Transcript of Grossman/ Melkonian 0 Chapter 4 and 5 Time Varying Circuits Capacitors Inductors Time Constant T C...
- Slide 1
- Grossman/ Melkonian 0 Chapter 4 and 5 Time Varying Circuits
Capacitors Inductors Time Constant T C Waveforms RC Circuits Step
Input RL Circuits Step Input
- Slide 2
- Grossman/ Melkonian 1 CAPACITORS: Where: C is the capacitance
in Farads. The farad is a large unit. Typical values of capacitance
are in F or pF. is the permittivity of the dielectric medium
(8.854x10 -12 F/m for air). A is the cross-sectional area of the
two parallel conducting plates. d is the distance between the two
plates. C = A d Dielectric Insulator = permittivity A d Metal
plates of area A Section 4.1
- Slide 3
- Grossman/ Melkonian 2 + When a voltage is applied to the
plates, an electric field is produced that causes an electric
charge q(t) to be produced on each plate. + - - Dielectric
Insulator = permittivity Metal plates of area A Standard Notations
(passive sign convention) + - + - vCvC vCvC VsVs i C (t)
CAPACITORS:
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- Grossman/ Melkonian 3 since: i(t) = dq(t)/dt and: q(t) = Cv C
(t) Differentiating: dq(t) = C dv C (t) i C (t) = dq(t)/dt = C(dv C
(t)/dt ) q (t) = Cv C (t) i C (t) = C[dv C (t)/dt] CAPACITORS:
Charge of a capacitor i-v relationship for a capacitor
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- Grossman/ Melkonian 4 Integrating the i-v relationship of the
capacitor: CAPACITORS: dv C (t) = 1/C i C (t)dt = v C (t) t t
Letting V C (0) = V 0 represent the initial voltage across the
capacitor at some time t = t 0 : v C (t) = V 0 + 1/C i C (t)dt t 0
t t0t0 i C (t) = C[dv C (t)/dt] Voltage across a capacitor Current
through a capacitor
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- Grossman/ Melkonian 5 CAPACITORS: Capacitor Power: p C (t) = i
C (t) v C (t) = C[dv C (t)/dt]v C (t) p C (t) = d/dt[Cv C 2 (t)]
Power associated with a capacitor Capacitor Energy: Energy is the
integral of power, therefore, w C (t) = Cv C 2 (t) Energy
associated with a capacitor
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- Grossman/ Melkonian 6 CAPACITORS: Example 1: Given i C (t) = I
o [e -t/Tc ] for t 0 and v C (0) = 0V, find the capacitors power
and energy. v c (t) = [(I o T c /C)(1-e -t/Tc )]V Power: p c (t) =
i C (t) v C (t) v C (t) = V 0 + 1/C i C (x)dx = 0V + 1/C I o [e
-x/Tc ]dx 0 t 0 t p C (t) = [I o e -t/Tc ][(I o T C /C)(1-e -t/Tc
)]W p c (t) = I o 2 T c /C (e -t/Tc - e -2t/Tc ) i C (t) v C (t)
Power can be positive or negative
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- Grossman/ Melkonian 7 CAPACITORS: Example 1 cont.: Energy: w C
(t) = Cv C 2 (t) w C (t) = [(I o T C ) 2 /2C](1-e -t/Tc ) 2 t IoIo
i C (t) 0 Waveforms: i C (t) = I o [e -t/Tc ] Energy is always
positive
- Slide 9
- Grossman/ Melkonian 8 t v C (t) 0 IoTc/CIoTc/C CAPACITORS:
Example 1 cont.: v c (t) = [(I o T c /C)(1-e -t/Tc )]V t p C (t) I
o 2 T c /4C 0 T c ln2 p c (t) = I o 2 T c /C (e -t/Tc - e -2t/Tc
)
- Slide 10
- Grossman/ Melkonian 9 CAPACITORS: Example 1 cont.: t w c (t) I
o 2 T c 2 /2C 0 w C (t) = [(I o T C ) 2 /2C](1-e -t/Tc ) 2
- Slide 11
- Grossman/ Melkonian 10 CAPACITORS: Series and Parallel
Capacitors: C1C1 Series: Capacitors connected in series combine
like resistors connected in parallel. C3C3 C7C7 C6C6 C5C5 C4C4 C2C2
v c (t) + - i(t) 1/C EQ = 1/C 1 + 1/C 2 + 1/C 3 + 1/C 4 +.+ 1/C
N
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- Grossman/ Melkonian 11 Series and Parallel Capacitors:
CAPACITORS: Parallel: Capacitors connected in parallel add like
resistors connected in series.... v c (t) C1C1 C2C2 C3C3 C4C4 C5C5
CNCN C EQ = C 1 + C 2 + C 3 + C 4 + + C N i C1 (t) i C2 (t) i C3
(t)
- Slide 13
- Grossman/ Melkonian 12 CAPACITORS: The voltage across a
capacitor cannot change instantaneously. Current through a
capacitor can change instantaneously. The current through a
capacitor is zero when the voltage across the capacitor is
constant. For example, when the capacitor is fully charged.
