CS计算机代考程序代写 Prepared by Dr. Majid MALEKPOUR

Prepared by Dr. Majid MALEKPOUR
In conjunction with School of Electrical Engineering and Telecommunication, UNSW, and McGraw-Hill Education
Electrical Engineering and Telecommunications EE1111
Topic 5: Capacitors and RC Circuits

Topic 5 Content
This lecture covers:
• The capacitors as linear circuit elements
• Circuit analysis with capacitors
• First-order circuits with resistors
and capacitors (RC circuits)
• Singularity functions
• Unit step
• Unit impulse • Unit ramp
• Source-free or natural response, forced response, and step response for RC circuits with switched sources
Corresponds to parts of Chapters 6 and 7 of your textbook
Page 1

Topic 4 recap
• Linear property in circuits • Homogeneity
• Additivity
• Superposition principle in linear circuits (circuit analysis simplification)
• The response of a linear circuit (a voltage or current of an element) with multiple sources is the algebraic sum of individual responses due to each independent source when the rest are turned off
• Dependent sources are left intact during all calculations • Source transformation (circuit analysis simplification)
• Replacing a voltage source 𝑣 (independent or dependent) in series with 𝑠
• 𝑖 = 𝑣𝑠 or 𝑣 = 𝑅𝑖 (essentially it is using Ohm’s Law) 𝑠𝑅𝑠𝑠
• The transformed resistor is considered to be the same
• Without a resistor, source transformation is not possible
a resistor with a current source 𝑖𝑠 (independent or dependent, respectively) in parallel with another resistor at the same terminals or vise versa
Page 2

Topic 4 recap II • Thevenin’s theorem
• A linear circuit can be replaced (modelled) with a voltage source 𝑽𝐓𝐡 in series with a resistor 𝑹𝑻𝒉 from a given terminal
• 𝑉 is equal to the open-circuit voltage across the terminals without the Th
load
•𝑉=𝑣
Th 𝑜𝑐
• There are 2 cases when it comes to obtaining 𝑅Th:
• Case 1: There is no dependent source in the circuit
• 𝑅Th = 𝑅eq = 𝑅in, equivalent/input resistance seen from the terminals without the load and all independent sources turned off
• Case 2: There are some dependent sources
• Attach a constant voltage source (e.g., 𝑣 = 1 V) to the terminals and
𝑜
find the current 𝑖𝑜 through the source while all independent sources
are set to zero and no load is attached • 𝑅Th = 𝑣𝑜 = 1 (𝑅Th can be negative!)
𝑖𝑜 𝑖𝑜
• Attach a constant current source (e.g., 𝑖𝑜 = 1 A) to the terminals and
find the voltage 𝑣 across the source while all independent sources 𝑜
are set to zero and no load attached
• 𝑅Th = 𝑣𝑜 = 𝑣𝑜 (𝑅Th can be negative!) Page 3 𝑖𝑜 1

Topic 4 recap III • Norton’s theorem
• A linear circuit can be replaced (modelled) with a current source 𝑰𝑵 in parallel with a resistor 𝑹𝑵 from a given terminal
• 𝐼 is equal to the short-circuit current at the terminals without the load 𝑁
•𝐼=𝑖 𝑁 𝑠𝑐
• 𝑅𝑁 is the same as 𝑅Th
• Thevenin-Norton transformation is exactly the same as source transformation
• 𝑉 = 𝑅 𝐼 or 𝐼 = 𝑉Th , Th 𝑁𝑁 𝑁 𝑅Th
and
• 𝑅Th =𝑅𝑁 =𝑣oc 𝑖sc
• Maximum power transfer
• Using Thevenin equivalent circuit if
• If 𝑅𝐿 = 𝑅𝑇h the power transferred to the load 𝑅𝐿 is maximum
• 𝑝max = 𝑉Th2, and 𝑣𝐿 = 𝑉Th
4𝑅Th 2
Page 4

Topic 4 recap IV • Source Modelling
• Practical sources can be modelled as Thevenin or Norton equivalent circuits
• The input resistance seen from the terminals of sources is known as internal resistance 𝑹𝒔 for voltage sources and 𝑹𝒑
for current sources
• The internal resistance limits the amount of power that can be transferred to the load
Page 5

Capacitors
• A capacitor is a passive element that
stores energy in its electric field.
• It consists of two conducting plates
separated by an insulator (or dielectric)
• The plates are typically aluminum foil
• The dielectric is often air, ceramic, paper, plastic, or mica.
• When a voltage source 𝑣 is connected to the capacitor, the source deposits a positive charge 𝑞 on one plate and a negative charge – 𝑞 on the other plate.
• The charges will be equal in magnitude on both plates.
• The amount of charge is proportional to the voltage.
𝐶 is the capacitance measured in farad (F)
1F = 1C/V(coulomb/volt)
𝑞 = 𝐶𝑣
Page 6

Capacitors II
• Capacitance 𝐶 depends on the physical dimensions and geometry of the capacitor.
• For parallel-plate capacitors, the capacitance is given as follows:
𝐶 = 𝜀𝐴 𝑑
• 𝐴 is the surface area of each plate
• 𝑑 is the distance between the plates • 𝜀 is the permittivity of the dielectric
Page 7

