15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative

15: Transients (A)

15: Transients (A)

• Differential Equation

• Piecewise steady stateinputs

• Step Input

• Negative exponentials

• Exponential Time Delays

• Inductor Transients

• Linearity

• Transient Amplitude

• Capacitor VoltageContinuity

• Summary

E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 1 / 11

Differential Equation

15: Transients (A)



• Step Input




• Linearity



• Summary


To find y(t):


15: Transients (A)



• Step Input




• Linearity



• Summary


To find y(t):x(t) constant: Nodal analysis


15: Transients (A)



• Step Input




• Linearity



• Summary


To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysis


15: Transients (A)



• Step Input




• Linearity



• Summary


To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation


15: Transients (A)



• Step Input




• Linearity



• Summary



i(t) = C dydt


15: Transients (A)



• Step Input




• Linearity



• Summary



i(t) = C dydt =

x−yR


15: Transients (A)



• Step Input




• Linearity



• Summary



i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x


15: Transients (A)



• Step Input




• Linearity



• Summary



i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x

General Solution: Particular Integral + Complementary Function


15: Transients (A)



• Step Input




• Linearity



• Summary



i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x


Particular Integral: Any solution to RC dydt + y = x


15: Transients (A)



• Step Input




• Linearity



• Summary



i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x



If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution for y(t).


15: Transients (A)



• Step Input




• Linearity



• Summary



i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x




Complementary Function: Solution to RC dydt + y = 0


15: Transients (A)



• Step Input




• Linearity



• Summary



i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x





Does not depend on x(t), only on the circuit.


15: Transients (A)



• Step Input




• Linearity



• Summary



i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x





Does not depend on x(t), only on the circuit.Solution is y(t) = Ae−

t/τ

where τ = RC is the time constant of the circuit.


15: Transients (A)



• Step Input




• Linearity



• Summary



i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x





Does not depend on x(t), only on the circuit.Solution is y(t) = Ae−

t/τ

where τ = RC is the time constant of the circuit.

The amplitude, A, is determined by the initial conditions at t = 0.

Piecewise steady state inputs

15: Transients (A)



• Step Input




• Linearity



• Summary


We will consider input signals that are sinusoidal or constant for a particulartime interval and then suddenly change in amplitude, phase or frequency.


15: Transients (A)



• Step Input




• Linearity



• Summary



Output is the sum of the steady state and a transient:y(t) = ySS(t) + yTr(t)


15: Transients (A)



• Step Input




• Linearity



• Summary




Steady state, ySS(t), is the same frequency as theinput; use phasors + nodal analysis.


15: Transients (A)



• Step Input




• Linearity



• Summary





Transient is always yTr(t) = Ae−t

τ at each change.


15: Transients (A)



• Step Input




• Linearity



• Summary





Transient is always yTr(t) = Ae−t

τ at each change.

Step Input

15: Transients (A)



• Step Input




• Linearity



• Summary


For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy

dt + y = x= 4Time Const: τ = RC = 1ms

-1 0 1 2 3

-2

0

2

4

t (ms)

Step Input

15: Transients (A)



• Step Input




• Linearity



• Summary




Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0

-1 0 1 2 3

-2

0

2

4

t (ms)

-1 0 1 2 3

-2

0

2

4

t (ms)

ySS

Step Input

15: Transients (A)



• Step Input




• Linearity



• Summary





Transient (Complementary Function)yTr(t) = Ae−

t/τ

-1 0 1 2 3

-2

0

2

4

t (ms)

-1 0 1 2 3

-2

0

2

4

t (ms)

ySS

Step Input

15: Transients (A)



• Step Input




• Linearity



• Summary






t/τ

Steady State + Transienty(t) = ySS + yTr = 4 +Ae−

t/τ

-1 0 1 2 3

-2

0

2

4

t (ms)

-1 0 1 2 3

-2

0

2

4

t (ms)

ySS

Step Input

15: Transients (A)



• Step Input




• Linearity



• Summary






t/τ


t/τ

To find A, use capacitor property:Capacitor voltage never changes abruptly

-1 0 1 2 3

-2

0

2

4

t (ms)

-1 0 1 2 3

-2

0

2

4

t (ms)

ySS

Step Input

15: Transients (A)



• Step Input




• Linearity



• Summary






t/τ


t/τ


-1 0 1 2 3

-2

0

2

4

t (ms)