Therefore, a capacitor acts like an open circuit when a DC voltage
is applied. Capacitors in parallel add. Capacitors in series
combine like resistors connected in parallel. Properties of the
Capacitor: The capacitor stores energy in its electric field.
- Slide 14
- Grossman/ Melkonian 13 INDUCTORS: + - i L (t) v L (t) Inductors
are typically made by winding a coil of wire around an insulator or
ferromagnetic material. Current flowing through an inductor creates
a magnetic field. i-v relationship for an inductor v L (t) = L[di L
(t)/dt] Magnetic Field Lines Where L is the inductance of the coil
measured in henrys (H). 1 H = 1 V-s/A
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- Grossman/ Melkonian 14 INDUCTORS: i L (t) = i L (0) + 1/L v L
(t) dt W L (t) = 1/2Li L 2 (t) p L (t) = i L (t) v L (t) = d/dt[W L
(t)] The basic results for the inductor can be derived in the same
way as we have done for the capacitor: t Current through an
inductor Power associated with an inductor Energy associated with
an inductor
- Slide 16
- Grossman/ Melkonian 15 INDUCTORS: Properties of the Inductor:
Inductor current is continuous. Cannot change instantaneously.
(Would require infinite power). The voltage across an inductor can
change instantaneously. When the current through an inductor is
constant, the voltage is zero. Therefore, the inductor acts like a
short circuit when a DC source is applied. Inductors connected in
series add like resistors connected in series. Inductors connected
in parallel combine like resistors connected in parallel
- Slide 17
- Grossman/ Melkonian 16 Section 4.2 TIME-DEPENDENT SIGNALS:
Sources that produce currents or voltages that vary with time are
called time-dependent signal sources. The sinusoidal waveform is
one of the most important time- dependent signals. + - + - V(t)
i(t) V(t), i(t) ~ Generalized time-dependent signals Sinusoidal
source
- Slide 18
- Grossman/ Melkonian 17 TIME-DEPENDENT SIGNALS: An important
class of time-dependent signals is a periodic signal. A periodic
signal satisfies the following equation: x(t) = x(t + nT) n = 1, 2,
3, where T is the period of the signal f = natural frequency = 1/T
cycles/s or Hz A generalized sinusiod is defined as follows: x(t) =
Acos( t + ) where = radian frequency = 2 f rad/s = 2 t/T = 360 t/T
rad
- Slide 19
- Grossman/ Melkonian 18 TIME-DEPENDENT SIGNALS: v(t) t t t t
Damped sinusoidSquare wave Pulse trainSawtooth wave Pulse
width
- Slide 20
- Grossman/ Melkonian 19 TIME-DEPENDENT SIGNALS: T A t x(t) =
Acos( t + ) Radial frequency Time Amplitude T A t Phase shift tt
Reference cosine Arbitrary sinusoid Sinusoidal Waveform
- Slide 21
- Grossman/ Melkonian 20 TIME-DEPENDENT SIGNALS: Root Mean Square
The operation of computing the Root-Mean-Square of a waveform is a
method for quantifying the strength of a time-varying signal. x rms
= 1/T x 2 (t) dt 0 T Root-mean-square The average value of a
sinusoidal signal is zero, independent of its amplitude and
frequency. Therefore, another method must be used to quantify the
strength of a time-varying signal.
- Slide 22
- Grossman/ Melkonian 21 TIME-DEPENDENT SIGNALS: Example 2:
Calculate the rms value of the sinusoidal voltage v(t) = Vsin( t).
v rms = 1/T v 2 (t) dt 0 T T = 1/f = 2 / v rms = /2 V 2 sin 2 ( t)
dt 0 v rms = V 2 /2 1/2 - 1/2cos(2 t) dt 0 2/2/ 2/2/
- Slide 23
- Grossman/ Melkonian 22 TIME-DEPENDENT SIGNALS: v rms = V 2 /2 [
t/2 - 1/4 sin(2 t) ] 0 2/2/ Example 2 cont.: v rms = V 2 /2 [( 2 /2
- /4sin(4 / ) ( 0 0 )] v rms = V 2 /2 ( / - /4sin(4 ) 0 ) 0 v rms =
2 2 V v rms = 0.707V
- Slide 24
- Grossman/ Melkonian 23 Sections 4.3 (pgs 161 & 162), 5.1,
5.2, 5.3 RC CIRCUIT, STEP RESPONSE: VTVT R t = 0 + - + - C v C (t)
Switch i C (t) Assume we have the following series circuit
containing a voltage source, a resistor, and a capacitor. How do we
write an expression for the voltage across the capacitor?
