Webnotes Lecture 10 Sinusoidal Steady State 2013

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Sinusoidal Steady StateSinusoidsPhasors

ImpedanceExamples

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Solve with matrix of differential equations, leading to a governing equation with is a third order diff eq

/Solve for the voltage across R1 for sinusoidal input V.

+V

R1

R2C1 C2

L

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Solve for the voltage across R1 for sinusoidal input V.

+V

Z1

Z5Z2 Z4

Z3

If V=costThenV1= V Z1/(Z1+Z2||(Z3+Z4||Z5) = A cos(t+)

Why Steady State?

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Steady state occurs after the natural response to the initial conditions has disappeared and the forced response has taken over…

We will now study this class of circuits, and in particular, the case where the forcing function is a pure sinusoid.

This is an important class of circuits!

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Why are sinusoids important?Sinsusoids are useful functions because ac power has this form (Tesla, AC

induction motor, transformers, high V transmission for low loss). Also, sinusoids form the basis of any waveform: any arbitrary function can always be represented perfectly by a sum of sinusoids of different frequencies and amplitudes. This is the “Fourier Transform.”

In order to mathematically manipulate sinusoids, it’s sometimes more convenient to first convert them to complex exponentials, which are easy to handle, and then convert back to sinusoids later. To do this, we use “Euler’s Identity.”

To make our calculations even simpler, we can even drop the time dependence in the exponential. We can then replace the complex exponential with a kind of vector notation that only depends on amplitude and phase angle. These vectors are called “Phasors.”

We can then easily apply all the techniques we’ve already developed for solving resistive circuits!

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Euler’s Identity

Define G(x) = cos(x) + j*sin(x) where j = (-1)1/2

Then dG = -sin(x)dx + j*cos(x)dx

Or dG/dx = j*[cos(x) + j*sin(x)]

dG/dx = j*G

dG/G = j*dx

Integrating both sides, ln(G) = j*x

G = e jx

e jx = cos(x) + j*sin(x)

This is a useful relation that allows us to convert sinusoidal functions to complex exponentials. The sinusoids, like exponentials, result in the same function when differentiated.

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Brief review of complex numbersDefinition of a complex number:“Ordered pairs (a,b) with +, defined.”

Rectangular form: z = a + jb where j = (-1)1/2

a = Re{z}b = Im{z}

Polar form:|z| = (a2 + b2)1/2 = r “ magnitude of z”a = r cos

b = r sin tan -1 (b/a) for a>0 = - tan -1 (b/a) a<0

Sub’ing in rect. form: z = r (cos j sin )From Euler: = r e j

Complex conjugate: z*= a - jb = r e -j

r

z*

.RectangularForm

PolarForm

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Solve circuit with cmplx exponentials

To simplify the math, we can replace cosines with complex exponentials, and then take the real part of the solution when we’re finished. This is the way it works:

1. Replace all the cosine functions by e jt

For sine functions, recall that sin(x) = cos(x - 90o) = cos(x - /2 )

and replace all sine functions by e j(t - /2 )

2. Solve the circuit using complex exponentials.

3. When the solution is obtained, get the cosines back

using Euler’s identity:

Re{e jt } = cos(t)

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Expressing the guessed forced response using exponentialsExpress the sinusoidal input in exponential form:

V(t) =V0sin(t) write in cosine form using /2 phase shift

=V0cos(t - /2) use Euler’s identity

= Re { V0e j( t - /2 ) } pull out the constant

= V0 * Re {e j( t - /2) } = V0 * Re {e j t e- j/2 }

Express the guessed forced response in exponential form:

VFR(t) = Acos(t) + Bsin(t) write in cosine form using trig identity

= (A2+B2) * cos(t - ) where, = tan -1 (B/A) .= C*Re {e j( t - ) } where C= (A2+B2)

= C*Re {e j te - j} response is also a complex exponential

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Forced response from a sinusoid

