Laplace Transforms, Linear Circuits and Bode Plots

Electronics – 96032

Alessandro Spinelli

Phone: (02 2399) 4001

[email protected] home.deib.polimi.it/spinelli

Slides are supplementary

material and are NOT a

replacement for textbooks

and/or lecture notes

• Laplace transforms

• Application to LTI circuits • Bode plots

𝐹𝐹 𝑠𝑠 = � 0 +∞ 𝑓𝑓 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑡𝑡𝑑𝑑𝑡𝑡 = ℒ(𝑓𝑓) ℒ is a linear operator: ℒ(𝛼𝛼𝑓𝑓 𝑡𝑡 + 𝛽𝛽𝛽𝛽(𝑡𝑡)) = 𝛼𝛼ℒ(𝑓𝑓) + 𝛽𝛽ℒ(𝛽𝛽)

Note: in the following, all signals are defined only for 𝑡𝑡 ≥ 0, i.e., they are (implicitely) multiplied by the unit step function 𝑢𝑢 𝑡𝑡

1. Time differentiation:

ℒ 𝑓𝑓′ 𝑡𝑡 = 𝑠𝑠𝐹𝐹 𝑠𝑠 − 𝑓𝑓(0)

 If we think in terms of distributional derivatives (e.g., Dirac delta as

derivative of step function), then _{𝑓𝑓(0) → 𝑓𝑓(0}−_{) (= 0 in our case)}

 If we think in terms of classical derivatives (e.g., zero as derivative of

step function), then _{𝑓𝑓(0) → 𝑓𝑓(0}+₎

2. “Frequency” differentiation:

ℒ 𝑡𝑡𝑓𝑓 𝑡𝑡 = − 𝑑𝑑𝐹𝐹 𝑠𝑠_{𝑑𝑑𝑠𝑠}

 More generally

ℒ 𝑡𝑡𝑛𝑛𝑓𝑓 𝑡𝑡 = −1 𝑛𝑛 𝑑𝑑𝑛𝑛_{𝑑𝑑𝑠𝑠}𝐹𝐹 𝑠𝑠_𝑛𝑛

3. Time integration: ℒ � 0 𝑡𝑡 𝑓𝑓 𝜏𝜏 𝑑𝑑𝜏𝜏 = 𝐹𝐹 𝑠𝑠_𝑠𝑠 4. Frequency integration: ℒ 𝑓𝑓 𝑡𝑡_𝑡𝑡 = � 𝑠𝑠 ∞ 𝐹𝐹 𝜎𝜎 𝑑𝑑𝜎𝜎

Properties

7

5. Time shifting: ℒ 𝑓𝑓 𝑡𝑡 − 𝑇𝑇 = 𝑒𝑒−𝑠𝑠𝑠𝑠𝐹𝐹(𝑠𝑠) 6. Frequency shifting: ℒ 𝑒𝑒𝑎𝑎𝑡𝑡𝑓𝑓 𝑡𝑡 = 𝐹𝐹(𝑠𝑠 − 𝑎𝑎) 7. Convolution: ℒ 𝑓𝑓 𝑡𝑡 ∗ 𝛽𝛽(𝑡𝑡) = 𝐹𝐹 𝑠𝑠 𝐺𝐺(𝑠𝑠)

 Dual properties is more complicated, involving integral in complex

domain

8. Initial value theorem:

𝑓𝑓 0+ = lim_{𝑠𝑠→∞} 𝑠𝑠𝐹𝐹(𝑠𝑠) 9. Final value theorem:

𝑓𝑓 +∞ = lim_𝑠𝑠→0 𝑠𝑠𝐹𝐹(𝑠𝑠)

Both can be extended to compute the values of the derivatives of 𝑓𝑓 just by recalling #1. For first derivatives, this means

𝑓𝑓′ 0+ = lim_{𝑠𝑠→∞} 𝑠𝑠2𝐹𝐹 𝑠𝑠 𝑓𝑓′ +∞ = lim_𝑠𝑠→0 𝑠𝑠2𝐹𝐹(𝑠𝑠)

