chapter 1

ECE 3205: Circuits and Systems II. 1–1

REVIEW OF CIRCUITS AND SYSTEMS

Complex Numbers

Cartesian Representation

A general complex number can be represented as:

z D x C jy

where x is the real part and y is the imaginary part. The imaginarynumber j is defined by:

j Dp

!1

Complex numbers in this form are easily represented in a rectangular, orcartesian, coordinate frame.

Polar Representation

Equivalently, we can express complex numbers in polar form, using theradial magnitude and rotation angle instead of rectangular coordinates:

z D r†!

We can relate polar to rectangular coordinates through the followingexpressions:

r Dp

x2 C y2

! D tan!1!y

x

"C

8̂<:̂

" if x < 0 and y " 00 if x " 0

!" if x < 0 and y < 0

Lecture notes prepared by M. Scott Trimboli. Copyright c# 2012, M. Scott Trimboli

ECE3205, 1–2

The complex conjugate of a complex number is defined by:

z$ D x ! jy

D r† ! !

Complex Exponentials

A complex exponential is defined by:z D r % ej!

where, using Euler’s formula we can equivalently write:z D r % .cos ! C j sin !/

From this expression we have:

Rfzg D r cos.!/

Ifzg D r sin.!/


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The complex conjugate of the complex exponential is written:

z$ D r % e!j!

Note we can obtain the inverse expressions:

cos.!/ D 1

2.ej! C e!j!/

sin.!/ D 1

2j.ej! ! e!j!/

Some Useful Properties of Complex Numbers

Rfzg D 1

2.z C z$/

Ifzg D 1

2j.z ! z$/

z % z$ D kzk2


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Example 1.1

Express the following complex numbers in Cartesian form:

1. z1 D jej.9"=4/

2. z2 D!p

3 ! j"

2p

2e!j.3"=4/


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Example 1.2

Express the following complex numbers in polar form:

1. z3 D .1 C j /.1 ! j 2/

2. z4 D j.2 C j /

.1 C j /.2 ! j /


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Example 1.3

Use the definition of the complex exponential or Euler’s relations todemonstrate the following trigonometric identities:

1. cos.2!/ D cos2.!/ ! sin2.!/

2. cos.!/ % cos.#/ D 1

2cos.! ! #/ C 1

2cos.! C #/


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Circuits

Let’s review the voltage-current (V-I) characteristics of primary circuitelements.

Resistor:

Vr.t/ D Rir.t/

Capacitor:

ic.t/ D Cdvc.t/

dt

vc.t/ D 1

C

Z t

0

ic.$/d$ C vc.0/

Inductor:

vl.t/ D Ldi`.t/

dt

What happens when we put these elements together?

Kirchoff’s Current Law:Xk2N

ik.t/ D 0

where N is the set of currents entering or leaving a node.


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Kirchoff’s Voltage Law:Xk2M

vk.t/ D 0

where M is the set of voltages around a closed path.

How do we solve circuit equations? There are a number of methodsavailable; we will highlight two: the node method and the mesh method.

Node Method

1. Create a supernode encircling each voltage source and the twonodes to which it is attached

2. Select one of the nodes of the circuit as the ground node.

3. Define n ! 1 node potential variables at the remaining nodes of thecircuit.

4. Set up KCL equations at n ! 1 of the nodes in the network. Thecurrents in these equations must be expressed in terms of the nodepotentials.

5. Solve the n ! 1 equations for the n ! 1 node potentials.

6. Calculate the element voltages and currents of interest from the nodepotentials.


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Example 1.4

For the circuit shown below,

use the Node Method to find the following quantities: e1.t/, e2.t/, i1.t/,i2.t/.


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Mesh Method

1. Create a supermesh from any pair of adjacent meshes that areseparated by a current source.

2. Define a mesh current for each mesh that does not contain a currentsource.

3. Write l KVL equations, each of which is written oer one of the pathsover which the mesh currents have been defined. the voltages inthese equations must be expressed in terms of the mesh currents.

4. Solve those l equations for the l mesh currents.

5. Compute the element currents and voltages of interest from the meshcurrents.


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Example 1.5


use the Mesh method to find the following quantities: e1.t/, e2.t/, i1.t/,i2.t/.


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Example 1.6


use the Mesh method to generate the system of equations needed tofind the currents, i1.t/ and i2.t/.


