4: Linearity and Superposition - Imperial College London Theorem 4: Linearity and Superposition •Linearity Theorem •Zero-value sources •Superposition •Superposition Calculation

4: Linearity and Superposition

4: Linearity andSuperposition

• Linearity Theorem

• Zero-value sources

• Superposition

• Superposition Calculation

• Superposition anddependent sources

• Single Variable Source

• Superposition and Power

• Proportionality

• Summary

E1.1 Analysis of Circuits (2018-10340) Linearity and Superposition: 4 – 1 / 10

Linearity Theorem




• Superposition





• Proportionality

• Summary


Suppose we use variables instead of fixed values for all of the independentvoltage and current sources. We can then use nodal analysis to find allnode voltages in terms of the source values.

Linearity Theorem




• Superposition





• Proportionality

• Summary



(1) Label all the nodes

Linearity Theorem




• Superposition





• Proportionality

• Summary



(1) Label all the nodes(2) KCL equations

X−U1

2 + X

1 + X−Y

3 = 0Y−X

3 + (−U2) = 0

Linearity Theorem




• Superposition





• Proportionality

• Summary




X−U1

2 + X

1 + X−Y

3 = 0Y−X

3 + (−U2) = 0(3) Solve for the node voltages

X = 13U1 +

23U2, Y = 1

3U1 +113 U2

Linearity Theorem




• Superposition





• Proportionality

• Summary




X−U1

2 + X

1 + X−Y

3 = 0Y−X


X = 13U1 +

23U2, Y = 1

3U1 +113 U2

Steps (2) and (3) never involve multiplying two source values together, so:

Linearity Theorem




• Superposition





• Proportionality

• Summary




X−U1

2 + X

1 + X−Y

3 = 0Y−X


X = 13U1 +

23U2, Y = 1

3U1 +113 U2


Linearity Theorem: For any circuit containing resistors and independentvoltage and current sources, every node voltage and branch current is alinear function of the source values and has the form

∑aiUi where the Ui

are the source values and the ai are suitably dimensioned constants.

Linearity Theorem




• Superposition





• Proportionality

• Summary




X−U1

2 + X

1 + X−Y

3 = 0Y−X


X = 13U1 +

23U2, Y = 1

3U1 +113 U2


Linearity Theorem: For any circuit containing resistors and independentvoltage and current sources, every node voltage and branch current is alinear function of the source values and has the form

∑aiUi where the Ui

are the source values and the ai are suitably dimensioned constants.

Also true for a circuit containing dependent sources whose values areproportional to voltages or currents elsewhere in the circuit.

Zero-value sources




• Superposition





• Proportionality

• Summary


A zero-valued voltage source has zero voltsbetween its terminals for any current. It isequivalent to a short-circuit or piece of wireor resistor of 0 Ω (or ∞ S).

Zero-value sources




• Superposition





• Proportionality

• Summary



Zero-value sources




• Superposition





• Proportionality

• Summary



A zero-valued current source has no currentflowing between its terminals. It is equivalentto an open-circuit or a broken wire or aresistor of ∞ Ω (or 0 S).

Zero-value sources




• Superposition





• Proportionality

• Summary



A zero-valued current source has no currentflowing between its terminals. It is equivalentto an open-circuit or a broken wire or aresistor of ∞ Ω (or 0 S).

Superposition




• Superposition





• Proportionality

• Summary


We can use nodal analysis to find X in terms of U , V and W .

Superposition




• Superposition





• Proportionality

• Summary



KCL: X−U

2 + X−V

6 + X

1 −W = 0

Superposition




• Superposition





• Proportionality

• Summary



KCL: X−U

2 + X−V

6 + X

1 −W = 0

10X − 3U − V − 6W = 0

Superposition




• Superposition





• Proportionality

• Summary



KCL: X−U

2 + X−V

6 + X

1 −W = 0

10X − 3U − V − 6W = 0

X = 0.3U + 0.1V + 0.6W

Superposition




• Superposition





• Proportionality

• Summary



KCL: X−U

2 + X−V

6 + X

1 −W = 0

10X − 3U − V − 6W = 0

X = 0.3U + 0.1V + 0.6W

From the linearity theorem, we know anyway that X = aU + bV + cW soall we need to do is find the values of a, b and c.

Superposition




• Superposition





• Proportionality

• Summary



KCL: X−U

2 + X−V

6 + X

1 −W = 0

10X − 3U − V − 6W = 0

X = 0.3U + 0.1V + 0.6W

From the linearity theorem, we know anyway that X = aU + bV + cW soall we need to do is find the values of a, b and c. We find each coefficient inturn by setting all the other sources to zero:

Superposition




• Superposition





• Proportionality

• Summary



KCL: X−U

2 + X−V

6 + X

1 −W = 0

10X − 3U − V − 6W = 0

X = 0.3U + 0.1V + 0.6W


We have XU = aU + b× 0 + c× 0 = aU .

