Transcript Ch. 8 Analysis of Continuous-Time Systems by Use of the

Ch. 8 Analysis of ContinuousTime Systems by Use of the
Transfer Function
8.1 Stability and the Impulse
Response
• Consider a causal linear time-invariant
continuous-time system with input x(t) and
output y(t).
• Let the system function, H(s), be finite
dimensional.
• The impulse response, h(t), is the inverse
Laplace transform of H(s).
8.1 (cont)
• A system is stable if all the poles of H(s)
are located in the open left-half plane.
• A system is marginally stable if its
impulse response is bounded.
• A system is unstable if the impulse
response grows without bound as t goes
to infinity.
8.2 Routh-Huritz Stability Test
• There are procedures for testing for
stability that do not require the
computation of the poles.
• Let A(s) = aNsN + aN-1SN-1+...+a1s + a0
• A necessary (but, in general, insufficient)
condition for stability is that all of the
above coefficients are greater than 0.
Routh-Huritz Test
• Gives necessary and sufficient conditions
for stability.
• Table 8.1 Routh Array
– N+1 rows, indexed on the power of s.
– (N/2) + 1 columns if N is even.
– (N+1)/2 columns if N is odd.
– First two rows contain coefficients as shown,
starting with aN.
Routh-Hurwitz Test (cont.)
• The other rows are computed using the two rows
above.
• Example 8.2: Second Order Case
– Let A(s) = s2 + a1s + a0.
– N = 2 is even, so there are (2/2 + 1) = 2 columns.
– Result: See Table 8.2
• Example 8.3: Third-Order Case
– Result: See Table 8.3.
• Example 8.4 Higher Order Case.
8.3 Analysis of the Step Response
• Given a system with Y(s) = B(s)/A(s) X(s).
• Now when x(t) = u(t), then X(s) = 1/s.
• So the Laplace Transform of the step response
is given by Y(s) = B(s)/A(s) (1/s).
• Using the residue formula
– Y(s) = E(s)/A(s) + c/s
– Where c = [sY(s)]s=0 = H(0)
• Then y(t) = y1(t) + H(0), t≥0
– Where y1(t) is the Inverse Laplace transform of
E(s)/A(s).
• Note: y1(t) is the transient response and H(0) is
the steady state value of the step response.
8.3.1 First Order Systems
• Let H(s) = k/(s-p), where k is a constant
and p is a pole.
• Then the step response is y(t)=(-k/p)(1-ept)
• The transient response is y1(t)= (k/p)(ept)
• The steady state value is H(0) =(- k/p)
• Fig. 8.3 Step response with p = 1,2,3
• Fig. 8.4 Step response with p = -1,-2,-3
• Fig. 8.5 --what is H(0)?what is p?
8.3.2 Second Order System
• Let H(s) = (k) / [s2 + 2ζωns + (ωn)2 ]
• Here we have
– ζ is the damping ratio
– ωn is the natural frequency
• The quadratic formula gives us the poles:
–
–
p1 = -ζωn+ ωn SQRT( ζ2 -1)
p1 = -ζωn - ωn SQRT( ζ2 -1)
Second Order Systems
• Case 1: Both poles are real
– ytr(t) = (k/p1p2) (k1ep1t + k2ep2t), t≥0
– H(0) = k/(p1p2)
• Case 2: Both poles are real and repeated
• Case 3: Poles are complex pairs—Fig. 8.8
• Comparing Parameters
–
–
–
–
Underdampled when ζ <1
Criticallly damped when ζ = 1
Overdampled when ζ > 1
Damped Natural Frequency = ωd = ωn SQRT( ζ2 -1)
8.4 Response to Sinusoids and
Arbitrary Inputs
• Let x(t) = C cos ωot , t≥0
• Using Laplace transforms it can be shown that
the output of system H(s) is
• y(t) = ytr(t) + yss(t)
• In section 5.1 the steady state response was
shown to be the input scaled by the magnitude
of the frequency response evaluated at
frequency of the sinusoid and with an additional
phase angle equal to the phase of the
frequency response evaluated at the frequency
of the sinusoid.
