Stanford University
Summer 2015-2016
Signal Processing and Linear Systems I
Lecture 5: Time Domain Analysis of Continuous Time
Systems
June 27, 2016
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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Time Domain Analysis of Continuous Time Systems
• Zero-input and zero-state responses of a system
• Impulse response
• Extended linearity
• Response of a linear time-invariant (LTI) system
• Superposition integral
• Convolution
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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System Equation
The System Equation relates the outputs of a system to its inputs.
Example from last time: the system described by the block diagram
x
+
+
-
Z
y
a
has a system equation
y 0 + ay = x.
In addition, the initial conditions must be given to uniquely specifiy a
solution.
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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Solutions for the System Equation
Solving the system equation tells us the output for a given input.
The output consists of two components:
• The zero-input response, which is what the system does with no input at
all. This is due to initial conditions, such as energy stored in capacitors
and inductors.
x(t) = 0
0
y(t)
0
t
t
H
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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• The zero-state response, which is the output of the system with all initial
conditions zero.
x(t)
0
y(t)
0
t
H
t
If H is a linear system, its zero-input response is zero. Homogeneity
states if y = F (ax), then y = aF (x). If a = 0 then a zero input requires
a zero output.
x(t) = 0
0
y(t) = 0
t
H
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
0
t
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Example: Solve for the voltage across the capacitor y(t) for an arbitrary
input voltage x(t), given an initial value y(0) = Y0.
i(t) R
x(t)
+
−
C
+
−
y(t)
From Kirchhoff’s voltage law
x(t) = Ri(t) + y(t)
Using i(t) = Cy 0(t)
RCy 0(t) + y(t) = x(t).
This is a first order LCCODE, which is linear with zero initial conditions.
First we solve for the homogeneous solution by setting the right side (the
input) to zero
RCy 0(t) + y(t) = 0.
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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The solution to this is
y(t) = Ae−t/RC
which can be verified by direct substitution.
To solve for the total response, we let the undetermined coefficient be a
function of time
y(t) = A(t)e−t/RC .
Substituting this into the differential equation
1
0
−t/RC
RC A (t)e
−
A(t)e−t/RC + A(t)e−t/RC = x(t)
RC
Simplying
1 t/RC
A (t) = x(t)
e
RC
which can be integrated from t = 0 to get
Z t
1 τ /RC
A(t) =
x(τ )
e
dτ + A(0)
RC
0
0
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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Then
y(t) = A(t)e−t/RC
Z t
1 τ /RC
x(τ )
= e−t/RC
e
dτ + A(0)e−t/RC
RC
0
Z t
1 −(t−τ )/RC
x(τ )
=
e
dτ + A(0)e−t/RC
RC
0
At t = 0, y(0) = Y0, so this gives A(0) = Y0
Z
y(t) =
t
x(τ )
|0
1 −(t−τ )/RC
−t/RC
e
dτ +
.
Y
e
0
|
{z
}
RC
{z
} zero−input response
zero−state response
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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Impulse Response
The impulse response of a linear system h(t, τ ) is the output of the system
at time t to an impulse at time τ . This can be written as
h(t, τ ) = H(δ(t − τ ))
Care is required in interpreting this expression!
δ(t)
0
h(t, 0)
t
0
H
δ(t − τ)
0 τ
t
t
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
h(t, τ)
0
t
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Note: Be aware of potential confusion here:
When you write
h(t, τ ) = H(δ(t − τ ))
the variable t serves different roles on each side of the equation.
• t on the left is a specific value for time, the time at which the output is
being sampled.
• t on the right is varying over all real numbers, it is not the same t as on
the left.
• The output at time specific time t on the left in general depends on the
input at all times t on the right (the entire input waveform).
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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• Assume the input impulse is at τ = 0,
h(t, 0) = H(δ(t)).
We want to know the impulse response at time t = 2. It doesn’t make
any sense to set t = 2, and write
h(2, 0) = H(δ(2))
⇐ No!
First, δ(2) is something like zero, so H(0) would be zero. Second, the
value of h(2, 0) depends on the entire input waveform, not just the value
at t = 2.
H
δ(t)
0
0
δ(2)
2
t
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
0
h(t, 0)
h(2, 0)
2
t
11
• Compare to an equation such as
y 0(t) + 2y(t) = x(t)
which holds for each t, so that y 0(1) + 2y(1) = x(1).
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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If H is time invariant, delaying the input and output both by a time τ
should produce the same response
h(t, τ ) = h(t − τ, τ − τ ) = h(t − τ, 0).
Hence h is only a function of t − τ . We suppress the second argument, and
define the impulse response of a linear time-invariant (LTI) system H to be
h(t) = H(δ(t))
δ(t)
0
h(t)
t
0
H
δ(t − τ)
0 τ
t
t
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
h(t − τ)
0
t
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RC Circuit example
i(t) R
x(t) +
−
C
+
−
y(t)
The solution for an input x(t) and initial y(0) = Y0 is
Z
y(t) =
0
t
1 −(t−τ )/RC
x(τ )
e
dτ + Y0e−t/RC
RC
The zero-state response is (Y0 = 0) is
Z
y(t) =
t
x(τ )
0
1 −(t−τ )/RC
e
dτ
RC
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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The impulse response is then
Z
t
δ(τ )
h(t) =
0−
=
1 −(t−τ )/RC
e
dτ
RC
1 −t/RC
e
RC
for t ≥ 0, and zero otherwise. We integrate from 0− to include the impulse.
This impulse response looks like:
1
RC
1 −t/RC
e
RC
RC
2RC
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
t
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Linearity and Extended Linearity
Linearity: A system S is linear if it satisfies both
• Homogeneity: If y = Sx, and a is a constant then
ay = S(ax).
