Lecture #10 EGR 261 – Signals and Systems

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Lecture #10 EGR 261 – Signals and Systems

Representation and Analysis of SystemsThe analysis of systems this course is limited to linear, time-invariant, continuous-time (LTIC) systems.

LTIC systems can be characterized (described) in three ways:

1) Differential equations – a differential equation (DE) for y(t) in terms of x(t) – found using standard D.E. methods (review of Ch. 7-8 in Nilsson text). The DE can be used to find the response to any input, x(t). Special cases include the unit step response (USR) and the impulse response, h(t).

2) Impulse response – the system output, y(t), can be determined using the impulse response, h(t), through evaluation of the convolution integral

3) Transfer function – The transfer function, H(s) can be determined through Laplace transform analysis. H(s) can be used to find the USR, h(t), or the output y(t) for any input x(t).

Read: Ch. 2, Sect. 1-8 in Linear Signals & Systems, 2nd Ed. by LathiCh. 13, Sect. 6 in Electric Circuits, 9th Ed. by Nilsson

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Example: Determine the differential equation for the output y(t) for any general input forcing function x(t).

¼ F y(t)

+

_

1

4/3 H

x(t)

+

_

Note that this differential equation completely characterizes the system.

Differential Equations System RepresentationAs we discussed the classification of systems, it was stated that a linear system can be represented by a linear differential equation of the form:

x(t)b dt

dxb

dt

xdb y(t)a

dt

yda

dt

yda N1-NM

M

M-NN1-N

1-N

1N

N

0

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Lecture #10 EGR 261 – Signals and SystemsUnit Step Response (USR) and Impulse Response, h(t)Recall that we earlier found the unit step response, USR, and the impulse response, h(t), using s-domain techniques involving the transfer function, H(s). To summarize, we found that:

We can also find h(t) and USR from the differential equation that characterizes a system.

Determining the USR and h(t) from a Differential EquationWe have just reviewed writing and solving differential equations.

Recall that if the input, x(t), to the differential equation is a unit step function, then y(t) is called the unit step response (USR).

So, USR = y(t) when x(t) = u(t)

The impulse response can then be found using the following relationship:

s

H(s) response stepunit USR

H(s) response impulse h(t)

1-

-1

L

L

Impulse response = h(t) = d/dt[USR(t)]

Also recall that (t) = d/dt[u(t)]

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Example:The USR for a system is y(t) = USR = [2 + 3e-4t]u(t).Find the impulse response, h(t). Hint: Use the product rule. Answer: h(t) = 5(t) – 12e-4tu(t)

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Example:A) Write the differential equation for y(t) that characterizes the system (circuit).B) Find the unit step response, USR, by solving the differential equation for x(t) = u(t).C) Find the impulse response, h(t), using the relationship h(t) = d/dt[USR(t)]

1

1 F y(t)

+

_

x(t)

+

_

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Example: (continued)

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Lecture #10 EGR 261 – Signals and SystemsDetermining a transfer function from a differential equationTaking the Laplace Transform of a differential equation will yield the transfer function if:

1)The differential equation uses a general input, x(t)

2)The initial conditions are zero

4- (0) x'2, x(0) where0,for t 20, 12y(t) dt

dy7

dt

yd2

2

Example: Find H(s) for the D.E. shown below.

Invalid. Input is not general.

Invalid. Initial conditions are not zero.

0 (0) x'0, x(0) where0,for t 4x(t), 12y(t) dt

dy7

dt

yd2

2

Example: Find H(s) for the D.E. shown below.

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Lecture #10 EGR 261 – Signals and SystemsThe Impulse Response and the Convolution IntegralRecall from our earlier study of systems using Laplace transforms that the transfer function, H(s), completely characterizes a system and that the output y(t) can be determined for any input x(t) using H(s).

H(s)Input = X(s) Y(s) = Output

s-domain representation of a circuit or system using a transfer function H(s)

response impulse h(t) H(s) 1H(s) y(t)

so 1, X(s) then (t), x(t)if that recall Also

X(s)H(s) y(t)or X(s)H(s) Y(s)

input(s)

output(s)

X(s)

Y(s) H(s) function Transfer

1-1-

1-

LL

L

so

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SystemInput = x(t) = (t) y(t) = h(t) = impulse response

Just as H(s) completely characterizes a system and can be used to determine the output of the system for any input, the impulse response h(t) also completely characterizes the system.

s-domain approach:The output of a system can be determined with H(s) and Laplace transform techniques.time-domain approaches:1) Differential equations (just covered)2) Convolution. The output of a system can be determined with h(t) and the

convolution integral.

h(t)x(t)y(t) can be found for any x(t) using h(t) and the convolution integral

y(t)

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Why use the convolution integral?• It allows us to work directly in the time-domain. This is particularly useful with

experimental data where using Laplace transforms might be difficult or impossible.• Insight can be gained as to how closely the output waveform replicates the input

waveform.

Development of the convolution integralThe following development is presented in section 13.6 of Electric Circuits, by Nilsson.

A LTIC system can be described by its impulse response, h(t), as shown below.

