bioe 162 final review 2014 ho
TRANSCRIPT
• Complex numbers and opera0ons • Addi0on of sinusoids -‐ phasor addi+on -‐ determine periodicity of the sum (review HW#2 & HW#3) • Signals and opera0ons
Final Reviews
• Representa)on and proper)es of LTIC systems -‐ Mathema+cal models/diff. eqs. for typical processes (first-‐order, second-‐order) -‐ System response from solu+on to diff. eq. zero-‐input + zero-‐state response • Time domain approach -‐ Impulse Response h(t) -‐ Convolu+on • Complex frequency domain (s-‐domain) approach -‐ Laplace Transform
-‐ Transfer func+on H(s) -‐ Zero-‐state response • Frequency domain approach -‐ Fourier Transform
-‐ Frequency Response H(jω)
)/sin( txA ωλ −
}Re{)( )( 0θω += tjAety
Re
Im y(t): Projec0on on Real axis
Represent sinusoids using phasors -‐ Euler formula:
A sinusoid:
0θ Re
Im ω (rad/s)
A
)(sin)(cos)( tjte tj ααα +=
Initial phase (at t=0):
)( 0θω +tjAeConsider as a rotating vector (at ω rad/s), a phasor,
0θ
)cos()( 0θω += tAtyUsing Euler’s formula
Time (t)
}Re{ )(2
)(1
21 θωθω ++ + tjtj eAeA
)cos()cos( 2211 θωθω +++ tAtA
}Re{ tjj eCe ωθ
}Re{ )( θω +tjCe
})Re{( 2121
tjjj eeAeA ωθθ +Where,
θθθ jjj CeeAeA =+ 2121
Addi+on of two sinusoids of same freq. using Euler’s formula
)cos( θω +tC
)/sin( txA ωλ −
)cos()cos()cos( 2211 θωθωθω +=+++ tCtAtA
Projec0on on Real axis
C
Re
Im
1A
2A
2θ θ 1θ
ω
At initial position (at t=0):
Phasor addi+on of two sinusoids of same freq.
Time (t)
θθθ jjj CeeAeA =+ 2121
}Re{}Re{ )()(2
)(1
21 θωθωθω +++ =+ tjtjtj CeeAeA
)/sin( txA ωλ −
)cos()cos()cos( 2211 θωθωθω +=+++ tCtAtA
Phasor addi+on of two sinusoids of same freq.
)cos(sincos θωωω +=+ tCtbta
a = A1 cosθ1 + A2 cosθ2
b = −(A1 sinθ1 + A2 sinθ2 )
Use trigonometric identities
)/sin( txA ωλ − )cos(sincos θωωω +=+ tCtbta
-‐ Phasor addi0on of sinusoids (of same frequency)
θθsincosCbCa=−
=Where, OR
abbaC−
=
+=
−1
22
tanθ
(Use trigonometric identities)
Re
Im
a
b− Cθ
Same as conversion of from Cartesian to Polar form
jbaZ −=θjCeZ =
Prac0ce Ques0on # 4: Use phasor addi;on to put the following into the form of x(t) = A cos(ωt +θ): a). x(t) = 2 cos(7t) + sin(7t)
A = 5
θ = tan−1 −12
Re
Im
2
−1
θ
Prac0ce Ques0on # 4: Use phasor addi;on to put the following into the form of x(t) = A cos(ωt +θ): b). x(t) = sin(2πt -‐ 45o ) + 3 sin(2πt)
acos2π t + bsin2π tStep 1: write x(t) in form of Step 2: find
a = − 22
b = 3+ 22
A = (0.707)2 + (3.707)2
θ = tan−1 −3.707−0.707
Re
Im
−0.707
−3.707
θ
Review: Linear Systems
• Linear system vs. non-‐linear system
linear system is characterized by linearity property that implies superposition (additive property); if several inputs are acting on a linear system, the total response is the sum of the responses from each input; time-‐shift property • Causal vs. non-‐causal signals/systems -‐ a causal signal (or response from a causal system) has zero values for t<0 -‐ unit step u(t) is often used to represent causal signals
Mathematical Models (Diff. Eqs.) are followed by
• Solutions of Diff. Equations - allow System Response and performance to be analyzed and designed
