review: laplace transform, linear circuits and bode...
TRANSCRIPT
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Review: Laplace Transform, Linear Circuits and Bode Plots
Alessandro SpinelliPhone: (02 2399) 4001
[email protected]/spinelli
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Electronics 96032 Alessandro Spinelli
Slides are supplementary material and are NOT a
replacement for textbooks and/or lecture notes
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Laplace transform
𝐹 𝑠 = න0
+∞
𝑓 𝑡 𝑒−𝑠𝑡𝑑𝑡 = ℒ(𝑓)
ℒ is a linear operator:ℒ(𝛼𝑓 𝑡 + 𝛽𝑔(𝑡)) = 𝛼ℒ(𝑓) + 𝛽ℒ(𝑔)
Note: in the following, all signals are defined only for 𝑡 ≥ 0, i.e., they are (implicitely) multiplied by the unit step function
Electronics 96032 Alessandro Spinelli
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Properties
1. Time differentiation:
ℒ 𝑓′ 𝑡 = 𝑠𝐹 𝑠 − 𝑓(0)
– If we think in terms of distributional derivatives(e.g., Dirac delta as derivative of step function), then 𝑓(0) → 𝑓(0−) (= 0 in our case)
– If we think in terms of classical derivatives (e.g., zero as derivative of step function), then 𝑓(0) →𝑓(0+)
Electronics 96032 Alessandro Spinelli
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Properties
2. “Frequency” differentiation:
−𝑑𝐹 𝑠
𝑑𝑠= ℒ(𝑡𝑓(𝑡))
– More generally
−1 𝑛𝑑𝑛𝐹 𝑠
𝑑𝑠𝑛= ℒ(𝑡𝑛𝑓(𝑡))
Electronics 96032 Alessandro Spinelli
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Properties
3. Time integration:
ℒ න0
𝑡
𝑓 𝜏 𝑑𝜏 =𝐹 𝑠
𝑠
4. Frequency integration:
ℒ𝑓 𝑡
𝑡= න
𝑠
∞
𝐹 𝜎 𝑑𝜎
Electronics 96032 Alessandro Spinelli
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Properties
5. Time shifting:
ℒ 𝑓 𝑡 − 𝑇 = 𝑒−𝑠𝑇𝐹(𝑠)
6. Frequency shifting:
ℒ 𝑒𝑎𝑡𝑓 𝑡 = 𝐹(𝑠 − 𝑎)
7. Convolution:ℒ 𝑓 𝑡 ∗ 𝑔(𝑡) = 𝐹 𝑠 𝐺(𝑠)
– Dual properties is more complicated, involvingintegral in complex domain
Electronics 96032 Alessandro Spinelli
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Properties
8. Initial value theorem:𝑓 0+ = lim
𝑠→∞𝑠𝐹(𝑠)
9. Final value theorem:𝑓 +∞ = lim
𝑠→0𝑠𝐹(𝑠)
Both can be extended to compute the values of the derivatives of 𝑓 just by recalling #1.
Electronics 96032 Alessandro Spinelli
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Elementary signals – 1
• Dirac delta function:
ℒ 𝛿 𝑡 = න0
+∞
𝛿 𝑡 𝑒−𝑠𝑡𝑑𝑡 = 1
• Heaviside step function (integral of 𝛿 𝑡 ):
ℒ 𝑢 𝑡 =1
𝑠• Ramp function (integral of 𝑢 𝑡 )
ℒ 𝑡 =1
𝑠2Electronics 96032 Alessandro Spinelli
Also obtainedfrom #2.
