EE305 Final

Time Dependent Signal Sources

A= amplitude f= frequency = 1/T T= period w= radian frequency = 2*pi*f phi= phase

An ideal current source is disconnected from any circuit. The following statement is true.

An ideal current source has no internal resistance.

An ideal voltage source is disconnected from any circuit. The following statement is true.

An ideal voltage source has no internal resistance.

The admittance of a capacitor is given by

YC = jωC

The admittance of a capacitor and a resistor in parallel is given by

YC||R = jωC + 1/R

The admittance of an inductor in parallel with a capacitor is given by

YL||C = 1/jωL + jωC

The admittance of a resistor in parallel with an inductor is given by

YR||L = 1/R + 1/jωL

There is an inductor L between the nodes i and j and while no other element connects these two nodes. For the dummy version of node voltage analysis, what is Yi,j

Yi,j = -1/(jωL)

There is a capacitor C between the nodes i and j and while no other element connects these two nodes. For the dummy version of node voltage analysis, what is Yi,j

Yi,j = -jωC

The impedance of a capacitor is given by

ZC = 1/jωC

The impedance of an inductor is given by

ZL = jωL

Active vs Passive Elements

active elements generate power (voltage, current) passive elements dissipate power (resistors) *active elements counted negative *passive elements counted positive

If you set up an equation with Kirchhoff's Voltage Law (KVL) the convention is to count

an element voltage positive if you pass the element voltage from + to − and count the element voltage negative if you pass the element voltage from − to +.

A practical amperemeter is described by

an ideal amperemeter with an internal resistance in series.

A practical voltage source is described by

an ideal voltage source with an internal resistance in series.

A capacitor in a DC circuit can be replaced by

an open circuit

To zero out a current source it must be replaced by

an open circuit

To zero out an ideal current source, it must be replaced by

an open circuit

To compute the Thevenin Resistance

we have to set all independent voltage and current sources to zero.

A capacitor is a device

which stores electric energy.

We can find the Norton Current

with any number of voltage and current sources present in the original circuit.

Voltage

work per unit charge

Power

work/time P= IV = I^2R = V^2/R

Capacitors in Series/Parallel

-capacitors in series all have the same current -capacitors in parallel have the same voltage but different currents

Integrating a sinusoidal signal over a full period yields

0

The unit of relative permittivity is

1

Ideal Current Source

provides the prescribed current at its terminals regardless of the circuit connected

Ideal Voltage Source

provides the prescribed voltage at its terminals regardless of the circuit connected

To find the Norton Current, we have to

replace the load with a short circuit and compute the short circuit current.

Req

series resistors: Req= sum of R's series parallel: Req= R1R2/R1+R2

You have a circuit with four sources embedded. To find the total current through one of the resistors you use the principle of superposition. For this you take the following steps:

step 1: turn off all sources step 2: turn one source on step 3: find the requested current with the single source in the circuit step 4: turn the one source off again step 5: if you haven't turned on all sources yet, go back to step 2 and repeat with next source, otherwise proceed to step 6 step 6: add all solutions together to find the total current through the resistor

A band stop filter is designed to

stop all frequencies within a band of frequencies, w1< w<w 2

Ideal Inductor

stores magnetic energy

Complex Number to Complex Conjugate

switch the sign of the j term ex. 150 + j15 ----> 150 -j15

When using the concept of "virtual zero" we assume

that no currents enter or leave the input terminals and there is no voltage difference between the two input terminals.

To measure the voltage across a circuit element

the voltmeter must be put in parallel to this element, while the element is still in the circuit.

energy stored =

total power which is put into the device over time

A one-port network has

two terminals.

You have a circuit with two sources. To find the total current through one specific resistor using the principle of superposition you need to find the current through the resistor

two times and add up the results.

Elements in parallel have the same

voltage

To compute the Norton Current

we do not have to change anything about the sources

Sinusoidal Signals

x(t)= Acos(wt + phi)

Power Rating

*power dissipated by a resistor is converted to heat Available power ratings are: 1/8W 1/4W 1/2W 1W

Color Code

digit, digit, multiplier, tolerance

E-Series

group resistors based on tolerance E3-50% E6-20% E12-10% E24-5% E48-2% E96-1%

Which relation is correct?

1/(1+j)=0.5-j0.5

The period of a sinusoidal signal is

2π

The impedance of a resistor and an inductor in series is given by

ZRLseries = R+jωL

Impedance (Z)

Zc= 1/jwc ZL= jwl ZR= R units: Ohms

Loops

any closed connection of branches

Branch

any portion of the circuit with two terminals connected to it -can contain more than element

Given is a parallel plate capacitor. It has a capacitance C1. The material between the parallel plates has a relative permittivity of εr1=1.5. If we this material with a new material with a relative permittivity of εr2=6 the new capacitance C2 is given by

C2 = 4C1

Given is a parallel plate capacitor. It has a capacitance C1. If we double the distance between the plates the new capacitance C2 is given by

C2 =C1/2

Parallel Plate Capacitor

C= EA/d E (epsilon)= permittivity A= area of cross section d= distance between plates

We have two series capacitors, C1 and C2. The equivalent capacitance is

Ceq = (C1C2)/(C1 + C2)