Capacitors III
• Capacitors are available in different values
and types.
• Most capacitors are rated in picofarad
(pF) to microfarad (𝜇F)
• They are described by two factors:
• Dielectric material (thin film capacitors with polyester (Fig. 1(a)), polystyrene, ceramic (Fig. 1(b)), or electrolytic (Fig. 1(c))
• Fixed type (Fig. 2(a)) or variable type (Fig. 2(b))
i
C v−
(a)
C
v− (b)
• Electrolytic capacitors produce very high capacitance
• They are used to block DC signal and pass AC signals, shift phase, store energy, suppress and filter noise, start motor, etc.
Circuit symbol for fixed capacitance
i
(a)
(b) (c) Fig. 1
Circuit symbol for variable capacitance
Fig. 2
Page 8
++

Capacitors IV
• Passive sign convention applies to
capacitors as well.
C
−
• If 𝒗 × 𝒊 > 𝟎, the capacitor is being charged (absorbing energy)
• If 𝒗 × 𝒊 < 𝟎, the capacitor is discharging (supplying energy) For capacitors in the range of 𝜇F, particularly electrolytic ones, the polarity is already assigned (the negative sign is marked on the capacitor) i C v− Circuit symbol i v Negative lead Page 9 + + Capacitor V • Current-voltage relationship • Recall that: 𝑖=𝑑𝑞 and 𝑞=𝐶𝑣 𝑑𝑡 • Differentiate both sides of 𝑞 = 𝐶𝑣 and use 𝑖 = 𝑑𝑞: 𝑑𝑡 • Voltage-current relationship • Integrate both side of 𝑖 = 𝑑𝑞 over the interval of 𝑡0 𝑡 : 𝑑𝑡 i C v− capacitor current 𝑖 is proportional to the rate of change of its voltage 𝑣 with capacitance 𝐶 as the constant of proportionality, assuming passive sign convention 𝑖 = 𝐶 𝑑𝑣 𝑑𝑡 1𝑡 𝑣𝑡=𝐶 𝑖𝜏𝑑𝜏+𝑣(𝑡0) 𝑡0 𝑣 𝑡0 = 𝑞 𝑡0 is called the 𝐶 initial voltage or initial conditions at time 𝑡0 Page 10 + Capacitor VI • Instantaneous power delivered to a capacitor: • The energy stored in a capacitor: 𝑡 𝑤𝐶(𝑡) = 𝑝 𝜏 𝑑𝜏 −∞ 𝑡𝑑𝑣𝜏𝑡 𝑣𝜏2𝑡 𝑤𝐶𝑡= 𝐶𝑣𝜏𝑑𝜏𝑑𝜏=𝐶 𝑣𝜏𝑑𝑣𝜏=𝐶2−∞ −∞ −∞ 𝑤𝐶 𝑡 =1𝐶(𝑣𝑡2−𝑣−∞2) 2 • We can always assume that capacitor was uncharged at the beginning of time, i.e., at 𝑡 = −∞, 𝑣 −∞ = 0 𝑝 = 𝑣𝑖 = 𝐶𝑣 𝑑𝑣 𝑑𝑡 𝑤𝐶 represents the energy stored in the electric field that exists between the plates of the capacitor 𝑤𝐶 𝑡 =1𝐶𝑣𝑡 2 2 Page 11 Properties of capacitors • Since in a capacitor 𝑖 = 𝐶 𝑑𝑣, the current is zero VDC −CVDC − for constant voltage. 𝑑𝑡 A capacitor acts like an open circuit to DC voltage • Similarly, a discontinuous voltage across the capacitor results in infinite current (not physically possible!) • If the applied voltage in a capacitive circuit suddenly changes, charge will flow until the voltage reaches the new applied voltage. Thus allowing current to change abruptly but not the voltage across the capacitor. The voltage across a capacitor cannot change abruptly This can be viewed as filtering ability of capacitor against high- frequency voltages Page 12 + + Properties of capacitors II • An ideal capacitor does not dissipate energy • It takes power from the circuit to charge and then it can discharge the stored energy to deliver power to the circuit. • In practice, it would take a short time for a capacitor to fully charge under DC input voltage and become open circuit if there is no other resistor attached to it. A capacitor can deliver its stored energy back to the circuit A real capacitor has a parallel- model leakage resistance, leading to a slow loss of the stored energy internally Page 13 Parallel capacitors • Similar to resistors, parallel or series capacitors can be combined to simplify the circuit. • In a circuit with two parallel capacitors, KVL confirms that both capacitors have the same voltage. • Apply KCL and current-voltage relation for capacitor, 𝑖𝐶 = 𝐶 𝑑𝑣𝐶: is ic1 ic2 vc C1 vc C2 −− KCL: Current-voltage relation: 𝑑𝑡 𝑖𝑠 = 𝑖𝐶1 + 𝑖𝐶2 1 𝑖 = 𝐶 𝑑𝑣𝐶 + 𝐶 𝑑𝑣𝐶 2 𝑠 1 𝑑𝑡 2 𝑑𝑡 is vc Ceq − 1 & 2 𝑖 = 𝐶 +𝐶 𝑑𝑣𝐶 → 𝑠 1 2 𝑑𝑡 Thus: is 𝑖=𝐶 𝑠 eq 𝑑𝑣𝐶 𝑑𝑡 𝐶=𝐶+𝐶 eq 1 2 Page 14 + + + Parallel capacitors II The equivalent capacitance of any number of parallel capacitors is the sum of the individual capacitances 𝐶=𝐶+𝐶+⋯+𝐶 eq12 𝑁 𝑁 𝐶=𝐶 eq 𝑛 𝑛=1 Parallel capacitors are combined in the same manner as series resistors Ceq C1C2CN Ceq Page 15 Series capacitors C1 vc1 − C2 vc2 − ic vs • In a circuit with two series capacitors, KCL confirms that both capacitors have the same current. • Apply KVL and voltage-current relation for capacitor,𝑣𝐶=1 𝑡𝑖𝐶 𝜏𝑑𝜏+𝑣𝐶(𝑡0). KVL: 𝐶 𝑡0 𝑣=𝑣+𝑣 𝑠 𝐶1 𝐶2 1 𝑣=1 𝑡𝑖𝜏𝑑𝜏+𝑣 𝑡 𝑠 𝐶1 𝑡0 𝐶 𝐶1 0 Voltage-current relation: 1&2 +1𝑡𝑖𝐶𝜏𝑑𝜏+𝑣𝐶𝑡0 2 𝐶2 𝑡0 2 ic 𝑣=+𝑖𝜏𝑑𝜏+𝑣𝑡+𝑣𝑡s s eq 11𝑡 𝑠 𝐶1 𝐶2 𝑡0 𝐶 𝐶1 0 𝐶2 0 vv−C 𝐶0𝐶0𝐶0 1𝑡 𝑣= 𝑖𝜏𝑑𝜏+𝑣𝑡 𝑠𝐶𝐶 𝐶eq0 eq 𝑡0 1=1+1 𝐶𝐶𝐶 eq 1 2 Thus: or 𝑣eq𝑡 =𝑣1𝑡 +𝑣2𝑡 Page 16 𝐶= eq 𝐶𝐶 12 𝐶+𝐶 12 + + + Series capacitors II The reciprocal of the equivalent capacitance of any number of series capacitors is the sum of the individual reciprocal capacitances 1=1+1⋯+1 𝐶𝐶𝐶𝐶 eq 1 2 𝑁 C1 C2 Ceq CN Ceq Series capacitors are combined in the same manner as parallel resistors Page 17 Example 1 • Obtain the energy stored in each capacitor in the circuit below under DC conditions • (Worked solutions of examples are given in class) Answer: 𝑤2mF = 16 mJ 𝑤4mF = 128 mJ Example1 (Web view) Page 18 Example 2 • Find the voltage and the charge for each capacitor in the given circuit. • (Worked solutions of examples are given in class) Example2 (Web view) Page 19 Answer: 𝑣1 = 15 V 𝑣2 = 10 V 𝑣3 = 5 V 𝑞1_20mF = 𝑞2_30mF = 0.3 C 𝑞3−20mF = 0.1 C 𝑞4−40mF = 0.2 C Practice problem 1 • Find the voltage and the charge for each capacitor in the given circuit. Answer: 𝑣1 =75V, 𝑣2 =75V 𝑣3 =25V, 𝑣4 =50V 𝑞1_40𝜇F = 3 mC 𝑞2−20𝜇F = 1.5 mC 𝑞3−60𝜇F = 𝑞3−30𝜇F = 1.5 mC Page 20 Source-free RC circuit • A circuit comprising of resistors and capacitors with sources is called RC circuit. • The response of an RC circuit is the way its output (any voltage or current) reacts to an input/excitation. • There are two ways to excite (power up) an RC circuit: • Initial conditions (initial capacitor’s voltage) • Independent sources • To find the response, one must solve the circuit equations obtained by applying KVL and KCL, and 𝒗-𝒊 or 𝒊-𝒗 relations of resistor (Ohm’s law) and capacitor • This results in a differential equation of the first order due to capacitor 𝑖-𝑣 relation𝑖𝐶 = 𝐶𝑑𝑣𝐶 𝑑𝑡 • Source-free response: • The response of a circuit to its initial conditions only • Forced-response: • The response of a circuit to independent sources only as external input/excitation A first order circuit is characterized by a first-order differential equation Page 21 Source-free RC circuit II • Consider a simple RC circuit after disconnecting the DC source (which has charged the capacitor) • The stored energy in the capacitor due to initially charged voltage is released to the resistor • Assume that at 𝑡 = 0: • Initialvoltage:𝑣 0 =𝑣 0 =𝑉, 𝐶𝐶0 • Initial stored energy: 𝑤 (0) = 1 𝐶𝑉2 𝑐20 ApplyKCL: 𝑖𝐶 +𝑖𝑅 =0 Use 𝑖𝐶 = 𝐶 𝑑𝑣𝐶 , 𝑖𝑅 = 𝑣𝐶: 𝐶 𝑑𝑣𝐶 + 𝑣𝐶 = 0 or Rearrange: 𝑑𝑣𝐶 = − 1 𝑑𝑡 Integration 𝑣𝐶 𝑅𝐶 𝑡 constant 𝑣𝐶 𝑡 Integratebothsides: ln𝑣𝐶 =−𝑅𝐶+ln𝐴 or ln 𝐴 =−𝑅𝐶 Take powers of 𝑒: Apply initial conditions: + 𝑣𝐶 − 𝑣 0+ =𝑉 𝐶0 First-order differential equation 𝑑𝑡 𝑅 𝑑𝑡𝑅 𝐶 𝑅𝐶 𝑣 𝑡 =𝐴𝑒−𝑡 𝑑𝑣𝐶 + 𝑣𝐶 = 0 𝑑𝑡 𝑅𝐶 Source-free response (homogeneous solution) 𝑣 𝑡 =𝑉𝑒− 𝑡 𝐶 0𝑅𝐶 𝑣0=𝐴=𝑉 𝐶0 Page 22 Source-free RC circuit III • The result shows that the voltage response of the RC circuit is an exponential decay of the initial voltage, known as Natural Response 𝑣𝐶(𝑡) 𝑉 0 𝑡=0− 𝑡=0+ 0 + 𝑣𝐶 − 𝑣 0+ =𝑉 𝐶0 The natural response of a circuit refers to the