VS(t) = 5sin(30t + /3) Guess: VFR(t) = Acos(30t + /3) + Bsin(30t + /3)

dVFR(t)/dt+ 5VFR(t) = 25sin(30t + /3) Now substitute for VFR(t)

Cumbersome! -30Asin(30t + /3) + 30Bcos(30t + /3) + 5Acos(30t + /3) + 5Bsin(30t + /3) = 25sin(30t + /3)

Equate coeff’s of cos & sin: -30A + 5B = 25 and 5A + 30B = 0A = -30/37 B = 5/37

VC(t) = (-30/37)cos(30t + /3) + (5/37)sin(30t + /3)

VC(t) = 0.8220cos(30t - 1.93) using trig identity

KVL: R[C(dVC(t)/dt)]+ VC(t) = VS(t)

or dVC(t)/dt + (1/RC)*VC(t) = (1/RC)VS(t)

VFR(t) = (A2+B2) cos(t - )

where = tan-1 (B/A)

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Forced response using cmplx exponentialsVS(t) = 5sin(30t + /3) Solve for VC(t)

In exponential form: VS(t) = Re{ 5 e j(30t - /6) } (note - /2 phase change from sin to cos)

KVL equation :

dVC(t)/dt + 5VC(t) = 5VS(t)Our guess will be: VC(t) = Ae j(30t - /6) Plug this in and solve for A:j30A e j(30t - /6) + 5A e j(30t - /6) = 25 e j(30t - /6) solving, A = 25 / (5 + 30j)

= 25 / [ (52 + 302)e j ] where = tan -1 (30/5) = 1.41 rad

And thus the complete answer is:

VC(t) = Re{ 0.8220*e j(30t - /6 - 1.41) }= 0.8220*cos(30t - 1.93)

(1st Quadrant)

Now convert to polar form

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Summary of the methodBecause the natural response has completely disappeared under steady state conditions, we only concern ourselves with the forced response.Natural response is transient, which dies out in time t= 5*(tau) at which time transient will equal about 0.0067 in amplitude.

The circuit has reached a ‘steady state’ when the transient has died away.

Write a sinusoidal input in cosine form: Vin = V0cos(t + ). Note: This does not apply to an arbitrary input as we had with the general forced response.

Use Euler’s equation to write the exponential form of that input: Vin = Re {V0e j(t + ) }

Guess the solution to be of the form VFR= Ae j(t + ) Then:

1.Plug VFR in gov. equation and solve for A

2. Convert to polar form

3. Combine phases

4. Find real part.

Example of finding the real part of the solution:

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Ex: Say the guessed solution is

VFR= Ae j(t + )

Then we solve for A by putting VFR in the governing equation and find

A = 10 e 5j

Then VFR = Re { 10 e 5je j(t + ) }

= Re { 10e j(t + + 5) } =10 cos(t + + 5)

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Simplify the symbols: Phasor notationA useful notation which provides further simplification in the final steps.

V(t) = V0sin(t) = V0cos(t - /2) write in exponential form

= Re{V0e j(t - /2) }

Now drop the notation “Re” and use V(t) = V0e j(t - /2) = V0e - j/2e jt

But the factor e jt gets carried through the calculations unchanged; let’s drop it too!

So V(t) = V0e - j/2

where the frequency factor and real part are understood, but not shown.

Or, in phasor notation V = V0(-/2) Phasors will be denoted in bold.

Example: V / I = (A ) / (B ) = (A / B) ( - )

Use phasors at the end of the calculation….

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RL example with phasorsFind the current across the inductor in steady state if

VS = 10cos(50t)

= Re { 10e j50t}

Assumed solution: i(t) =Ae j(50t + ).