10.Scaling

ℒ 𝑓𝑓 𝑎𝑎𝑡𝑡 = _{|𝑎𝑎| 𝐹𝐹}1 _𝑎𝑎𝑠𝑠

In particular, for 𝑎𝑎 = −1 (time reversal) we get: ℒ 𝑓𝑓 −𝑡𝑡 = 𝐹𝐹 −𝑠𝑠

• Dirac delta function:

ℒ 𝛿𝛿 𝑡𝑡 = �

0

+∞

𝛿𝛿 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑡𝑡𝑑𝑑𝑡𝑡 = 1 • Heaviside step function (integral of 𝛿𝛿 𝑡𝑡 ):

ℒ 𝑢𝑢 𝑡𝑡 = 1_𝑠𝑠 • Ramp function (integral of 𝑢𝑢 𝑡𝑡 )

ℒ 𝑡𝑡 = _𝑠𝑠1₂

Elementary signals – 1

11

Also obtained from #2

• Rectangle function:

ℒ rect 0, 𝑇𝑇 = ℒ 𝑢𝑢 𝑡𝑡 − 𝑢𝑢 𝑡𝑡 − 𝑇𝑇 = 1 − 𝑒𝑒_𝑠𝑠 −𝑠𝑠𝑠𝑠

• Decreasing exponential function (from #6, with 𝑓𝑓 𝑡𝑡 = 𝑢𝑢 𝑡𝑡 and 𝑎𝑎 = −1/𝜏𝜏):

ℒ 𝑒𝑒−𝑡𝑡/𝜏𝜏𝑢𝑢(𝑡𝑡) = _{𝑠𝑠 + 1/𝜏𝜏 =}1 _{1 + 𝑠𝑠𝜏𝜏}𝜏𝜏

𝑓𝑓 𝑡𝑡 = 𝑒𝑒−𝑡𝑡/𝜏𝜏 ⇒ 𝐹𝐹 𝑠𝑠 = _{1 + 𝑠𝑠𝜏𝜏}𝜏𝜏 ℒ 𝑓𝑓′ 𝑡𝑡 = 𝑠𝑠𝐹𝐹 𝑠𝑠 = _{1 + 𝑠𝑠𝜏𝜏}𝑠𝑠𝜏𝜏 In fact, in a distribution frame:

𝑓𝑓′ 𝑡𝑡 = 𝛿𝛿 𝑡𝑡 − 1_𝜏𝜏 𝑒𝑒−𝑡𝑡/𝜏𝜏

ℒ 𝑓𝑓′ 𝑡𝑡 = 1 − 1_{𝜏𝜏 1 + 𝑠𝑠𝜏𝜏}𝜏𝜏 = _{1 + 𝑠𝑠𝜏𝜏}𝑠𝑠𝜏𝜏

• Sine function

ℒ sin 𝜔𝜔𝑡𝑡 = _{𝑠𝑠2 + 𝜔𝜔}𝜔𝜔 ₂ • Cosine function (from #1)

ℒ cos 𝜔𝜔𝑡𝑡 = _{𝑠𝑠2 + 𝜔𝜔}𝑠𝑠 ₂

• Laplace transforms

• Application to LTI circuits

• Bode plots

Application to LTI systems

16 𝐻𝐻(𝑠𝑠) ℎ(𝑡𝑡) 𝑥𝑥(𝑡𝑡) 𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 ∗ ℎ(𝑡𝑡) 𝑋𝑋(𝑠𝑠) 𝑌𝑌 𝑠𝑠 = 𝑋𝑋 𝑠𝑠 𝐻𝐻(𝑠𝑠) ℒ ℒ ℒ−1

