Chapter 2: part a Convolution for C-T systems • Impulse Response: h(t) is what “comes out”when δ(t) “goes in”

• How do we know or get the impulse response h(t)? 1. It is given to us by the designer of the C-T system. 2. It is measured experimentally But, we cannot just “put in δ(t)”

-There are other ways to get h(t) but we need the later chapters information first

3. Mathematically analyze the C-T system

Signal and System Dis EELE 3040 Eng. Abdo Salah

Islamic University of Gaza Faculty of Engineering Electrical Engineering Dep.

• In what form will we know h(t)? 1. h(t) known analytically as a function

h(t) = 𝑒𝑒−𝛾𝛾𝛾𝛾𝑢𝑢(𝛾𝛾) most common in this chapter • Use h(t) to find the zero-state response of the system for an input Convolution is a powerful tool for determining the output of a system to any input. The Convolution Theorem is developed here in a completely mathematical way. On this page we will derive the convolution theorem: If the input to a system is x(t), and the impulse response of that system is h(t), then we can determine the output of the system, y(t), from the integral:

• Start with a linear time-invariant (LTI) system (in box). There is an input to the system, x(t), and an output from the system,

• Now consider the same system with the input as an impulse, δ(t). The output is then, by definition, the impulse response, h(t).

• If we use a delayed impulse into the system, the output is just the impulse response with the same delay.(Time invariant )

• Because of linearity, if we scale the input by any factor, the output will be scaled

by the same factor. In particular, we can scale the input (and hence the output) by the factor x(λ)dλ.

• If we integrate the input, the output is also integrated. In this case we'll take the

upper limit of the integration to be t+; but it could just as well have been positive infinity. We could also take the lower lower limit to be negative infinity.

• By the sifting property of the impulse response we know that:

• Therefore the input to the system is x(t)

• If the input is x(t), then the output must be y(t) (by definition, from the first

diagram).

• Equating the outputs of the last two diagrams leads directly to the convolution

theorem:

'

ومعروف الريسبونس LTI system : لو كان عندي convolutionاالن باختصار نشرح مفهوم متوفرة بامكاني اجيب الريسبونس تبع شغالت ، بيحكيلك انو لو كانت هاي الثالث (h(t تبع االمبلس

أي انبت من خالل الكونفوليوشن كالتالي : بقسم االنبوت لمجموعة من االمبلسات االن بما انو السيستم عندي تايم انفارينت فكل امبلس مزاحة سيقابلها نفس االزاحة في االوت بوت(االمبلس

ريسبونس ) وبما انو السيتم عندي لينير وانا قسمت االنبوت لمجموعة من االمبلسات فاالوت بوت عندي هو مجموع الريسبونس تبع كل امبلسة .

Convolution Properties These are things you can exploit to make it easier to solve convolution problems. 1.Commutativity

⇒You can choose which signal to “flip” 2. Associativity

Can change order →sometimes one order is easier than another 3. Distributivity ⇒may be easier to split complicated system h[t] into sum of simple ones ⇒we can split complicated input into sum of simple ones (nothing more than “linearity”)

4. Convolution with impulses

“Computing”CT Convolution

If we are analyzing a given system (e.g., a circuit) we may need to compute a convolution to determine how it behaves in response to various different input signals. If we are designing a system (e.g., a circuit) we may need to be able to visualize how convolution works in order to choose the correct type of system impulse response to make the system work the way we want it to.

We’ll learn how to perform “Graphical Convolution ,”which is nothing more than steps that help you use graphical insight to evaluate the convolution integral.

Steps for Graphical Convolution x(t)*h(t)

1. Re-Write the signals as functions of τ: x(τ) and h(τ)

2. Flip just one of the signals around t = 0 to get either x(-τ) or h(-τ)

a. It is usually best to flip the signal with shorter duration b. For notational purposes here: we’ll flip h(τ) to get h(-τ)

3. Shift h(-τ) by an arbitrary value of t to get h(t-τ) and get the edges of the tow functions : we mean by edge the points where the functions expressions change

