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THIS CHAPTER DEALS WITH CONVOLUTION ALONG WITH A SET OF SOLVED PROBLEMS
Typology: Exercises
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Prof. Paul Cuff Slides courtesy of John Pauly (Stanford)
Princeton University
Fall 2011-
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 1 / 55
Today’s topics
Impulse response Extended linearity Response of a linear time-invariant (LTI) system Convolution Zero-input and zero-state responses of a system
The impulse response of a linear system hτ (t) is the output of the system at time t to an impulse at time τ. This can be written as
hτ = H(δτ )
Care is required in interpreting this expression!
0 0 t
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 3 / 55
Note: Be aware of potential confusion here: When you write hτ (t) = H(δτ (t))
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).
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 y 1 = Sx 1 and y 2 = Sx 2 , then
y 1 + y 2 = S(x 1 + x 2 ).
Combined Homogeneity and Superposition: If y 1 = Sx 1 and y 2 = Sx 2 , and a and b are constants,
ay 1 + by 2 = S(ax 1 + bx 2 )
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 7 / 55
Extended Linearity
Summation: If yn = Sxn for all n, an integer from (−∞ < n < ∞), and an are constants
∑
n
anyn = S
n
anxn
Summation and the system operator commute, and can be interchanged. Integration (Simple Example) : If y = Sx, ∫ (^) ∞
−∞
a(τ )y (t − τ ) dτ = S
−∞
a(τ )x(t − τ )dτ
Integration and the system operator commute, and can be interchanged.
We would like to determine an expression for the output y (t) of an linear time invariant system, given an input x(t)
x y H
We can write a signal x(t) as a sample of itself
x(t) =
−∞
x(τ )δτ (t) dτ
This means that x(t) can be written as a weighted integral of δ functions.
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 9 / 55
Applying the system H to the input x(t),
y (t) = H (x(t))
= H
−∞
x(τ )δτ (t)dτ
If the system obeys extended linearity we can interchange the order of the system operator and the integration
y (t) =
−∞
x(τ )H (δτ (t)) dτ.
The impulse response is
hτ (t) = H(δτ (t)).
The System Equation relates the outputs of a system to its inputs.
Example from last time: the system described by the block diagram
++
Z
a
has a system equation y ′^ + ay = x. In addition, the initial conditions must be given to uniquely specify a solution.
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 13 / 55
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.
t
0 0 t
The zero-state response, which is the output of the system with all initial conditions zero.
t
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.
t
(^00)
t
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 15 / 55
Example: Solve for the voltage across the capacitor y (t) for an arbitrary input voltage x(t), given an initial value y (0) = Y 0.
C y ( t )
i ( t )
Then
y (t) = A(t)e−t/RC
= e−t/RC
∫ (^) t
0
x(τ )
eτ /RC
dτ + A(0)e−t/RC
∫ (^) t
0
x(τ )
e−(t−τ^ )/RC
dτ + A(0)e−t/RC
At t = 0, y (0) = Y 0 , so this gives A(0) = Y 0
y (t) =
∫ (^) t
0
x(τ )
e−(t−τ^ )/RC
dτ ︸ ︷︷ ︸ zero−state response
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 19 / 55
RC Circuit example
The impulse response of the RC circuit example is
h(t) =
e−t/RC
The response of this system to an input x(t) is then
y (t) =
∫ (^) t
0
x(τ )hτ (t)dτ
∫ (^) t
0
x(τ )
e−(t−τ^ )/RC
dτ
which is the zero state solution we found earlier.
Example: High energy photon detectors can be modeled as having a simple exponential decay impulse response.