behaviour of the circuit (in terms of voltage or current) due to its initial stored energy and physical characteristics with no external source or excitation 𝑣 𝑡 =𝑉𝑒− 𝑡 𝐶 0𝑅𝐶 0 𝑅𝐶 𝑉𝑒− 𝑡 Note: The voltage across capacitor cannot change immediately: 𝑣𝐶(0−) = 𝑣𝐶(0+) = 𝑣𝐶(0) • 𝑡=0− meansjustbefore𝑡=0 • 𝑡 = 0+ means immediately after 𝑡 = 0 Page 23 Time constant • The speed at which the voltage decays depends on the coefficient of 𝑡 in the power of exponential function, which expressed in terms of what is known as time constant. + 𝑣𝐶 − 𝑣 0+ =𝑉 𝐶0 The time constant of a circuit is the time required for the response to decay to a factor of 1/𝑒 or 36.8% of its initial value (or to reduce by 63.2%), denoted by 𝝉, (lowercase Greek letter Tau) 𝑣 𝑡 =𝑉𝑒− 𝑡 𝐶 0𝑅𝐶 • This implies that at 𝑡 = 𝜏, the voltage should be 0.368𝑉 : 0 𝑉𝑒−𝜏 =𝑉𝑒−1=0.368𝑉 0𝑅𝐶0 0 𝑣𝐶 𝜏 = 𝑅𝐶 Page 24 Time constant II • The source-free response or natural response of an RC circuit is written in terms of time constant as below. 𝜏 = 𝑅𝐶 • Notice that for every time interval of 𝜏, the voltage is reduced to 36.8 percent of its previous value. • After 5 time constants, the voltage 𝑣𝐶(𝑡) on the capacitor is less that one percent of its initial value 𝑉 . 0 + 𝑣𝐶 − 𝑣 0+ =𝑉 𝐶0 −𝑡 𝑣𝑡=𝑉𝑒𝜏 𝐶0 𝑣𝐶 𝑡 −𝑡 Values of 𝑉 = 𝑒 𝜏 0 𝑡 𝑣𝐶 𝑡 𝑉 0 𝜏 2𝜏 3𝜏 4𝜏 𝟓𝝉 0.36788 0.13534 0.04974 0.01832 0.00674 It takes 5 time constants (𝟓𝝉) for an RC circuit to reach its final state or steady-state (either fully charged or fully discharged) Page 25 Time constant III • A circuit with a small time constant has a fast response and vice versa. • This means it can dissipate the stored energy and reach steady-state much faster with smaller time constant. • It also means that for a capacitor to be discharged quicker, either the resistance 𝑅 or the capacitance 𝐶 must be reduced. + 𝑣𝐶 − 𝑣 0+ =𝑉 𝐶0 𝑣𝐶 𝑡 =𝑒−𝑡 𝑉𝜏 0 𝑣𝑡=𝑉𝑒 𝐶0 −𝑡 𝜏 𝜏 = 𝑅𝐶 Page 26 Other important variables in RC circuit • Knowing the voltage across capacitor, other circuit variables can be obtained as follows: • Resistor and capacitor currents: 𝑣 𝑡 𝑉 −𝑡 Note: The actual current in the + 𝑣𝐶 − 𝑖𝑅 𝑡 = 𝐶 = 0𝑒𝜏 circuitis𝑖 =–𝑖 ,capacitoractsas 𝑅𝑅𝑅𝐶 𝑖𝐶𝑡=𝐶 𝐶=−𝐶0𝑒𝜏=−0𝑒𝜏=−𝑖𝑅(𝑡) 𝑑𝑡 𝜏 𝑅 a source and discharges it energy. 𝑣 0+ 𝐶0 𝜏 = 𝑅𝐶 lim 𝑤 (𝑡) = 1 𝐶𝑉2 = 𝑤 (0) 𝑡→∞𝑅 20 𝑐 = 𝑉 𝑑𝑣 𝑉 −𝑡 𝑉 −𝑡 • Dissipated power in the resistor : 𝑉2 −2𝑡 𝑝 𝑡 = 𝑣𝐶 𝑡 𝑖𝑅 𝑡 = 0 𝑒 𝜏 𝑅 −𝑡 𝑣𝑡=𝑉𝑒𝜏 𝐶0 • Absorbed energy by the resistor up to time 𝑡: 𝑡 12−2𝑡 𝑤𝑡= 𝑝(𝜆)𝑑𝜆=𝐶𝑉(1−𝑒𝜏) 𝑅20 0 Note: The total energy dissipated by the resistor is equal to the initial stored energy in the capacitor Page 27 Solving source-free RC circuits • Follow these steps to find the source-free response of RC circuits: 1. Find the initial voltage 𝑣 0− = 𝑉 across the 𝐶0 capacitor before any changes/switching in thecircuit,then𝑣 0− =𝑣 0+ =𝑣 0 =𝑉 𝐶𝐶𝐶0 • The capacitor is assumed to be fully charged just before the changes and can be replaced with open circuit 2. Find the time constant 𝜏 = 𝑅𝐶 after the changes in the circuit • Note: 𝑅 is the Thevenin equivalent resistance as seen from terminals of the capacitor after the changes/switching in the circuit, 𝑅 = 𝑅Th • 𝜏 = 𝑅Th𝐶 3. The voltage across capacitor is 𝑣 𝑡 = 𝑉 𝑒 𝜏 −𝑡 𝐶0 4. Find all other circuit variables using capacitor voltage𝑣𝐶 𝑡 The changes in the circuit can be a switch which opens or closes and the action can either remove part of the circuit or add a new part to the circuit Page 28 Example 3 • Find the voltages 𝑣𝐶, 𝑣𝑥, and the current 𝑖𝑥 for 𝑡 > 0 if the initial
voltage is 𝑣𝐶 0 = 15 V.
• (Worked solutions of examples are given in class)
Answer:
𝑣𝐶 = 15𝑒−2.5𝑡 V 𝑣𝑥 = 9𝑒−2.5𝑡 V
𝑖𝑥 = 0.75𝑒−2.5𝑡 A
Example3 (Web view) Page 29

