Sinusoidal Steady-State

responseresponse

EE3301

Kamran Kiasaleh

Learning Objectives

1. Be able to obtain the steady-state response of RLC circuits (in all forms) to a sinusoidal input

2. Be able to represent currents and voltages in “Phasor” format

3. Be able to obtain circuit impedance and admittance. 3. Be able to obtain circuit impedance and admittance.

4. Be able to obtain Thevenin and Norton equivalent circuits for steady-state sinusoidal circuits

5. Be able to write mesh, node, KVL, and KCL equations for sinusoidal steady-state circuit

6. Be able to conduct steady-state sinusoidal analysis of circuits with transformers

A sinusoidal voltage

What are the key parameters of a sinusoid

T;period(sec); f0 =1

T; frequency(Hz)

ω 0 = 2πf

v t( ) =Vm cos ω 0t +θ( )( ) m 0( )

Vrms =1

Tv t( )

0

T∫2

dt = lim T → ∞

1

Tv t( )

0

T∫2

dt

P =Vrms

2

R;

Vrms =Vm

2

Only for sinusoidal

signals

Phase shift (does not change the

frequency)-moves signal in time!

What is the rms of a triangular wave?

We have to average i2

Irms =IP

3

How do we assess the response of circuits to a sinusoidal signal (direct substitution)

1. First, write the differential equation that relates the desired output to the input

2. Ignore all initial conditions (this includes switches, etc.). All initial energies (initial conditions) are assumed to have dissipated in the resistive part of the circuit. circuit.

3. Assume that the response (in this case, the particular response) is also sinusoidal with different amplitude and phase, but the same frequency (linear circuit)

4. Plug in the proposed response in the differential equation and solve for the unknown amplitude and phase

An RL circuit excited by a sinusoidal voltage

source.

Differential Equation

1. Using KVL,

Ldi

dt+ Ri =V =Vm cos ωt + φ( )

This will disappear

Ldt

+ Ri =V =Vm cos ωt + φ( )

i t( ) = Im cos ωt + φ −θ( )− Im cos φ −θ( )e−R

Lt

i 0+( )= 0 = i 0−( )iss t( ) = Im cos ωt + φ −θ( )

Steady State Response

1. Using KVL,

Ldi

dt+ Ri =V =Vm cos ωt + φ( )

i t( ) = Im cos ωt + φ −θ( )m

−LImω sin ωt + φ −θ( )+ RIm cos ωt + φ −θ( )=

Vm cos ωt + φ( )RIm sin θ( )− LImω cos θ( ){ }sin ωt + φ( )+

RIm cos θ( )+ LImω sin θ( ){ }cos ωt + φ( )=

Vm cos ωt + φ( )

Steady State Response

RIm sin θ( )− LImω cos θ( ){ }sin ωt + φ( )+

RIm cos θ( )+ LImω sin θ( ){ }cos ωt + φ( )=

V cos ωt + φ( )

Is there an

easier and

more

intuitive way

to get this?

Vm cos ωt + φ( )

RIm sin θ( )− LImω cos θ( ) = 0⇒ θ = tan−1 ωLR

RIm cos θ( )+ LImω sin θ( ) =Vm ⇒ Im =Vm

ω 2L2 + R2

Phasors

1. Phasors are actually vector representation of sinusoidal signals

2. They suppress the element of time (if you know phase and amplitude, you can reconstruct the signal assuming a known frequency)

3. The length of the vector is the amplitude of the signal 3. The length of the vector is the amplitude of the signal (fixed) and the direction of the phasor at t=0 is the phase of the sinusoid

4. We can combine (add and subtract) phasors using vector addition

5. Sinusoidal signals may be related to phasors by observing the projection of the vector onto x and y axis

Asin θ t( )( )

Acos θ t( )( )

A

θ t( ) = ωt + φ

θ t( )

Acos θ t( )( )

e± jθ = cos θ( )± j sin θ( )Ae± jθ = Acos θ( )± jAsin θ( )Ae± j ωt+φ( ) = Acos ωt + φ( )± jAsin ωt + φ( )

How to add two phasors (22.43

degrees is the phase difference)

Can we use phasors to represent current and voltages of a passive devices?