We have two parallel capacitors, C1 and C2. The equivalent capacitance is

Ceq = C1 + C2

Units

Current (I) in Amperes Voltage (V) in Volts Resistance (R) in Ohms Power (P) in Watts Resistivity (Rho) in Ohms*meters Conductivity (sigma) in S/m Charge (q) in Coulombs Conductance (G) in Siemens

Relative permittivity

E= Er*E0

Constant Current

I = delta(q)/delta(t)

Ohmeter

Measures resistance -is connected to the two terminals of the resistor -must disconnect the resistor from the circuit to measure its individual resistance

Integrator vs Differentiator

Integrator has Zf = Cf and Zs = Rs Diff. has Zf = Rf and Zs = Cs

2 Element Current Divider

Iout= (G2/(G1+G2))*Iinput

Given is a long cylindrical coil. It has an inductance L1. If we reduce the cross section of the coil to half of the old one the new inductance L2 is given by

L2 = L1/2

Inductance for long cylindrical coil

L= uN^2A/l u= permeability of core N= number of windings A= cross sectional area l= length

We have two parallel inductors, L1 and L2. The equivalent inductance is

Leq = (L1L2)/(L1 + L2)

We have two parallel inductors L1 and L2. The equivalent inductance is given by

Leq = L1L2/(L1+L2)

Can we use the dummy version of mesh current analysis to find the unknown mesh currents?

No when there is an ideal current source in the circuit. Yes when there is NO ideal current source in the circuit

Short Circuit

R -> 0 V=0

Open Circuit

R -> infinity i=0

If we replace the element(s) in a branch with an open circuit, the following statement is true:

R → ∞ , i = 0 for any v.

Cylindrical Resistor

R= sigma(l)/A upside down sigma= resistivity l= length of resistor A= cross sectional area

Conductance

Reciprocal of resistance G=1/R unit: S (Siemens)

To find the Thevenin Voltage, we have to

Replace the load with an open circuit and compute the open circuit voltage.

Complex Power

S= P + jQ P= avg power [W] Q= reactive power [VAR] S= apparent power [VA]

A practical voltage source is disconnected from any circuit. The following statement is true.

There will be no current through the internal resistance. The disconnected voltage source has the same voltage at its terminals as the ideal voltage source it contains.

A practical voltage source is disconnected from any circuit. The following statement is true:

The disconnected voltage source has the same voltage at its terminals as the ideal voltage source it contains

A practical current source is disconnected from any circuit. The following statement is true.

There is a current flowing through the internal resistance and therefore power used at all times.

The Thevenin Resistance is the inverse of the Norton Conductance

True

Voltage Divider Rule

V1= (R1/Req)*Vs

Ohm's Law

V=IR

Complex power is measured in

VA

Reactive power is measured in

VAR

2 Element Voltage Divider

Vout= (R2/(R1+R2))*Vin

Average power is measured in

W

Mesh

a loop that does not contain other loops

An ideal capacitor is

a passive element

An ideal inductor is

a passive element

An inductor in a DC circuit can be replaced by

a short circuit

To zero out a voltage source it must be replaced by

a short circuit

To zero out an ideal voltage source, it must be replaced by

a short circuit

an amplifier circuit, that combines several inputs and produces an output that is the weighted sum of the inputs is called

a summing amplifier.

An operational amplifier is

an active element

A practical voltmeter is described by

an ideal voltmeter with an internal resistance in parallel.

Current

charge/time

A quantity that contains all the power information in a given load is the

complex power.

225VA∠0.62 The power factor is given by

cos(θ)= 0.81

Elements in series have the same

current

Current Sign Convention

current entering a node: negative current leaving a node: positive

For an ideal operational amplifier as given in the figure usually the following assumption is made:

i1 = 0 and i2 = 0

Admittance (Y)

inverse of impedance Yc= jwc YL= 1/jwl YR= 1/R

The two input terminals of an operational amplifier are labeled as:

inverting and noninverting.

Active Filters

inverting: A= -Zf/Zs non-inverting: A= 1 + Zf/Zs (Zf is element above op-amp, Zs is element to the left of op-amp near source)

Complex Numbers

j = SQRT (-1) z = a + jb = Acos(phi) + jAsin(phi) = Ae^(j*phi) a= Acosphi b= Asinphi A= IzI= SQRT (a^2 + b^2) phi= inverse tangent (b/a)

Node

junction of two or more branches

The base units in the SI system are

m, kg, s, A

IV characteristic

many materials have linear relationship between voltage and current for a constant temp.

Amperemeter

measures current -must be placed in series to the element whose current is to be measured

Voltmeter

measures voltage -placed in parallel to element you want to measure, element stays connected to circuit

Usually the feedback loop connects the output terminal with the negative input terminal. This is called

negative feedback.

An active filter consists

of an op-amp and the elements R,L, and C.

A low pass filter is designed to

only pass frequencies from DC (0Hz) up to the cutoff frequency wc.

Superposition

only works with linear circuits can be used to find voltage or current, cannot be used to find power

Current Divider Rule

parallel resistors: Geq= sum of G's and Req= 1/(sum of 1/R)

A band pass filter is designed to

pass all frequencies within a band of frequencies, w1< w<w2

Voltage Sign Convention

positive current flows out of the positive terminal, negative current flows out of the negative terminal voltage is positive when going from + to - and negative when going from - to +

To measure a current flowing through a circuit element

the amperemeter must be put in series to this element

An amplifier is called inverting when

the source is connected to the inverting terminal.