Practice problem 2
• Find the voltages 𝑣𝐶, 𝑣𝑥, and the current 𝑖𝑜 for 𝑡 ≥ 0 if the initial voltage is 𝑣𝐶 0 = 60 V.
Answer:
𝑣𝐶 = 60𝑒−0.25𝑡 V 𝑣𝑥 = 20𝑒−0.25𝑡 V 𝑖𝑜 = −5𝑒−0.25𝑡 A
Page 30

Example 4
• The switch in the circuit has been closed for a long time, and it is opened at 𝑡 = 0, Find the voltage 𝑣 𝑡 for 𝑡 ≥ 0. Calculate the initial energy stored in the capacitor. How much energy is dissipated in the circuit in 100 ms, then find the remaining energy in capacitor after 100 ms.
• (Worked solutions of examples are given in class)
Answer:
𝑣 = 15𝑒−5𝑡 V 𝑤𝐶 0 =2.25J
𝑤𝑅Th 𝑡=0.1 s = 1.422 J 𝑤𝐶 𝑡=0.1 s = 0.828 J
Example4 (Web view) Page 31

Practice problem 3
• If the switch opens at 𝑡 = 0 after a long time, find the voltage 𝑣 𝑡 for 𝑡 ≥ 0, initial energy 𝑤𝐶 0 , and the dissipated energy in the circuit after 500 ms.
Answer:
𝑣 = 8𝑒−2𝑡 V
𝑤𝐶 0 =16=5.333J 3
𝑤𝑅Th 𝑡=0.5 s = 4.611 J
Page 32

Singularity functions – I. Unit step function
• Singularity functions are functions that are either discontinuous or have discontinuous derivatives.
• They serve as good approximations to switching signals representing a sudden change in voltage source or current source.
• An important singular function is known as unit step function.
𝑢𝑡=
0, 𝑡<0 1, 𝑡>0
A unit step function, denoted by 𝑢(𝑡), is zero for negative values of time 𝑡 and one or positive values of time 𝑡
• The unit step function is undefined at 𝑡 = 0
• It is a dimensionless function
Page 33

Time shift in unit step function • If the sudden change occurs at 𝑡 = 𝑡0 (𝑡0 > 0),
it is said that 𝑢(𝑡) is delayed by 𝑡0 seconds.
• If the sudden change occurs at 𝑡 = −𝑡0 (𝑡0 > 0), it is said that 𝑢(𝑡) is advanced by 𝑡0 seconds.
𝑢𝑡−𝑡0 =
0, 𝑡<𝑡0 1, 𝑡>𝑡0
𝑢𝑡+𝑡0 =
0, 𝑡<−𝑡0 1, 𝑡>−𝑡0
Page 34

Equivalent circuits with switch
• Unit step function 𝑢(𝑡) is used to represent sudden changes in sources when a switch in the circuit is opened or closed at 𝑡 = 𝑡0.
𝑡 = 𝑡0
𝑉 𝑢(𝑡 − 𝑡 ) 00
𝑣(𝑡) =
0, 𝑡<𝑡0 𝑉, 𝑡>𝑡 0 0
𝑣𝑡=𝑉𝑢𝑡−𝑡 00
𝐼 𝑢(𝑡 − 𝑡 ) 00
𝑖(𝑡) =
𝑡 = 𝑡0
0, 𝑡<𝑡0 𝐼, 𝑡>𝑡 0 0
𝑖𝑡=𝐼𝑢𝑡−𝑡 00
Page 35

