Time Domain Analysis of Continuous Time Systems

Lecture 3
ELE 301: Signals and Systems Today’s topics

Impulse response
Prof. Paul Cuff Extended linearity
Slides courtesy of John Pauly (Stanford) Response of a linear time-invariant (LTI) system
Convolution
Princeton University
Zero-input and zero-state responses of a system

Fall 2011-12

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Impulse Response

The impulse response of a linear system hτ (t) is the output of the system Note: Be aware of potential confusion here:
at time t to an impulse at time τ . This can be written as When you write
hτ (t) = H(δτ (t))
hτ = H(δτ )
the variable t serves different roles on each side of the equation.
Care is required in interpreting this expression!
t on the left is a specific value for time, the time at which the output
δ(t) is being sampled.
h(t, 0)
t on the right is varying over all real numbers, it is not the same t as
0 t 0 t on the left.
H The output at time specific time t on the left in general depends on
δ(t − τ) h(t, τ) the input at all times t on the right (the entire input waveform).

0 τ t 0 t

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 Assume the input impulse is at τ = 0,
Time-invariance
If H is time invariant, delaying the input and output both by a time τ
h = h0 = H(δ0 ). should produce the same response

We want to know the impulse response at time t = 2. It doesn’t hτ (t) = h(t − τ ).

make any sense to set t = 2, and write

h(2) = H(δ(2)) ⇐ No! In this case, we don’t need to worry about hτ because it is just h shifted in
time.
First, δ(2) is something like zero, so H(0) would be zero. Second, the
value of h(2) depends on the entire input waveform, not just the δ(t)
value at t = 2. h(t)
0 t 0 t
H 0 H
δ(t) δ(t − τ)
δ(2) h(t, 0) h(2, 0) h(t − τ)
0 2 t 0 2 t 0 τ t 0 t

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Linearity and Extended Linearity

Extended Linearity

Linearity: A system S is linear if it satisfies both Summation: If yn = Sxn for all n, an integer from (−∞ < n < ∞),
and an are constants
Homogeneity: If y = Sx, and a is a constant then !
X X
an yn = S an x n
ay = S(ax).
n n

Superposition: If y1 = Sx1 and y2 = Sx2 , then Summation and the system operator commute, and can be
interchanged.
y1 + y2 = S(x1 + x2 ). Integration (Simple Example) : If y = Sx,
Combined Homogeneity and Superposition: Z ∞ Z ∞ 
If y1 = Sx1 and y2 = Sx2 , and a and b are constants, a(τ )y (t − τ ) dτ = S a(τ )x(t − τ )dτ
−∞ −∞

ay1 + by2 = S(ax1 + bx2 ) Integration and the system operator commute, and can be
interchanged.

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Output of an LTI System
Applying the system H to the input x(t),
We would like to determine an expression for the output y (t) of an linear y (t) = H (x(t))
time invariant system, given an input x(t) Z ∞ 
= H x(τ )δτ (t)dτ
−∞
x y
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 (δτ (t)) dτ.
We can write a signal x(t) as a sample of itself −∞
Z ∞
The impulse response is
x(t) = x(τ )δτ (t) dτ
−∞
hτ (t) = H(δτ (t)).
This means that x(t) can be written as a weighted integral of δ functions.

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Graphically, this can be represented as:

Substituting for the impulse response gives

Input Output
Z ∞
y (t) = x(τ )hτ (t)dτ. δ(t) h(t)
−∞
0 t 0 t
This is a superposition integral. The values of x(τ )h(t, τ )dτ are δ(t − τ) h(t − τ)
superimposed (added up) for each input time τ .
0 τ t 0 τ t
If H is time invariant, this written more simply as
(x(τ)dτ)δ(t − τ) (x(τ)dτ)h(t − τ)
Z ∞ x(t)
y (t) = x(τ )hτ (t)dτ. t
−∞ 0 τ 0τ t
y(t)
x(t)
This is in the form of a convolution integral, which will be the subject of
the next class. 0 τ t 0 t
Z ∞ Z ∞
x(t) = x(τ)δ(t − τ)dτ y(t) = x(τ)h(t − τ)dτ
−∞ −∞

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System Equation Solutions for the 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 Solving the system equation tells us the output for a given input.
The output consists of two components:
x +
Z y
+ 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.
a
x(t) = 0 y(t)
0 t 0 t
has a system equation H
0
y + ay = x.
In addition, the initial conditions must be given to uniquely specify a
solution.

