revision lecture 4: transients & lines · e1.1 analysis of circuits (2014-4649) revision...

Revision Lecture 4: Transients &Lines

Revision Lecture 4:Transients & Lines

• Transients: Basic Ideas

• Steady States

• Determining TimeConstant• Determining TransientAmplitude

• Transmission Lines Basics

• Reflections

• Sinewaves and Phasors

• Standing Waves

E1.1 Analysis of Circuits (2014-4649) Revision Lecture 4 – 1 / 9

Transients: Basic Ideas



• Steady States



• Reflections


• Standing Waves


• Transients happen in response to a sudden change

◦ Input voltage/current abruptly changes its magnitude,frequency or phase

◦ A switch alters the circuit




• Steady States



• Reflections


• Standing Waves





• 1st order circuits only: one capacitor/inductor




• Steady States



• Reflections


• Standing Waves






• All voltage/current waveforms are: Steady State + Transient

◦ Steady States: find with nodal analysis or transfer function

• Note: Steady State is not the same as DC Level• Need steady states before and after the sudden change




• Steady States



• Reflections


• Standing Waves









◦ Transient: Always a negative exponential: Ae−tτ




• Steady States



• Reflections


• Standing Waves










• Time Constant: τ = RC or LR where R is the Thévenin

resistance at the terminals of C or L




• Steady States



• Reflections


• Standing Waves











resistance at the terminals of C or L• Find transient amplitude, A, from continuity since VC orIL cannot change instantly.




• Steady States



• Reflections


• Standing Waves











resistance at the terminals of C or L• Find transient amplitude, A, from continuity since VC orIL cannot change instantly.

• τ and A can also be found from the transfer function.

Steady States



• Steady States



• Reflections


• Standing Waves


A steady-state output assumes the input frequency, phase and amplitudeare constant forever. You need to determine two ySS(t) steady stateoutputs: one for before the transient (t < 0) and one after (t ≥ 0).

-5 0 5 10 15 200

2

4

6

t (µs)

Steady States



• Steady States



• Reflections


• Standing Waves


A steady-state output assumes the input frequency, phase and amplitudeare constant forever. You need to determine two ySS(t) steady stateoutputs: one for before the transient (t < 0) and one after (t ≥ 0).At t = 0, ySS(0−) means the first one and ySS(0+) means the second.

-5 0 5 10 15 200

2

4

6

t (µs)

Steady States



• Steady States



• Reflections


• Standing Waves



Method 1: Nodal analysisInput voltage is DC (ω = 0)⇒ ZL = 0 (for capacitor: ZC =∞)So L acts as a short citcuit

-5 0 5 10 15 200

2

4

6

t (µs)

Steady States



• Steady States



• Reflections


• Standing Waves




Potential divider: ySS = 12x

-5 0 5 10 15 200

2

4

6

t (µs)

Steady States



• Steady States



• Reflections


• Standing Waves





ySS(0−) = 1, ySS(0+) = 3

-5 0 5 10 15 200

2

4

6

t (µs)

Steady States



• Steady States



• Reflections


• Standing Waves





ySS(0−) = 1, ySS(0+) = 3

Method 2: Transfer functionYX (jω) = R+jωL

2R+jωL

-5 0 5 10 15 200

2

4

6

t (µs)

Steady States



• Steady States



• Reflections


• Standing Waves





ySS(0−) = 1, ySS(0+) = 3


2R+jωL

set ω = 0: YX (0) = 1

2

-5 0 5 10 15 200

2

4

6

t (µs)

Steady States



• Steady States



• Reflections


• Standing Waves





ySS(0−) = 1, ySS(0+) = 3


2R+jωL

set ω = 0: YX (0) = 1

2ySS(0−) = 1, ySS(0+) = 3

-5 0 5 10 15 200

2

4

6

t (µs)

Steady States



• Steady States



• Reflections


• Standing Waves





ySS(0−) = 1, ySS(0+) = 3


2R+jωL

set ω = 0: YX (0) = 1

2ySS(0−) = 1, ySS(0+) = 3

-5 0 5 10 15 200

2

4

6

t (µs)

Sinusoidal input⇒ Sinusoidal steady state⇒ use phasors.Then convert phasors to time waveforms to calculate the actual outputvoltages ySS(0−) and ySS(0+) at t = 0.