Photon
Light
Scintillating Crystal
ing the PMT socket containing the dynode resistor chain network, is 3 cm long, 3 cm wide, and 9.75 cm long. ETHODS—DETECTOR CHARACTERIZATION lood source histogram detector module was uniformly irradiated with a 68 Ge nt source !2.6 "Ci#. The signals from the PS-PMT were ted and digitized as described above in Sec. II D. The er energy threshold was set to approximately $100 keV h the aid of the threshold on the constant fraction dis- inator and no upper energy threshold was applied. nergy spectra oundaries were drawn on the 2D position map to define k-up table !LUT# which relates position in the 2D his- am to the appropriate element in the LSO array. The raw mode data were then resorted and a histogram of total e amplitudes !sum of the four position outputs# gener- for each crystal in the array. These energy spectra were lyzed to determine the FWHM and the location of the keV photopeak of each crystal. These two parameters sure the energy resolution and light collection efficiency, ectively. iming resolution wo detectors were aligned facing each other, 15 cm rt, and connected in coincidence. A 22 Na point source "Ci# encapsulated in a 25.4 mm diameter, 3 mm thick r plastic disc with the activity in the central 1 mm was ed in the center of the two detectors. For each detected cidence event, the sum of the four position signals for h detector was sent to a constant fraction discriminator ch generated timing pulses. The lower energy threshold he CFD was set to approximately 100 keV. These two ing pulses !one for each module# were in turn fed into a brated time-to-amplitude converter !TAC# module. The put from the TAC was then digitized to produce the tim- pectrum. oincidence point spread function lood source histograms of both detectors were obtained
were defined. The detectors were then configured in coinci- dence, 15 cm apart, and list-mode data was acquired by step- ping a 1 mm diameter 22 Na point source !same as used in Sec. III C# between the detectors in 0.254 mm steps. The point source was scanned across the fifth row of the detector. For each opposing crystal pair, the counts were recorded as a function of the point source position. A lower energy win- dow of $100 keV was applied. The FWHM of the resulting distribution for each crystal pair was determined to give the intrinsic spatial resolution of the detectors. E. Detector efficiency A measure of the absolute detector efficiency was ob- tained. A 18 F point source with known activity! 68 "Ci# was placed 15 cm away from the face of the detector module. The actual gamma-ray flux impinging on the detector face was calculated from the solid angle subtended by the detector module at the source. The constant fraction discriminator was set to eliminate electronic noise ($100 keV# and the full energy spectrum was obtained for each crystal over a fixed time. A background measurement without the 18 F point source was also obtained to subtract the LSO background from the measurement. A lower energy window of 350 keV was applied to all of the crystals and the number of counts falling under the photopeak was calculated. The number of counts detected was then divided by the total number of gamma rays impinging on the detector module to obtain the detector efficiency. IV. RESULTS—DETECTOR CHARACTERIZATION A. Flood source histogram results An image of the flood histogram from one detector mod- ule is shown in Fig. 6. All 81 crystals from the 9!9 LSO array are clearly visible. An average peak-to-valley ratio of 3.5 was obtained over the central row of nine crystals. Not all crystals are uniformly spaced in the flood histogram after applying Anger logic. This may be a result of the nonuni- form tapering of the optical fiber taper, the nonuniform pack- ing of the reflectance powder between the crystals, or most
LE ts. I. Summary results from the various lightguide configuration experi-
oupler
Energy !^ resolutionFWHN %#
Light efficiency^ collection !%#
Average valley ratio^ peak-to-
Number of crystals clearly resolved ect LSO htguideaa 13.019.9 100.040.6 10.02.5 (^97) V lens 27.2 28.0 2.5 7 er er tapera^ 35.019.5 12.627.0 6.07.5 (^69) rgy resolution and light collection efficiency were measured with single tguide elements. F array of 3IG. 5. A picture of the assembled detector module consisting of a 9! 3 !20 mm (^3) LSO crystals coupled through a tapered optical fiber! 9 bundle to a Hamamatsu R5900-C8 PS-PMT.
0 Doshi et al.: LSO PET detector 1540
Crystal
Light Fibers
Photomultiplier
From: Doshi et al, Med Phys. 27(7), p1535 July 2000
These are used in Positiron Emmision Tomography (PET) systems. Input is a sequence of impulses (photons).
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 21 / 55
Output is superposition of impulse responses (light).
For a causal system h(t) = 0 for t < 0, and
y (t) =
−∞
x(τ )hτ^ (t) dτ.
Since hτ^ (t) = 0 for t < τ , we can replace the upper limit of the integral by t y (t) =
∫ (^) t
−∞
x(τ )hτ (t) dτ.
Only past and present values of x(τ ) contribute to y (t).
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 25 / 55
A LTI system has a real impulse response h(t). A sinusoidal input
x(t) = A cos(2πf 1 t + θ)
produces an output
y (t) =
−∞
h(τ ) [A cos(2πf 1 (t − τ ) + θ)] dτ.