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The zero-state response, which is the output of the system with all
initial conditions zero.

x(t) y(t) Example: Solve for the voltage across the capacitor y (t) for an arbitrary
input voltage x(t), given an initial value y (0) = Y0 .
0 t 0 t
H
i(t) R
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 x(t) +
−
C y(t)
−
requires a zero output.

x(t) = 0 y(t) = 0
0 t 0 t
H

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 To solve for the total response, we let the undetermined coefficient be a
From Kirchhoff’s voltage law
function of time
x(t) = Ri(t) + y (t) y (t) = A(t)e −t/RC .
Substituting this into the differential equation
Using i(t) = Cy 0 (t)
 
RCy 0 (t) + y (t) = x(t). 1
RC A0 (t)e −t/RC − A(t)e −t/RC + A(t)e −t/RC = x(t)
This is a first order LCCODE, which is linear with zero initial conditions. RC

First we solve for the homogeneous solution by setting the right side (the Simplifying  
1 t/RC
input) to zero A0 (t) = x(t) e
RCy 0 (t) + y (t) = 0. RC
which can be integrated from t = 0 to get
The solution to this is
y (t) = Ae −t/RC Z t  
1 τ /RC
A(t) = x(τ ) e dτ + A(0)
which can be verified by direct substitution. 0 RC

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Then RC Circuit example

y (t) = A(t)e −t/RC The impulse response of the RC circuit example is
Z t  
1 τ /RC
= e −t/RC x(τ ) e dτ + A(0)e −t/RC h(t) =
1 −t/RC
e
RC
Z t 0  RC
1 −(t−τ )/RC
= x(τ ) e dτ + A(0)e −t/RC The response of this system to an input x(t) is then
0 RC
Z t
At t = 0, y (0) = Y0 , so this gives A(0) = Y0 y (t) = x(τ )hτ (t)dτ
Z t   Z0 t  
1 −(t−τ )/RC 1 −(t−τ )/RC
y (t) = x(τ ) e dτ + Y e −t/RC . = x(τ ) e dτ
RC | 0 {z } RC
|0 {z } zero−input response 0
zero−state response
which is the zero state solution we found earlier.

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 Example: Output is superposition of impulse responses (light).
High energy photon detectors can be modeled as having a simple
exponential decay impulse response.
Input: Photons Output: Light
Doshi et al.: LSO PET detector 1540

ummary results from the various lightguide configuration experi-

Photomultiplier
Energy Light Average
Scintillating
Number of
resolution collection peak-to- Crystal
crystals clearly
t t
r !FWHN %# efficiency !%# valley ratio resolved
Light Fibers
Oa 13.0
ea 19.9 Light 100.0
40.6
10.0
2.5
9
7
27.2 28.0 2.5 7
35.0 12.6 6.0 6
Crystal
r 19.5 27.0 7.5 9