Determining Time Constant



• Steady States



• Reflections


• Standing Waves


Method 1: Thévenin




• Steady States



• Reflections


• Standing Waves


Method 1: Thévenin(a) Remove the capacitor/inductor




• Steady States



• Reflections


• Standing Waves


Method 1: Thévenin(a) Remove the capacitor/inductor(b) Set all sources to zero (including the inputvoltage source). Leave output unconnected.




• Steady States



• Reflections


• Standing Waves


Method 1: Thévenin(a) Remove the capacitor/inductor(b) Set all sources to zero (including the inputvoltage source). Leave output unconnected.(c) Calculate the Thévenin resistancebetween the capacitor/inductor terminals:RTh = 8R||4R||(6R + 2R) = 2R




• Steady States



• Reflections


• Standing Waves



(d) Time constant: = RThC or LRTh

τ = RThC = 2RC




• Steady States



• Reflections


• Standing Waves




τ = RThC = 2RC

Method 2: Transfer function




• Steady States



• Reflections


• Standing Waves




τ = RThC = 2RC

Method 2: Transfer function(a) Calculate transfer function using nodal analysis




• Steady States



• Reflections


• Standing Waves




τ = RThC = 2RC


KCL @ V: V−X4R + V

8R + jωCV + V−Y2R = 0

KCL @ Y: Y−V2R + Y−X

6R = 0




• Steady States



• Reflections


• Standing Waves




τ = RThC = 2RC



8R + jωCV + V−Y2R = 0


6R = 0

→ Eliminate V to get transfer Function: YX = 8jωRC+13

32jωRC+16




• Steady States



• Reflections


• Standing Waves




τ = RThC = 2RC



8R + jωCV + V−Y2R = 0


6R = 0


32jωRC+16

(b) Time Constant = 1Denominator corner frequency




• Steady States



• Reflections


• Standing Waves




τ = RThC = 2RC



8R + jωCV + V−Y2R = 0


6R = 0


32jωRC+16

(b) Time Constant = 1Denominator corner frequency

ωd = 1632RC ⇒ τ = 1

ωd= 2RC

Determining Transient Amplitude



• Steady States



• Reflections


• Standing Waves


After an input change at t = 0, y(t) = ySS(t) +Ae−tτ .

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves



⇒ y(0+) = ySS(0+) +A⇒ A = y(0+)− ySS(0+)

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves



⇒ y(0+) = ySS(0+) +A⇒ A = y(0+)− ySS(0+)Method: (a) calculate true output y(0+), (b) subtract ySS(0+) to get A

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves




(i) Version 1: vC or iL continuity

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves




(i) Version 1: vC or iL continuityx(0−) = 2

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves




(i) Version 1: vC or iL continuityx(0−) = 2⇒ iL(0−) = 1mA

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves




(i) Version 1: vC or iL continuityx(0−) = 2⇒ iL(0−) = 1mAContinuity⇒ iL(0+) = iL(0−)

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves




(i) Version 1: vC or iL continuityx(0−) = 2⇒ iL(0−) = 1mAContinuity⇒ iL(0+) = iL(0−)Replace L with a 1mA current source

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves




(i) Version 1: vC or iL continuityx(0−) = 2⇒ iL(0−) = 1mAContinuity⇒ iL(0+) = iL(0−)Replace L with a 1mA current sourcey(0+) = x(0+)− iR = 6− 1 = 5

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves





(i) Version 2: Transfer function

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves





(i) Version 2: Transfer functionH(jω) = Y

X (jω) = R+jωL2R+jωL

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves







Input step, ∆x = x(0+)− x(0−) = +4-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves







Input step, ∆x = x(0+)− x(0−) = +4y(0+) = y(0−) +H(j∞)×∆x

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves








= 1 + 1× 4 = 5

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

ySS




• Steady States



• Reflections


• Standing Waves








= 1 + 1× 4 = 5

(ii) A = y(0+)− ySS(0+) = 5− 3 = 2

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

yTr

ySS




• Steady States



• Reflections


• Standing Waves








= 1 + 1× 4 = 5

(ii) A = y(0+)− ySS(0+) = 5− 3 = 2

(iii) y(t) = ySS(t) +Ae−t/τ

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

yTr

ySS




• Steady States



• Reflections


• Standing Waves








= 1 + 1× 4 = 5

(ii) A = y(0+)− ySS(0+) = 5− 3 = 2

(iii) y(t) = ySS(t) +Ae−t/τ

= 3 + 2e−t/2µ

-5 0 5 10 15 200

2

4

6

t (µs)

-5 0 5 10 15 200

2

4

6

t (µs)

yTr

ySS

y

Transmission Lines Basics



• Steady States



• Reflections


• Standing Waves


Transmission Line: constant L0 and C0 : inductance/capacitance permetre.Forward wave travels along the line: fx(t) = f0

(

t− xu

)

.

Velocity u =√

1L0C0

≈ 12c = 15 cm/ns




• Steady States



• Reflections


• Standing Waves



(

t− xu

)

.

Velocity u =√

1L0C0

≈ 12c = 15 cm/ns

fx(t) equals f0 (t)but delayed by x

u .




• Steady States



• Reflections


• Standing Waves



(

t− xu

)

.

Velocity u =√

1L0C0

≈ 12c = 15 cm/ns


u .

0 2 4 6 8 10Time (ns)

f(t-0/u)




• Steady States



• Reflections


• Standing Waves



(

t− xu

)

.

Velocity u =√

1L0C0

≈ 12c = 15 cm/ns


u .

0 2 4 6 8 10Time (ns)

f(t-0/u) f(t-45/u)




• Steady States



• Reflections


• Standing Waves



(

t− xu

)

.

Velocity u =√

1L0C0

≈ 12c = 15 cm/ns


u .

Knowing fx(t) forx = x0 fixes it for allother x.

0 2 4 6 8 10Time (ns)

f(t-0/u) f(t-45/u) f(t-90/u)




• Steady States



• Reflections


• Standing Waves



(

t− xu

)

.

Velocity u =√

1L0C0

≈ 12c = 15 cm/ns


u .


0 2 4 6 8 10Time (ns)

f(t-0/u) f(t-45/u) f(t-90/u)

Backward wave: gx(t) is the same but travelling←: gx(t) = g0(

t+ xu

)

.




• Steady States



• Reflections


• Standing Waves



(

t− xu

)

.

Velocity u =√

1L0C0

≈ 12c = 15 cm/ns


u .


0 2 4 6 8 10Time (ns)

f(t-0/u) f(t-45/u) f(t-90/u)


t+ xu

)

.

Voltage and current are: vx = fx + gx and ix = fx−gxZ0

where ix is

positive in the +x direction (→) and Z0 =√

L0

C0




• Steady States



• Reflections


• Standing Waves



(

t− xu

)

.

Velocity u =√

1L0C0

≈ 12c = 15 cm/ns


u .


0 2 4 6 8 10Time (ns)

f(t-0/u) f(t-45/u) f(t-90/u)


t+ xu

)

.

Voltage and current are: vx = fx + gx and ix = fx−gxZ0

where ix is

positive in the +x direction (→) and Z0 =√

L0

C0

Waveforms of fx and gx are determined by the connections at both ends.

Reflections



• Steady States



• Reflections


• Standing Waves


vx = fx + gxix = fx−gx

Z0

Reflections



• Steady States



• Reflections


• Standing Waves



Z0

At x = L, Ohm’s law⇒ vL(t)iL(t) = RL ⇒ gL (t) = RL−Z0

RL+Z0

× fL (t).