-1 0 1 2 3

-2

0

2

4

t (ms)

ySS

y(0+) = 4 +A and y(0−) = 1

Step Input

15: Transients (A)



• Step Input




• Linearity



• Summary






t/τ


t/τ


-1 0 1 2 3

-2

0

2

4

t (ms)

-1 0 1 2 3

-2

0

2

4

t (ms)

ySS

y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1

Step Input

15: Transients (A)



• Step Input




• Linearity



• Summary






t/τ


t/τ


-1 0 1 2 3

-2

0

2

4

t (ms)

-1 0 1 2 3

-2

0

2

4

t (ms)

ySS

y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3

Step Input

15: Transients (A)



• Step Input




• Linearity



• Summary






t/τ


t/τ


-1 0 1 2 3

-2

0

2

4

t (ms)

-1 0 1 2 3

-2

0

2

4

t (ms)

ySS

yTr

y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3

So transient: yTr(t) = −3e−t/τ

Step Input

15: Transients (A)



• Step Input




• Linearity



• Summary






t/τ


t/τ


-1 0 1 2 3

-2

0

2

4

t (ms)

-1 0 1 2 3

-2

0

2

4

t (ms)

ySS

yTr

y

y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3

So transient: yTr(t) = −3e−t/τ and total y(t) = 4− 3e−t/τ

Step Input

15: Transients (A)



• Step Input




• Linearity



• Summary






t/τ


t/τ


-1 0 1 2 3

-2

0

2

4

t (ms)

-1 0 1 2 3

-2

0

2

4

t (ms)

ySS

yTr

y

y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3

So transient: yTr(t) = −3e−t/τ and total y(t) = 4− 3e−t/τ

Transient amplitude ⇐ capacitor voltage continuity: vC(0+) = vC(0−)

Negative exponentials

15: Transients (A)



• Step Input




• Linearity



• Summary


Positive exponentials grow to ±∞:et

0 1 2 3 4 5

-10

0

10 et

t


15: Transients (A)



• Step Input




• Linearity



• Summary


Positive exponentials grow to ±∞:et, 3et/4

0 1 2 3 4 5

-10

0

10 et3e¼t

t


15: Transients (A)



• Step Input




• Linearity



• Summary


Positive exponentials grow to ±∞:et, 3et/4, −2et/2

0 1 2 3 4 5

-10

0

10 et3e¼t

-2e½t

t


15: Transients (A)



• Step Input




• Linearity



• Summary



0 1 2 3 4 5

-10

0

10 et3e¼t

-2e½t

t

Negative exponentials decay to 0:2e−t

0 2 4 6 8-2

0

22e-t

t


15: Transients (A)



• Step Input




• Linearity



• Summary



0 1 2 3 4 5

-10

0

10 et3e¼t

-2e½t

t

Negative exponentials decay to 0:2e−t, e

−t/4

0 2 4 6 8-2

0

22e-t

e-¼t

t


15: Transients (A)



• Step Input




• Linearity



• Summary



0 1 2 3 4 5

-10

0

10 et3e¼t

-2e½t

t


−t/4, −2e−t/2

0 2 4 6 8-2

0

22e-t

e-¼t

-2e-½t

t


15: Transients (A)



• Step Input




• Linearity



• Summary



0 1 2 3 4 5

-10

0

10 et3e¼t

-2e½t

t


−t/4, −2e−t/2

Transients are negative exponentials.

0 2 4 6 8-2

0

22e-t

e-¼t

-2e-½t

t


15: Transients (A)



• Step Input




• Linearity



• Summary



0 1 2 3 4 5

-10

0

10 et3e¼t

-2e½t

t


−t/4, −2e−t/2


0 2 4 6 8-2

0

22e-t

e-¼t

-2e-½t

t

Decay rate of e−t/a

a 2a 3a 4a 5a0

0.5

1

t


15: Transients (A)



• Step Input




• Linearity



• Summary



0 1 2 3 4 5

-10

0

10 et3e¼t

-2e½t

t


−t/4, −2e−t/2


0 2 4 6 8-2

0

22e-t

e-¼t

-2e-½t

t


37% after 1 time constant

a 2a 3a 4a 5a0

0.5

1

0.37

t


15: Transients (A)