solution and light collection efficiency were measured with single

t t
elements. Photon
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
he PMT socket containing the dynode resistor chain bundle to a Hamamatsu R5900-C8 PS-PMT.
work, is 3 cm long, 3 cm wide, and 9.75 cm long.
These are used in Positiron Emmision
HODS—DETECTOR CHARACTERIZATION
Tomography (PET) systems.
were defined. The detectors were then configured in coinci-
t t
dence, 15 cm apart, and list-mode data was acquired by step-
Input is a sequence of impulses
d source histogram ping (photons).
a 1 mm diameter 22Na point source !same as used in
ector module was uniformly irradiated with a 68Ge Sec. III C# between the detectors in 0.254 mm steps. The
urce !2.6 " Ci#. The signals from the PS-PMT were point source was scanned across the fifth row of the detector.
nd digitized as described above in Sec. II D. The For each opposing crystal pair, the counts were recorded as a
ergy threshold was set (Lecture
Cuff to approximately
3) $100 keV function
ELE 301: of the point
Signals source position. A lower energyFall
and Systems win-
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aid of the threshold on the constant fraction dis- dow of $100 keV was applied. The FWHM of the resulting
or and no upper energy threshold was applied. distribution for each crystal pair was determined to give the
intrinsic spatial resolution of the detectors.
gy spectra
E. Detector efficiency
daries were drawn on the 2D position map to define
p table !LUT# which relates position in the 2D his- A measure of the absolute detector efficiency was ob-
o the appropriate element in the LSO array. The raw tained. A 18F point source with known activity !68 " Ci# was
e data were then resorted and a histogram of total placed 15 cm away from the face of the detector module. The
mplitudes !sum of the four position outputs# gener- actual gamma-ray flux impinging on the detector face was
each crystal in the array. These energy spectra were calculated from the solid angle subtended by the detector
to determine the FWHM and the location of the module at the source. The constant fraction discriminator
photopeak of each crystal. These two parameters was set to eliminate electronic noise ($100 keV# and the full
the energy resolution and light collection efficiency, energy spectrum was obtained for each crystal over a fixed
ely. time. A background measurement without the 18F point
source was also obtained to subtract the LSO background
ng resolution from the measurement. A lower energy window of 350 keV
was applied to all of the crystals and the number of counts
detectors were aligned facing each other, 15 cm
falling under the photopeak was calculated. The number of
nd connected in coincidence. A 22Na point source
counts detected was then divided by the total number of
i# encapsulated in a 25.4 mm diameter, 3 mm thick
gamma rays impinging on the detector module to obtain the
stic disc with the activity in the central 1 mm was
detector efficiency.
n the center of the two detectors. For each detected
nce event, the sum of the four position signals for
ector was sent to a constant fraction discriminator IV. RESULTS—DETECTOR CHARACTERIZATION
enerated timing pulses. The lower energy threshold A. Flood source histogram results
CFD was set to approximately 100 keV. These two
An image of the flood histogram from one detector mod-
ulses !one for each module# were in turn fed into a
ule is shown in Fig. 6. All 81 crystals from the 9!9 LSO
d time-to-amplitude converter !TAC# module. The
array are clearly visible. An average peak-to-valley ratio of
om the TAC was then digitized to produce the tim-
3.5 was obtained over the central row of nine crystals. Not
trum.
all crystals are uniformly spaced in the flood histogram after
applying Anger logic. This may be a result of the nonuni-
Summary
cidence point spread function
form tapering of the optical fiber taper, the nonuniform pack-
Another expression for the superposition integral can be found by
source histograms of both detectors were obtained ing of the reflectance powder between the crystals, or most
ibed in Sec. III A from which the position LUTs likely, the nonuniform spacing of the anode plates in the substituting for τ = t − τ1 . Then dτ = −dτ1 and τ1 = t − τ ,
Physics, Vol. 27, No. 7, July 2000 Z ∞
For an input x(t), the output of an linear system is given by the y (t) = x(τ )h(t − τ )dτ
superposition integral −∞
Z−∞
Z ∞
= x(t − τ1 )h(t − (t − τ1 ))d(−τ1 )
y (t) = x(τ )hτ (t) dτ
−∞
Z∞∞
= x(t − τ1 )h(τ1 )dτ1 .
If the system is also time invariant, the result is a convolution integral −∞
Z ∞
y (t) = x(τ )h(t − τ ) dτ The block diagrams for a system using the impulse response:
−∞

The response of an LTI system is completely characterized by its x(t) y(t) x(t) y(t)
impulse response h(t). ∗h(t) ∗h(t)