𝑣𝑣 𝑡𝑡 = 𝑅𝑅𝑅𝑅 𝑡𝑡 𝑉𝑉 𝑠𝑠 = 𝑅𝑅𝑅𝑅(𝑠𝑠) 𝑉𝑉(𝑠𝑠) 𝑅𝑅(𝑠𝑠) = 𝑅𝑅

R, L, C

17 𝑅𝑅 𝑡𝑡 𝑣𝑣 𝑡𝑡 𝑅𝑅 𝑡𝑡 = 𝐶𝐶 𝑑𝑑𝑣𝑣(𝑡𝑡)_{𝑑𝑑𝑡𝑡} 𝑅𝑅 𝑠𝑠 = 𝑠𝑠𝐶𝐶𝑉𝑉(𝑠𝑠) 𝑉𝑉(𝑠𝑠) 𝑅𝑅(𝑠𝑠) = 1 𝑠𝑠𝐶𝐶 = 𝑍𝑍𝐶𝐶 𝑣𝑣 𝑡𝑡 = 𝐿𝐿 𝑑𝑑𝑅𝑅(𝑡𝑡)_{𝑑𝑑𝑡𝑡} 𝑉𝑉 𝑠𝑠 = 𝑠𝑠𝐿𝐿𝑅𝑅(𝑠𝑠) 𝑉𝑉(𝑠𝑠) 𝑅𝑅(𝑠𝑠) = 𝑠𝑠𝐿𝐿 = 𝑍𝑍𝐿𝐿 𝑅𝑅 𝑡𝑡 𝑣𝑣 𝑡𝑡 𝑅𝑅 𝑡𝑡 𝑣𝑣 𝑡𝑡

𝑅𝑅𝐶𝐶𝑅𝑅 + 1 = 0 ⇒ 𝑅𝑅 = − _{𝑅𝑅𝐶𝐶 ⇒ 𝑣𝑣}1 ₀ = 𝐾𝐾𝑒𝑒− 𝑡𝑡𝑅𝑅𝐶𝐶

𝑣𝑣_𝑖𝑖 = 𝐴𝐴𝑢𝑢 𝑡𝑡 ⇒ 𝑣𝑣_𝑜𝑜 = 𝐴𝐴 (particular solution) 𝑣𝑣 0 = 0 ⇒ 𝐾𝐾 = −𝐴𝐴 ⇒ 𝑣𝑣 = 𝐴𝐴 1 − 𝑒𝑒− 𝑡𝑡𝑅𝑅𝐶𝐶

RC network – time domain

18

𝑣𝑣_𝑖𝑖 𝑡𝑡 _{𝑣𝑣𝑜𝑜 𝑡𝑡} 𝐶𝐶 𝑅𝑅 𝑅𝑅 𝑡𝑡 �𝑣𝑣𝑖𝑖 = 𝑣𝑣𝑜𝑜 + 𝑅𝑅𝑅𝑅 𝑅𝑅 = 𝐶𝐶 𝑑𝑑𝑣𝑣_{𝑑𝑑𝑡𝑡}𝑜𝑜 ⇒ 𝑅𝑅𝐶𝐶 𝑑𝑑𝑣𝑣_𝑜𝑜 𝑑𝑑𝑡𝑡 + 𝑣𝑣𝑜𝑜 = 𝑣𝑣𝑖𝑖 (homogeneous sol.)

Poles are the eigenvalues of the characteristic equation ⇒ they represent the time constants of the output

RC network – Laplace domain

19

𝑉𝑉_𝑜𝑜(𝑠𝑠) 𝑉𝑉_𝑖𝑖(𝑠𝑠) = 1/𝑠𝑠𝐶𝐶 𝑅𝑅 + 1/𝑠𝑠𝐶𝐶 = 1 1 + 𝑠𝑠𝐶𝐶𝑅𝑅

Single, real pole in 𝑠𝑠 = −1/𝐶𝐶𝑅𝑅

𝑣𝑣_𝑖𝑖 𝑡𝑡 _{𝑣𝑣𝑜𝑜 𝑡𝑡}

𝐶𝐶

𝑅𝑅

• Delta function response, 𝑣𝑣_𝑖𝑖 𝑡𝑡 = 𝐴𝐴𝛿𝛿(𝑡𝑡)