Ø Diff. Equation for LTIC (Linear Time-‐Invariant Continuous) System: f(t)-‐input; y(t)-‐output Ø Solution to Diff. Eq. Total response = zero-‐input response + zero-‐state response
)()()()( tfDPtyDQ =
011
1
012
21
1
.....)(
.....)(
bDbDbDbDPaDaDaDaDDQ
mm
mm
nn
nn
n
+++=
++++=−
−
−−
−−
)( nm ≤
Review of Lecture 9 • Impulse response is the zero-‐input or natural
response with specific ini;al condi;ons
• Given Impulse response of a system, h(t), the system’s response to any input f(t) can be obtained from the convolution of h(t) and f(t)
• Convolu;on proper;es and the given table
τττ∫ −=t
dthfthtf0
)()()(*)( (For causal systems)
)(tf )(ty)(th
Arbitrary Input System Response
h(t): Impulse Response
Time-‐domain Representation of System Input-‐Output
)(*)()( thtfty =
The Convolution Integral
)( τ−tf )( τ−ty)(th
Time-‐domain Representation of System Input-‐Output
)(*)()( thtfty ττ −=−
The Convolution shift property
)(tf )(ty)(th
Arbitrary Input System Response
h(t): Impulse Response
6
Prac;ce problem 6
g(t) = e−tu(t)Consider function Time-delayed by 1: Let y1(t) = g(t)*h(t) Then y1(t-1) = g(t-1)*h(t) and y(t)=y1(t-1)/e
g(t −1) = e−(t−1)u(t −1)
f (t) = e−tu(t −1) = 1eg(t −1)
g(t) y1(t))(th
g(t −1) y1(t −1))(th
f (t) = 1eg(t −1) 1
ey(t −1) = y(t)
)(thInput Response
• Transfer Function of a system is the LT of the Impulse Response Definition of Laplace Transform: Inverse transform:
H (s) = L[h(t)]
L[h(t)]= h(t)e−st0
∞
∫ dt
h(t) = L−1[H (s)]
• Transfer Function of a system is the LT of the Impulse Response • Transfer Function: the ratio of the LT of the output y(t) to the LT of the input x(t)
)()()(sXsYsH =
H (s) = L[h(t)]
Assume ini+al condi+ons = 0 when determine transfer func+on !
Ø Diff. Equation for LTIC (Linear Time-‐Invariant Continuous) System: x(t)-‐input; y(t)-‐output
)()()()( txDPtyDQ =
011
1
012
21
1
.....)(
.....)(
bDbDbDbDPaDaDaDaDDQ
mm
mm
nn
nn
n
+++=
++++=−
−
−−
−−
)( nm ≤
Apply Laplace Transform to the diff. eq. and obtain: or Transfer Function:
)()()()( sYsPsYsQ =
)()(
)()()(
sQsP
sXsYsH ==
Common System Types
• First-‐order
• Second-‐order
• Integra;on • Time-‐Delay
)()()( tKxtydttdy
=+τ
)()( tKxdttdy=
)()()(2)(12
2
2tKxty
dttdy
dttyd
nn
=++ως
ω
)()( DtKxty −=
Transfer Func;ons for Common System Types
• First-‐order
• Second-‐order
• Integra;on • Time-‐Delay
1)()(
+=sK
sXsY
τ
)121(
1)()(
22 ++
=sssX
sY
nn ως
ω
sK
sXsY
=)()(
sDesXsY −=)()(
Example: determine the impulse response of a second-‐order system:
1. Transfer function form:
2. Find inverse LT of H(s)
)()4()()2( 2 txDtyDD +=+
sss
sXsYsH
24
)()()(
2 +
+==
Partial fraction expansion & L.T. table
h(t) = L−1[H (s)]= L−1[ s+ 4s2 + 2s
]
Example: determine the impulse response of a second-‐order system:
1. Transfer function form:
2. Find inverse LT of H(s)
)()4()()2( 2 txDtyDD +=+
sss
sXsYsH
24
)()()(
2 +
+==
h(t) = L−1[H (s)]= L−1[ s+ 4s2 + 2s
] = L−1[2s−
1(s+ 2)
]
Poles (roots of the denominator): 0, -‐2 Factors: s, s+2
h(t) = (2− e−2t )u(t)
In case of complex conjugate factors
Non-‐repeated or repeated real factors
Laplace Transform Table
)()()()( txDPtyDQ =
)()()()( sXsPsYsQ =
)()()( sXsHsY =
Diff. Eq. Algebraic eq.
