ES250:Electrical Science

HW8: Complete Response

of RL and RC Circuitsof RL and RC Circuits

• RL and RC circuits are called first‐order circuits. In thisIntroduction

RL and RC circuits are called first order circuits. In this chapter we will do the following:– develop vocabulary that will help us talk about the

response of a first‐order circuit– analyze first‐order circuits with inputs that are constant

ft ti l ti t t i ll t 0after some particular time, t0, typically t0 = 0– analyze first‐order circuits that experience more than one

abrupt change e g when a switch opens or closesabrupt change, e.g., when a switch opens or closes– introduce the step function and use it to determine the

step response of a first‐order circuit

• Circuits that contain only one inductor or capacitor can beFirst‐Order Circuits

• Circuits that contain only one inductor or capacitor can be represented by a first‐order differential equation

– these circuits are called first‐order circuitsthese circuits are called first order circuits

– Thévenin and Norton equivalent circuits simplify the analysis of first‐order circuits by permitting us toanalysis of first order circuits by permitting us to represent all first‐order circuits as one of two possible simple equivalent first‐order circuits, as shown:

First‐Order Circuits

• Consider the first‐order circuit with input voltage vs(t); theFirst‐Order Circuits

Consider the first order circuit with input voltage vs(t); the output, or response, is the voltage across the capacitor:

• Assume the circuit is in steady state before the switch is• Assume the circuit is in steady state before the switch is closed at time t = 0, then closing the switch disturbs the circuit; eventually, the disturbance dies out and the ; y,resulting circuit assumes a new steady state condition, as shown on the next slide

First‐Order Circuits

6.28ms2 rad1000 sec

T

Tπω

≈

⇒ = ≈6.28ms

secTT =

• When the input to a circuit is sinusoidal the steady‐stateFirst‐Order Circuits

• When the input to a circuit is sinusoidal, the steady‐state response is also sinusoidal; furthermore, the frequency of the response sinusoid must be the same as the frequency of p q ythe input sinusoid

• If the prior circuit is at steady state before the switch is closed, the capacitor voltage will be of the form:

• The switch closes at time t = 0, the capacitor voltage is:

• After the switch closes, the response will consist of two parts: a transient part that eventually dies out and a steady‐state part, as shown:

• The steady‐state part of the circuit response to a sinusoidalFirst‐Order Circuits

• The steady‐state part of the circuit response to a sinusoidal input will also be sinusoidal at the same frequency as the input, while the transient part of the response of a first‐p , p porder circuit is exponential of the form Ke−t/τ

• Note, the transient part of the response goes to zero as tbecomes large; when this part of the response “dies out,” the steady‐state response remains, e.g., M cos(1000t + δ)

• The complete response of a first‐order circuit can be represented in several ways, e.g.:

• Alternatively the complete response can be written as:First‐Order Circuits

• Alternatively, the complete response can be written as:

• The natural response is the part of the circuit response• The natural response is the part of the circuit response solely due to initial conditions, such as a capacitor voltage or inductor current, when the input is zero; while the forcedinductor current, when the input is zero; while the forced response is the part of the circuit response due to a particular input, with zero initial conditions, e.g.:

• In the case when the input is a constant or a sinusoid, the forced response is the same as the steady‐state response and the natural response is the same as the transient response

• Steps to find the complete response of first‐order circuits:First‐Order Circuits

• Steps to find the complete response of first‐order circuits:

– Step 1: Find the forced response before the disturbance, e g a switch change; evaluate this response at time t = t0e.g., a switch change; evaluate this response at time t t0to obtain the initial condition of the energy storage element

– Step 2: Find the forced response after the disturbance

– Step 3: Add the natural response = Ke−t/τ to the forced p p /response to get the complete response; use the initial condition to evaluate the constant K

Questions?Questions?

• Find the complete response of a first‐order circuit shown

Complete Response to a Constant Input• Find the complete response of a first‐order circuit shown

below for time t0 > 0 when the input is constant:

• Note the circuit contains a single capacitor and no

Complete Response to a Constant Input• Note, the circuit contains a single capacitor and no

inductors, so its response is first order in nature

• Assume the circuit is at steady state before the switch closesAssume the circuit is at steady state before the switch closes at t0 = 0 disturbing the steady state condition for t0 < 0

• Closing the switch at t0 = 0 removes resistor R1 from theClosing the switch at t0 0 removes resistor R1 from the circuit; after the switch closes the circuit can be represented with all elements except the capacitor replaced by its Thévenin equivalent circuit, as shown:

• The capacitor current is given by:

Complete Response to a Constant Input• The capacitor current is given by:

• The same current, i(t), passes through the resistor Rt, Appling KVL to the circuit yields:Appling KVL to the circuit yields:

C bi i th lt i ld th fi t d diff• Combining these results yields the first‐order diff. eqn.:

• Find the complete response of a first‐order circuit shown

Complete Response to a Constant Input• Find the complete response of a first‐order circuit shown

below for time t0 > 0 when the input is constant:

• Closing the switch at t = 0 removes resistor R from the

Complete Response to a Constant Input• Closing the switch at t0 = 0 removes resistor R1 from the

circuit; after the switch closes the circuit can be represented with all elements except the capacitor replaced by its p p p yNorton equivalent circuit, as shown:

≡Element law:

KCL:

⇒

• Both of these circuits have eqns. of the form:Complete Response to a Constant Input

Both of these circuits have eqns. of the form:

h th t i ll d th ti t t

⇒where the parameter τ is called the time constant

• Separating the variables and forming an indefinite integral, e ha ewe have:

where D is a constant of integrationwhere D is a constant of integration

• Performing the integration and solving for x yields:

⇒where A = e D, which is determined from the IC x(0)

⇒

• To find A, let t = 0, then:

⇒

• Therefore, we obtain:Complete Response to a Constant Input

Therefore, we obtain:

h th t i ll d th ti t twhere the parameter τ is called the time constant

• Since the solution can be written as:

• The circuit time constant can be measured from a plot ofComplete Response to a Constant Input

The circuit time constant can be measured from a plot of x(t) versus t, as shown:

Note: The circuit settles toNote: The circuit settles to steady state within 5τ

• Applying these results to the RC circuit yields the solution:Complete Response to a Constant Input

Applying these results to the RC circuit yields the solution:

⇒

⇒

• Applying these results to the RL circuit yields the solution:Complete Response to a Constant Input

Applying these results to the RL circuit yields the solution:

⇒

⇒

• The circuit below is at steady state before the switch opens;Example 8.3‐5: First‐Order Circuit

The circuit below is at steady state before the switch opens; find the current i(t) for t > 0:

• Note:

• The figures below show circuit after the switch opens (left)Solution

The figures below show circuit after the switch opens (left) and its the Thévenin equivalent circuit (right):

• The parameters of the Thévenin equivalent circuit are:• The parameters of the Thévenin equivalent circuit are:

• The time constant is:Solution

The time constant is:

• Substituting these values into the standard RC solution:Substituting these values into the standard RC solution:

where t is expressed in units of ms

• Now that the capacitor voltage is known, node voltage l d d “ ” h f h ldapplied to node “a” at the top of the circuit yields:

• Substituting the expression for the capacitor voltage yields:

• Solving for va(t) yields:Solution

Solving for va(t) yields:

• Finally, we calculating i(t) using Ohm's law yields:

Solution

>> tau=120e‐3;>> t=0:(5*tau)/100:5*tau;>> i=66.7e‐6‐16.7e‐6*exp(‐t/120e‐3);>> plot(t,i,'LineWidth',4)>> plot(t,i, LineWidth ,4)>> xlabel('t [s]')>> ylabel('i [A]')>> title('Plot of i(t)')

Questions?Questions?

• The application of a constant source, e.g., a battery, byThe Unit Step Source

The application of a constant source, e.g., a battery, by means of switches may be considered equivalent to a source that is zero up to t0 and equal to the voltage V0thereafter, as shown below:

≡

• We can represent voltage v(t) using the unit step as shown:• We can represent voltage v(t) using the unit step, as shown:

• Where the unit step forcing function is defined as:The Unit Step Source

• Where the unit step forcing function is defined as:

Note: That value of u(t0) is undefined

• Consider the pulse source v(t) = V0u(t − t0)−V0u(t − t1)The Unit Step Source

Consider the pulse source v(t) V0u(t t0) V0u(t t1) defined:

W t lt (t) i th it t h• We can represent voltage v(t) using the unit step, as shown:

≡

• The pulse source v(t) can schematically as:The Unit Step Source

The pulse source v(t) can schematically as:

• Recognize that the unit step function is an ideal model No• Recognize that the unit step function is an ideal model. No real element can switch instantaneously at t = t0; However, if it switches in a very short time (say, 1 ns), we can consider y ( y, ),the switching as instantaneous for medium‐speed circuits

• As long as the switching time is small compared to the time g g pconstant of the circuit, it can be ignored.

• Consider the application of a pulse source to an RL circuit asEx: Pulse Source Driving an RL Circuit

Consider the application of a pulse source to an RL circuit as shown below with t0 = 0, implying a pulse duration of t1 sec:

• Assume the pulse is applied to the RL circuit when i(0) = 0; since the circuit is linear, we may use the principle of

iti th t i i i h i i th tsuperposition, so that i = i1 + i2 where i1 is the response to V0u(t) and i2 is the response to V0u(t − t1)

• We recall that the response of an RL circuit to a constantEx: Pulse Source Driving an RL Circuit

We recall that the response of an RL circuit to a constant forcing function applied at t = tn with i(0) = 0, Isc = V0/R, andwhere τ = L/R is given by:

• Consequently we may add the two solutions to the two• Consequently, we may add the two solutions to the two‐step sources, carefully noting t0 = 0 and t1 as the start of each response, respectively, as shown:p , p y,

• Adding the responses provides the complete response of Ex: Pulse Source Driving an RL Circuit

the RL circuit shown:

Questions?Questions?