Using the identity cos(a − b) = cos a cos b + sin a sin b,
y (t) = A cos(2πf 1 t + θ)
−∞
h(τ ) cos(2πf 1 τ )dτ
+A sin(2πf 1 t + θ)
−∞
h(τ ) sin(2πf 1 τ )dτ.
Since h(t) is real,
y (t) = Hc (f 1 )A cos(2πf 1 t + θ) + Hs (f 1 )A sin(2πf 1 t + θ).
where
Hc (f 1 ) =
−∞
h(τ ) cos(2πf 1 τ )dτ
Hs (f 1 ) =
−∞
h(τ ) sin(2πf 1 τ )dτ
are real constants.
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 27 / 55
We can then write the output as
y (t) = |H(f 1 )|A cos (2πf 1 t + θ + ∠H(f 1 ))
(using the same trigonometric identity in reverse), where
|H(f )| =
H^2 c (f 1 ) + H s^2 (f 1 ) ∠H(f 1 ) = tan−^1 (−Hs (f 1 )/Hc (f 1 ))
Note that the response to a sinusoidal input is determined by a single complex number H(f 1 ), which determines the magnitude of the output, and the phase shift.
A sinusoidal input is scaled and delayed by an LTI system, but is otherwise unchanged.
Review: response of an LTI system Representation of convolution Graphical interpretation Examples Properties of convolution
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 31 / 55
The convolution of an input signal x(t) with and impulse response h(t) is
y (t) =
−∞
x(τ )h(t − τ ) dτ
= (x ∗ h)(t)
or y = x ∗ h. This is also often written as
y (t) = x(t) ∗ h(t)
which is potentially confusing, since the t’s have different interpretations on the left and right sides of the equation (your book does this).
For a causal system h(t) = 0 for t < 0, and
y (t) =
−∞
x(τ )h(t − τ ) dτ.
Since h(t − τ ) = 0 for t < τ , the upper limit of the integral is t
y (t) =
∫ (^) t
−∞
x(τ )h(t − τ )dτ.
Only past and present values of x(τ ) contribute to y (t).
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 33 / 55
0 t (^0) t
y ( t )
τ
x ( t )
x ( t ) =
Z (^) ∞ −∞
x (τ)δ( t − τ) d τ
τ < t τ > t
y(t) =
∫ (^) t
−∞
x(τ )h(t − τ )dτ
Does not contribute to y(t)
If x(t) is also causal, x(t) = 0 for t < 0, and the integral further simplifies
y (t) =
∫ (^) t
0
x(τ )h(t − τ ) dτ.
0 t (^0) t
y ( t )
τ
x ( t )
Does not contribute to y(t)
Does not contribute to y(t)
x(t) =
∫ (^) t 0
x(τ )δ(t − τ )dτ y(t) =
∫ (^) t 0
x(τ )h(t − τ )dτ
Simple Example
-1 0 1 2 3
1
2
-1 0 1 2 3
1
2 h (τ)
τ
h ( t − τ)
h (−τ)
-1 0 1 2 3
1
(^2) x (τ)
τ
-1 0 1 2 3
1
2
τ
τ
Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 37 / 55
-1 0 1 2 3
1
(^2) x (τ)
τ
h ( t − τ) t < 0
-1 0 1 2 3
1
x (τ)
τ
h ( t − τ) 0 < t < 1
-1 0 1 2 3
1 h ( t − τ) x (τ)^1 <^ t^ <^2
-1 0 1 2 3
1
(^2) x (τ)
τ
h ( t − τ)^2 <^ t^ <^3
-1 0 1 2 3
1
(^2) x (τ)
τ
h ( t − τ) t^ >^3
-1 0 1 2 3
1
2
τ
y ( t ) = ( x ∗ h )( t )
Communication channel, e.g., twisted pair cable
x ( t ) y ( t ) ∗ h ( t )
Impulse response:
(^00 2 4 6 8 )
1
t
h
h ( t )
t
This is a delay ≈ 1, plus smoothing. Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 39 / 55
Simple signaling at 0.5 bit/sec; Boolean signal 0, 1 , 0 , 1 , 1 ,...
0 2 4 6 8 10
0
1
t
u
0 2 4 6 8 10
0
1
t
y
t
x ( t )
y ( t )
t
Output is delayed, smoothed version of input.
1’s & 0’s easily distinguished in y