Superposition




• Superposition





• Proportionality

• Summary



KCL: X−U

2 + X−V

6 + X

1 −W = 0

10X − 3U − V − 6W = 0

X = 0.3U + 0.1V + 0.6W


We have XU = aU + b× 0 + c× 0 = aU .

Superposition




• Superposition





• Proportionality

• Summary



KCL: X−U

2 + X−V

6 + X

1 −W = 0

10X − 3U − V − 6W = 0

X = 0.3U + 0.1V + 0.6W


We have XU = aU + b× 0 + c× 0 = aU .

Superposition




• Superposition





• Proportionality

• Summary



KCL: X−U

2 + X−V

6 + X

1 −W = 0

10X − 3U − V − 6W = 0

X = 0.3U + 0.1V + 0.6W


We have XU = aU + b× 0 + c× 0 = aU .Similarly, XV = bV and XW = cW ⇒ X = XU +XV +XW .

Superposition Calculation




• Superposition





• Proportionality

• Summary


Superposition:

Find the effect of each source on its ownby setting all other sources to zero. Thenadd up the results.





• Superposition





• Proportionality

• Summary


Superposition:


XU =6

7

2+ 6

7

U = 620U = 0.3U





• Superposition





• Proportionality

• Summary


Superposition:


XU =6

7

2+ 6

7

U = 620U = 0.3U

XV =2

3

6+ 2

3

V = 220V = 0.1V





• Superposition





• Proportionality

• Summary


Superposition:


XU =6

7

2+ 6

7

U = 620U = 0.3U

XV =2

3

6+ 2

3

V = 220V = 0.1V

XW = 66+ 2

3

W × 23 = 12

20W = 0.6W





• Superposition





• Proportionality

• Summary


Superposition:


XU =6

7

2+ 6

7

U = 620U = 0.3U

XV =2

3

6+ 2

3

V = 220V = 0.1V

XW = 66+ 2

3

W × 23 = 12

20W = 0.6W

Adding them up: X = XU +XV +XW





• Superposition





• Proportionality

• Summary


Superposition:


XU =6

7

2+ 6

7

U = 620U = 0.3U

XV =2

3

6+ 2

3

V = 220V = 0.1V

XW = 66+ 2

3

W × 23 = 12

20W = 0.6W

Adding them up: X = XU +XV +XW = 0.3U + 0.1V + 0.6W

Superposition and dependent sources




• Superposition





• Proportionality

• Summary


A dependent source is one that is determined by the voltage and/or currentelsewhere in the circuit via a known equation. Here V , Y −X .





• Superposition





• Proportionality

• Summary



Step 1: Pretend all sources are independentand use superposition to find expressions forthe node voltages:





• Superposition





• Proportionality

• Summary




X = 103 U1

Y = 2U1





• Superposition





• Proportionality

• Summary




X = 103 U1 + 2U2

Y = 2U1 + 6U2





• Superposition





• Proportionality

• Summary




X = 103 U1 + 2U2 +

16V

Y = 2U1 + 6U2 +12V





• Superposition





• Proportionality

• Summary




X = 103 U1 + 2U2 +

16V

Y = 2U1 + 6U2 +12V

Step 2: Express the dependent source values in terms of node voltages:V = Y −X





• Superposition





• Proportionality

• Summary




X = 103 U1 + 2U2 +

16V

Y = 2U1 + 6U2 +12V


Step 3: Eliminate the dependent source values from the node voltageequations:X = 10

3 U1 + 2U2 +16 (Y −X)

Y = 2U1 + 6U2 +12 (Y −X))





• Superposition





• Proportionality

• Summary




X = 103 U1 + 2U2 +

16V

Y = 2U1 + 6U2 +12V



3 U1 + 2U2 +16 (Y −X) ⇒ 7

6X − 16Y = 10

3 U1 + 2U2

Y = 2U1 + 6U2 +12 (Y −X))





• Superposition





• Proportionality

• Summary




X = 103 U1 + 2U2 +

16V

Y = 2U1 + 6U2 +12V



3 U1 + 2U2 +16 (Y −X) ⇒ 7

6X − 16Y = 10

3 U1 + 2U2

Y = 2U1 + 6U2 +12 (Y −X)) ⇒ 1

2X + 12Y = 2U1 + 6U2





• Superposition





• Proportionality

• Summary




X = 103 U1 + 2U2 +

16V

Y = 2U1 + 6U2 +12V



3 U1 + 2U2 +16 (Y −X) ⇒ 7

6X − 16Y = 10

3 U1 + 2U2

Y = 2U1 + 6U2 +12 (Y −X)) ⇒ 1

2X + 12Y = 2U1 + 6U2

X = 3U1 + 3U2

Y = U1 + 9U2





• Superposition





• Proportionality

• Summary




X = 103 U1 + 2U2 +

16V

Y = 2U1 + 6U2 +12V



3 U1 + 2U2 +16 (Y −X) ⇒ 7

6X − 16Y = 10

3 U1 + 2U2

Y = 2U1 + 6U2 +12 (Y −X)) ⇒ 1

2X + 12Y = 2U1 + 6U2

X = 3U1 + 3U2

Y = U1 + 9U2

Note: This is an alternative to nodal anlysis: you get the same answer.