- Slide 25
- Grossman/ Melkonian 24 VSVS R t = 0 + - + - C v C (t) Switch i
C (t) RC CIRCUIT, STEP RESPONSE: Writing KVL for the circuit: RC(dv
C (t)/dt) + v C (t) = V S V S + Ri c (t) + v C (t) = 0 i c (t) =
C(dv c (t)/dt ) RC(dv C (t)/dt) + v C (t) = V S for t 0 First order
linear differential equation.
- Slide 26
- Grossman/ Melkonian 25 Since the circuit is linear, we can
separate the solution v C (t) into two components: v C (t) = v N
(t) + v F (t) RC CIRCUIT, STEP RESPONSE: 1. RC(dv N (t)/dt) + v N
(t) = 0 t 0 Natural Response: V S = 0 Natural Response: The
classical approach to solving this type of an equation is to try a
solution of the following form: v N (t) = Ke st where K and s are
constants
- Slide 27
- Grossman/ Melkonian 26 Substituting v N (t) = Ke st into the
above equation: RCsKe st + Ke st = 0 Ke st (RCs + 1) = 0 K = 0 is
trivial solution RCs + 1 = 0 Characteristic Equation RC CIRCUIT,
STEP RESPONSE: Solving the Characteristic Equation: s = -1/RC
Therefore, the Natural Response has the form: v N (t) = Ke -t/RC t
> 0
- Slide 28
- Grossman/ Melkonian 27 RC CIRCUIT, STEP RESPONSE: 2. RC(dv F
(t)/dt) + v F (t) = V S t 0 Forced Response Forced Response:
Equation 2 requires the linear combination of v F (t) and its
derivative equal a constant V S. Setting v F (t) = V S satisfies
this condition. Combining the Natural and Forced Responses: v C (t)
= v N (t) + v F (t) = Ke -t/RC + V S t > 0 (General Solution)
Evaluating K using initial conditions (at t=0, the voltage across
the capacitor): v(0) = V 0 = Ke 0 + V S = K + V S This requires K =
(V 0 V S )
- Slide 29
- Grossman/ Melkonian 28 v C (t) = (V 0 V S )e -t/RC + V S t >
0 RC CIRCUIT, STEP RESPONSE: Substituting into the general
solution: Decaying exponential Constant RC: Defined as the time
constant, sometimes written as T C or . The time constant depends
only on fixed circuit parameters. V 0 : The initial condition or
initial voltage across the capacitor. Sometimes written as V i or v
C (0) Important Values: V S : The Thevenin voltage seen by the
capacitor as t . Sometimes written as V f or v C ( ).
- Slide 30
- Grossman/ Melkonian 29 Important Values cont.: RC CIRCUIT, STEP
RESPONSE: R: The Thevenin resistance or equivalent resistance seen
by the capacitor. Sometimes written as R T. t(0 - ): The instant of
time just prior to t=0 or the switch changing its position. t(0):
The time t=0 or the instant the switch changes its position. t(0 +
): The instant of time just after t=0 or just after the switch
changes its position.
- Slide 31
- Grossman/ Melkonian 30 Important Concepts: The voltage across a
capacitor cannot change instantaneously. Therefore, v C (0 - ) = v
C (0) = v C (0 + ). A capacitor acts like a short circuit the
instant a voltage is applied across its terminals. When a DC source
is applied to a capacitor, it becomes an open circuit as t . RC
CIRCUIT, STEP RESPONSE: v C (t) = (V i V f )e -t/T C + V f t > 0
The current through a capacitor can change instantaneously.
- Slide 32
- Grossman/ Melkonian 31 RL CIRCUIT, STEP RESPONSE: Assume we
have the following series circuit containing a voltage source, a
resistor, and an inductor. We can write an expression for the
current through the inductor using a similar procedure as that
applied previously for finding the voltage across a capacitor. VTVT
R t = 0 + - + - L v L (t) Switch i L (t)
- Slide 33
- Grossman/ Melkonian 32 RL CIRCUIT, STEP RESPONSE: i L (t) = (I
0 I S )e -Rt/L + I S t > 0 Decaying exponential Constant
Applying a similar procedure as that applied to finding the voltage
across a capacitor, we obtain the equation for the current through
an inductor when the input is a step function. L/R: Defined as the
time constant, sometimes written as T C or . The time constant
depends only on fixed circuit parameters. I 0 : The initial
condition or initial current through the inductor. Sometimes
written as I i or i L (0) Important Values: I S : The Norton
current seen by the inductor as t . Sometimes written as I f or i L
( ).