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Elementary signals – 2
• Rectangle function:
ℒ rect 0, 𝑇 = ℒ 𝑢 𝑡 − 𝑢 𝑡 − 𝑇
=1 − 𝑒−𝑠𝑇
𝑠• Decreasing exponential function (from #6.):
ℒ 𝑒−𝑡/𝜏𝑢(𝑡) =1
𝑠 + 1/𝜏=
𝜏
1 + 𝑠𝜏
Electronics 96032 Alessandro Spinelli
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Example
• Let 𝑓 𝑡 = 𝑒−𝑡/𝜏 ⇒ 𝐹 𝑠 = 𝜏/(1 + 𝑠𝜏)
• Then ℒ 𝑓′ 𝑡 = 𝑠𝐹 𝑠 = 𝑠𝜏/(1 + 𝑠𝜏)
• In fact, in a distribution frame:
𝑓′ 𝑡 = 𝛿 𝑡 −1
𝜏𝑒−𝑡/𝜏
ℒ 𝑓′ 𝑡 = 1 −1
𝜏
𝜏
1 + 𝑠𝜏=
𝑠𝜏
1 + 𝑠𝜏
Electronics 96032 Alessandro Spinelli
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Elementary signals – 3
• Sine function
ℒ sin 𝜔𝑡 =𝜔
𝑠2 +𝜔2
• Cosine function (from #1.)
ℒ cos 𝜔𝑡 =𝑠
𝑠2 +𝜔2
Electronics 96032 Alessandro Spinelli
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Application to LTI systems
Electronics 96032 Alessandro Spinelli
𝐻(𝑠)
ℎ(𝑡)𝑥(𝑡) 𝑦 𝑡 = 𝑥 𝑡 ∗ ℎ(𝑡)
𝑋(𝑠) 𝑌 𝑠 = 𝑋 𝑠 𝐻(𝑠)
ℒ ℒ ℒ−1
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Resistors
Electronics 96032 Alessandro Spinelli
𝑖 𝑡
𝑣 𝑡
𝑣 𝑡 = 𝑅𝑖 𝑡
𝑉 𝑠 = 𝑅𝐼(𝑠)
𝑉(𝑠)
𝐼(𝑠)= 𝑅
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Capacitors
𝑖 𝑡 = 𝐶𝑑𝑣(𝑡)
𝑑𝑡
𝐼 𝑠 = 𝑠𝐶𝑉(𝑠)
𝑉(𝑠)
𝐼(𝑠)=
1
𝑠𝐶= 𝑍𝐶
Electronics 96032 Alessandro Spinelli
𝑖 𝑡
𝑣 𝑡
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Inductors
𝑣 𝑡 = 𝐿𝑑𝑖(𝑡)
𝑑𝑡
𝑉 𝑠 = 𝑠𝐿𝐼(𝑠)
𝑉(𝑠)
𝐼(𝑠)= 𝑠𝐿 = 𝑍𝐿
Electronics 96032 Alessandro Spinelli
𝑖 𝑡
𝑣 𝑡
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RC network
Electronics 96032 Alessandro Spinelli
𝑣𝑖 𝑡 𝑣𝑜 𝑡
𝐶𝑅
𝑖 𝑡
ቐ
𝑣𝑖 = 𝑣𝑜 + 𝑅𝑖
𝑖 = 𝐶𝑑𝑣𝑜𝑑𝑡
⇒ 𝑅𝐶𝑑𝑣𝑜𝑑𝑡
+ 𝑣𝑜 = 𝑣𝑖
𝑉𝑜(𝑠)
𝑉𝑖(𝑠)=
1/𝑠𝐶
𝑅 + 1/𝑠𝐶=
1
1 + 𝑠𝐶𝑅
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Output signals
• Delta function response, 𝑣𝑖 𝑡 = 𝐴𝛿(𝑡)
𝑉𝑖 𝑠 = 𝐴 ⇒ 𝑉𝑜 =𝐴
1 + 𝑠𝐶𝑅⇒ 𝑣𝑜 =
𝐴
𝜏𝑒−𝑡/𝜏
• Step function response, 𝑣𝑖 𝑡 = 𝐴𝑢(𝑡)
𝑉𝑖 𝑠 =𝐴
𝑠⇒ 𝑉𝑜 =
𝐴
1 + 𝑠𝜏
1
𝑠= 𝐴
1
𝑠−
𝜏
1 + 𝑠𝜏
⇒ 𝑣𝑜 = 𝐴 𝑢 𝑡 − 𝑒−𝑡/𝜏𝑢 𝑡 = 𝐴(1 − 𝑒−𝑡/𝜏)
Electronics 96032 Alessandro Spinelli
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Ex: Lag network (step response)