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Superposition Integral for Causal Systems LTI System Response to a Sinusoidal Input
A LTI system has a real impulse response h(t). A sinusoidal input

For a causal system h(t) = 0 for t < 0, and x(t) = A cos(2πf1 t + θ)

Z ∞
produces an output
y (t) = x(τ )hτ (t) dτ.
−∞ Z ∞
y (t) = h(τ ) [A cos(2πf1 (t − τ ) + θ)] dτ.
Since hτ (t) = 0 for t < τ , we can replace the upper limit of the integral −∞
by t Z t Using the identity cos(a − b) = cos a cos b + sin a sin b,
y (t) = x(τ )hτ (t) dτ. Z ∞
−∞
y (t) = A cos(2πf1 t + θ) h(τ ) cos(2πf1 τ )dτ
Only past and present values of x(τ ) contribute to y (t). −∞
Z ∞
+A sin(2πf1 t + θ) h(τ ) sin(2πf1 τ )dτ.
−∞

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We can then write the output as

Since h(t) is real, y (t) = |H(f1 )|A cos (2πf1 t + θ + ∠H(f1 ))
y (t) = Hc (f1 )A cos(2πf1 t + θ) + Hs (f1 )A sin(2πf1 t + θ). (using the same trigonometric identity in reverse), where
where q
Z |H(f )| = Hc2 (f1 ) + Hs2 (f1 )
∞
Hc (f1 ) = h(τ ) cos(2πf1 τ )dτ ∠H(f1 ) = tan−1 (−Hs (f1 )/Hc (f1 ))
Z−∞
∞
Hs (f1 ) = h(τ ) sin(2πf1 τ )dτ Note that the response to a sinusoidal input is determined by a single
−∞ complex number H(f1 ), which determines the magnitude of the output,
and the phase shift.
are real constants.
A sinusoidal input is scaled and delayed by an LTI system, but is otherwise
unchanged.

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Summary

The response of an LTI system is completely characterized by its

impulse response h(t). For a sinusoidal input at frequency f ,the output is
For an input x(t), the output of an linear system is given by the I a sinusoid at the same frequency,
superposition integral I scaled in amplitude, and
Z ∞ I phase shifted.
y (t) = x(τ )hτ (t) dτ This can be represented by a single complex number H(f ).
−∞

If the system is also time invariant, the result is a convolution integral

Z ∞
y (t) = x(τ )h(t − τ ) dτ
−∞

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Convolution Evaluation and Properties Convolution Integral

The convolution of an input signal x(t) with and impulse response h(t) is
Z ∞
y (t) = x(τ )h(t − τ ) dτ
−∞
Review: response of an LTI system = (x ∗ h)(t)
Representation of convolution
Graphical interpretation or
Examples y = x ∗ h.
Properties of convolution
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).

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Convolution Integral for Causal Systems τ <t τ >t
Does not
x(t) y(t) contribute to y(t)

0 τ t 0 t
Z ∞
! t
For a causal system h(t) = 0 for t < 0, and x(t) = x(τ)δ(t − τ)dτ y(t) = x(τ )h(t − τ )dτ
−∞ −∞
Z ∞
y (t) = x(τ )h(t − τ ) dτ.
−∞ If x(t) is also causal, x(t) = 0 for t < 0, and the integral further simplifies
Z t
Since h(t − τ ) = 0 for t < τ , the upper limit of the integral is t
y (t) = x(τ )h(t − τ ) dτ.
Z t 0
y (t) = x(τ )h(t − τ )dτ.
−∞ Does not
contribute to y(t) τ <t τ >t
Does not
Only past and present values of x(τ ) contribute to y (t). x(t) y(t) contribute to y(t)