- Slide 34
- Grossman/ Melkonian 33 Important Values cont.: R: The Norton
resistance or equivalent resistance seen by the inductor. Sometimes
written as R N. t(0 - ): The instant of time just prior to t=0 or
the switch changing its position. t(0): The time t=0 or the instant
the switch changes its position. t(0 + ): The instant of time just
after t=0 or just after the switch changes its position. RL
CIRCUIT, STEP RESPONSE:
- Slide 35
- Grossman/ Melkonian 34 Important Concepts: The current through
an inductor cannot change instantaneously. Therefore, i L (0 - ) =
i L (0) = i L (0 + ). A inductor acts like an open circuit the
instant a voltage or current is applied to its terminals. When a DC
source is applied to a inductor, it becomes a short circuit as t .
RL CIRCUIT, STEP RESPONSE: i L (t) = (I i I f )e -t/T C + I f t
> 0 The voltage across an inductor can change
instantaneously.
- Slide 36
- Grossman/ Melkonian 35 RC CIRCUIT, STEP RESPONSE: Example 3:
16V 1.5K t = 0 + - + - 1.5 F v C (t) Switch i C (t) Calculate the
RC time constant (T C ), V i, V f, the voltage across the
capacitor, and the current through the capacitor for t > 0. The
switch has been open for a long time before closing. 1.5K
- Slide 37
- Grossman/ Melkonian 36 Example 3 cont.: RC CIRCUIT, STEP
RESPONSE: 16V 1.5K t = 0 + - + - 1.5 F v C (t) Switch i C (t) 1.5K
RC time constant (T C ): T C = R TH C EQ = 1.5K 1.5K 1.5 F T C =
1.125ms
- Slide 38
- Grossman/ Melkonian 37 Example 3 cont.: RC CIRCUIT, STEP
RESPONSE: 16V 1.5K t = 0 + - + - 1.5 F v C (t) Switch i C (t) 1.5K
V i = 0V The capacitor has had time to discharge. V f = 16V 1.5K
1.5K + 1.5K V f = 8V Voltage divider
- Slide 39
- Grossman/ Melkonian 38 16V 1.5K t = 0 + - + - 1.5 F v C (t)
Switch i C (t) 1.5K Example 3 cont.: RC CIRCUIT, STEP RESPONSE: v C
(t) t > 0 v C (t) = (V i V f )e -t/T C + V f t > 0 v C (t) =
(0V - 8V)e -t/T C + 8V t > 0 v C (t) = 8 - 8e -t/1.125ms V t
> 0
- Slide 40
- Grossman/ Melkonian 39 16V 1.5K t = 0 + - + - 1.5 F v C (t)
Switch i C (t) 1.5K Example 3 cont.: RC CIRCUIT, STEP RESPONSE: i C
(t) t > 0 i C (t) = C[dv C (t)/dt] i C (t) = 1.5 F d/dt[8 - 8e
-t/1.125ms ]A t > 0 i C (t) = 10.67e -t/1.125ms mA t > 0
- Slide 41
- Grossman/ Melkonian 40 Example 4: RL CIRCUIT, STEP RESPONSE:
14V 5K t = 0 + - + - v L (t) Switch i L (t) 3k 5mH Calculate the
L/R time constant (T C ), I i, I f, the current through the
inductor, and the voltage across the inductor for t > 0. The
switch has been closed for a long time. 1K
- Slide 42
- Grossman/ Melkonian 41 Example 4 cont.: RL CIRCUIT, STEP
RESPONSE: 14V 5K t = 0 + - + - v L (t) Switch i L (t) 3k 5mH 1K L/R
time constant (T C ): T C = L EQ /R N = 5mH 3k + 1K T C = 1.25
s
- Slide 43
- Grossman/ Melkonian 42 Example 4 cont.: RL CIRCUIT, STEP
RESPONSE: 14V 5K t = 0 + - + - v L (t) Switch i L (t) 3k 5mH 1K I i
= 1/5K + 1/3k + 1/1K 2.8mA I i = 1.83mA I f = 0A 1/1K Source
transformation and current divider
- Slide 44
- Grossman/ Melkonian 43 Example 4 cont.: RL CIRCUIT, STEP
RESPONSE: 14V 5K t = 0 + - + - v L (t) Switch i L (t) 3k 5mH 1K i L
(t) = (I i I f )e -t/T C + I f t > 0 i L (t): i L (t) = (1.83 -
0)e -t/1.25 s + 0 mA t > 0 i L (t) = 1.83e -t/1.25 s mA t >
0
- Slide 45
- Grossman/ Melkonian 44 Example 4 cont.: RL CIRCUIT, STEP
RESPONSE: 14V 5K t = 0 + - + - v L (t) Switch i L (t) 3k 5mH 1K v L
(t): v L (t) = L[di L (t)/dt] v L (t) = 5mH d/dt[1.83x10 -3 e
-t/1.25 s ] V v L (t) = - 7.32e -t/1.25 s V t > 0
- Slide 46
- Grossman/ Melkonian 45 Example 5 : RC CIRCUIT, STEP RESPONSE:
10V 800 t = 0 + - + - 2F2F v C (t) i C (t) 400 + - A B 6V i R (t)
The switch has been in position A for a long time. At t=0, the
switch moves to position B. Calculate V i, V f, T C, and v C (t), i
C (t), and i R (t) for t > 0.
