• Slides borrowed from Evans, U. of Texas @ Austin
Zero-State Response • Linear constant coefficient differential equation Input x(t) and output y (t ) = y zero−input (t ) + y zero− state (t ) Zero-state response: all initial conditions are zero y(t ) ↔ Y (s ) x(t ) ↔ X (s ) dr r ( ) y t ↔ s Y (s ) r dt
dk k ( ) x t ↔ s X (s ) k dt
Laplace transform both sides of differential equation with all initial conditions being zero and solve for Y(s)/X(s) y ' (t ) + y (t ) = x(t )
sY ( s ) + Y ( s ) = X ( s )
y (0 − ) = 0
H (s) =
Y ( s) 1 = X (s) s + 1
Transfer Function • H(s) is called the transfer function because it describes how input is transferred to the output in a transform domain (s-domain in this case) Y(s) = H(s) X(s) y(t) = L-1{H(s) X(s)} = h(t) * x(t) ⇒ H(s) = L{h(t)}
• Transfer function is Laplace transform of impulse response
Transfer Function Examples • Laplace transform ∞
X (s ) = ∫ − x(t ) e − s t dt 0
• Assume input x(t) and output y(t) are causal • Ideal delay of T seconds Initial conditions (initial voltages in delay buffer) are zero y (t ) = x(t − T ) Y (s ) = X (s ) e − s T Y (s ) H (s ) = = e−s T X (s )
x(t)
T
y(t)
Transfer Function Examples • Ideal integrator with y(0-) = 0 t
y (t ) = ∫ − x(τ )dτ 0
1 1 Y (s ) = X (s ) + y 0 − s s Y (s) 1 H (s ) = = X (s) s
( )
x(t)
t
∫ (•) dt 0−
y(t)
• Ideal differentiator with x(0-) = 0 d y (t ) = x(t ) dt Y (s ) = s X (s ) − x 0 − = s X ( s ) Y (s ) H (s ) = =s X (s )
( )
x(t)
d y(t) (•) dt
Cascaded Systems • Assume input x(t) and output y(t) are causal x(t) • Integrator first, then differentiator X(s)
• Differentiator first, x(t) then integrator
X(s)
t
1/s
s
∫ x(τ )dτ 0−
X(s)/s d x(t ) dt
s X(s)
s
1/s
• Common transfer functions A constant (finite impulse response) A polynomial (finite impulse response) Ratio of two polynomials (infinite impulse response)
x(t) X(s) x(t) X(s)
Block Diagrams X(s)
X(s)
H1(s)
W(s)
H(s)
Y(s)
H2(s)
Y(s)
=
X(s)
H1(s)H2(s)
Y(s)
Σ
Y(s)
=
X(s)
H1(s) + H2(s)
Y(s)
Y(s)
=
X(s)
G(s) 1 + G(s)H(s)
Y(s)
H1(s) X(s) H2(s)
X(s)
-
Σ
E(s)
G(s) H(s)
Cascade and Parallel Connections • Cascade
X(s)
W(s) = H1(s) X(s) Y(s) = H2(s)W(s) Y(s) = H1(s) H2(s) X(s) ⇒ Y(s)/X(s) = H1(s)H2(s) H1(s) H2(s) Y(s) ⇔ X(s) H2(s) H1(s) Y(s) One can switch the order of the cascade of two LTI systems if both LTI systems compute to exact precision
Feedback Connection • Governing equations E (s ) = F (s ) − H (s ) Y (s ) Y (s ) = G(s ) E (s )
• What happens if H(s) is a constant K? Choice of K controls all poles in transfer function
• Combining equations Y (s ) = G (s ) [ F (s ) − H (s ) Y (s ) ] Y (s ) + G (s ) H (s ) Y (s ) = G (s ) F (s ) G (s ) Y (s ) = F (s ) 1 + G (s ) H (s ) F(s)
-
Σ
E(s)
G(s) H(s)
Y(s)
=
F(s)
G(s) 1 + G(s)H(s)
Y(s) 19 - 9
External Stability Conditions • Bounded-input bounded-output stability Zero-state response given by h(t) * x(t) Two choices: BIBO stable or BIBO unstable
• Remove common factors in transfer function H(s) • If all poles of H(s) in left-hand plane, All terms in h(t) are decaying exponentials h(t) is absolutely integrable and system is BIBO stable
