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# Linear Circuit Analysis PowerPoint PPT Presentation

Linear Circuit Analysis. In an oscilloscope a timing signal called a horizontal sweep acts as a time base, which allows one to view measured input signals as a function of time.

Linear Circuit Analysis

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Linear Circuit Analysis

In an oscilloscope a timing signal called a horizontal sweep acts as a time base, which allows one to view measured input signals as a function of time.

In practice, the linear increase in voltage is approximated by the linear part of an exponential response of an RC circuit.

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First-Order RL and RC Circuits

1. What is a first-order circuit?

A first-order circuit is characterized by a first-order differential equation. It consists of resistors and the equivalent of one energy storage element.

Typical examples of first-order circuits:

(a) First-Order RL circuit

(b) First-Order RC circuit

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Interconnections of sources, resistors, capacitors, and inductors lead to new and fascinating circuit behavior.

since

First-Order RC circuit

or , equivalently,

This equation is called Constant-coefficient first-order linear differential equation

Apply duality principle,

2. Some mathematical preliminaries

RC circuit first-order differential equation

RL circuit first-order differential equation

The first-order RL and RC circuits have differential equation models of the form

or , equivalently,

valid for tt0, where x(t0)=x0 is the initial condition. The term f(t) denotes a forcing function. Usually, f(t) is a linear function of the input excitations to the circuit.

The parameter denotes a natural frequency of the circuit.

or

The main purpose of this chapter is to find a solution to the differential equation.

The solution to the equation for tt0 has the form

(1) Satisfies the differential equation

(2) Satisfies the correct initial condition, x(t0)=x0

Integrating factor method

First step, multiply both sides of the equation by a so-called integrating factor et.

By the product rule for differentiation,

Integrate both sides of the equation from t0 to t as follows

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3. Source-free or zero-input response

A parallel connection of a resistor with an inductor or capacitor without a source.

In these circuits, one assumes the presence of an initial inductor current or initial capacitor voltage.

(a) KCL implies

(b) KVL implies

Both differential equation models have the same general form,

The solution to the equation for tt0 has the form

Where is a special constant called the time constant of the circuit. The response for tt0 of the undriven RL and RC circuit are, respectively, given by

RL circuit

RC circuit

The time constant of the circuit is the time it takes for the source-free circuit response to drop to e1=0.368 of its initial value.

Linear

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For more general circuits, those containing multiple resistors and dependent sources, it is necessary to use the Thevenin equivalent resistance seen by the inductor or capacitor in place of the R.

Thevenin equivalent

Thevenin equivalent

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Example 1. For the circuit of the figure, find iL(t) and uL(t) for t 0 given that iL(0)=10A and the switch S closes at t=0.4s. Then compute the energy dissipated in the 5 resistor over the time interval [0.4, ).

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transient state

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0 uC(0) iL(0)

uC(0+) iL(0+)

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uC(0)

iL(0)

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delayed

piecewise-constant signal

pulse

pulse train

Example 8.2. For the circuit of the figure, find iL(t) and uL(t) for t 0 given that iL(0)=10A and the switch S closes at t=0.4s. Then compute the energy dissipated in the 5 resistor over the time interval [0.4, ).

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Solution

Step 1. With switch S open, compute the response for 0t 0.4s.

From the continuity property of the inductor current,

Step 2. With switch S closed, compute the response for t0.4s.

Step 3. Write the complete response as a single expression.

Step 4. Plot the complete response.

The 0.4s time constant has a much faster rate of decay than the lengthy 2s time constant.

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Step 5. Compute uL(t).

for 0t 0.4s,

in particular,

hence,

for t 0.4s,

in particular,

Step 6. Compute energy dissipated in the 5 resistor over the time interval [0.4, ).

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Example 8.3. Find uC(t) for t 0 for the circuit of the figure given that uC(0)=9V.

Solution

Step 1. Compute the response for 0t 1s. By the continuity of the capacitor voltage,

hence,

Step 2. Compute the response for t 1s.

Step 3. Use step functions to specify the complete response.

Step 4. Obtain a plot of the response.

Here the part of the response with the 0.3s time constant shows a greater rate of decay than the longer 0.8s time constant.

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Exercise. Suppose that in example 2 the switch moves to the 4.5 resistor at t=0.5s instead of 1s. Compute the value uC(t) at t=1.2s.

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Example 8.4. Find uC(t) for the circuit of the figure, assuming that gm=0.75S and uC(0)=10V.

Solution

It is straightforward to show that the Thevenin equivalent seen by the capacitor is a negative resistance,

Thevenin equivalent

hence,

Because of the negative resistance, this response grows exponentially, as shown in the figure.

A circuit having a response that increases without bound is said to be unstable.

Negative resistance causes capacitor voltage to increase without bound.

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Exercise. In example 3, let gm=0.125S. Find the equivalent resistance seen by the capacitor and uC(t), t 0.

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24V K

RC

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t = 0 K12

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4. DC or step response of first-order circuits

This section takes up the calculation of voltage and current responses when constant voltage or constant current sources are present.

Thevenin equivalent

Thevenin equivalent

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Deriving the differential equation models characterizing each circuits voltage and current responses.

By KVL and Ohm law,

By KCL and Ohm law,

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Exercise. Constant differential equation models for the parallel RL and RC circuits of the figure. Note that these circuits are Norton equivalents of those in the figure. Again choose iL(t) as the response for the RL circuit and uC(t) as the response for the RC circuit.