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Resistor-Only Subnetworks

In Series:

In Parallel:

Voltage Divider:

Current Divider:


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Example 1.7

Find the equivalent resistance for the following circuit:


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Example 1.8

Find the equivalent resistance for the following circuit:


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Thevenin Equivalent Circuits

How do we compute the Thevenin voltage and resistance?

1. Find the open-circuit voltage, voc.t/ (this is the Thevenin sourcevoltage).

2. Find the short-circuit current, isc.t/ .

3. The Thevenin equivalent resistance is then given by:

RT D !voc.t/

isc.t/


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Example 1.9

Find the Thevenin equivalent of the following two-terminal network.


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Norton Equivalent Circuits

Example 1.10

Find the Norton equivalent of the following two-terminal network.


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Laplace Transforms

Definition: [single-sided Laplace Transform]

L fx.t/g D X.s/ DZ 1

0

x.t/e!stdt

The Laplace transform allows us to transition between the time domainand the complex (frequency) domain.

Example 1.11

Find the Laplace transform of

x.t/ D e!2tu.t/ C e!t cos.3t/u.t/

Example 1.12

Find the Laplace transform of

x.t/ D ı.t ! 1/ C ı.t/ C e!2.tC3/u.t ! 1/


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Transfer Functions

Definition:

For a dynamic system described by linear, ordinary, constantcoefficient differential equations, a transfer function is the ratio ofthe Laplace transform of the input to the Laplace transform of theoutput with all initial conditions set to zero.

We will derive transfer functions from differential equations later. For themoment, we simply note that a transfer function describes theinput-output relationship for a system and is expressed as a ratio ofpolynomials:

H.s/ D B.s/

A.s/D bmsm C bm!1s

m!1 C % % % C b1s C b0

sn C an!1sn!1 C % % % C a1s C a0

The roots of the numerator B.s/ D 0 are called the zeros of H.s/.

The roots of the denominator A.s/ D 0 are called the poles of H.s/.

Inverse Laplace Transform

The inverse Laplace transform provides a way to convert Laplacetransformed expressions back into the time domain.

Definition:

x.t/ D 1

2"j

Z cCj 1

c!j 1X.s/estdt

In practice, we rarely use this integral to find the inverse transform.Instead, we’ll rely on algebraic techniques and tables of commonLaplace transform pairs.


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Example 1.13

Find the inverse Laplace transform of

H.s/ D s C 2

s2 C 4s C 3


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Example 1.14

Find the inverse Laplace transform of

H.s/ D s C 2

.s C 1/2.s C 3/


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Circuit Analysis in the Laplace Domain

Laplace transforms make circuit analysis easy by replacing often difficultdifferential equations with simple algebraic expressions.We’ll begin this section by developing the Laplace transforms for the v-icharacteristic of fundamental circuit elements.

Resistor:vr.t/ D ir.t/R ) Vr.s/ D Ir.s/ % R

Capacitor:

ic.t/ D Cdvc.t/

dt) C ŒsVc.s/ ! vc.0/%

) 1

C sIc.s/ C 1

svc.0/

Inductor:

v`.t/ D Ldi`.t/

dt) V`.s/ D L ŒsI`.s/ ! i`.0/%

For the case where we have zero initial conditions, these relationshipsreduce to:

Vr.s/ D Ir.s/ % R

Vc.s/ D Ic.s/ % 1

C s

V`.s/ D I`.s/ % Ls

And when we connect the elements together in circuits !


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Kirchoff’s Current Law:

Xk2N

ik.t/ D 0 )Xk2N

Ik.s/ D 0

Kirchoff’s Voltage Law:

Xk2M

vk.t/ D 0 )Xk2M

Vk.s/ D 0

Impedance

The impedance, Z.s/; of a fundamental circuit element can be thought ofas the transfer function relationship between the voltage output and thecurrent input:

Z.s/ D V.s/

I.s/

Working with impedences allows us to use algebraic expressions tosimplify circuit analysis.

Series Connection:

Parallel Connection:


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Voltage Divider:

Current Divider:


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Example 1.15

Find the Laplace domain expression for the voltage across the capacitor,Vc.s/. Assume that the capacitor has an initial charge of vc.0/.


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Example 1.16

Find the Laplace domain expression for the voltage across the capacitor,Vc.s/. Assume that the circuit is initially at rest.


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Example 1.17

Find the Laplace domain expression for the current through the inductor,IL.s/. Assume that both the capacitor and inductor have initial conditions.


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