8.4 (cont.)
• Let x(t) be an arbitrary input.
• Let X(s) = C(s)/D(s) be the Laplace transform of the
input.
• Then if the system function is H(s)=B(s)/A(s) the Laplace
transform of the output is Y(s) = B(s) C(s)/A(s)D(s)
• Using partial fraction expansion we have Y(s)= E(s)/A(s)
and F(s)/D(s).
• And finally y(t) = ytr(t) + yss(t).
• Note that the form of the transient response depends on
the poles of the system function and the form of the
steady state response depends on the poles of the input
(assuming stability).
8.5 Frequency Response Function
• The magnitude function is sometimes
given in decibels.
• |H(ω)| db = 20 log10 |H(ω)|
• Bode Plots — plots of the magnitude and
phase angle of the frequency response as
a function of frequency, ω, on a
logarithmic scale.
• 8.5.3 discusses the construction of Bode
Plots.
8.5.3 Bode Plots
• H(s) = A(s+C1)…(s+CM)/s(s+B1)…(s+BN-1)
• Let s=jω
• H(ω)= A(jω+C1)…(jω +CM)/jω(j ω +B1)…(j
ω+BN-1)
• Let K = AC1C2…CM/B1B2…BN-1
• H(ω)= K(jω/C1+1)…(jω /CM+1)
/jω(jω/B1+1)…(jω/BN-1+1)
• Log(AB) = Log(A) + Log(B)
8.5.3 Bode Plots (cont.)
• ǀH(ω)ǀdb= 20 logǀKǀ +20 logǀ(jω/C1+1)ǀ…
-20logǀjωǀ - 20logǀ(jω/B1+1)ǀ…
• The phase angle is shown on page 455.
• Constant factors: K
• (jωT + 1) factors: the magnitude db can be
approximated as 0 for values less than the
corner frequency and an increase of 20 db
per decade for frequencies greater than
the corner frequency (1/T).
8.6 Causal Filters
• In real-time filtering, ideal filters cannot be
implemented because they are not causal.
• The magnitude functions of the causal
approximations have gradual transitions from
the passband to the stopband.
• See figure 8.32.
• When the peak value is normalized to 1, the 3dB
point (|H(ω)| = .707) marks the passband.
• Usually the stopbands are down 40 or 50 dB
and the region between passbands and
stopbands is called the transition region.
8.6.1 Butterworth Filters
• Two pole lowpass Butterworth filter:
• H(s) = (ωn2) /(s2 + sqrt(2)ωns + ω2)
• Maximally flat —the magnitude variation
is as small as possible.
• The cutoff frequency (ωn) is the -3dB
point of the magnitude function.
• Figures 8.33 and 8.35 illustrates the
difference between one, two, and three
poles.
8.6.2 Chebyshev Filters
• Monotone—means that the magnitude curve is
gradually decreasing over the pass band or stop
band.
• The Butterworth filter is monotone in both the
pass and stop bands.
• Type 1 Chebyshev is monotone decreasing in
the stop band, but has equal ripple in the pass
band.
• Type 2 Chebyshev is the opposite.
• Figures 8.37 and 8.38 compares Butterworth
and Chebyshev filters.
8.6.3 Frequency Transformation
• Starting with any lowpass filter having a transfer
function H(s), we can modify the cutoff
frequency of the filter or construct other types of
filters (high-pass, etc.)
• If the cut-off frequency is ωc=ω1 and we want to
change it to ω2, then H(s) we replace every s by
s ω1/ω2 .
• To convert a lowpass with cut-off frequency is
ω1 to a highpass filter with a passband starting
at ω2, we replace s in H(s) with ω1ω2 /s.
• Page 473 gives transformations for bandpass
and bandstop filters.