𝑉𝑉_𝑖𝑖 𝑠𝑠 = 𝐴𝐴 ⇒ 𝑉𝑉_𝑜𝑜 = _{1 + 𝑠𝑠𝐶𝐶𝑅𝑅 =}𝐴𝐴 𝐴𝐴_{𝜏𝜏 1 + 𝑠𝑠𝜏𝜏 ⇒ 𝑣𝑣}𝜏𝜏 _𝑜𝑜 = 𝐴𝐴_𝜏𝜏 𝑒𝑒−𝑡𝑡/𝜏𝜏 • Step function response, 𝑣𝑣_𝑖𝑖 𝑡𝑡 = 𝐴𝐴𝑢𝑢(𝑡𝑡)

𝑉𝑉_𝑖𝑖 𝑠𝑠 = 𝐴𝐴_{𝑠𝑠 ⇒ 𝑉𝑉𝑜𝑜} = _{1 + 𝑠𝑠𝜏𝜏}𝐴𝐴 1_{𝑠𝑠 = 𝐴𝐴} 1_{𝑠𝑠 − 1 + 𝑠𝑠𝜏𝜏 ⇒}𝜏𝜏 𝑣𝑣_𝑜𝑜 = 𝐴𝐴 𝑢𝑢 𝑡𝑡 − 𝑒𝑒−𝑡𝑡/𝜏𝜏𝑢𝑢 𝑡𝑡 = 𝐴𝐴(1 − 𝑒𝑒−𝑡𝑡/𝜏𝜏)

Ex: Lag network (step response)

21 𝑉𝑉_𝑖𝑖 𝑉𝑉_𝑜𝑜 𝐶𝐶 𝑅𝑅₁ 𝑅𝑅₂ 𝑉𝑉_𝑜𝑜 𝑉𝑉_𝑖𝑖 = 𝑅𝑅₂ + 1/𝑠𝑠𝐶𝐶 𝑅𝑅₁ + 𝑅𝑅₂ + 1/𝑠𝑠𝐶𝐶 = 1 + 𝑠𝑠𝐶𝐶𝑅𝑅₂ 1 + 𝑠𝑠𝐶𝐶(𝑅𝑅₁ + 𝑅𝑅₂) 𝑉𝑉_𝑜𝑜 = _{1 + 𝑠𝑠𝐶𝐶(𝑅𝑅}1 + 𝑠𝑠𝐶𝐶𝑅𝑅2 1 + 𝑅𝑅2) 𝐴𝐴 𝑠𝑠 𝑉𝑉_𝑜𝑜 = 𝐴𝐴 1_{𝑠𝑠 − 1 + 𝑠𝑠𝐶𝐶}𝐶𝐶𝑅𝑅_𝑅𝑅1 1 + 𝑅𝑅2 ⇒ 𝑣𝑣𝑜𝑜(𝑡𝑡) = 𝐴𝐴 1 − 𝑅𝑅₁ 𝑅𝑅₁ + 𝑅𝑅₂ 𝑒𝑒−𝑡𝑡/𝜏𝜏

1) Compute:

𝑣𝑣_𝑜𝑜 0 = lim_{𝑠𝑠→∞} 𝑠𝑠𝑉𝑉_𝑜𝑜(𝑠𝑠) = 𝐴𝐴 _𝑅𝑅 𝑅𝑅2

1 + 𝑅𝑅2

𝑣𝑣_𝑜𝑜 ∞ = lim_𝑠𝑠→0 𝑠𝑠𝑉𝑉_𝑜𝑜(𝑠𝑠) = 𝐴𝐴 2) Connect extremes with an

exponential curve having time constant determined by the single pole

Alternative (rough) method

22

One more approach

Voltage across a capacitor

cannot change abruptly (unless you are applying a 𝛿𝛿-function current)

