Chapter 15 Circuit Analysis in the...

Chapter 15 Circuit Analysis in the s-Domain

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Resistors in the Frequency Domain

Ohm’s law specifies that

v(t) = Ri(t)

Taking the Laplace transform of both sides

V(s) = RI(s)

The impedance Z(s) is defined as

and the admittance Y(s)=I(s)/V(s) is 1/R

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Z(s)V(s)

I(s) R

Modeling Inductors in the s-Domain

Since v(t) = L di/dt taking the Laplace transform of both sides of this equation yields

V(s) = L[sI(s) − i (0−)]

The impedance is Z(s)=sL and the initial condition is modeled as a voltage source in series.

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Example: Inductors

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Find the voltage v(t) given an initial current i(0−) =1 A.

Answer: v(t) = [3.2e−8t − 1.2e−0.5t ]u(t) volts

Solution

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Modeling Capacitors in the s-Domain

Using i=C dv/dt, we can find Z(s)=1/sC and

we model the initial condition as in (b) and (c)

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Example: Mesh Currents

Determine the mesh currents i1(t) and i2(t) (assume zero initial energies)

Answer: (using a computer for calculations)

i1(t) =−96.39e−2t − 344.8e−t +

841.2e−0.15t cos 0.8529t + 197.7e−0.15t sin 0.8529t mA

i2(t) =−481.9e−2t − 241.4e−t +

723.3e−0.15t cos 0.8529t + 472.8e−0.15t sin 0.8529t mA

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Example: Nodal Analysis

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Find vx

Answer: vx = [4 + 6.864e−1.707t − 5.864e−0.2929t ]u(t)

Solution

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Example: Source Transformations

Find v(t) using source transformation (The initial conditions can be determined to be zero).

Answer:

v(t) =[5.590 ×10−5e−0.1023t + 2.098 cos(3t + 3.912◦)

+ 0.1017e−0.04885t cos(0.6573t + 157.9◦)]u(t)

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Example: Thévenin Equivalent

Find the frequency-domain Thévenin equivalent of the highlighted network:

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Method: apply a test current:

The three impedance in parallel on the left and on the right can all be combined.

Solution

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The Transfer Function H(s)

H(s) is the transfer function of the circuit, defined as the ratio of the output to the input.

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H(s)Vout

Vin

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1 sRC1/RC

s1/RC

Impulse Response

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𝛿(𝑡) ℎ(𝑡) Network

Network

Network

Network

𝛿(𝑡 − 𝜆) ℎ(𝑡 − 𝜆)

𝑥 𝜆 𝛿(𝑡 − 𝜆) 𝑥 𝜆 ℎ(𝑡 − 𝜆)

𝑥 𝑡 = 𝑥 𝜆 𝛿(𝑡 − 𝜆)∞

−∞

𝑑𝜆 𝑦 𝑡 = 𝑥 𝜆 ℎ(𝑡 − 𝜆)𝑑𝜆∞

−∞

Convolution

If we define the impulse response as h(t), then the output y(t) is related to the input x(t) via the convolution integral:

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If x(t)=vi(t)=u(t)-u(t-1) and h(t)=2e-tu(t), then by flip/slide/integrate, we find vo(t):

Graphical Convolution

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t <1 t >1

Example: Convolution

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Convolution and Laplace

Convolution in the time domain is multiplication in the frequency domain:

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Visualizing Laplace: the s Plane

We can explore the properties of F(s) by graphing |F(s)| on the s plane.

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Interpreting the Complex Frequency (s) Plane

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Example: Y(s) = 1 / (s+3)

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Pole-Zero Constellations

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Z(s) ks2

s2 2s26

s2

(s1)2 52

Natural Response from the s Plane

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When Vs=0, the output is the

natural response, and it can only be

non-zero at the pole s=−R/L.

Hence in(t)=Ae-Rt/L.

Implementing a Pole

H (s) R f

R1

1/R fC f

s1/R fC f

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Implementing a Zero

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H (s) R fC1 s1

R1C1

Designing a Circuit to Achieve a Particular H(s)

Synthesize a circuit that will yield the transfer function H(s) =Vout/Vin =10(s+2)/(s+5).

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