0 τ t 0 t
! t ! t
x(t) = x(τ )δ(t − τ )dτ y(t) = x(τ )h(t − τ )dτ
0 0

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Graphical Interpretation
This is multiplied point by point with the input,
An increment in input x(τ )δτ (t)dτ produces an impulse response x(τ )
h(t − τ ) x(τ )h(t − τ )
x(τ )hτ (t)dτ . The output is the integral of all of these responses
Z ∞
y (t) = x(τ )hτ (t) dτ t τ t τ
−∞
Then integrate over τ to find y (t) for this t.
Another perspective is just to look at the integral.
Graphically, to find y (t):
hτ (t) = h(t − τ ) is the impulse response delayed to time τ
If we consider h(t − τ ) to be a function of τ , then h(t − τ ) is delayed flip impulse response h(τ ) backwards in time (yields h(−τ ))
to time t, and reversed. drag to the right over t (yields h(−(τ − t)))
h(t − τ ) h(t − τ ) multiply pointwise by x (yields x(τ )h(t − τ ))
Z ∞
integrate over τ to get y (t) = x(τ )h(t − τ ) dτ
τ t t τ −∞

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Simple Example

2 2
x(τ) x(τ) x(τ)
h(t − τ) t <0 h(t − τ)
1 1 1 0<t <1

-1 0 1 2 3 τ 0 3 τ 0 3 τ
-1 1 2 -1 1 2

2
2
1 h(τ) x(τ) x(τ)
h(t − τ) 1<t <2 2<t <3
1 1 h(t − τ)
-1 0 1 2 3 τ
-1 0 1 2 3 -1 0 1 2 3 τ
2
h(−τ) y(t) = (x ∗ h)(t)
1 2
2
x(τ)
1 h(t − τ) t >3 1
-1 0 1 2 3τ

2 -1 0 1 2 3 τ -1 0 1 2 3 τ
h(t − τ)
1

-1 0 1 2 3τ

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Communication channel, e.g., twisted pair cable Simple signaling at 0.5 bit/sec; Boolean signal 0, 1, 0, 1, 1, . . .

x(t) y(t)
∗h(t)
1
x(t)
0.5
u

Impulse response: 0
0 2 4 6 8 10
1.5
tt
1
y(t)
1 h(t)
0.5
y
h

0
0.5
0 2 4 6 8 10
tt

0
0 2 4 6 8 10
tt Output is delayed, smoothed version of input.

This is a delay ≈ 1, plus smoothing. 1’s & 0’s easily distinguished in y

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Simple signalling at 4 bit/sec; same Boolean signal Examples: Try these:

x(t) h(t) (x ∗ h)(t)

x(t) 1 1 1
1

0.5 0 1 2
u

0 1 2 0 1 2

0
1 1 1
0 2 4 6 8 10
t 0 1 2 0 1 2 0 1 2
1 y(t)
δ(t − 1)
1 1 1
0.5
y

0 1 2 0 1 2 0 1 2
0
0 2 4 6 8 10 1 1 1
tt
0 1 2 0 1 2 0 1 2
Smoothing makes 1’s & 0’s very hard to distinguish in y .
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Properties of Convolution
Simple Example (x*h)
For any two functions f and g the convolution is
2 2
Z ∞ x(τ)
1 1 h(τ)
(f ∗ g )(t) = f (τ )g (t − τ ) dτ
−∞
-1 0 1 2 3 τ -1 0 1 2 3 τ
If we make the substitution τ1 = t − τ , then τ = t − τ1 , and dτ = −dτ1 .
Z −∞ 2 h∗x 2
x∗h x(τ)
(f ∗ g )(t) = f (t − τ1 )g (τ1 ) (−dτ1 ) h(t − τ) 1
x(t − τ)
1
Z∞∞ h(τ)

= g (τ )f (t − τ1 ) dτ1 -1 0 1 2 3 τ -1 0 1 2 3 τ
−∞
= (g ∗ f )(t) y(t) = (x ∗ h)(t)
2

1
This means that convolution is commutative.
Practically, If we have two signals to convolve, we can choose either to be -1 0 1 2 3 τ

the signal we hold constant and the other to ”flip and drag.”

