Structure of the transfer function, H(s), of the complex frequency, s, and its relation to Vout(t) for delta function input through the Laplace transform.
Draft 2/3/99
Introduction | Underdamped | Critical Damping
LRC low pass filter (nodes given as in SPICE):
Vin 1 0
L 1 2
R 2 3
C 3 0
Vout is measured at node 3. Node 0 is ground.
The transfer function for this LRC low-pass filter circuit is:*
H(s)=1/(1+(s/w0)^2+s/(Q*w0)),
where
w0=1/(LC)^.5, Q=(1/R)*(L/C)^.5.
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* Find H(j*w)=Vout/Vin using network analysis with complex impedance, then substitute s for j*w. (Note that w represents the angular frequency, omega).
Choose some parameter values (underdamped case):
w0=5, Q=5, and
H(s)=1/(1+(s/5)^2+s/25).
For these values, the poles (the points where the denominator of H(s) vanishes) are at
s1 = -1/2+j*(3/2)*(11)^.5, s2 = -1/2-j*(3/2)*(11)^.5.
The magnitude of H(s) in the neighborhood of the pole, s1, is shown below. The real part of s (sigma) is plotted along x and the imaginary part (omega) along y.
The value of Abs(H(y, x=0)) (i.e., along the imaginary axis) is the magnitude of the output voltage as a function of angular frequency for unit input AC (sinusoid) amplitude (here, y is the angular frequency). Notice the resonance peak.
The linearity of the Laplace transform and its transformation of ordinary differential equations with constant coefficients to algebraic equations allows a direct connection between H(s) and f(t). Essentially, f(t), the inverse Laplace transform of H(s), represents the output voltage of the circuit for a unit impulse function (delta function) input. This also assumes the voltages and currents are zero at t=0 (zero initial conditions). The Laplace transform method is studied in 116B.
For this circuit, the inverse Laplace transform for H(s) is a damped sinusoid (see Table 5.1 in Bobrow or use Mathematica):
V(t)=(50/(3*11^.5))*exp(-t/2)*sin((3*11^.5)*t/2)*u(t),
where u(t) is the unit step function.
Note that the expression is a linear combination of exp(s1*t) and exp(s2*t) where s1 and s2 are the pole locations (complex conjugate poles in the left half of the complex plane). The real part of s1 and s2 determines the time constant of the exponential falloff of the amplitude while the imaginary part sets the angular frequency of oscillation.
Now we repeat this analysis for the critically damped case, Q=0.5, keeping w0=5.
H(s)=1/(1+(s/5)^2+s/2.5).
The denominator has a double root at s=-5, leading to a double pole.
The magnitude of H(s) in the neighborhood of the double pole is shown below. Again, the real part of s (sigma) is plotted along x and the imaginary part (omega) along y.
Again, observe that the value of Abs(H(y, x=0)) (i.e., along the imaginary axis) is the magnitude of the output voltage as a function of angular frequency for unit input AC amplitude (here, y is the angular frequency).
Now, the inverse Laplace transform for H(s) is:
V(t)=25*exp(-5*t)*t*u(t),
as expected for the pole of order 2 (i.e., double root of denominator) on the negative real axis. The pole location determines the time constant of the exponential.
Again, this represents the output voltage of the circuit for a unit impulse function (delta function) input and zero initial conditions.
Q=0.5, w0=5