• Example: BIBO stable but asymptotically unstable ⎛ s − 1 ⎞ ⎛ 1 ⎞⎛ s − 1 ⎞ ⎛ 1 ⎞ H ( s) = ⎜ ⎟ = ⎜ ⎟⎜ ⎟ = ⎜ ⎟ 2 s − 1 s + 1 s + 1 ⎠⎝ ⎠ ⎝ ⎠ ⎝ s − 1 ⎠ ⎝
Based on slide by Prof. Adnan Kavak
Internal Stability Conditions • Stability based on zero-input solution • Asymptotically stable if and only if Characteristic roots are in left-hand plane (LHP) Roots may be repeated or non-repeated
• Unstable if and only if (i) at least characteristic root in right-hand plane and/or (ii) repeated characteristic roots are on imaginary axis
• Marginally stable if and only if There are no characteristic roots in right-hand plane and Some non-repeated roots are on imaginary axis Based on slide by Prof. Adnan Kavak
Frequency-Domain Interpretation est
h(t)
y(t)
y (t ) = h(t )∗ e s t ∞
• y(t) = H(s) e s t for a particular value of s
= ∫ h(τ )e s (t −τ )dτ −∞
=e
st
∞
−s τ ( ) h τ e ∫−∞dτ H (s )
• Recall definition of frequency response: ej 2π f t
h(t)
y(t)
y (t ) = h(t )∗ e j 2π ∞
ft
= ∫ h(τ )e j 2π
f (t −τ )
−∞
=e
j 2π f t
dτ
∞
j 2π f τ ( ) h τ e ∫−∞dτ H(f
)
Frequency-Domain Interpretation • Generalized frequency: s = σ + j 2 π f • We may convert transfer function into frequency response by if and only if region of convergence of H(s) includes the imaginary axis H freq ( f ) = H (s )
s = j 2πf
• What about h(t) = u(t)?
1 H (s ) = s
for Re{s} > 0
We cannot convert H(s) to a frequency response However, this system has a frequency response
• What about h(t) = δ(t)? H (s ) = 1 for all s ⇒ H freq ( f ) = 1
Frequency Selectivity in Filters • Lowpass filter |Hfreq(f)|
• Bandpass filter |Hfreq(f)|
1
1
• Highpass filter
f
|Hfreq(f)|
f
• Bandstop filter |Hfreq(f)|
f
Linear time-invariant filters are BIBO stable
f
Passive Circuit Elements • Laplace transforms with zero-valued initial conditions • Capacitor + dv v(t) i (t ) = C dt – I (s ) = C s V (s ) 1 ( ) V s = I (s ) Cs V (s ) 1 H (s ) = = I (s ) C s
• Inductor
+
v(t) di v(t ) = L dt – V (s ) = L s I (s ) V (s ) H (s ) = =Ls I (s )
• Resistor v(t ) = R i (t ) V (s ) = R I (s ) V (s ) H (s ) = =R I (s )
+
v(t) –
First-Order RC Lowpass Filter R + x(t)
C
i(t)
+ y(t)
Time domain R + X(s)
I(s)
1 Cs
+ Y(s)
X ( s) 1 R+ Cs 1 Y (s) = I ( s) Cs 1 Y ( s) = C s X ( s) 1 R+ Cs 1 Y (s) = RC X (s) s + 1 RC I ( s) =
X ( s) 1 R+ Cs Y (s) = R I (s) R Y (s) = X ( s) 1 R+ Cs Y (s) s = X (s) s + 1 RC I ( s) =
Frequency response is also an example of a notch filter
Passive Circuit Elements • Laplace transforms with non-zero initial conditions • Capacitor i (t ) = C
[
[
( )] = L s I (s ) − L i (0 ) ⎡ i (0 )⎤ = L s I (s ) −
V (s ) = L s I (s ) − i 0 −
−
( )] v(0 ) I (s ) + −
− 1 V (s ) = Cs s 1 = I (s ) + C v 0 − Cs
[
di V (t ) = L dt −
dv dt
I (s ) = C s V (s ) − v 0
• Inductor
( )]
⎢ ⎣
⎥ s ⎦
Operational Amplifier • Ideal case: model this nonlinear circuit as linear and time-invariant Input impedance is extremely high (considered infinite) vx(t) is very small (considered zero) + vx(t) _
_ +
+ y(t)
_
Operational Amplifier Circuit • Assuming that Vx(s) = 0, Y (s ) = − I (s ) Z f (s ) F (s ) Z (s ) F(s) Z f (s ) Y (s ) = − F (s ) + _ Z (s ) Z f (s ) Y (s ) H (s ) = =− F (s ) Z (s )
H(s)
I (s ) =
Z(s)
• How to realize a gain of –1? • How to realize a gain of 10?