 𝑣𝑣_𝑐𝑐 0+ = 𝑣𝑣_𝑐𝑐 0− = 0  𝑣𝑣₀ 0+ = 𝐴𝐴𝑅𝑅1

𝑅𝑅₁+𝑅𝑅₂

For 𝑡𝑡 → ∞, the capacitor behaves as an open circuit

 𝑣𝑣_𝑜𝑜 ∞ = 𝐴𝐴 23 𝑉𝑉_𝑖𝑖 𝑉𝑉_𝑜𝑜 𝐶𝐶 𝑅𝑅₁ 𝑅𝑅₂ 𝑉𝑉_𝑖𝑖 𝑉𝑉_𝑜𝑜 𝐶𝐶 𝑅𝑅₁ 𝑅𝑅₂

• Laplace transforms

• Application to LTI circuits • Bode plots

• The output of a stable LTI system to a sinusoidal input contains a transient and a steady-state term

• The steady-state term is also sinusoidal at the same frequency as the input, but with different amplitude and phase:

where

𝐵𝐵 = 𝑇𝑇 𝑗𝑗𝜔𝜔 𝜙𝜙 = ∡(𝑇𝑇 𝑗𝑗𝜔𝜔 )

LTI systems and sinusoidal inputs

25

• Provide an efficient way to plot the frequency response (magnitude and phase) of LTI systems

• We consider asymptotic Bode plots, i.e., piecewise linear approximations on a suitable scale:

 Magnitude: dB = 20 log₁₀ | � |  Phase: linear scale

 Frequency: log scale

𝑇𝑇 𝑠𝑠 = _{1 + 𝑠𝑠𝜏𝜏 ⇒ 𝑇𝑇 𝑗𝑗𝜔𝜔}1 _{𝑑𝑑𝑑𝑑} = 20 log₁₀ 1

1 + 𝜔𝜔𝜏𝜏 2

= −20 log₁₀ 1 + 𝜔𝜔𝜏𝜏 2

≈ �_{−20 log}0 𝜔𝜔𝜏𝜏 ≪ 1

10 𝜔𝜔𝜏𝜏 𝜔𝜔𝜏𝜏 ≫ 1

Single real pole: magnitude

27

log₁₀𝑓𝑓 Slope = −20 dB/dec 1 2𝜋𝜋𝜏𝜏 Bode plot of a real zero

∡ 𝑇𝑇 𝑗𝑗𝜔𝜔 = ∡ _{1 + 𝑗𝑗𝜔𝜔𝜏𝜏 = −arctan(𝜔𝜔𝜏𝜏)}1

Single real pole: phase

28

log₁₀ 𝑓𝑓 1 2𝜋𝜋𝜏𝜏 ∡ ⋅ Bode plot of a real zero

𝑇𝑇 𝑠𝑠 = _{𝑠𝑠𝜏𝜏 ⇒ 𝑇𝑇 𝑗𝑗𝜔𝜔}1 _{𝑑𝑑𝑑𝑑} = 20 log_{10 𝜔𝜔𝜏𝜏 = −20 log}1 ₁₀ 𝜔𝜔𝜏𝜏 𝑥𝑥-axis intercept: 𝑇𝑇 𝑗𝑗𝜔𝜔 = 1 ⇒ 𝜔𝜔 = 1_𝜏𝜏

Pole (zero) in 𝒔𝒔 = 𝟎𝟎

29 Bode plot of a zero in s = 0 Slope = −20 dB/dec log₁₀ 𝑓𝑓 1 2𝜋𝜋𝜏𝜏

• 𝑇𝑇 𝑠𝑠 can always be expressed as

𝑇𝑇 𝑠𝑠 = 𝐺𝐺 1 + 𝑠𝑠𝜏𝜏𝑧𝑧1 1 + 𝑠𝑠𝜏𝜏𝑧𝑧2 … (1 + 𝑠𝑠𝜏𝜏𝑧𝑧𝑛𝑛) 1 + 𝑠𝑠𝜏𝜏_𝑝𝑝1 1 + 𝑠𝑠𝜏𝜏_𝑝𝑝2 … (1 + 𝑠𝑠𝜏𝜏_{𝑝𝑝𝑝𝑝}) • Thanks to log and arctan properties:

𝑇𝑇 _{𝑑𝑑𝑑𝑑} = 𝐺𝐺_{𝑑𝑑𝑑𝑑} + � 𝑖𝑖=1 𝑛𝑛 1 + 𝑗𝑗𝜔𝜔𝜏𝜏_{𝑧𝑧𝑖𝑖 𝑑𝑑𝑑𝑑} − � 𝑗𝑗=1 𝑝𝑝 1 + 𝑗𝑗𝜔𝜔𝜏𝜏_{𝑝𝑝𝑗𝑗 𝑑𝑑𝑑𝑑} ∡(𝑇𝑇) = � 𝑛𝑛 ∡(1 + 𝑗𝑗𝜔𝜔𝜏𝜏 ) − � 𝑝𝑝 ∡(1 + 𝑗𝑗𝜔𝜔𝜏𝜏 )

Ex: Lag network

31 log10 𝑓𝑓 1 2𝜋𝜋𝜏𝜏𝑧𝑧 1 2𝜋𝜋𝜏𝜏_𝑝𝑝 𝜏𝜏_𝑧𝑧 𝜏𝜏_𝑝𝑝 1 𝑅𝑅₂ 𝑅𝑅₁ + 𝑅𝑅₂ 𝑇𝑇 _{𝑑𝑑𝑑𝑑} 𝑇𝑇(𝑠𝑠) = _{1 + 𝑠𝑠𝐶𝐶}1 + 𝑠𝑠𝐶𝐶_(𝑅𝑅 𝑅𝑅2 1 + 𝑅𝑅2)

• Magnitude:

 At every zero frequency, the slope is increased by 1 (20 dB/dec)  At every pole frequency, the slope is reduced by 1 (20 dB/dec)

• Phase:

 Composition is a bit more complicated, but a rough view can be

obtained by abruptly adding ±90° at every zero/pole frequency

𝑇𝑇(𝑠𝑠) = _{1 + 𝑠𝑠𝐶𝐶}1 + 𝑠𝑠𝐶𝐶_(𝑅𝑅 𝑅𝑅2 1 + 𝑅𝑅2)

Asymptotic values

33 𝑉𝑉_𝑖𝑖 𝑉𝑉_𝑜𝑜 𝐶𝐶 𝑅𝑅₁ 𝑅𝑅₂ log₁₀ 𝑓𝑓 log₁₀ 𝑓𝑓 𝑉𝑉_𝑖𝑖 𝑉𝑉_𝑜𝑜 𝐶𝐶 𝑅𝑅₁ 𝑅𝑅₂ 1 2𝜋𝜋𝜏𝜏_𝑧𝑧 1 2𝜋𝜋𝜏𝜏_𝑝𝑝 𝜏𝜏_𝑧𝑧 𝜏𝜏_𝑝𝑝 1 𝑅𝑅₂ 𝑅𝑅₁ + 𝑅𝑅₂ ∡ 𝑇𝑇 𝑇𝑇 _{𝑑𝑑𝑑𝑑}

1. Compute the step response of a CR filter

2. Design a simple network (made with resistors and capacitors only) having a zero and a pole with 𝑓𝑓_𝑧𝑧 < 𝑓𝑓_𝑝𝑝 (a lead network) 3. Consider a BP filter having

𝑇𝑇 𝑠𝑠 = _{(1 + 𝑠𝑠𝜏𝜏}𝐴𝐴𝑠𝑠𝜏𝜏1

1)(1 + 𝑠𝑠𝜏𝜏2) (𝜏𝜏1≫ 𝜏𝜏2)

Draw its quantitative Bode diagram and work out the step response