Grossman/ Melkonian 0 Chapter 4 and 5 Time Varying Circuits Capacitors Inductors Time Constant T C...

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

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

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:

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

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

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

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

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

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 )

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

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

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)

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.

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

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

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

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

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

Grossman/ Melkonian 18 TIME-DEPENDENT SIGNALS: v(t) t t t t Damped sinusoidSquare wave Pulse trainSawtooth wave Pulse width

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

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.

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/

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

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?

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.

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

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

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 )

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 ( ).

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.

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.

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)

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 ( ).

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:

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.

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

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

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

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

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

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

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

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

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

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

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.

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

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)

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)

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)

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

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:

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

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

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

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

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

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

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

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.

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

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

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:

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 .

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

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.

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

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

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.

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

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

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

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.

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

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

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

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

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

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

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

Grossman/ Melkonian 80 Transient Analysis of Second Order Circuit KCL:

Grossman/ Melkonian 81 Transient Analysis of Second Order Circuit KVL: v C (t) = v L (t)

Grossman/ Melkonian 82 Transient Analysis of Second Order Circuit Knowing i L (t):

Grossman/ Melkonian 83 Transient Analysis of Second Order Circuit Another formulation:

Grossman/ Melkonian 84 Solution of Second Order Circuit Generalized second order DEQ: Natural Frequency: Damping Ratio DC gain

Grossman/ Melkonian 85 Natural Response of a Second- Order System Solution is known to be of the form: x N (t) = e st

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

Grossman/ Melkonian 87 Response of the Second Order System as function of

Grossman/ Melkonian 88 Roots of Second Order System Real and Distinct Roots: > 1 Over-damped Solution:

Grossman/ Melkonian 89 Overdamped Solution Natural response of overdamped second-order system for 1 = 2 = 1; = 1.5; n = 1

Grossman/ Melkonian 90 Roots of Second Order System Real and Repeated Roots: = 1 Critically-damped Solution:

Grossman/ Melkonian 91 Critically Damped Solution Natural response of a critically damped second-order system for 1 = 2 = 1; = 1; n = 1

Grossman/ Melkonian 92 Roots of Second Order System Complex Conjugate Roots: < 1 Under-damped Solution: Damped Natural Frequency

Grossman/ Melkonian 93 Underdamped Solution Natural response of an underdamped second-order system for 1 = 2 = 1; = 0.2; n = 1

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

Grossman/ Melkonian 95 Complete Solution Overdamped case ( > 1): Critically Damped case ( = 1): Underdamped case ( < 1)

Grossman/ Melkonian 0 Chapter 4 and 5 Time Varying Circuits Capacitors Inductors Time Constant T C...

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Transcript of Grossman/ Melkonian 0 Chapter 4 and 5 Time Varying Circuits Capacitors Inductors Time Constant T C...