Step response of an RC circuit
• Step response is the reaction of the circuit due to a sudden application of a DC voltage or current source, which can be modeled as a switch as shown in Fig. (a) or with a step function for the source as shown in Fig. (b).
+
𝑣𝐶 −
The step response of a circuit is its behaviour when the excitation/input is the step function, which may be a voltage or a current source
(a)
• If the initial voltage of capacitor is 𝑉 , right 0
+
𝑣𝐶 −
after switching at 𝑡 = 0, the capacitor voltage cannot immediately change.
Vs
0t
𝑣 0− =𝑣 0+ =𝑉=𝑣(0) 𝐶𝐶0𝐶
• 𝑣𝐶 0− is the capacitor voltage just before switching
• 𝑣𝐶 0+ is the capacitor voltage just after switching
Note: An exception to this is when the
switch adds a voltage source in parallel
to capacitor (𝑣 0+ = 𝑉 ≠ 𝑣 (0−)), or if it 𝐶𝑠𝐶
adds a short circuit in parallel to the capacitor (𝑣𝐶 0+ = 0 V ≠ 𝑣𝐶(0−)). They are
not practical through.
(b)
Page 36

Step response of an RC circuit II • Consider a simple RC circuit with step function as its
input voltage source with initial condition 𝑣 0 = 𝑉 . 𝐶0
iRi
• Apply KCL at the node between 𝑅 and 𝐶:
𝑑𝑣 𝑉𝑢𝑡−𝑣 𝑖𝐶=𝑖𝑅 →𝐶 𝐶=𝑠 𝐶
Vs
iC
+ 𝑣𝐶
𝑑𝑡𝑅0t − • First-order differential equation for 𝑣𝐶:
𝑑𝑣𝐶 + 𝑣𝐶 = 1 𝑉 , 𝑡 > 0
(𝑢(𝑡) = 1 for 𝑡 > 0) • Solve the differential equation for 𝑣𝐶 by finding its
homogenous and particular solutions (or responses) : 𝑣𝐶 𝑡 =𝑣𝐶𝑝 𝑡 +𝑣𝐶h 𝑡 =𝐵+𝐴𝑒𝑠𝑡 where and
• Apply initial condition to find 𝐴: 𝑉 = 𝐴𝑒0 + 𝑉 → 0𝑠
Thus:
Root of characteristic equation𝑠+ 1 =0
𝑅𝐶
𝑑𝑡 𝑅𝐶 𝑅𝐶
𝑠
𝑠=−1 𝑅𝐶
𝐵=𝑉 𝑠
𝐴=𝑉 −𝑉 0𝑠
𝑉, 𝑡≤0 0
𝑣𝐶𝑡=𝑉+(𝑉−𝑉)𝑒−𝑡, 𝑡≥0 𝑠0𝑠𝜏
𝜏 = 𝑅𝐶
𝜏 is the circuit time constant
Page 37

Step response of an RC circuit III • Depending on the initial condition value of 𝑉 and
0 voltage source 𝑉 , the capacitor can be charged or
discharged.
𝑠
• After 5 time constants (5𝜏), the capacitors is again considered as open circuit as its voltage can no longer change beyond the source voltage.
+
𝑣𝐶 −
𝑣 v ( 𝑡t ) 𝐶
Vs
𝑉, 𝑡≤0 0
𝑣𝐶𝑡=𝑉+(𝑉−𝑉)𝑒−𝑡, 𝑡≥0 𝑠0𝑠𝜏
𝑉>𝑉 𝑠0
𝑣𝐶 𝑡
V0 0t
V0
𝑉<𝑉 𝜏=𝑅𝐶 In practice, capacitors have a limit on the maximum voltage that can be applied across them known as breakdown or nominal voltage (they cannot be overcharged physically) Vs 𝑠0 0 t Page 38 Forced response of an RC circuit • The response of an RC circuit to a sudden application of an external source when the capacitor is initially uncharged (i.e., 𝑣𝐶 (0) = 0 V) is known as Forced Response. or iC + 𝑣𝐶 − Forced response of an RC circuit is the response to external source only 𝜏 = 𝑅𝐶 𝑣𝐶 𝑡 = 0, 𝑡≤0 −𝑡 𝑉(1−𝑒𝜏), 𝑡≥0 𝑠 𝑣𝐶 𝑡 −𝑡 𝑣 𝑡 =𝑉(1−𝑒𝜏)𝑢(𝑡) 𝐶𝑠 • The current through capacitor is also exponentially decreasing as capacitor is charged leading to open 𝑖𝐶 𝑡 circuit: 𝑑𝑣𝐶 𝑖𝐶 𝑡 = 𝐶 → Beware of the derivate of step function!!! 𝑉 −𝑡 𝑖𝐶 𝑡 = 𝑠𝑒𝜏 𝑢(𝑡) 𝑅 𝑑𝑡 • Note: An uncharged capacitor is a short circuit since𝑣𝐶 0 =0V. • Note: Capacitor current CAN change immediately after switching, i.e., 𝑖𝐶 0+ ≠ 𝑖𝐶 0− (a jump at 𝑡 = 0) Page 39 Complete response of an RC circuit • The response of an RC circuit to a sudden application of an external source as well as its initial condition is known as Complete Response or Total Response. + • We can express the complete response in 𝑣𝐶 terms of natural response (source-free response) and forced response. − −𝑡 𝑣𝑡=𝑉+𝑉−𝑉𝑒𝜏,𝑡>0
𝐶𝑠0𝑠
𝑣 𝑡 =𝑉𝑒−𝑡+𝑉(1−𝑒−𝑡), 𝑡>0
𝐶0𝜏𝑠𝜏 Natural Forced
Complete response = natural response + forced response
𝑣𝐶 𝑡 =𝑣𝐶𝑛(𝑡)+𝑣𝐶𝑓(𝑡), 𝑡>0
response response
𝑣𝑣 𝑛𝑓
This can be viewed as superposition principle in RC circuits with two input sources of energy:
1. Initial condition (initial stored energy) 2. Independent sources
Due to
initial stored energy (source-free response)
Due to Independent sources
Page 40

