time domain analysis a very superficial...
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TIME DOMAIN ANALYSIS (A VERY SUPERFICIAL APPROACH)
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Laplace Transform (LT)
§ The Laplace Transform (LT) is an integral transform similar to the Fourier Transform
§ The LT of a function f(t) is defined as
• s is a complex variable • This integral does not necessarily exist for all possible f(t) and s
(If s has a real part >0, f(t) must not grow faster than C eRe (s) t)
§ The Inverse Transform is more complicated:
• where γ > Re(all singularities of f). • This is a line integral in the complex s-plane, ‘right’ of all
singularities
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§ The Residue Theorem states that the line integral of a function f(z) along a closed curve γ in the complex z-plane is 2πi × (the sum of the residues at the singularities ak of f):
§ The residue is a characteristic of a singularity ak (or c below) • For a first order (simple) pole at c (where f behaves ~ like 1/z):
• More generally, for a pole of order n:
Reminder (hopefully..): Integration with Residues
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Wikipedia
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§ Assume we want to find .
§ The function has poles and
§ The residue at the ‘simple pole’ is:
§ The line integral along green curve C is
§ This is independent of R! With increasing R, the contribution of the upper arc vanishes (the length of the arc rises ~R, but f falls as 1/R2)
§ Therefore
Example for Integration with Residues
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i
-i Re
Im
R C
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Example 1: LT of 1
§ For f(t) = 1:
§ Valid for Re[s] > 0 to make sure this vanishes
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Example 1 (inverse): Inverse LT of 1/s
§ For F(s) = :
§ The integral has just one pole at s = 0. § The Residuum is:
§ So the Integral is
§ And we just have (the arc contribution at infinity vanishes again…)
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Re
Im s
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Properties of Laplace Transforms
§ For we have:
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Linearity
Function Laplace Transform
Derivative
Integration
Convolution
Time Shift
u(t): Step function u(t): Step function
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Example 2 ('frequency shift'):
§ For F(s) = :
§ The pole is now at s = -k.
§ The Residuum is:
§ And we just have
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Re
Im
- k
s
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Why is Laplace Transform so Useful ?
§ Differential / Integral equations in t can be converted to Analytical equations in s, where they can be solved
§ EQ(t) → transform to H(s) → Solve in s → Transform back
§ Example: Radioactive Decay • f[t]: Number of atoms at time t • The # of decaying atoms is prop. to # of atoms:
• With F[s] = LT(f[t]): (f[0] = N0 is initial number of atoms)
• This can be solved in s-domain:
• Transforming back (see example) gives:
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LAPLACE TRANSFORM AND TRANSFER FUNCTION
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Time Response
§ The transfer function tells us how sine inputs are modified by the system, i.e. what happen in the frequency domain
§ How can be get the time response for an arbitrary input?
§ For a linear, time invariant (LTI) system, we can use:
• The response of a k × larger input pulse is just k × larger • The response for a time shifted input is time shifted
§ For such a system we can • express the input signal as a superposition of 'simple' signals • Calculate the output for each 'simple' component • Superimpose the outputs
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in H(s) out
t
in(t)
t
out(t)
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Illustration
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in H(s) out t
in(t)
t
out(t)
a(t) A(t)
in H(s) out t
in(t)
t
out(t)
a(t) + 0.5 a(t-t0)
t0 A(t) + 0.5 A(t-t0)
in H(s) out t
in(t)
t
out(t)
Note that the integrals are CONVOLUTIONS (Faltung) of two functions!
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Clever Choice of the 'nominal input' a[t]
§ To make the convolutions as simple as possible, it is best to chose a[t] to be Dirac Delta 'function'
§ For any input function we can write
§ The output is then just
where Δ[t] is the response of the circuit to a δ[t] input, the 'delta response':
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in H(s) out Δ[t] δ[t]
Note: I am a bit sloppy here with integration limits..
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What is the Delta Response Δ[t] ?
§ We do not know Δ[t], but: it turns out that its LT is just the transfer function!
§ Knowing that LT(Δ[t]) = H[s], what is Δ[t] ?
It's the Inverse LT:
§ Why is this?
• If we write down Kirchhoff's rules in the time domain, we get differential / integral equations.
• The 'topology' of the equations is the same as using complex impedances.
• If we transform this, we can get the impulse response
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The Laplace Transform of the Delta Response of a circuit is just given by its transfer function H[s]
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General Time Response
§ Start from
§ Laplace transform both sides and use Convolution rule:
§ Use our knowledge that :
§ And transform back:
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To calculate the time response of a circuit to an arbitrary input f[t]: 1. Laplace Transform f[t], yielding F[s] 2. Multiply with the Transfer function H[s] 3. Laplace Transform back
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§ The most important input to test a circuit is the Unit step: • It is often called u[t], Heaviside Step function, UnitStep,…
§ For a Shifted Step, use Time Shift rule:
§ A rectangular Pulse is just the difference of two Unit Steps § For very short input signals (charge deposition in detector),
input is the Dirac Delta, with LT = 1.
Important Input Functions
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t
u(t)
LT
t
u(t)
LT
τ
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§ Consider
§ This is the step response
Example 1: Step Response of Low Pass
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R
C t
u(t)
LT H[s]
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Watch out!
§ What is LT-1[H[s]] ???, e.g.
§ That is the output for a Dirac pulse at the input (with LT[δ[t]] = 1), i.e. that is the impulse response
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§ Now Consider a linear input ramp in[t] = k t
§ The LT is
§ So our response is
Example 2: Response of Low Pass to Slope
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R
C t
in[t]
k=1, τ=1
in[t]
f[t]
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Example 3: Step Response of RLC
§ Consider the RLC circuit
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R
vin vout
R=1, L=1, C=1,2,5