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Some general input x(t) is shown.

x(t) can be expressed as a series of rectangular pulses. So,

x(t) = x0(t) + x0(t) + … + xi(t) + …

Each pulse could be expressed in terms of unit step functions (i.e., window functions) as:

xi(t) = x(i)[u(t - i) - u(t – (i + )]

As the width of each pulse approaches zero, x(t) can be expressed as a series of impulse functions as follows:

x (t) = x(0) (t - 0) + x(1) (t - 1) + …

+ x(i) (t - i)

Area under pulse

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-

)d -)h(t x( y(t)

This integral is referred to as the convolution integral and is generally expressed as:

integral)on (convoluti )d -)h(t x( h(t)* x(t) y(t)-

The output y(t) consists of a sum of uniformly delayed impulse responses where the strength of each responses depends on the strength of the impulse driving the circuit.

y (t) = x(0) h(t - 0) + x(1) h(t - 1) + …

+ x(i) h(t - i) or

y (t) = x(i) h(t - i)

As approaches zero, this summation approaches a continuous integration, or

Recall that ifx(t) = K(t),then y(t) = Kh(t)

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Properties of the convolution integralCommutative property:

Distributive property:

Associative property:

Shift property:

--

)d -)x(t h( )d -)h(t x( x(t)*h(t) h(t)*x(t)

(t)y* x(t) (t)y* x(t) (t)y (t)y*x(t) 2121

(t) x* (t)x*(t)x (t) x* (t)x*(t)x 321321

)T-T-y(t )T-(t x* )T-(tx

and

T)-y(t (t) x* T)-(t x T)-(t x* (t)x

then

y(t) (t) x* (t) xIf

212211

2121

21

y(t)" and x(t)ofn convolutio the" y(t)*x(t)

or

y(t)" with convolved x(t)" y(t)*x(t)

)d -)h(t x( y(t)* x(t) y(t)-

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Convolution can be understood by examining the graphical interpretation of the convolution integral. The graphical approach is helpful for the following reasons:• It is helpful in evaluating the convolution of complicated signals• It allows us to grasp visually the convolution integral’s result• It allows us to perform convolution with signals that can only be described graphically.

--

)d -)x(t h( )d -)h(t x( x(t)*h(t) h(t)* x(t) y(t)

Why use graphical evaluation of the convolution integral?The convolution integral can be evaluated directly, but this can be very difficult (especially for piecewise-continuous functions), so a graphical approach is more commonly used. However, even the graphical approach is quite challenging. A detailed example will follow shortly to illustrate the procedure.

Convolution Integral:

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Procedure for evaluating the convolution integral graphically1. Graph x() (with on the horizontal-axis). (see note below)2. Invert h() to form h(-). Then shift h(-) along the axis t

seconds for form h(t - ). Note that as the time shift t varies, the waveform h(t - ) will “slide” across x() and in some cases the waveforms will overlap.

3. Determine the different ranges of t which result in unique overlapping portions of x() and h(t - ). For each range determine the area under the product of x() and h(to - ). This area is y(t) for the range.

4. Compile the results of y(t) for each range and graph y(t).

Note: x(t) and h(t) can be reversed since x(t)*h(t) = h(t)*x(t), i.e., commutative property. In general, it is easiest to invert and delay the simplest function.

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Example: Find y(t) = x(t)*h(t) for x(t) and h(t) shown below using the graphical method. Follow the procedure listed on the previous slide.

0 1

1

x(t)

0 1

2

h(t)

3tt

Solution: 1. Form x():

0 1

1

x()

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Lecture #10 EGR 261 – Signals and Systems2. Form h(t - ):

0 1

2

h()

3

0-1

2

h(-)

-3

0-1

2

h(t-)

-3

t

t-1t-3

Note: Remember that is the independent variable, not t. t is simply a constant. h(t - ) = h(- + t), so t is the amount of time shift (to the left).

3. Consider different ranges of t (time-shift):Each range of t selected should result in different portions of the waveforms overlapping. The convolution y(t) = x(t)*h(t) is the area under the product of the overlapping portions.

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Lecture #10 EGR 261 – Signals and Systems3A. First range selected: (- <t < 1):

This range results in no overlap for all values of (i.e., - < < ).

0

2

h(t-)

t-1t-3

h(t-) sliding as t varies over the range - < t < 1

1

1

No overlap

x()

1 t -for 0 y(t) so

d0 )-h(t)x( y(t) and

) allfor overlap (no - 0, )-h(t)x(

--

d

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3B. Second range selected: (1 <t < 2):This range results in an overlap from = 0 to = t-1.

0

2

h(t-)

t-1t-3

h(t-) sliding as t varies over the range 1 < t < 2

1

1

Area of overlap between = 0 and = t-1

x()

2 t 1for 1)-2(t 2 y(t)

d0 d2 d0 )-h(t)x( y(t) and

1-t0

1-t0221

00

)-h(t)x(

1t

0

0

-

1-t

0 1t-

d

20


3C. Third range selected: (2 <t < 3):This range results in an overlap from = 0 to = 1.

0

2

h(t-)

t-1t-3


1

1

Area of overlap between = 0 and = 1

x()

3 t 2for 2 0)-2(1 2 y(t)


10

10221

00

)-h(t)x(

1

0

1

0-

d

21


3D. Fourth range selected: (3 <t < 4):This range results in an overlap from = t-3 to = 1.

0

2

h(t-)

t-1t-3


1

1

Area of overlap between = t-3 and = 1

x()

4 t 3for 2t -8 t-42 3-t-12 2 y(t)


10

13-t221

3t0

)-h(t)x(

1

3t

1

3t-

d

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3E. Fifth range selected: (4 <t):This range results in no overlap for all values of (i.e., - < < ).

0

2

h(t-)

t-1t-3

h(t-) sliding as t varies over the

range t > 4

1

1

No overlap

x()

t 4for 0 y(t) so


) allfor overlap (no - 0, )-h(t)x(

--

d

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5. Compile the results of y(t) over the five ranges. Also graph y(t).

t 40

4 t 382t-

3 t 22

2 t 12-2t

1 t 0

h(t)* x(t) y(t)

0

2

t431

y(t)

2

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Example: Find y(t) = x(t)*h(t) for x(t) and h(t) shown below using the graphical method.

0 3-1

x(t)

0 4

20

h(t)

6tt

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Lecture #10 EGR 261 – Signals and Systems