Complete response of an RC circuit II
• From another perspective, the natural
response eventually decays to zero along
with the transient component of the
forced response, leaving only the steady-
state component of the forced response 𝑣𝐶
−𝑡 𝑣𝑡=𝑉+𝑉−𝑉𝑒𝜏,𝑡>0
+ −
𝐶𝑠0𝑠
Complete response = transient response
+ steady-state response
𝑣𝐶 𝑡 =𝑣𝐶𝑡(𝑡)+𝑣𝐶𝑠𝑠(𝑡), 𝑡>0
Steady-state Transient
response
𝑣 𝑠𝑠
Permanent Temporary component component
response
𝑣𝑡
• Transient response is the circuit‘s temporary response that will die out with time.
• Steady-state response is the circuit’s permanent response a long time after an external input/excitation is applied (after 5 time constants, 5𝜏).
Page 41

Analyzing an RC circuit
• We can describe the overall behavior of a simple RC circuit before and after a sudden application of a DC constant source into two stages of steady-state behavior and one stage of transient behavior
−𝑡
𝑣𝑡=𝑉+𝑉−𝑉𝑒𝜏,𝑡>0 𝐶𝑠0𝑠
• First stage of steady-state 𝑡 < 0: • There has been no change in the circuit for a long time and the capacitor is open circuit with 𝑣0=𝑉 𝐶0 • Transient stage 0 < 𝑡 < 5𝜏: • The capacitor voltage changes exponentially (charging 𝑉 > 𝑉
or discharging 𝑉 < 𝑉 ) 𝑠0 • Second stage of steady-state 𝑡 > 5𝜏:
• The capacitor voltage reaches its final value or steady-state value and becomes open circuit
againas𝑡→∞with𝑣 ∞ = 𝑉 𝐶𝑠
𝑠0
𝑣𝐶 𝑡
Vs
V0
𝑡 < 0 0 0<𝑡<5𝜏 Transient stage t 𝑡 > 5𝜏
Second stage of steady-state
𝑉>𝑉 𝑠0
First stage of steady-state
Page 42

Analyzing an RC circuit II
• The main question here is how to find the voltage across a capacitor in a complicated circuit with multiple resistors, switches, independent and dependent sources!!!!!!!
a
i
v
−
b
C
Resistive circuit with a switch
or
• The standard approach is to use KVL (Mesh analysis), KCL (Nodal analysis), Ohm’s law, and capacitor current-voltage relation to obtain and solve circuit differential equation. But there is a short-cut for first-order circuits.
• Recall that any resistive circuit can be replaced with its Thevenin equivalent circuit.
• For each stage of steady-state (before and after changes in the circuit), there is a Thevenin equivalent circuit from capacitor terminals.
RTh_0
a
i=0
v(0) C VTh_∞
RTh_∞ a
i=0 v(∞) C
VTh_0
−−
bb
Thevenin equivalent circuit before the switch changes (first stage of steady-state)
Thevenin equivalent circuit after the switch changes (second stage of steady state)
Page 43
+
+
+

Analyzing an RC circuit III
• Once the Thevenin equivalent circuit is obtained
after switching, we can still use the same
VTh_0
RTh_0 a
i=0
v(0) −
b
formula for step response of the capacitor −𝑡
C
withaninitialvoltage,𝑣 𝑡 =𝑉 + 𝑉 −𝑉 𝑒𝜏 ,
𝑡 > 0, as below:
𝑠 0 𝑠
Thevenin equivalent circuit at first stage of steady-state)
𝜏 𝑣𝑡=𝑣∞+𝑣0+ −𝑣∞𝑒−𝑡, 𝑡>0
VTh_∞
a
i=0 C
RTh_∞
b
v(∞) −
• 𝑣(0+): Initial voltage at 𝑡 = 0+ (which is the same as 𝑣(0−) or before the switch changes at 𝑡 < 0 with capacitor being open circuit) • 𝑣 ∞ : Final or steady-state voltage at 𝑡 → ∞ after the switch changes (𝑡 > 0 ) with capacitor being open circuit again
• 𝜏 = 𝑅Th_∞𝐶: Time constant at 𝑡 → ∞ after the switch changes (𝑡 > 0 ) with 𝑅Th_∞ as the Thevenin equivalent resistance after the switch changes
Thevenin equivalent circuit at second stage of steady state)
Note: The assumption here is that the output of the circuit is capacitor voltage, that is why 𝑣 0+ =𝑣 0− .Allothercircuit variables can be found using capacitor voltage
Page 44
++