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Convolution is associative Linearity
Convolution is also distributive,

f ∗ (g + h) = f ∗ g + f ∗ h
If we convolve three functions f , g , and h
which is easily shown by writing out the convolution integral,
(f ∗ (g ∗ h))(t) = ((f ∗ g ) ∗ h)(t) Z ∞
(f ∗ (g + h))(t) = f (τ ) [g (t − τ ) + h(t − τ )] dτ
which means that convolution is associative. Z−∞ Z ∞
∞
Combining the commutative and associate properties, = f (τ )g (t − τ ) dτ + f (τ )h(t − τ ) dτ
−∞ −∞
f ∗ g ∗ h = f ∗ h ∗ g = ··· = h ∗ g ∗ f = (f ∗ g )(t) + (f ∗ h)(t)

We can perform the convolutions in any order. Together, the commutative, associative, and distributive properties mean
that there is an “algebra of signals” where

addition is like arithmetic or ordinary algebra, and

multiplication is replaced by convolution.
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Time-invariant Properties of Convolution Systems

The properties of the convolution integral have important consequences
for systems described by convolution:

Convolution systems are linear: for all signals x1 , x2 and all α, β ∈ <,

Convolution with a delayed signal gives a delayed output. h ∗ (αx1 + βx2 ) = α(h ∗ x1 ) + β(h ∗ x2 )

(f ∗ gτ )(t) = (fτ ∗ g )(t) = (f ∗ g )(t − τ ) Convolution systems are time-invariant: if we shift the input signal x
by T , i.e., apply the input

x1 (t) = x(t − T )

to the system, the output is

y1 (t) = y (t − T ).

In other words: convolution systems commute with delay.

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 Composition of convolution systems corresponds to convolution of Since convolution is commutative, the convolution systems are also
impulse responses. commutative. These two cascade connections have the same response
The cascade connection of two convolution systems y = (x ∗ f ) ∗ g
Composition y
x w
∗f ∗g
x w y
∗f ∗g
x v y
∗g ∗f
is the same as a single system with an impulse response h = f ∗ g

x y
∗( f ∗ g) Many operations can be written as convolutions, and these all commute
(integration, differentiation, delay, ...)

Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 49 / 55 Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 50 / 55

Example: Measuring the impulse response of an LTI system.

The impulse response is determined by differentiating the step response,
We would like to measure the impulse response of an LTI system,
described by the impulse response h(t) s(t)
u(t) h(t)
δ(t)
h(t)
0 t 0 t 0 t
0 t 0 t
∗h d
∗h
dt

This can be practically difficult because input amplitude is often limited. A

very short pulse then has very little energy. To show this, commute the convolution system and the differentiator to
produce a system with the same overall impulse response
A common alternative is to measure the step response s(t), the response
to a unit step input u(t) δ(t)
u(t) h(t)

s(t) 0 t 0 t 0 t
u(t) d
∗h
0 t 0 t dt
∗h

Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 51 / 55 Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 52 / 55
Convolution Systems with Complex Exponential Inputs
If we have a convolution system with an impulse response h(t), and
and input e st where s = σ + jω
Z ∞
H(s) is the transfer function of the system.
y (t) = h(τ )e s(t−τ ) dτ
−∞ If the input is a complex sinusoid e jωt ,
Z ∞
Z ∞
= e st h(τ )e −sτ dτ
−∞ H(jω) = h(τ )e −jωτ dτ
−∞
We get the complex exponential back, with a complex constant y (t) = e jωt H(jω)
multiplier
Z ∞
H(s) = h(τ )e −sτ dτ
−∞
y (t) = e st H(s)

provided the integral converges.

Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 53 / 55 Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 54 / 55

Summary

LTI systems can be represented as a the convolution of the input with

an impulse response.
Convolution has many useful properties (associative, commutative,
etc).
These carry over to LTI systems
I Composition of system blocks
I Order of system blocks
Useful both practically, and for understanding.
While convolution is conceptually simple, it can be practically difficult.
It can be tedious to convolve your way through a complex system.
There has to be a better way ...

Cuff (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 55 / 55