)]([)( 1 sYLty −=
LT
Inverse LT
S-domain
Assuming zero initial conditions
)()(
)()()(
sQsP
sXsYsH ==
Substitute transform of input X(s) and solve for Y(s) Obtain time-domain Solution y(t) (zero-state response)
Find zero-‐state response: Example: determine the unit step response of a second-‐order system: (assuming zero initial conditions) 1. Transfer function form:
2. Find LT of the input: X(s) = L[u(t)]=1/ s
)()4()()2( 2 txDtyDD +=+
Y (s) = s+ 4s2 + 2s
X(s) = s+ 4s(s2 + 2s)
sss
sXsYsH
24
)()()(
2 +
+==
3. Find Y(s):
Example: determine the impulse response of a second-‐order system: (assuming zero initial conditions) 4. Find y(t) from inverse L.T. and partial fraction expansion
)()4()()2( 2 txDtyDD +=+
y(t) = L−1[ s+ 4s(s2 + 2s)
]= L−1[− 0.5s+2s2+0.5(s+ 2)
]
Poles (roots of the denominator): 0,0, -‐2 Factors: s repeated, s+2
y(t) = (−0.5+ 2t + 0.5e−2t )u(t)
(Continue)
aa
Prac0ce Ques0on # 8:
Define: τ = t – D
LT of a time delayed function
for t < D: f(t-‐D) = 0
t
f(t)
f(t-‐D)
0
L[ f (t −D)]= f (t −D)e−st0
∞
∫ dt
L[ f (t −D)]= e−sD f (τ )e−sτ0
∞
∫ dτ
L[ f (t −D)]= e−sDL[ f (t)]
aa
Prac0ce Ques0on # 9:
L[x(t − 2)]= e−2sL[x(t)]
X(s)
L−1[F(s)]= x(t − 2)
X(s) = 2s+ 5s2 + 5s+ 6
x(t) = L−1[X(s)]
L[x(t − 2)]= F(s)
Step 1
Step 2
Lecture 10 Review: • Fourier theory: any periodic signal can be
represented as a sum of sine and cosine waves with harmonic frequencies
Fourier Transform (FT)
• A non-‐periodic func;on f(t) can be represented as a sum of sin’s and cos’s of (possibly) all frequencies:
∫∞
∞−= ωω
πω deFtf tj)(
21)(
∫∞
∞−
−= dtetfF tjωω )()(Fourier Transform (FT) :
Inverse Fourier Transform (IFT)
Prac0ce ques0on #2 Let x(t) denotes a rectangular pulse shown in the figure below, find the Fourier Transform of this signal, X(ω).
X(ω) = x(t)e− jωt dt−∞
∞
∫ = e− jωt dt−1
1∫
=1− jω
(e− jω − e jω )
=2sinωω
Sampling Theorem • If a signal contains the highest frequency fmax, the
sampling frequency (rate) (fs) >= 2 fmax
• Nyquist frequency: fNyquist = 2 fmax
• The sampling theory insures that the digi;zed signal truly represents the original analog signal (No loss of informa+on, No aliasing)
Prac0ce Ques0on # 9: