"xn(t)" = Aest S1 = -a₁ + √²-4a₂a0 2a2 xn(t) = A₁eS₁t + A₂eszt (a₂s² + a₁s + ao) = 0 S2 = -a₁-√²-4a₂a⁰ 2a2 NATURAL RESPONSE, Xn Alesit +A₂e21 (A₁+A₂t)e (A₁ cos wat+A₂ sin wat)e -αl

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The complete response is the sum of the natural response and the forced response
X = Xn+Xf
Natural response of a first-order circuit
Natural response of a second-order circuit
CASE
Overdamped
Critically damped
Underdamped
az
S1 =
FORCING FUNCTION
K
Kt
K₁²
K sin cot
Ke at
Forced response of a first-order, or a second-order circuit
+ A₁
NATURAL FREQUENCIES
$1,52 = -α± √√² - 0²
$1,$₂=-α
S1, S2 = -x±j√√/0²-a² -α±jood
dx
dt
d²x
dt²
Solution of the Second-Order Differential Equation
d²x
x(t) = xn(t) + xf (t)
a2
dt²
+ a。x = f(t)
+ A1
dx
dt
"xn(t)" = Aest
Xn(t) = Ke-t/t
+ a。x = 0
-a₁ + √²-4a₂a0
242
-
Xn(t) = A₁e³₁ª + A₂e³₂t
ASSUMED RESPONSE
A
At + B
At² +Bt+C
A sin cot + B cos cot
Ae-at
NATURAL RESPONSE, Xn
Ale+Azer
(A₁+A₂t)e-at
(A, cos coat+A₂ sin coat)e
xn(t) = ?
(a₂s² + a₁s+ao) = 0
-α₁ - √²-4a₂a0
242
NATURAL RESPONSE, Xn
A₁ est + A₂e21
-α1
(A₁+A₂t)e
(A₁ cos wat+A₂ sin wat)ext
Transcribed Image Text:The complete response is the sum of the natural response and the forced response X = Xn+Xf Natural response of a first-order circuit Natural response of a second-order circuit CASE Overdamped Critically damped Underdamped az S1 = FORCING FUNCTION K Kt K₁² K sin cot Ke at Forced response of a first-order, or a second-order circuit + A₁ NATURAL FREQUENCIES $1,52 = -α± √√² - 0² $1,$₂=-α S1, S2 = -x±j√√/0²-a² -α±jood dx dt d²x dt² Solution of the Second-Order Differential Equation d²x x(t) = xn(t) + xf (t) a2 dt² + a。x = f(t) + A1 dx dt "xn(t)" = Aest Xn(t) = Ke-t/t + a。x = 0 -a₁ + √²-4a₂a0 242 - Xn(t) = A₁e³₁ª + A₂e³₂t ASSUMED RESPONSE A At + B At² +Bt+C A sin cot + B cos cot Ae-at NATURAL RESPONSE, Xn Ale+Azer (A₁+A₂t)e-at (A, cos coat+A₂ sin coat)e xn(t) = ? (a₂s² + a₁s+ao) = 0 -α₁ - √²-4a₂a0 242 NATURAL RESPONSE, Xn A₁ est + A₂e21 -α1 (A₁+A₂t)e (A₁ cos wat+A₂ sin wat)ext
• Problem 5. The Circuit shown is at the steady state, before the switch is closed at time t=0
Assume steady-state conditions exists.
(For your knowledge, this might be an example, when a device is plugged into the power outlet
at the wall in the house.)
vs (t) = 220 sin(100 t) (v)
π
Vin m
vs(t)
R₁
V10M
Un
(a) Derive the differential equation of i(t) over t for t≥ 0.
(b) Find the forced response of i(t) over t for t≥ 0.
(a) Find out the complete response of i(t) over t for t≥ 0.
▪ Brief catalog of the Operational Amplifier circuits
(Based on the ideal operational amplifier)
(a) Inverting amplifier
R₁
U₂0-W
R₂
R₁
www
R₁
Vout=-
R₁ R₁
Xo
t = 0
R₁
R₁
Ovout R₁
U₁+ ひnt
Vin
- Vout
R₁
Run)
(d) Summing amplifier
1H
ww
Vina
i(t)
2H
R₁.
R₁.
U10
out
(b) Noninverting amplifier
U20
R₂K1
R₂/K₂
R.K3
+
U30-W
R₂/(1-(K₁ + K₂ + K3))
5Ω
ww.
Vin
Ro(K4-1)
0.5F
Rp
Vin O
(c) Voltage follower (buffer amplifier)
Ovout Vin
Ovout K4(K₁v1 + K₂V₂ + K3V3)
(e) Noninverting summing amplifier
Transcribed Image Text:• Problem 5. The Circuit shown is at the steady state, before the switch is closed at time t=0 Assume steady-state conditions exists. (For your knowledge, this might be an example, when a device is plugged into the power outlet at the wall in the house.) vs (t) = 220 sin(100 t) (v) π Vin m vs(t) R₁ V10M Un (a) Derive the differential equation of i(t) over t for t≥ 0. (b) Find the forced response of i(t) over t for t≥ 0. (a) Find out the complete response of i(t) over t for t≥ 0. ▪ Brief catalog of the Operational Amplifier circuits (Based on the ideal operational amplifier) (a) Inverting amplifier R₁ U₂0-W R₂ R₁ www R₁ Vout=- R₁ R₁ Xo t = 0 R₁ R₁ Ovout R₁ U₁+ ひnt Vin - Vout R₁ Run) (d) Summing amplifier 1H ww Vina i(t) 2H R₁. R₁. U10 out (b) Noninverting amplifier U20 R₂K1 R₂/K₂ R.K3 + U30-W R₂/(1-(K₁ + K₂ + K3)) 5Ω ww. Vin Ro(K4-1) 0.5F Rp Vin O (c) Voltage follower (buffer amplifier) Ovout Vin Ovout K4(K₁v1 + K₂V₂ + K3V3) (e) Noninverting summing amplifier
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