1. If we add (subtract) two sinusoidal signals,

the resulting phasors add (subtract)

2. KVL and KCL still applies for sinusoidal signals

3. This implies that we can apply KVL and KCL 3. This implies that we can apply KVL and KCL

for circuits using phasors assuming that we

have access to the relationship between

current and voltage of all components

4. Let us consider passive components

Resistors

v = Ri

V sin ωt + φ( )= RIsin ωt + φ( )⇒V = RI

V = RI

Figure 9.9 A plot showing that the

voltage and current at the terminals of

a resistor are in phase.

Inductor

( )sin

div L

dt

i I tω φ

=

= +( )( )

( )cos

sin 90

v I L t

v V t

VV j LI j L

I

ω ω φ

ω φ

ω ω

= +

= + +

= ⇒ =

Inductor response

Capacitor

i = Cdv

dt

v =V sin ωt + φ( )i =VωCcos ωt + φ( )i =VωCcos ωt + φ( )i =VωC sin ωt + φ + 90( )i = I sin ωt + φ( )

I =VjωC ⇒V

I=1

jωC

Capacitor response

What do previous observations imply

1. Capacitors, inductor, and resistors may be looked as having “Impedance”

2. Impedance of a resistor is real and is called resistance resistance

3. The impedance of a capacitors or an inductor is purely imaginary (is called Reactance)

4. The reactance is frequency dependent

5. Given KVL and KCL, we can treat R, L, and C as we would treat a simple resistors through

V = ZI

Z = impedance = R + jXReactance

KVL

Vab = Z1 + Z2 + ...+ Zn( )IVab = ZI

Z = Z1 + Z2 + ...+ Zn

Example 9.6.

The circuit at the 800 Hz

KCL

Vab = Z1I1 = Z2I2 = ...= ZnIn

I = I1 + I2 + ...+ In

Vab = ZI

Z = Z1 || Z2 || ... || Zn

A parallel circuit

Previous circuit at ω=200,000

We can use ∆∆∆∆-Y transformation

Example

Simplified circuit

source transformation

Thevenin equivalent

Norton equivalent

Example

Use of Thevenin to solve circuit

Compute output voltage (no

load)

Thevenin Impedance

Final circuit

Writing Node Equation

Mesh Current

Linear TransformersMutual Inductance

Vs = R1 + jωL1 + ZS( )I1 − jωMI2

0 = − jωMI1 + R2 + jωL2 + ZL( )I2

Transformer as a 2-port networkZ11 = Zs + R1 + jωL1Z22 = ZL + R2 + jωL2Zab = R1 + jωL1 + Zr2

Zcd = R2 + jωL2 + Zr1

=ω 2M 2

Self-impedance primary

Self-impedance secondary

Zr2 =ω 2M 2

R2 + RL + j ωL2 + XL( )

Zr1 =ω 2M 2

R1 + Rs + j ωL1 + X s( )

k1 =Zr2

Z22*

=ω 2M 2

| Z22 |2

k2 =Zr1

Z11*

=ω 2M 2

| Z11 |2

Reflected impedance

Scaling factor

How do you model an ideal

transformer?

M

Coefficient of coupling

k =M

L1L2

L1

L2=

N1

N2

2

Ideal transformer

k =1

L1 →∞

L →∞

V1

V2=N1

N2

I1 =N2

No power

loss

L2 →∞

L1

L2=

N1

N2

2

1

I2= 2

N1

V1I1 =V2I2

Zin =N1N2

2

ZL

Ideal Transformer

Input/Output relationship

Can we use phasors to solve a

circuit?

The complex number -7 – j3 = 7.62

-156.80°.

Example

The phasor diagram

Example

Impedance model

Phasor diagram for the circuit

addition of a capacitor

One more phasor

Final picture