Governing equation: L si + Ri = VS or si + (R/L)i = (1/L) 10e j50t

Substituting for i: (j50 + 10)Ae j(50t + ) = (10)10e j50t

I = 10*10 0 / [(2600) 1.3734 ] = 1.9612 -1.3734

Now reinsert frequency factor and take real part: i(t) =Re{1.9612 e j(50t -1.3734 )} = 1.9612 cos (50t - 1.3734)

Phasor form: I = Ae j =A and VS = 10e j(0) =100

Rewrite using the phasor notation: ( j50 + 10) I = 10*VS

I = 10*VS / (10 + 50j)

(Boldface=phasors)

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Resistive ImpedanceFor a sinusoidal input voltage, use exponential form v = Ve j(t + )

Current will be of the form i=Ie j(t + )

From Ohm’s Law

Ve j(t + ) = R*Ie j(t + )

Ve j = R*Ie j so = resistor does not store energy or change phase

V=R I (phasors-bold face) For a resistor, current & voltage are in phase.

Impedance is defined as Z = V / I (phasors representing sinusoidal values)

For a resistor, the impedance Z = V/I (phasors) = R

The impedance of a resistor to a sinusoidal input is R. The resistance is purely lossy.

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Inductive ImpedanceFor an inductor, v= L*di/dt

Let i = Ie j(t + )

v = Ve j(t + )

Vej(t + ) = L*j*I ej(t + )

Vej = L*j*I ej

or V = jL I (phasors)

since j = e j for an inductor, voltage leads current by

Z = V/I = jL

in response to a sinusoidal input.

All of the analysis techniques we developed for resistive circuits apply to impedances.

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Capacitive ImpedanceFor a capacitor, i= C*dv/dt

Let i = Ie j(t + )

v = Ve j(t + )

Iej(t + ) = C*j*Vej(t + )

Iej = C*j*Vej phase of current is shifted compared to voltage or I = jCV (phasors)

since j = e j for a capacitor, current leads voltage by

Z = V/I = 1 / (jC)

in response to a sinusoidal input.

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Go to RL impedance example

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Go to RLC example, impedance

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The Advantage of SimplicityUsing impedance we do not need the differential equations.

Use Impedance to find I(t) when

VS(t) = 10cos(3t).

I = VS/(1 + Z).Z is the series combination of the resistor and inductor, in parallel with the capacitor.

Z = (3 + j) // [ 1 / j(1/9)] = (3 + 3j)(-3j) / [ (3 + 3j) + (-3j) ] = 3 - 3j

I(t) = VS(t) / (4 - 3j) = VS(t) / (5e j ) where = tan -1 (-3/4) = -36.9°

OR: I(t) = (10/5)cos(3t + 36.9° */180)

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Dependent Source ExampleThis example shows how to use impedance with normal circuit techniques. Node and Mesh analysis should be usable after this. Solve for VX(t). Redraw the circuit.

KCL: I1 + I2 + I3 = 10cos(20t).

Express the currents in terms of VX(t).

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Dependent Source Example:ba) VX = 30*I1 I1 = (1/30)VX

b) jL*I2 + 2VX = VX

or I2 = (1/10)*j*VX

c) VX = 1/ (jC)*I3

or I3 = (1/5)*j*VX

I1 + I2 + I3 = 10cos(20 t) = [(1/30) + j(9/30)] *VX(t)

OR: VX(t) = { 10 / [ (1/30) + j(9/30) ] } cos(20 t)

= 94.87cos(20 t – tan-19) volts

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Op-Amp ExampleFind VOUT / VS (as phasors) for the given circuit.

v - = R2 / (R2 + R3) *VOUT

OR: VOUT = (1 + R3 / R2)* v -

v - = v+

v+ = {R1 / [ R1 + (1/ jC) ] }* VS OR: v+ =[ jR1C / (1 + jR1C) ]* VS

VOUT = (1 + R3 / R2) * v+ = [jR1C( 1 + R3 / R2) / ( 1 + jR1C)]* VS

OR: VOUT / VS = jR1C( 1 + R3 / R2) / ( 1 + jR1C)

Then plug in component values, convert to polar form, write in cosine form.