I(s)
Zf(s)
+ _ Vx(s) _ +
+ Y(s)
_
Differentiator • A differentiator amplifies high frequencies, e.g. high-frequency components of noise: H(s) = s for all values of s (see next slide) Frequency response is H(f) = j 2 π f ⇒ | H( f ) |= 2 π | f |
• Noise has equal amounts of low and high frequencies up to a physical limit • A differentiator may amplify noise to drown out a signal of interest • In analog circuit design, one would generally use integrators instead of differentiators
Ø The DECIBEL value is a logarithmic measurement of the ratio of one variable to another of the same type. Ø Decibel value has no dimension. Ø It is used for voltage, current and power gains.
Magnitude H
Decibel Value HdB
0.001
-60
0.01
-40
0.1
-20
0.5
-6
Some Properties of Logarithms
1/√2
-3
log P1 P2 = log P1 + log P2
1
0
√2
3
2
6
10
20
20
26
⎛ P1 ⎞ log ⎜ ⎟ = log P1 − log P2 ⎝ P2 ⎠ log P n = n log P log 1 = 0
100
40
H dB
V2 = 20 log10 H = 20 log10 V1
Typical Sound Levels and Their Decibel Levels.
EXAMPLE 1 Construct Bode plots for
H (ω ) =
200 jω ( jω + 2)( jω + 10)
• Express transfer function in Standard form. STANDARD FORM H (ω ) =
10 jω (1 + jω 2)(1 + jω 10)
• Express the magnitude and phase responses. H db = 20log10 10 + 20log10 jω − 20log10 1 + jω 2 − 20log10 1 + jω 10
φ = 90° − tan -1 (ω 2) − tan -1 (ω 10) • Two corner frequencies at ω=2, 10 and a zero at the origin ω=0. • Sketch each term and add to find the total response.
EXAMPLE 1 Construct Bode plots for
H (ω ) =
200 jω ( jω + 2)( jω + 10)
H db = 20log10 10 + 20log10 jω − 20log10 1 + jω 2 − 20log10 1 + jω 10 26 dB
20log10 2 = 6dB
X
X
φ = 90° − tan -1 (ω 2) − tan -1 (ω 10)
X
X
EXAMPLE 2 Continued: Let us calculate |H| and φ at ω=50 rad/sec graphically.
• Express transfer function in Standard form. STANDARD FORM 0.4(1 + jω 10) H (ω ) = jω (1 + jω 5) 2
• Express the magnitude and phase responses. H db = 20log10 0.4 + 20log10 1 + jω 10 − 20log10 jω − 40log10 1 + jω 5
φ = 0° + tan -1 (ω 10) − 90° − 2 tan -1 (ω 5) • Two corner frequencies at ω=5, 10 and a zero at ω=10. • The pole at ω=5 is a double pole. The slope of the magnitude is -40 dB/decade and phase has slope -90 degree/decade. • Sketch each term and add to find the total response.
EXAMPLE 3 Construct Bode plots for
H (ω ) =
jω + 10 jω ( jω + 5 )
2
H db = 20 log10 0.4 + 20 log10 1 + jω 10 −
X
O
X
O
20 log10 jω − 40 log10 1 + jω 5
φ = 0° + tan -1 (ω 10) -1
− 90° − 2 tan (ω 5)
PRACTICE PROBLEM 4 Obtain the transfer function for the Bode plot given.
A zero at ω = 0.5, 1 + jω 0.5 1 A pole at ω = 1, 1 + jω 1 Two poles at