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).

Reflection coefficient: ρL = gL(t)fL(t) =

RL−Z0

RL+Z0

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0

ρL ∈ [−1, +1] and increases with RL

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0


Knowing fx(t) for x = x0 now tells you fx, gx, vx, ix ∀x

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0



At x = 0: f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t)

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0



At x = 0: f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t) = τ0vS(t) + ρ0g0(t)

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0



At x = 0: f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t) = τ0vS(t) + ρ0g0(t)

Wave bounces back and forth getting smaller with each reflection:

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0



At x = 0: f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t) = τ0vS(t) + ρ0g0(t)


vS(t)×τ0−→ f0(t)

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0



At x = 0: f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t) = τ0vS(t) + ρ0g0(t)


vS(t)×τ0−→ f0(t)

×ρL−→ g0(t+

2Lu )

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0



At x = 0: f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t) = τ0vS(t) + ρ0g0(t)


vS(t)×τ0−→ f0(t)

×ρL−→ g0(t+

2Lu )

×ρ0

−→ f0(t+2Lu )

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0



At x = 0: f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t) = τ0vS(t) + ρ0g0(t)


vS(t)×τ0−→ f0(t)

×ρL−→ g0(t+

2Lu )

×ρ0

−→ f0(t+2Lu )

×ρL−→ g0(t+

4Lu )

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0



At x = 0: f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t) = τ0vS(t) + ρ0g0(t)


vS(t)×τ0−→ f0(t)

×ρL−→ g0(t+

2Lu )

×ρ0

−→ f0(t+2Lu )

×ρL−→ g0(t+

4Lu )

×ρ0

−→ · · ·

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0



At x = 0: f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t) = τ0vS(t) + ρ0g0(t)


vS(t)×τ0−→ f0(t)

×ρL−→ g0(t+

2Lu )

×ρ0

−→ f0(t+2Lu )

×ρL−→ g0(t+

4Lu )

×ρ0

−→ · · ·

Infinite sum:f0(t) = τ0vS(t)+τ0ρLρ0vS(t−

2Lu )+. . .

Reflections



• Steady States



• Reflections


• Standing Waves



Z0


RL+Z0

× fL (t).


RL−Z0

RL+Z0



At x = 0: f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t) = τ0vS(t) + ρ0g0(t)


vS(t)×τ0−→ f0(t)

×ρL−→ g0(t+

2Lu )

×ρ0

−→ f0(t+2Lu )

×ρL−→ g0(t+

4Lu )

×ρ0

−→ · · ·

Infinite sum:f0(t) = τ0vS(t)+τ0ρLρ0vS(t−

2Lu )+. . .=

∑

∞

i=0 τ0ρiLρ

i0vS

(

t− 2Liu

)

Sinewaves and Phasors



• Steady States



• Reflections


• Standing Waves


Sinewaves are easier because:

1. Use phasors to eliminate t:

2. Time delays are just phase shifts:




• Steady States



• Reflections


• Standing Waves



1. Use phasors to eliminate t: f0(t) = A cos (ωt+ φ)⇔ F0 = Aejφ





• Steady States



• Reflections


• Standing Waves





fx(t) = A cos(

ω(

t− xu

)

+ φ)

⇔ Fx = Aej(φ−ωux)= F0e

−jkx




• Steady States



• Reflections


• Standing Waves





fx(t) = A cos(

ω(

t− xu

)

+ φ)


−jkx

k is the wavenumber: radians per metre (c.f. ω in rad per second)




• Steady States



• Reflections


• Standing Waves





fx(t) = A cos(

ω(

t− xu

)

+ φ)


−jkx


As before: Vx = Fx +Gx and Ix = Fx−Gx

Z0




• Steady States



• Reflections


• Standing Waves





fx(t) = A cos(

ω(

t− xu

)

+ φ)


−jkx



Z0

As before:GL = ρLFL

F0 = τ0VS + ρ0G0




• Steady States



• Reflections


• Standing Waves





fx(t) = A cos(

ω(

t− xu

)

+ φ)


−jkx



Z0


F0 = τ0VS + ρ0G0

But G0 = F0ρLe−2jkL : roundtrip delay of 2L

u + reflection at x = L.