• Step Input




• Linearity



• Summary



0 1 2 3 4 5

-10

0

10 et3e¼t

-2e½t

t


−t/4, −2e−t/2


0 2 4 6 8-2

0

22e-t

e-¼t

-2e-½t

t


37% after 1 time constant5% after 3

a 2a 3a 4a 5a0

0.5

1

0.37

0.05

t


15: Transients (A)



• Step Input




• Linearity



• Summary



0 1 2 3 4 5

-10

0

10 et3e¼t

-2e½t

t


−t/4, −2e−t/2


0 2 4 6 8-2

0

22e-t

e-¼t

-2e-½t

t


37% after 1 time constant5% after 3, <1% after 5

a 2a 3a 4a 5a0

0.5

1

0.37

0.05 0.01

t


15: Transients (A)



• Step Input




• Linearity



• Summary



0 1 2 3 4 5

-10

0

10 et3e¼t

-2e½t

t


−t/4, −2e−t/2


0 2 4 6 8-2

0

22e-t

e-¼t

-2e-½t

t



a 2a 3a 4a 5a0

0.5

1

0.37

0.05 0.01

t

Gradient of e−t/a

Gradient at t hits zero at t+ a.

a 2a 3a0

0.5

1

t


15: Transients (A)



• Step Input




• Linearity



• Summary



0 1 2 3 4 5

-10

0

10 et3e¼t

-2e½t

t


−t/4, −2e−t/2


0 2 4 6 8-2

0

22e-t

e-¼t

-2e-½t

t



a 2a 3a 4a 5a0

0.5

1

0.37

0.05 0.01

t

Gradient of e−t/a

Gradient at t hits zero at t+ a.True for any t.

a 2a 3a0

0.5

1

t

Exponential Time Delays

15: Transients (A)



• Step Input




• Linearity



• Summary


Negative exponential with a finalvalue of F .

y(t) = F + (A− F ) e−(t−TA)/τ


15: Transients (A)



• Step Input




• Linearity



• Summary



y(t) = F + (A− F ) e−(t−TA)/τ

How long does it take to go from A to B ?


15: Transients (A)



• Step Input




• Linearity



• Summary



y(t) = F + (A− F ) e−(t−TA)/τ


At t = TB :y(TB) = B = F + (A− F ) e−(TB−TA)/τ


15: Transients (A)



• Step Input




• Linearity



• Summary



y(t) = F + (A− F ) e−(t−TA)/τ



B−FA−F = e

−(TB−TA)/τ


15: Transients (A)



• Step Input




• Linearity



• Summary



y(t) = F + (A− F ) e−(t−TA)/τ



B−FA−F = e

−(TB−TA)/τ

Hence TB − TA = τ ln(

A−FB−F

)

= τ ln(

initial distance toFfinal distance toF

)


15: Transients (A)



• Step Input




• Linearity



• Summary



y(t) = F + (A− F ) e−(t−TA)/τ



B−FA−F = e

−(TB−TA)/τ

Hence TB − TA = τ ln(

A−FB−F

)

= τ ln(

initial distance toFfinal distance toF

)

Useful formula - worth remembering.

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt =

LR ×

d(x−y)dt

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt =

LR ×

d(x−y)dt = L

Rdxdt − L

Rdydt

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt =

LR ×

d(x−y)dt = L

Rdxdt − L

Rdydt

⇒ LR

dydt + y = L

Rdxdt

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt =

LR ×

d(x−y)dt = L

Rdxdt − L

Rdydt

⇒ LR

dydt + y = L

Rdxdt

Solution: Particular Integral + Complementary Function

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt =

LR ×

d(x−y)dt = L

Rdxdt − L

Rdydt

⇒ LR

dydt + y = L

Rdxdt


Particular Integral: Any solution to LR

dydt + y = L

Rdxdt

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt =

LR ×

d(x−y)dt = L

Rdxdt − L

Rdydt

⇒ LR

dydt + y = L

Rdxdt



dydt + y = L

Rdxdt

If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution, ySS(t).

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt =

LR ×

d(x−y)dt = L

Rdxdt − L

Rdydt

⇒ LR

dydt + y = L

Rdxdt



dydt + y = L

Rdxdt


Complementary Function: Solution to LR

dydt + y = 0

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt =

LR ×

d(x−y)dt = L

Rdxdt − L

Rdydt

⇒ LR

dydt + y = L

Rdxdt



dydt + y = L

Rdxdt



dydt + y = 0

Does not depend on x(t), only on the circuit.

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt =

LR ×

d(x−y)dt = L

Rdxdt − L

Rdydt

⇒ LR

dydt + y = L

Rdxdt



dydt + y = L

Rdxdt



dydt + y = 0

Does not depend on x(t), only on the circuit.Solution is yTr(t) = Ae−

t/τ

where τ = LR is the time constant of the circuit.

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt =

LR ×

d(x−y)dt = L

Rdxdt − L

Rdydt

⇒ LR

dydt + y = L

Rdxdt



dydt + y = L

Rdxdt



dydt + y = 0


t/τ


1st order transient is always yTr(t) = Ae−t/τ where τ = RC or L

R

Inductor Transients

15: Transients (A)



• Step Input




• Linearity



• Summary


We know i = x−yR

y(t) = L didt =

LR ×

d(x−y)dt = L

Rdxdt − L

Rdydt

⇒ LR

dydt + y = L

Rdxdt



dydt + y = L

Rdxdt



dydt + y = 0


t/τ


1st order transient is always yTr(t) = Ae−t/τ where τ = RC or L

RAmplitude A ⇐ no abrupt change in capacitor voltage or inductor current.

Linearity

15: Transients (A)



• Step Input




• Linearity



• Summary


1st order circuit has only one C or L.

Linearity

15: Transients (A)



• Step Input




• Linearity



• Summary


1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C .

Linearity

15: Transients (A)



• Step Input




• Linearity



• Summary


1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C . Rememberx is a voltage source but y is not.

Linearity

15: Transients (A)



• Step Input




• Linearity



• Summary



Linearity

15: Transients (A)



• Step Input




• Linearity



• Summary



Now v(t) = vSS(t) + vTr(t)= vSS(t) +Ae

−t/τ

Linearity

15: Transients (A)



• Step Input




• Linearity



• Summary




−t/τ

Time constant is τ = RThC

where RTh is the Thévenin resistance.

Linearity

15: Transients (A)



• Step Input




• Linearity



• Summary




−t/τ



Replace the capacitor with a voltage sourcev(t); all voltages and currents in the circuitwill remain unchanged.

Linearity

15: Transients (A)



• Step Input




• Linearity



• Summary




−t/τ




Linearity: y = ax+ bv = ax+ bvSS + bvTr

Linearity

15: Transients (A)



• Step Input




• Linearity



• Summary




−t/τ




Linearity: y = ax+ bv = ax+ bvSS + bvTr = ySS + bvTr

Linearity

15: Transients (A)



• Step Input




• Linearity



• Summary




−t/τ





All voltages and currents in a circuit have the same transient (but scaled).

Linearity

15: Transients (A)



• Step Input




• Linearity



• Summary




−t/τ





All voltages and currents in a circuit have the same transient (but scaled).

The circuit’s time constant is τ = RThC or LRTh

where RTh is theThévenin resistance of the network connected to C or L.

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary


Find Steady State (DC ⇒ ZL = 0)

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary


Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1

2x

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3

Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3


At t = 0+

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3


At t = 0+x− y = 1mA× 1 k = 1

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3


At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3


At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

Time Constant

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3


At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

Time ConstantSet x ≡ 0 → RTh = 2k

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3


At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

Time ConstantSet x ≡ 0 → RTh = 2kτ = L

RTh= 2µs

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3


At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5


RTh= 2µs

Result

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3


At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5


RTh= 2µs

Resulty = ySS + (y (0+)− ySS (0+)) e−t/τ

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3


At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5


RTh= 2µs


= 3 + (5− 3) e−t/τ

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3


At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5


RTh= 2µs


= 3 + (5− 3) e−t/τ

= 3 + 2e−t/τ

Transient Amplitude

15: Transients (A)



• Step Input




• Linearity



• Summary



2x

ySS(0−) = 1, ySS(0+) = 3


At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5


RTh= 2µs


= 3 + (5− 3) e−t/τ

= 3 + 2e−t/τ

Capacitor Voltage Continuity

15: Transients (A)



• Step Input




• Linearity



• Summary


Find Steady State (DC ⇒ ZC = ∞)

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t


15: Transients (A)



