Impedance - Engineering Electrical Circuits - Lecture Slides, Slides of Electrical Circuit Analysis

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Impedance

KCL & KVL

Review → V-I in Phasor Space

• Resistors

 Inductors

 Capacitors

R

V

I =

ω L

V

I

I = ω C V ∠ 90 °

No Phase Shift

i(t) LAGS

i(t) LEADS

Impedance cont.

• Since V & I are COMPLEX, Then Z is also Complex

 However, Z IS a COMPLEX NUMBER that can be written in polar or Cartesian form.

• In general, its value DOES depend on the  Impedance is NOT a Sinusoidal frequency Phasor
• It’s Magnitude and Phase Do Not Change regardless of the Location within The Circuit

v i z M

M M i

M v Z I

V I

V θ θ θ θ

θ (^) = ∠ − = ∠ ∠

= = ∠ ( ) | | I

Z V

( ) REACTive component

RESISTivecomponent

( ) ( )

• =
• =

= +

ω

ω ω

X

R

Z R jX

• Note that the REACTANCE, X , is a function of ω

Impedance cont.

• Thus

 The Magnitude and Phase

Z ∠θ z = R + jX

R

X

Z R X

z

1

2 2

=tan−

= +

θ

 Where

z

z

X Z

R Z

sin

cos

j C

Z

Z j L

j C

j L

C

L

R R Z R

1 1

=

=

= =

V I

V I

V I

Element PhasorEq. Impedance

 Summary Of Passive- Element Impedance

 Examine ZC

C

j jj C

j j C

Z (^) C ω ω 1 ω

1 −

= = =

C

X C

Z (^) C j C ω ω

1 − 1 ⇒ =

− ∴ =

Series & Parallel Impedances

• Impedances (which have units of Ω) Combine as do RESISTANCES - The SERIES Case - The Parallel Case

I +^ V^1 − Z 1

• V 2 − Z 2

I Z (^) s = Z 1 + Z 2

Zs = (^) ∑ k Zk

Z 1 Z 2 −

V

I I

−

V 1 2

1 2 Z Z

Z ZZ p =^ + =^ ∑ k Zp Zk

1 1

Admittance

• The Frequency Domain Analog of CONDUCTANCE is ADMITTANCE - Admittance is Thus Inverse Impedance

 Multiply Denominator by the Complex Conjugate

 Find G & B In terms of Resistance, R, and Reactance, X

(Siemens)

G jB

Z

Y = = +

Z R jX

Y

2 2

2 2

2 2

R X

X

B

R X

R

G

R X

R jX

R jX

R jX

R jX

Y

 Note that G & R and X & B are NOT Reciprocals

• G ≡ CONDUCTance
• B ≡ SUSCEPTance

Complex Numbers in MATLAB

• MATLAB recognizes

complex numbers these in these forms

• Rectangular
• Exponential
• Can Use “i” or “j” for √(-

• MATLAB Always returns

“i” for √(-1)

• Sometimes need “*”

>> phiR = 23*pi/180 % 23deg in Rads phiR =

>> Z1 = 7 + i*23 % if i or j BEFORE, then need * Z1 = 7.0000 +23.0000i >> Z2 = 11 - 13j Z2 = 11.0000 -13.0000i

>> Z3 = 43exp(jphiR) % Need * Z3 = 39.5817 +16.8014i >> Z4 = 37*exp(0.61j) Z4 = 30.3270 +21.1961i

Phasors in MATLAB

• MATLAB Does NOT

Recognize Phasor NOTATION

• But it DOES handle Complex Exponentials
• e.g.:





53 17

29 43

8

7

= − ∠

= ∠ −

Z

Z

>> phi7 = -43pi/ phi7 = -0. >> phi8 = 17pi/ phi8 = 0. >> Z7 = 29exp(jphi7)** Z7 = 21.2093 -19.7780i >> Z8 = -53exp(jphi8)** Z8 = -50.6842 -15.4957i >> Zsum = Z7 + Z Zsum = -29.4749 -35.2737i >> Zdif = Z7 - Z Zdif = 71.8934 - 4.2823i >> Zprod = Z7*Z Zprod = -1.3814e+003 +6.7378e+002i >> Zquo = Z7/Z Zquo = -0.2736 + 0.4739i

MATLAB: a+jb ↔ A∟φ

• BMayer MATLAB Functions function Zrectd = Rectab(Mag, phi_deg) % B. Mayer 22Apr09 * ENGR % finds for POLAR COMPLEX number Z the Rectangular Equivalet %% note that phi is in DEGREES % a = Magcosd(phi_deg); b = Magsind(phi_deg); Zrectd = a + j*b

function Phasor = MagPh(Zr) % B. Mayer 22Apr09 * ENGR % finds for RECTANGULAR COMPLEX number Z %% Magnitude %% Phase Angle in DEGREES Magnitude = abs(Zr); Phase_deg = angle(Zr)*180/pi; Phasor = [Magnitude, Phase_deg];

 Example >> Z1r = 13 - 19j Z1r = 13.0000 -19.0000i >> Phasor1 = MagPh(Z1r) Phasor1 = 23.0217 -55. >> Phasor2 = [43 -127] Phasor2 = 43 - >> Zr2 = Rectab(Phasor2(1), Phasor2(2)) Zr2 = -25.8780 -34.3413i

MATLAB Equivalent Functions

 Rectangular to Polar  Polar to Rectangular

 Both use RADIANS only

Vector Addition

• Parallelogram Rule

For Vector Addition

• Examine Top & Bottom of

The Parallelogram

• Triangle Rule For Vector Addition
• Vector Addition is Commutative
• Vector Subtraction → Reverse Direction of The Subtrahend

B

B

C

C

P − Q = P + ( − Q )

P + Q = Q + P

Example  Phasor Diagram

• For The Ckt at Right, Draw the Phasor Diagrams as a function of Frequency
• First Write KCL

 That is, we Can Select ONE Phasor to have a ZERO Phase Angle

• In this Case Choose V  Next Examine Frequency Sensitivity of the Admittances

 Now we can Select ANY Phasor Quantity, I or V , as the BaseLine

= ∑ ⇒



 

 = + +

= + +

S k k

S

S

Y

j C R j L

j C R j L

Admittance s

1 1

I V

I V

I V V V

ω ω

ω ω

Example  Phasor Diagram cont.

• Case-I: ω=Med so That
  • YL ≈ YC

 Case-III: ω=Hi so That

• YC ≈ 2YL  The Circuit is Basically

 Case-II: ω=Low so That CAPACITIVE

• YL ≈ 2YC

 The Circuit is Basically INDUCTIVE

I (^) C = j ω C V

L j ω L I =^ V

| I (^) L |>| I C |

| I (^) L |<| I C |

I C + I L ≈ 0

∴ I (^) S ≈ I R

KCL & KVL for AC Analysis

• Simple-Circuit Analysis
  • AC Version of Ohm’s Law → V = IZ
  • Rules for Combining Z and/or Y
  • KCL & KVL
  • Current and/or Voltage Dividers
• More Complex Circuits
  • Nodal Analysis
  • Loop or Mesh Analysis
  • SuperPosition