Single Variable Source




• Superposition





• Proportionality

• Summary


Any current or voltage can be written X = a1U1 + a2U2 + a3U3 + . . ..





• Superposition





• Proportionality

• Summary



Using nodal analysis (slide 4-2) or elsesuperposition:

X = 13U1 +

23U2.





• Superposition





• Proportionality

• Summary




X = 13U1 +

23U2.

Suppose we know U2 = 6 mA





• Superposition





• Proportionality

• Summary




X = 13U1 +

23U2.

Suppose we know U2 = 6 mA, then

X = 13U1 +

23U2 = 1

3U1 + 4.





• Superposition





• Proportionality

• Summary




X = 13U1 +

23U2.


X = 13U1 +

23U2 = 1

3U1 + 4.

If all the independent sources except for U1

have known fixed values, then

X = a1U1 + b

where b = a2U2 + a3U3 + . . . .





• Superposition





• Proportionality

• Summary




X = 13U1 +

23U2.


X = 13U1 +

23U2 = 1

3U1 + 4.

If all the independent sources except for U1

have known fixed values, then

X = a1U1 + b

where b = a2U2 + a3U3 + . . . .

This has a straight line graph.

Superposition and Power




• Superposition





• Proportionality

• Summary


The power absorbed (or dissipated) by a component always equals V I

where the measurement directions of V and I follow the passive signconvention.

For a resistor V I = V2

R= I2R.





• Superposition





• Proportionality

• Summary





R= I2R.

Power in resistor is P = (U1+U2)2

10 = 6.4W





• Superposition





• Proportionality

• Summary





R= I2R.


10 = 6.4W

Power due to U1 alone is P1 =U

2

1

10 = 0.9W





• Superposition





• Proportionality

• Summary





R= I2R.


10 = 6.4W


2

1

10 = 0.9W


2

2

10 = 2.5W





• Superposition





• Proportionality

• Summary





R= I2R.


10 = 6.4W


2

1

10 = 0.9W


2

2

10 = 2.5W

P 6= P1 + P2 ⇒ Power does not obey superposition.





• Superposition





• Proportionality

• Summary





R= I2R.


10 = 6.4W


2

1

10 = 0.9W


2

2

10 = 2.5W

P 6= P1 + P2 ⇒ Power does not obey superposition.

You must use superposition to calculate the total V and/or the total I andthen calculate the power.

Proportionality




• Superposition





• Proportionality

• Summary


From the linearity theorem, all voltages and currents have the form∑

aiUi

where the Ui are the values of the independent sources.

If you multiply all the independent sources by the same factor, k, then allvoltages and currents in the circuit will be multiplied by k.

The power dissipated in any component will be multiplied by k2.

Proportionality




• Superposition





• Proportionality

• Summary


From the linearity theorem, all voltages and currents have the form∑

aiUi

where the Ui are the values of the independent sources.

If you multiply all the independent sources by the same factor, k, then allvoltages and currents in the circuit will be multiplied by k.

The power dissipated in any component will be multiplied by k2.

Special Case:If there is only one independent source, U , then all voltages and currentsare proportional to U and all power dissipations are proportional to U2.

Summary




• Superposition





• Proportionality

• Summary


• Linearity Theorem: X =∑

iaiUi over all independent sources Ui

Summary




• Superposition





• Proportionality

• Summary




• Superposition: sometimes simpler than nodal analysis, often moreinsight. Zero-value voltage and current sources Dependent sources - treat as independent and add dependency

as an extra equation

Summary




• Superposition





• Proportionality

• Summary






• If all sources are fixed except for U1 then all voltages and currents inthe circuit have the form aU1 + b.

Summary




• Superposition





• Proportionality

• Summary







• Power does not obey superposition.

Summary




• Superposition





• Proportionality

• Summary








• Proportionality: multiplying all sources by k multiplies all voltages andcurrents by k and all powers by k2.

Summary




• Superposition





• Proportionality

• Summary








• Proportionality: multiplying all sources by k multiplies all voltages andcurrents by k and all powers by k2.

For further details see Hayt Ch 5 or Irwin Ch 5.

4: Linearity and Superposition - Imperial College London Theorem 4: Linearity and Superposition •Linearity Theorem •Zero-value sources •Superposition •Superposition Calculation

Documents