• Steady States



• Reflections


• Standing Waves





fx(t) = A cos(

ω(

t− xu

)

+ φ)


−jkx



Z0


F0 = τ0VS + ρ0G0


u + reflection at x = L.Substituting for G0 in source end equation: F0 = τ0VS + ρ0F0ρLe

−2jkL




• Steady States



• Reflections


• Standing Waves





fx(t) = A cos(

ω(

t− xu

)

+ φ)


−jkx



Z0


F0 = τ0VS + ρ0G0


u + reflection at x = L.Substituting for G0 in source end equation: F0 = τ0VS + ρ0F0ρLe

−2jkL

⇒ F0 = τ01−ρ0ρL exp(−2jkL)VS so no infinite sums needed ,

Standing Waves



• Steady States



• Reflections


• Standing Waves


Standing waves arise whenever a wave meets its reflection:at places where the two waves are in phase their amplitudes addbut where they are anti-phase their amplitudes subtract.

Standing Waves



• Steady States



• Reflections


• Standing Waves



At any point x,delay of x

u ⇒Fx = F0e

−jkx

Standing Waves



• Steady States



• Reflections


• Standing Waves




u ⇒Fx = F0e

−jkx

Backward wave: Gx = ρLFxe−2jk(L−x)

Standing Waves



• Steady States



• Reflections


• Standing Waves




u ⇒Fx = F0e

−jkx

Backward wave: Gx = ρLFxe−2jk(L−x): reflection + delay of 2L−x

u

Standing Waves



• Steady States



• Reflections


• Standing Waves




u ⇒Fx = F0e

−jkx


u

Voltage at x: Vx = Fx +Gx

Standing Waves



• Steady States



• Reflections


• Standing Waves




u ⇒Fx = F0e

−jkx


u

Voltage at x: Vx = Fx +Gx = F0e−jkx

(

1 + ρLe−2jk(L−x)

)

Standing Waves



• Steady States



• Reflections


• Standing Waves




u ⇒Fx = F0e

−jkx


u


(


)

Voltage Magnitude : |Vx| = |F0|∣

∣1 + ρLe−2jk(L−x)

∣

∣

Standing Waves



• Steady States



• Reflections


• Standing Waves




u ⇒Fx = F0e

−jkx


u


(


)



∣

∣: depends on x

Standing Waves



• Steady States



• Reflections


• Standing Waves




u ⇒Fx = F0e

−jkx


u


(


)



∣

∣: depends on x

If ρL ≥ 0, max magnitude is (1 + ρL) |F0| whenever e−2jk(L−x) = +1

Standing Waves



• Steady States



• Reflections


• Standing Waves




u ⇒Fx = F0e

−jkx


u


(


)



∣

∣: depends on x

If ρL ≥ 0, max magnitude is (1 + ρL) |F0| whenever e−2jk(L−x) = +1⇒ x = L or x = L− π

k or x = L− 2πk or . . .

Standing Waves



• Steady States



• Reflections


• Standing Waves




u ⇒Fx = F0e

−jkx


u


(


)



∣

∣: depends on x


k or x = L− 2πk or . . .

Min magnitude is (1− ρL) |F0| whenever e−2jk(L−x) = −1

Standing Waves



• Steady States



• Reflections


• Standing Waves




u ⇒Fx = F0e

−jkx


u


(


)



∣

∣: depends on x


k or x = L− 2πk or . . .

Min magnitude is (1− ρL) |F0| whenever e−2jk(L−x) = −1⇒ x = L− π

2k or x = L− 3π2k or x = L− 5π

2k or . . .

revision lecture 4: transients & lines · e1.1 analysis of circuits (2014-4649) revision...

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