• Superposition: If y1 = Sx1 and y2 = Sx2, then
y1 + y2 = S(x1 + x2).
Combined Homogeneity and Superposition:
If y1 = Sx1 and y2 = Sx2, and a and b are constants,
ay1 + by2 = S(ax1 + bx2)
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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Extended Linearity
• Summation: If yn = Sxn for all n, an integer from (−∞ < n < ∞),
and an are constants
!
X
anyn = S
X
n
anxn
n
Summation and the system operator commute, and can be interchanged.
• Integration (Simple Example) : If y = Sx,
Z
∞
Z
a(τ )y(t − τ ) dτ = S
−∞
∞
a(τ )x(t − τ )dτ
−∞
Integration and the system operator commute, and can be interchanged.
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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Output of an LTI System
We would like to determine an expression for the output y(t) of an linear
time invariant system, given an input x(t)
y
x
H
We can write a signal x(t) as a sample of itself
Z
x(t) =
∞
x(τ )δ(t − τ ) dτ
−∞
This means that x(t) can be written as a weighted integral of δ functions.
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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Applying the system H to the input x(t),
y(t) = H (x(t))
Z ∞
x(τ )δ(t − τ )dτ
= H
−∞
If the system obeys extended linearity we can interchange the order of the
system operator and the integration
Z
y(t) =
∞
x(τ )H (δ(t − τ )) dτ.
−∞
The impulse response is
h(t, τ ) = H(δ(t − τ )).
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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Substituting for the impulse response gives
Z
y(t) =
∞
x(τ )h(t, τ )dτ.
−∞
This is a superposition integral.
The values of x(τ )h(t, τ )dτ are
superimposed (added up) for each input time τ .
If H is time invariant, this written more simply as
Z
y(t) =
∞
x(τ )h(t − τ )dτ.
−∞
This is in the form of a convolution integral, which will be the subject of
the next class.
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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Graphically, this can be represented as:
Input
Output
h(t)
δ(t)
0
t
t
0
δ(t − τ)
0 τ
x(t)
h(t − τ)
t
(x(τ)dτ)δ(t − τ)
0 τ
t
0 τ
t
(x(τ)dτ)h(t − τ)
0τ
t
y(t)
x(t)
0 τ
x(t) =
Z
∞
−∞
t
x(τ)δ(t − τ)dτ
t
0
y(t) =
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
Z
∞
−∞
x(τ)h(t − τ)dτ
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RC Circuit example, again
The impulse response of the RC circuit example is
1 −t/RC
h(t) =
e
RC
The response of this system to an input x(t) is then
Z
t
x(τ )h(t − τ )dτ
y(t) =
0
Z
=
t
x(τ )
0
1 −(t−τ )/RC
e
dτ
RC
which is the zero state solution we found earlier.
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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Example:
1540
High energy photon detectors can be modeled as having a simple exponential
decay impulse response.
1540
Doshi et al.: LSO PET detector
TABLE I. Summary results from the various lightguide configuration experiments.
Coupler
Direct LSOa
Lightguidea
PCV lens
Fibera
Fiber taper
Average
Light
Energy
peak-toresolution
collection
!FWHN %# efficiency !%# valley ratio
13.0
19.9
27.2
35.0
19.5
Light
100.0
40.6
28.0
12.6
27.0
10.0
2.5
2.5
6.0
7.5
Photomultiplier
Scintillating
Number of
crystals
clearly
Crystal
resolved
Light Fibers
9
7
7
6
9
Crystal
Energy resolution and light collection efficiency were measured with single
lightguide elements.
Photon
luding the PMT socket containing the dynode resistor chain
bias network, is 3 cm long, 3 cm wide, and 9.75 cm long.
From: Doshi et al, Med Phys. 27(7), p1535 July 2000
FIG. 5. A picture of the assembled detector module consisting of a 9!9
array of 3!3!20 mm3 LSO crystals coupled through a tapered optical fiber
bundle to a Hamamatsu R5900-C8 PS-PMT.
These are used in positiron emmision tomography (PET) systems.
were defined. The detectors were then configured in coincidence, 15 cm apart, and list-mode data was acquired by stepA. Flood source histogram
ping a 1 mm diameter 22Na point source !same as used in
Sec. III C# between the detectors in 0.254 mm steps. The
A detector module was uniformly irradiated with a 68Ge
point source was scanned across the fifth row of the detector.
point source !2.6 " Ci#. The signals from the PS-PMT were
For each opposing crystal pair, the counts were recorded as a
reated and digitized as described above in Sec. II D. The
function of the point source position. A lower energy winower energy threshold was set to approximately $100 keV
dow of $100 keV was applied. The FWHM of the resulting
with the aid of the threshold on the constant fraction disdistribution
each crystal pair was determined to give the
Processing
Systems I; Summer
15-16,forGibbons
riminator and EE102A:Signal
no upper energy
thresholdand
wasLinear
applied.
intrinsic spatial resolution of the detectors.
II. METHODS—DETECTOR CHARACTERIZATION
B. Energy spectra
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Input is a sequence of impulses (photons).
Output is superposition of impulse responses (light).
Input: Photons
Output: Light
t
t
t
t
t
t
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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Summary
• For an input x(t), the output of an linear system is given by the
superposition integral
Z
y(t) =
∞
x(τ )h(t, τ ) dτ
−∞
• If the system is also time invariant, the result is a convolution integral
Z
y(t) =
∞
x(τ )h(t − τ ) dτ
−∞
• The response of an LTI system is completely characterized by its impulse
response h(t).
EE102A:Signal Processing and Linear Systems I; Summer 15-16, Gibbons
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