- Slide 47
- Grossman/ Melkonian 46 Example 5 cont.: RC CIRCUIT, STEP
RESPONSE: 10V 800 t = 0 + - + - 2F2F v C (t) i C (t) 400 + - A B 6V
i R (t) V f = 6V 400 800 + 400 Voltage divider V i = 10V 400 800 +
400 Voltage divider V i = 3.34V V f = 2V
- Slide 48
- Grossman/ Melkonian 47 Example 5 cont.: RC CIRCUIT, STEP
RESPONSE: T C : T C = R TH C EQ = 800 400 2 F T C = 533.4 s 10V 800
t = 0 + - + - 2F2F v C (t) i C (t) 400 + - A B 6V i R (t)
- Slide 49
- Grossman/ Melkonian 48 Example 5 cont.: RC CIRCUIT, STEP
RESPONSE: v C (t) t > 0 v C (t) = (V i V f )e -t/T C + V f t
> 0 v C (t) = (3.34V - 2V)e -1875t + 2V t > 0 v C (t) = 2 +
1.34e -1875t V t > 0 10V 800 t = 0 + - + - 2F2F v C (t) i C (t)
400 + - A B 6V i R (t)
- Slide 50
- Grossman/ Melkonian 49 i C (t) t > 0 i C (t) = C[dv C
(t)/dt] i C (t) = 2 F d/dt[2 + 1.34e -1875t ]A t > 0 i C (t) =
-5.01e -1875t mA t > 0 Example 5 cont.: RC CIRCUIT, STEP
RESPONSE: 10V 800 t = 0 + - + - 2F2F v C (t) i C (t) 400 + - A B 6V
i R (t)
- Slide 51
- Grossman/ Melkonian 50 i R (t) t > 0 i x (t) = v C (t)/400 =
i R (t) = i x (t) + i C (t) = 5 + 3.35e -1875t - 5.01e -1875t mA t
> 0 i R (t) = 5 1.66e -1875t mA t > 0 Example 5 cont.: RC
CIRCUIT, STEP RESPONSE: 10V 800 t = 0 + - + - 2F2F v C (t) i C (t)
400 + - A B 6V i R (t) i x (t) 2 + 1.34e -1875t 400 = 5 + 3.35e
-1875t mA t > 0
- Slide 52
- Grossman/ Melkonian 51 Section 4.3 Forced Response of Circuits
Excited by Sinusoidal Sources: v S (t) R + - + - C v C (t) i C (t)
Consider the following circuit. If v S (t) = Vcos( t) we obtain the
following differential equation: ~ RC(dv C (t)/dt) + v C (t) =
Vcos( t) t > 0 This equation is similar to the first order
differential equation found previously for a step input. RC
CIRCUIT, SINUSOIDAL INPUT:
- Slide 53
- Grossman/ Melkonian 52 v S (t) R + - + - C v C (t) i C (t) ~ RC
CIRCUIT, SINUSOIDAL INPUT: As with the step response, we find the
natural and forced response: v N (t) = Ke -t/RC t > 0 Natural
Response
- Slide 54
- Grossman/ Melkonian 53 RC CIRCUIT, SINUSOIDAL INPUT: Forced
Response The forced response depends on both the circuit and the
nature of the forcing function. It is the particular solution to
the following equation: RC(dv F (t)/dt) + v F (t) = Vcos( t) t >
0 This equation requires v F (t) plus RC times its first derivative
add to produce a cosine function. Try a solution of a general
sinusoid. v F (t) = Acos( t) + Bsin( t) Substituting this into the
differential equation we obtain: RCd/dt[Acos( t) + Bsin( t)] +
Acos( t) + Bsin( t) = Vcos( t) t > 0 S
- Slide 55
- Grossman/ Melkonian 54 RC CIRCUIT, SINUSOIDAL INPUT: RC[-A sin(
t) + Bcos( t)] + Acos( t) + Bsin( t) = Vcos( t) t > 0 Performing
the differentiation: Rearranging the equation: [RC B + A V]cos t +
[-RC A + B]sin t = 0 The left side of the equation is zero for all
t > 0 when the coefficients of the cosine and sine terms are
zero. A + (RC )B = V and -(RC )A + B = 0 Solving the two equations
yields: A = V 1 + ( RC) 2 andB = RCV 1 + ( RC) 2
- Slide 56
- Grossman/ Melkonian 55 RC CIRCUIT, SINUSOIDAL INPUT: Combining
the forced and natural responses: v c (t) = Ke -t/RC + V 1 + ( RC)
2 (cos t + RCsin t) t > 0 The initial condition requires: v(0) =
V 0 = K + V 1 + ( RC) 2 K = V 0 - V 1 + ( RC) 2 Substituting this
value for K into the equation for the voltage across the capacitor:
v c (t) = V 0 - V 1 + ( RC) 2 (cos t + RCsin t) t > 0 e -t/RC +
V 1 + ( RC) 2 Natural Response Forced Response
- Slide 57
- Grossman/ Melkonian 56 v S (t) = 5cos(2000t)V 500 t = 0 + - + -
2F2F v C (t) Switch i C (t) Example 6: RC CIRCUIT, SINUSOIDAL
INPUT: Calculate the voltage across the capacitor for t > 0. V 0
= 0V. ~ v c (t) = V 0 - V 1 + ( RC) 2 (cos t + RCsin t) t > 0 e
-t/RC + V 1 + ( RC) 2
- Slide 58
- Grossman/ Melkonian 57 Example 6 cont.: RC CIRCUIT, SINUSOIDAL
INPUT: v S (t) = 5cos(2000t)V 500 t = 0 + - + - 2F2F v C (t) Switch
i C (t) ~ V 0 = 0V, V = 5V, = 2000, R = 500 , C = 2 F, T C = 1ms v
c (t) = 0 - 1 + (2) 2 (cos2000t + 2sin2000t) t > 0 e -1000t + 5
1 + (2) 2 5 v c (t) = -e -1000t + (cos2000t + 2sin2000t)V t >
0
- Slide 59
- Grossman/ Melkonian 58 PHASORS: Section 4.4 A phasor is the
representation of a sinusoidal signal in the frequency domain. This
phasor representation eliminates the need for solving differential
equations. Eulers Identity is the basis of phasor notation. e j =
cos( ) + jsin( ) Eulers Identity is a trigonometric relationship in
the complex plain. e j = cos + jsin Re 1 Im j -j 1 cos sin
- Slide 60
- Grossman/ Melkonian 59 Ae j = Acos + jAsin = A PHASORS: The
relationship between rectangular and polar forms is as follows:
Acos( t + ) V(j ) = Ae j A Mathematical Representations:
Time-domain Frequency-domain Polar form Rectangular form
Multiplication and Division is performed in polar form. Addition
and Subtraction is performed in rectangular form.
- Slide 61
- Grossman/ Melkonian 60 Example 7: PHASORS: Construct the
phasors for the following signals: v 1 (t) = 10cos(1000t - 45 ) v 2
(t) = 5cos(1000t + 30 ) v 1 (t) = 10e -j45 = 10cos(- 45 ) +
j10sin(-45 ) Solution: Polar form 10 -45 V Rectangular form 7.07 -
j7.07 V
- Slide 62
- Grossman/ Melkonian 61 Example 7 cont.: PHASORS: v 2 (t) = 5e
j30 = 5cos(30 ) + j5sin(30 ) Polar form 5 30 V Rectangular form
4.33 + j2.5 V Use the additive property of phasors to find the sum
of v 1 (t) and v 2 (t). v(t) = v 1 (t) + v 2 (t) = (7.07 - j7.07) +
(4.33 + j2.5) V v(t) = 11.4 - j4.57 V Rectangular form = 12.28
-21.8 V Polar form = 12.28cos(1000t - 21.8 ) V Time-domain
- Slide 63
- Grossman/ Melkonian 62 Use the derivative property of phasors
to find the time derivative of v(t) = 15cos(200t - 30 ) V. Example
8: Solution: PHASORS:
- Slide 64
- Grossman/ Melkonian 63 PHASORS: Solution (cont.) dv(t)/dt = j
15 -30 = j200 15 -30 = 200 90 15 -30 = 3000 60 V/s dv(t)/dt =
3000cos(200t + 60 ) V/s The phasor form of the sinusoid is V(j ) =
15 -30 V. The time derivative is found by multiplying V(j ) by j
.
- Slide 65
- Grossman/ Melkonian 64 IMPEDANCE: Using phasor notation we will
analyze the ideal resistor, inductor and capacitor. When dealing
with AC signals, we will define a new parameter called impedance.
Impedance can be viewed as complex resistance. The concept of
impedance shows that certain parameters of inductors and capacitors
are frequency-dependent. The Resistor: Ohms law dictates the
relationship v = iR. If the source is sinusoidal, v S ( t) =
Acos(t), then the current through the resistor is: i(t) = v S (t) /
R = (A / R)cos t Converting v S (t) and i(t) to phasor notation: V
Z (j ) = A 0 I(j ) = (A / R) 0
- Slide 66
- Grossman/ Melkonian 65 IMPEDANCE: The relationship between V Z
and I viewed in the complex plane is shown below: Therefore, the
impedance of a resistor is found to be the ratio of the phasor
voltage across it to the phasor current flowing through it. Z R (j
) = V Z (j ) I(j ) = R Real Imaginary I V Relationship between V Z
& I for the resistor.
- Slide 67
- Grossman/ Melkonian 66 IMPEDANCE: The Inductor: + - v L (t)
i(t) L v S (t) + - ~ v L (t) = L[di L (t) / dt] i L (t) = 1 / L v L
(t)dt v L (t) = v S (t) i L (t) = i(t) We can write: i L (t) = i(t)
= 1 / L v L (t)dt if v S (t) = Acos( t) then, i L (t) = 1 / L Acos(
t)dt = (A / L)sin( t) For the circuit below
- Slide 68
- Grossman/ Melkonian 67 IMPEDANCE: v S (t) = v L (t) = Acos( t)
i(t) = i L (t) = (A / L)sin( t) = (A / L)cos( t - / 2) Note that
the inductor current is dependent on the radian frequency, , of the
source and is shifted in phase by 90 with respect to the voltage.
Converting to phasor notation: V Z (j ) = A 0 I(j ) = (A / L) - / 2
Z L (j ) = V Z (j ) I(j ) = L / 2 = j L Therefore: Relationship
between V Z & I viewed in the complex plane (for the inductor).
Real Imaginary I V -/2-/2
- Slide 69
- Grossman/ Melkonian 68 IMPEDANCE: The Capacitor: + - v C (t)
i(t) C v S (t) + - ~ i C (t) = C[dv C (t) / dt] v C (t) = 1 / C i C
(t)dt i C (t) = i(t) v C (t) = v S (t) We can write: i C (t) = C[dv
C (t) / dt] = C[dAcos( t) / dt] {for v S (t) = Acos( t)} = -C(A sin
t) = CAcos( t + / 2) For the circuit below Note that the capacitor
current is dependent on the radian frequency, , of the source and
is shifted in phase by 90 with respect to the voltage.
- Slide 70
- Grossman/ Melkonian 69 IMPEDANCE: Converting to phasor
notation: V Z (j ) = A 0 I(j ) = CA / 2 Z C (j ) = V Z (j ) I(j ) =
1 / ( C) - / 2 = -j / ( C) = Therefore: Relationship between V Z
& I viewed in the complex plane (for the capacitor). Real
Imaginary I V /2/2 (j C) 1
- Slide 71
- Grossman/ Melkonian 70 IMPEDANCE: Impedance of a circuit
element is defined as the sum of the real and imaginary parts. Z(j
) = R(j ) + jX(j ) Impedance Re Im ZLZL ZCZC ZRZR LL CC 1 R - /2/2
-/2-/2 Z R = R Z L = j L Z C = 1 jCjC
- Slide 72
- Grossman/ Melkonian 71 IMPEDANCE: KCL, KVL, voltage division,
current division, series connections, and parallel connections
apply to circuits in the frequency domain or in phasor form. Z1Z1
Z2Z2 ZNZN Rest of the circuit V + - I + V 1 - + V 2 - + V N - For
the circuit below, the elements are connected in series, therefore:
1. The same phasor current I flows through each impedance. 2. The
voltage across the series connection can be written as: V = V 1 + V
2 +... + V N = Z 1 I + Z 2 I +... + Z N I
- Slide 73
- Grossman/ Melkonian 72 Z1Z1 Z2Z2 ZNZN Rest of the circuit V + -
I + V 1 - + V 2 - + V N - IMPEDANCE: 3. The equivalent impedance is
Z EQ = Z 1 + Z 2 +... + Z N In general, the equivalent impedance is
a complex quantity of the form: Z EQ = R + jX Where R is the real
part and X is the imaginary part. The real part is called
resistance and the imaginary part is called reactance.
- Slide 74
- Grossman/ Melkonian 73 IMPEDANCE: Z1Z1 Z2Z2 ZNZN Rest of the
circuit V + - I + V 1 - + V 2 - + V N - For passive circuits:
Resistance is always positive Capacitive reactance is always
negative X C = -1 / C ohms ( ) Inductive reactance is always
positive X L = L ohms ( ) The phasor voltage across the k th
element in a series connection is: V k = Z k I k = ZkZk Z EQ V
Phasor version of Voltage Divider Rule
- Slide 75
- Grossman/ Melkonian 74 IMPEDANCE: Z1Z1 Z2Z2 ZNZN Rest of the
circuit V + - I ININ I1I1 I2I2 For the circuit below, the elements
are connected in parallel, therefore: 1. The same phasor voltage V
is across each impedance. 2. The phasor current I can be written
as: I = I 1 + I 2 +... + I N = V / Z 1 + V / Z 2 +... + V / Z
N
- Slide 76
- Grossman/ Melkonian 75 IMPEDANCE: Z1Z1 Z2Z2 ZNZN Rest of the
circuit V + - I ININ I1I1 I2I2 1 / Z EQ = 1 / Z 1 + 1 / Z 2 +... +
1 / Z N The equivalent impedance of the parallel connection is: The
phasor current through the k th element in a parallel connection
is: I k = 1/Zk1/Zk 1 / Z 1 + 1 / Z 2 +... + 1 / Z N I
- Slide 77
- Grossman/ Melkonian 76 IMPEDANCE: Example 9: Find the impedance
(Z) of the elements in the rectangular box. The circuit is in
steady-state. v s (t) = 50cos(4000t - 20 )V and i 1 (t) =
0.5cos(4000t)A. 50 i 3 (t) 100 + - i 1 (t) i 2 (t) ~ + - v 2 (t) v
1 (t) + - v s (t) Phasor representation of signals: Solution: V s =
50 -20 V and I 1 = 0.5 0 A
- Slide 78
- Grossman/ Melkonian 77 50 i 3 (t) + - i 1 (t) i 2 (t) ~ + - v 2
(t) v 1 (t) + - v s (t) IMPEDANCE: Example 9 cont.: Z = V 2 / I 2 V
2 = V s - V 1 and V 1 = I 1 R s = 0.5 0 50 = 25 0 V V 2 = 50 -20 -
25 0 = (46.98 - j17.10) - (25 + j0) V 2 = 21.98 - j17.10 = 27.85
-37.88 100
- Slide 79
- Grossman/ Melkonian 78 IMPEDANCE: Example 9 cont.: 50 i 3 (t) +
- i 1 (t) i 2 (t) ~ + - v 2 (t) v 1 (t) + - v s (t) 100 V L = V 2 =
27.85 -37.88 I 3 = V L / 100 0 = [27.85 -37.88 ] / [100 0 ] = 278.5
-37.88 mA + - v L (t) I 2 = I 1 - I 3 = (0.5 + j0) - (0.2198 -
j0.1710) = 0.2802 +j0.1710 A I 2 =0.328 31.39
- Slide 80
- Grossman/ Melkonian 79 Z = V 2 / I 2 = IMPEDANCE: Example 9
cont.: 50 i 3 (t) + - i 1 (t) i 2 (t) ~ + - v 2 (t) v 1 (t) + - v s
(t) 100 + - v L (t) 27.85 -37.88 V 0.328 31.39 A Z = 84.9 -69.3 =
30.1 -j79.4
- Slide 81
- Grossman/ Melkonian 80 Transient Analysis of Second Order
Circuit KCL:
- Slide 82
- Grossman/ Melkonian 81 Transient Analysis of Second Order
Circuit KVL: v C (t) = v L (t)
- Slide 83
- Grossman/ Melkonian 82 Transient Analysis of Second Order
Circuit Knowing i L (t):
- Slide 84
- Grossman/ Melkonian 83 Transient Analysis of Second Order
Circuit Another formulation:
- Slide 85
- Grossman/ Melkonian 84 Solution of Second Order Circuit
Generalized second order DEQ: Natural Frequency: Damping Ratio DC
gain
- Slide 86
- Grossman/ Melkonian 85 Natural Response of a Second- Order
System Solution is known to be of the form: x N (t) = e st
- Slide 87
- Grossman/ Melkonian 86 Roots of Second Order System 1.Real and
Distinct Roots: > 1 Over-damped Response 2.Real and Repeated
Roots: = 1 Critically Damped Response 3.Complex Conjugate Roots:
< 1 Under-damped Response
- Slide 88
- Grossman/ Melkonian 87 Response of the Second Order System as
function of
- Slide 89
- Grossman/ Melkonian 88 Roots of Second Order System Real and
Distinct Roots: > 1 Over-damped Solution:
- Slide 90
- Grossman/ Melkonian 89 Overdamped Solution Natural response of
overdamped second-order system for 1 = 2 = 1; = 1.5; n = 1
- Slide 91
- Grossman/ Melkonian 90 Roots of Second Order System Real and
Repeated Roots: = 1 Critically-damped Solution:
- Slide 92
- Grossman/ Melkonian 91 Critically Damped Solution Natural
response of a critically damped second-order system for 1 = 2 = 1;
= 1; n = 1
- Slide 93
- Grossman/ Melkonian 92 Roots of Second Order System Complex
Conjugate Roots: < 1 Under-damped Solution: Damped Natural
Frequency
- Slide 94
- Grossman/ Melkonian 93 Underdamped Solution Natural response of
an underdamped second-order system for 1 = 2 = 1; = 0.2; n = 1
- Slide 95
- Grossman/ Melkonian 94 Forced Response Solution to the
equation: 1.Constant Input: f(t) = F for t 0 x F (t) = K s F t 0 DC
steady-state solution
- Slide 96
- Grossman/ Melkonian 95 Complete Solution Overdamped case ( >
1): Critically Damped case ( = 1): Underdamped case ( < 1)