Electronics 96032 Alessandro Spinelli
𝑉𝑖 𝑉𝑜𝐶
𝑅1
𝑅2
𝑉𝑜𝑉𝑖=
𝑅2 + 1/𝑠𝐶
𝑅1 + 𝑅2 + 1/𝑠𝐶=
1 + 𝑠𝐶𝑅21 + 𝑠𝐶(𝑅1 + 𝑅2)
𝑉𝑜 =1 + 𝑠𝐶𝑅2
1 + 𝑠𝐶(𝑅1 + 𝑅2)
𝐴
𝑠
𝑉𝑜 = 𝐴1
𝑠−
𝐶𝑅11 + 𝑠𝐶 𝑅1 + 𝑅2
⇒ 𝑣𝑜(𝑡) = 𝐴 1 −𝑅1
𝑅1 + 𝑅2𝑒−𝑡/𝜏
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Alternative method
• Compute:
𝑣𝑜 0 = lim𝑠→∞
𝑠𝑉𝑜(𝑠) = 𝐴𝑅2
𝑅1 + 𝑅2𝑣𝑜 ∞ = lim
𝑠→0𝑠𝑉𝑜(𝑠) = 𝐴
• Connect extremes withan exponential curvehaving time constantdetermined by the singlepole
Electronics 96032 Alessandro Spinelli
𝑡
𝑣𝑜
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One more approach
• Voltage across a capacitors cannot changeabruptly (unless you are applying a 𝛿-functioncurrent) 𝑣𝑐 0+ = 𝑣𝑐 0− = 0
– 𝑣0 0+ = 𝐴𝑅1/(𝑅1 + 𝑅2)
• For 𝑡 → ∞, the capacitor behaves as an open circuit
– 𝑣𝑜 ∞ = 𝐴
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LTI systems and sinusoidal inputs
• The output of a stable LTI system to a sinusoidalinput contains a transient and a steady-state term
• The steady-state term is also sinusoidal at the same frequency as the input, but with differentamplitude and phase:
where𝐵 = 𝑇 𝑗𝜔 𝜙 = ∡(𝑇 𝑗𝜔 )
Electronics 96032 Alessandro Spinelli
𝐴 sin(𝜔𝑡) 𝑇(𝑠) 𝐴𝐵 sin(𝜔𝑡 + 𝜙)
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Bode plots
• Provide an efficient way to plot the frequencyresponse (magnitude and phase) of LTI systems
• We consider asymptotic Bode plots, i.e., piecewise linear approximations on a suitablescale:
– Magnitude: dB = 20 log10 | ∙ |
– Phase: linear scale
– Frequency: log scale
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Single real pole: magnitude
𝑇 𝑠 =1
1 + 𝑠𝜏⇒ 𝑇 𝑗𝜔 𝑑𝐵 = 20 log10
1
1 + 𝜔𝜏 2
= −20 log10 1 + 𝜔𝜏 2 ≈ ቊ0 𝜔𝜏 ≪ 1
−20 log10𝜔𝜏 𝜔𝜏 ≫ 1
Electronics 96032 Alessandro Spinelli
log10 𝑓
Slope = −20 dB/dec
1
2𝜋𝜏
Bode plot of a real zero
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Single real pole: phase
Electronics 96032 Alessandro Spinelli
∡ 𝑇 𝑗𝜔 = ∡1
1 + 𝑗𝜔𝜏= −arctan(𝜔𝜏)
log10 𝑓
1
2𝜋𝜏∡ ⋅
Bode plot of a real zero
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General case (real singularities)
• 𝑇 𝑠 can always be expressed as
𝑇 𝑠 = 𝐺1 + 𝑠𝜏𝑧1 1 + 𝑠𝜏𝑧2 …(1 + 𝑠𝜏𝑧𝑛)
1 + 𝑠𝜏𝑝1 1 + 𝑠𝜏𝑝2 …(1 + 𝑠𝜏𝑝𝑚)
• Thanks to log and arctan properties:
𝑇 𝑑𝐵 =
𝑖=1
𝑛
1 + 𝑗𝜔𝜏𝑧𝑖 𝑑𝐵 −
𝑗=1
𝑚
1 + 𝑗𝜔𝜏𝑝𝑗 𝑑𝐵
∡(𝑇) =
𝑖=1
𝑛
∡(1 + 𝑗𝜔𝜏𝑧𝑖) −
𝑗=1
𝑚
∡(1 + 𝑗𝜔𝜏𝑝𝑗)
Electronics 96032 Alessandro Spinelli
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Ex: Lag network
𝑇(𝑠) =1 + 𝑠𝐶𝑅2
1 + 𝑠𝐶(𝑅1 + 𝑅2)
Electronics 96032 Alessandro Spinelli
log10 𝑓
1
2𝜋𝜏𝑧
1
2𝜋𝜏𝑝
𝜏𝑧
𝜏𝑝
1
𝑅2𝑅1 + 𝑅2
𝑇 𝑑𝐵
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In practice…
• 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
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Asymptotic values
𝑇(𝑠) =1 + 𝑠𝐶𝑅2
1 + 𝑠𝐶(𝑅1 + 𝑅2)
Electronics 96032 Alessandro Spinelli
𝑉𝑖 𝑉𝑜𝐶
𝑅1
𝑅2
log10 𝑓
log10 𝑓
𝑉𝑖 𝑉𝑜𝐶
𝑅1
𝑅21
2𝜋𝜏𝑧
1
2𝜋𝜏𝑝
𝜏𝑧
𝜏𝑝
1
𝑅2𝑅1 + 𝑅2
∡ 𝑇
𝑇 𝑑𝐵
![Page 30: Review: Laplace Transform, Linear Circuits and Bode Plotshome.deib.polimi.it/spinelli/corsi/ele/E01.pdf · Review: Laplace Transform, Linear Circuits and Bode Plots ... Dirac delta](https://reader030.vdocuments.site/reader030/viewer/2022021505/5ac18b7d7f8b9a5a4e8d4120/html5/thumbnails/30.jpg)
Ex: BP filter
𝑇 𝑠 =𝐴𝑠𝜏0
(1 + 𝑠𝜏1)(1 + 𝑠𝜏2)𝜏1 ≫ 𝜏2
Electronics 96032 Alessandro Spinelli
log10 𝑓
log10 𝑓
1
2𝜋𝜏2
1
2𝜋𝜏1
90
∡ 𝑇
𝑇 𝑑𝐵
−90
![Page 31: Review: Laplace Transform, Linear Circuits and Bode Plotshome.deib.polimi.it/spinelli/corsi/ele/E01.pdf · Review: Laplace Transform, Linear Circuits and Bode Plots ... Dirac delta](https://reader030.vdocuments.site/reader030/viewer/2022021505/5ac18b7d7f8b9a5a4e8d4120/html5/thumbnails/31.jpg)
Homework
1. Compute the step response of a CR filter
2. Design a simple network (made with 𝑅s and 𝐶s) with a zero and a pole having 𝑓𝑧 < 𝑓𝑝 (a
lead network)
3. Quote the Bode diagram of the previous slide
4. Work out the step response of the BP filter of the previous slide
Electronics 96032 Alessandro Spinelli