Observe that four differential equations have the same structure:

for RL case

where

for RC case

And the general formula for solving such a differential equation:

where

as long as x(t) is a capacitor

voltage or inductor current, and f()=F is a constant (nonimpulsive) forcing function.

Which is valid for tt0. After some interpretation, this formula will serve as a basis for computing the response to RL and RC circuits driven by constant sources.

then

if

for RL case

for RC case

Hence, the solution of the differential equation given constant or dc excitation becomes

for RL case

for RC case

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Example 8.5. For the circuit of the figure, suppose a 10V unit step excitation is applied at t=1 when it is found that the inductor current is iL(1)=1A. The 10V excitation is represented mathematically as uin(t)=10u(t1)V for t1. Find iL(t) and uL(t) for t1.

Solution

Step 1. Determine the circuits differential equation model.

where the time constant

Step 2. Determine the form of the response.

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Step 3. Compute iL() and set forth the final expression for iL(t).

replace the inductor by a short circuit,

and since

It follows that,

Step 4. Plot iL(t).

Exercise. Verify that in example 4, uL(t) can be obtained without differentiation by

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Specifically, we need only compute , , and the time constant or .

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Step 5. Compute uL(t).

Exercise. In example 4, suppose R is changed to 4. Find iL(t) at t=2s.

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Example 8.6. The source in the circuit of the figure furnishes a 12V excitation for t<0 and a 24V excitation for t0, denoted by uin(t)=12u(t)+24u(t)V. The switch in the circuit closes at t=10s. First determine the value of the capacitor voltage at t=0, which by continuity equals uC(0+). Next determine uC(t) for all t0.

Solution

Step 1. Compute initial capacitor voltage

Step 2. Obtain uC(t) for 0t 10s.

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Step 3. Compute the initial condition for the interval t>10.

Step 4. Find uC(t) for t>10.

Thevenin equivalent

Step 5. Set forth the complete response using step functions.

Step 6. Plot uC(t).

Exercise. Suppose the switch in example 5 opens again at t=20s. Find uC(t) at t=25s.

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t = 0 , K uC(0) = 0V12uC80V t

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t = 0 , K t > 0 iCuC u uC(0+) = 1V

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t = 0 , uC(t)

uC

iL

is .

5. Superposition and linearity

Provided one properly accounts for initial conditions, superposition still apply when capacitors and inductors are added to the circuit.

for a capacitor

suppose

and

By the same arguments, the current due to the input excitation

One the other hand, suppose

and

Thus linearity and, hence, superposition hold.

Arguments analogous to the preceding imply that a relaxed inductor satisfies a linear relationship, and thus superposition is valid, whether the inductor is excited by currents or by voltages.

For a general linear circuit, one can view each initial condition as being set up by an input which shuts off the moment the initial condition is established.

This mean that when using superposition on a circuit, one first looks at the effect of each independent source on a circuit having no initial conditions.

Then one sets all independent sources to zero and computes the response due to each initial condition with all other initial conditions set to zero.

The sum of all the responses to each of the independent sources plus the individual initial condition responses yields the complete circuit response, by the principle of superposition.

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Example 8.8. The linear circuit of the figure has two source excitations applied at t=0, as indicated by the presence of the step functions. The initial condition on the inductor current is iL(0)=1A. Determine the response iL(t) for t0 using superposition.

Solution

Step 1. Compute the part of the circuit response due only to the initial condition, with all independent sources set to zero.

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Step 2. Determine the response due only to US=10u(t)V.

Step 3. Compute the response due only to the current source IS=2u(t)A.

Step 4. Apply the principle of superposition.

due to source US

due to source IS

due to initial condition

The question still remains as to why the superposition principle holds an advantage over the Thevenin equivalent method.

Question 1. What is the new response if the initial condition is changed to iL(0)=5A?

Question 2. What is the new response if the voltage source US is changed to 5u(t)V, with all other parameters held constant at their original values?

Question 3. What is the new response if the initial condition is changed to 5A, the voltage source US is changed to 5u(t)V, and the current source IS is changed to 8u(t)A?

This allows one to explore easily a circuits behavior over a wide range of excitations and initial conditions.

6. Response classifications

Zero-input response The zero-input response of a circuit is the response to the initial conditions when the input is set to zero.

Zero-state response The zero-state response of a circuit is the response to a specified input signal or set of input signals given that the initial conditions are all set to zero.

Complete response By linearity, the sum of the zero-input and zero-state responses is the complete response of the circuit.

Natural response The natural response is that portion of the complete response that has the same exponents as the zero-input response.

Forced response The forced response is that portion of the complete response that has the same exponents as the input excitation.

7. Further points of analysis and theory

Not only a capacitor voltage or an inductor current, it turns that any voltage or current in an RC or RL first-order linear circuit with constant input has the form

by linearity, any voltage or current in the circuit has the form

and

which implies that

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Example 8.9. For the circuit of the figure, let uin(t)=18u(t )+9u(t)V. Find iin(t) for t>0.

Solution

Step 1. Compute iin(0+).

0+ circuit

Step 2. Computeand iin().

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Step 3.Set forth the complete response iin(t).

Exercise. In example 7, find iC(0), iC(0+), and iC(t) for t>0.

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t = 0 K1t =0.2sK2