• Step Input




• Linearity



• Summary


Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x

4R + v8R + v−y

2R = 0

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t


15: Transients (A)



• Step Input




• Linearity



• Summary



4R + v8R + v−y

2R = 0

KCL @ Y: y−v2R + y−x

6R = 0

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t


15: Transients (A)



• Step Input




• Linearity



• Summary



4R + v8R + v−y

2R = 0


6R = 0

vSS = 34x, ySS = 13

16x

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

ySS


15: Transients (A)



• Step Input




• Linearity



• Summary



4R + v8R + v−y

2R = 0


6R = 0

vSS = 34x, ySS = 13

16x

Capacitor Voltage ContinuityvSS(0−) = −3 ⇒ v(0+) = −3

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

ySS


15: Transients (A)



• Step Input




• Linearity



• Summary



4R + v8R + v−y

2R = 0


6R = 0

vSS = 34x, ySS = 13

16x


At t = 0+: x = 4 and v = −3

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

ySS


15: Transients (A)



• Step Input




• Linearity



• Summary



4R + v8R + v−y

2R = 0


6R = 0

vSS = 34x, ySS = 13

16x


At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0 -RC 0 RC 2RC 3RC

-4

-2

0

2

4

t

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

ySS


15: Transients (A)



• Step Input




• Linearity



• Summary



4R + v8R + v−y

2R = 0


6R = 0

vSS = 34x, ySS = 13

16x


At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

4

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

ySS


15: Transients (A)



• Step Input




• Linearity



• Summary



4R + v8R + v−y

2R = 0


6R = 0

vSS = 34x, ySS = 13

16x


At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

4

Time Constantτ = RThC = 2RC (from earlier slide)

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

ySS


15: Transients (A)



• Step Input




• Linearity



• Summary



4R + v8R + v−y

2R = 0


6R = 0

vSS = 34x, ySS = 13

16x


At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

4



-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

ySS


15: Transients (A)



• Step Input




• Linearity



• Summary



4R + v8R + v−y

2R = 0


6R = 0

vSS = 34x, ySS = 13

16x


At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

4



= 134 +

(

− 54 − 13

4

)

e−t/τ

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

ySS


15: Transients (A)



• Step Input




• Linearity



• Summary



4R + v8R + v−y

2R = 0


6R = 0

vSS = 34x, ySS = 13

16x


At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

4



= 134 +

(

− 54 − 13

4

)

e−t/τ

= 134 − 18

4 e−t/τ = 3 1

4 − 4 12e

−t/2RC

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

yTr

ySS


15: Transients (A)



• Step Input




• Linearity



• Summary



4R + v8R + v−y

2R = 0


6R = 0

vSS = 34x, ySS = 13

16x


At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

4



= 134 +

(

− 54 − 13

4

)

e−t/τ

= 134 − 18

4 e−t/τ = 3 1

4 − 4 12e

−t/2RC

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

-RC 0 RC 2RC 3RC-4

-2

0

2

4

t

yTr

ySS

y

Summary

15: Transients (A)



• Step Input




• Linearity



• Summary


• 1st order circuits: include one C or one L.◦ vC or iL never change abruptly. The output, y, is not necessarily

continuous unless it equals vC .

Summary

15: Transients (A)



• Step Input




• Linearity



• Summary




• Circuit time constant: τ = RThC or LRTh

◦ RTh is the Thévenin resistance seen by C or L.◦ Same τ for all voltages and currents.

Summary

15: Transients (A)



• Step Input




• Linearity



• Summary






• Output = Steady State + Transient◦ Steady State: use nodal/Phasor analysis when input is piecewise

constant or piecewise sinusoidal. The steady state has the samefrequency as the input signal.

Summary

15: Transients (A)



• Step Input




• Linearity



• Summary








◦ Transient: Find vC(0−) or iL(0−): unchanged at t = 0+Find y(0+) assuming source of vC(0+) or iL(0+)Amplitude never complex, never depends on t.

Summary

15: Transients (A)



• Step Input




• Linearity



• Summary









◦ y(t) = ySS(t) + (y(0+)− ySS(0+)) e−t/τ

Summary

15: Transients (A)



• Step Input




• Linearity



• Summary









◦ y(t) = ySS(t) + (y(0+)− ySS(0+)) e−t/τ

See Hayt Ch 8 or Irwin Ch 7.

15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative

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