Solving for step response of RC circuits
• Follow these steps to find complete or step response of an RC circuit:
1. Find the initial voltage 𝑣 0− = 𝑉 across the 𝐶0
capacitor before any changes/switching in thecircuit,then𝑣 0+ =𝑣 0− =𝑣 0 =𝑉
𝐶𝐶𝐶0
• The capacitor is assumed to be open circuit
2. Find the final voltage 𝑣𝐶(∞) at 𝑡 → ∞ across the capacitor after the changes/switching in the circuit (𝑡 > 0)
• The capacitor is assumed to be open circuit
3. Find the time constant 𝜏 = 𝑅Th_∞𝐶 after the changes in the circuit
• 𝑅Th_∞ is Thevenin Equivalent resistance after the changes
4. The voltage across capacitor is
𝑣𝐶 𝑡 =𝑣𝐶 ∞ + 𝑣𝐶 0 −𝑣𝐶 ∞ 𝑒𝜏
5. Find all other circuit variables using capacitor voltage 𝑣𝐶 𝑡
−𝑡
The changes in the circuit can be a switch which opens or closes and the action can either remove part of the circuit or add a new part to the circuit
Page 45

Time shift in step response of an RC circuit • Note that if the switch changes at time 𝑡 = 𝑡0 instead of 𝑡 = 0, there is a
time delay in the response.
• Since RC circuits are linear time-invariant circuits (LTI), so time delay
can be expressed as time shift in the original response as follows:
• 𝑣 𝑡 is the initial voltage at 𝑡 = 𝑡+ 𝐶00
𝑣 𝑡 =𝑣 ∞ + 𝑣 𝑡 −𝑣 ∞ 𝑒−(𝑡−𝑡0), 𝑡>𝑡 𝐶𝐶𝐶0𝐶𝜏0
• Timeinvarianceisa property of a system where the output of the system does not depend on the time of application of the input.
• This means that if the input 𝑥(𝑡) generates output 𝑦(𝑡), applying the same input at𝑡=𝑡0,i.e.,𝑥 𝑡−𝑡0 ,will generate the same output at𝑡=𝑡0,i.e.,𝑦 𝑡−𝑡0
• This method can also be used for multiple switching at different times
• Initial voltage 𝑣𝐶 𝑡0 for each time interval after the first switching is calculated using the voltage function of the capacitor in the previous time interval.
• Final value 𝑣𝐶 ∞ for each time interval is calculated in the same manner as before by replacing the capacitor with an open circuit assuming that the time interval is long enough before the next switching
Page 46

Example 5
• The switch in the circuit below has been in position A for a long time. At 𝑡 = 0, the switch moves to B. Determine 𝑣(𝑡) for 𝑡 > 0 and calculate its value at 𝑡 = 1 s and 𝑡 = 4 s.
• (Worked solutions of examples are given in class)
Answer:
𝑣𝑡 =30−15𝑒−0.5𝑡V, 𝑡>0 𝑣1 =20.9V
𝑣4 =27.97V
Example5 (Web view) Page 47

Example 6
• In the circuit below, the witch has been closed for a long time and
it is opened at 𝑡 = 0, find 𝑣 and 𝑖 for all time. • (Worked solutions of examples are given in class)
Answer:
𝑣𝑡= 𝑖𝑡=
10 V, 𝑡 ≤ 0 20−10𝑒−0.6𝑡V, 𝑡≥0
−1 A, 𝑡 < 0 1+𝑒−0.6𝑡A, 𝑡>0
Example6 (Web view) Page 48

Practice problem 5
• Find 𝑣(𝑡) for 𝑡 > 0 in the circuit below. Assume the switch has been open for a long time and it is closed at 𝑡 = 0. Calculate 𝑣(𝑡) at 𝑡 = 0.5 s for all time.
Answer:
𝑣𝑡 =9.375+5.625𝑒−2𝑡V, 𝑡>0 𝑣0.5 =11.444V
Page 49

Practice problem 6
• the switch in the circuit below is closed at 𝑡 = 0. Find 𝑣(𝑡) and 𝑖(𝑡) for all time. Note that 𝑢(−𝑡) = 1 for 𝑡 < 0 and it is zero for 𝑡 > 0. Also, you can express it as 𝑢 −𝑡 = 1 − 𝑢(𝑡).
Answer:
𝑣𝑡= 𝑖𝑡=
20 V,
10(1 + 𝑒−1.5𝑡) V,
0 A,
−2(1 + 𝑒−1.5𝑡) A,
𝑡 ≤ 0 𝑡 ≥ 0
𝑡 < 0 𝑡 > 0
Page 50

Questions!?
Page 51