =1
2𝑅𝐶, 𝜔0 =
1
𝐿𝐶
• 𝛼 (Np/s)
• 𝜔0 (rad/s)
The Source-Free Parallel RLC
Three type of solution
• If α > ω0 overdamped case
• If α = ω0 critically damped case
• If α < ω0 underdamped case
The Source-Free Parallel RLC
Overdamped case (α>ω0)
• Both roots S1 and S2 are negative and real
• The response is
𝑣 𝑡 = 𝐴1𝑒𝑠1𝑡 + 𝐴2𝑒
𝑠2𝑡
The Source-Free Parallel RLC
Critically damped case (α= ω0)
• The roots are real and equal so the response is
𝑣 𝑡 = (𝐴1+𝐴2𝑡)𝑒−𝛼𝑡
The Source-Free Parallel RLC
Underdamped case(α<ω0)
• Roots
𝑠1,2 = −𝛼 ± j𝜔𝑑
where 𝜔𝑑 = 𝜔02 − 𝛼2
• The response is
𝑣 𝑡 = 𝑒−𝛼𝑡(𝐴1 cos𝜔𝑑𝑡 + 𝐴2 sin𝜔𝑑𝑡)
Example
Find v(t) for t>0 in the RLC circuit shown
below
Step Response of a Series RLC Circuit
• Applying KVL around the
loop for t>0
𝐿𝑑𝑖
𝑑𝑡+ 𝑅𝑖 + 𝑣 = 𝑉𝑠
but 𝑖 = 𝐶𝑑𝑣
𝑑𝑡
substitute i in equation above
𝑑2𝑣
𝑑𝑡2+𝑅
𝐿
𝑑𝑣
𝑑𝑡+𝑣
𝐿𝐶=𝑉𝑠𝐿𝐶
Step Response of a Series RLC Circuit
• There is two components in the equation (i) transient
response 𝑣𝑡 𝑡 (ii) steady-state response 𝑣𝑠𝑠 𝑡
𝑣 𝑡 = 𝑣𝑡 𝑡 + 𝑣𝑠𝑠 𝑡
• The transient response 𝑣𝑡 𝑡 is similar as discussed in
source-free circuit.
• The final value of the capacitor voltage is the same as
the source voltage Vs
𝑣𝑠𝑠 𝑡 = 𝑣 ∞ = 𝑉𝑠
Step Response of a Series RLC Circuit
• The complete response solution are:-
𝑣 𝑡 = 𝑉𝑠 + 𝐴1𝑒𝑠1𝑡 + 𝐴2𝑒
𝑠2𝑡 (Overdamped)
𝑣 𝑡 = 𝑉𝑠 + (𝐴1+𝐴2𝑡)𝑒−𝛼𝑡 (Critically damped)
𝑣 𝑡 = 𝑉𝑠 + (𝐴1𝑐𝑜𝑠𝜔𝑑𝑡 + 𝐴2𝑠𝑖𝑛𝜔𝑑𝑡)𝑒−𝛼𝑡 (Underdamped)
Example
For the circuit shown in figure below, find
v(t) and i(t) for t>0.
Given R = 5 Ω, C = 0.25 F
Step Response of a Parallel RLC Circuit
• Applying KCL at the top
node for t > 0,𝑣
𝑅+ 𝑖 + 𝐶
𝑑𝑣
𝑑𝑡= 𝐼𝑠
but 𝑣 = 𝐿𝑑𝑖
𝑑𝑡
substitute vin equation above
and dividing by LC:
𝑑2𝑖
𝑑𝑡2+1
𝑅𝐶
𝑑𝑖
𝑑𝑡+𝑖
𝐿𝐶=𝐼𝑠𝐿𝐶
Step Response of a Parallel RLC Circuit
• There is two components in the equation (i) transient
response 𝑖𝑡 𝑡 (ii) steady-state response 𝑖𝑠𝑠 𝑡
𝑖 𝑡 = 𝑖𝑡 𝑡 + 𝑖𝑠𝑠 𝑡
• The transient response 𝑖𝑡 𝑡 is similar as discussed in
source-free circuit.
• The final value of the current through the inductor is the
same as the source current Is
Step Response of a Parallel RLC Circuit
• The complete response solution are:-
𝑖 𝑡 = 𝐼𝑠 + 𝐴1𝑒𝑠1𝑡 + 𝐴2𝑒
𝑠2𝑡 (Overdamped)
𝑖 𝑡 = 𝐼𝑠 + (𝐴1+𝐴2𝑡)𝑒−𝛼𝑡 (Critically damped)
𝑖 𝑡 = 𝐼𝑠 + (𝐴1𝑐𝑜𝑠𝜔𝑑𝑡 + 𝐴2𝑠𝑖𝑛𝜔𝑑𝑡)𝑒−𝛼𝑡 (Underdamped)
Example
Find i(t) and v(t) for t > 0
END