4. Find Regions of τ- Overlap

a.What you are trying to do here is find intervals of t over which the product x(τ) h(t-τ) has a single mathematical form in terms of τ b.In each region find: Interval of t that makes the identified overlap happen c.Working examples is the best way to learn how this is done d.To find the regions equal the changing poin of the signals with each others

5. For Each Region: Form the Product x(τ) h(t-τ) and Integrate

a. Form product x(τ) h(t-τ) b. Find the Limits of Integration by finding the interval of τ over which the product is nonzero

i. Found by seeing where the edges of x(τ) and h(t-τ) lie ii. Recall that the edges of h(t-τ) depend on the value of t

So…the limits of integration may depend on t c. Integrate the product x(τ) h(t-τ) over the limits found in 5b i. The result is generally a function of t, but is only valid for the interval of t found for the current region ii. Think of the result as a “time-section” of the output y(t)

6. “Assemble” the output from the output time-sections for all the regions a. Note: you do NOT add the sections together b. You define the output “piecewise” c. Finally , if possible, look for a way to write the output in a simpler form

Examples1: Graphically Convolve Two Signals

Convolve these two signals:

Step #1: Write as Function of τ

Step #2: Flip h(τ) to get h(-τ)

Step #3: Shift by t to get h(t-τ) & get Edges

For ( )h t τ− changing point (t-1),(t)

For ( )x τ changing point 0,2

Step #4: Find Regions of τ-Overlap

t-1=0 t-1=2 t=0 t=2

t=1 t=3 t=0 t=2

Y(t)=

s.th t

Step #5: Form Product & Integrate For Each Region

Step #6: “Assemble ” Output Signal

Example 2:

• Convolve the following two functions:

• Replace t with τ in f(t) and g(t) • Choose to flip and slide g(τ) since it is simpler

and symmetric . • Functions overlap like this:

For ( )g t τ− changing point (-2+t),(2+t)

For ( )f τ changing point 0,2

-2+t =0 -2+t =2 2+t =0 2+t =2

t=2 t=4 t= -2 t=0

Y(t)=

s.th t

Example 3

t=0 t=4

Y(t)=

s.th t

( )

( )

2 2 2

0 0

2 2

0

2

( ) 15 15

1152

( ) 7.5 1

t tt t

tt

t

y t e d e e d

e e

y t e

τ τ

τ

τ τ− − −

−

−

= =

=

→ = −

∫ ∫

( )4

2

0

( ) 15 ty t e dτ τ− −= ∫

( )

( )

4 42 2 2

0 04

2 2 2 8

0

( ) 15 15

115 7.5 12

t t

t t

y t e d e e d

e e e e

τ τ

τ

τ τ− − −

− −

= =

= = −

∫ ∫

( )( )

2

2 8

0 0

( ) 7.5 1 0 4

7.5 1 4

t

t

t

y t e t

e e t

−

−

The actions of flipping and shifting can be applied to EITHER function

Repeat example 3 by flipping and shifting x(t) rather than h(t)

( ) ( )

( ) ( )

( )* ( )

( )* ( )

x t h t x h t d

h x t d h t x t

τ τ τ

τ τ τ

∞

−∞

∞

−∞

= −

= − =

∫

∫

( )

2 2

00

2

( ) 15 7

for

.5

0 4

( ) 7.5 1

tt

t

y t e d e

y t e

t

τ ττ− −

−

= = −

= −

< <

∫

Same result as before

( )( ) ( )

2 2

44

2 4 2 8 2

for 4

1( ) 15 152

( ) 7.5 7.5 1

tt

tt

t t t

t

y t e d e

y t e e e e

τ ττ− −−−

− − − −

>

− = =

→ = − − = −

∫

( )( )

2

2 8

0 0

( ) 7.5 1 0 4

7.5 1 4

t

t

t

y t e t

e e t

−

−

3

Length of the response y(t) = x(t)*h(t)In general if the length of the input x(t) is T1 and the length of the impulse response h(t) is T2, then the length of y(t) = x(t)*h(t) will be T1 + T2.

T1

x(t) h(t)

ttT2

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

tT1 +T2

0 1

1

x(t)

0 1

2

h(t)

3tt

ExampleRecall the example worked in the last presentation where we determined y(t) = x(t)*h(t) as shown below.

0

2

t431

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

2

Length = 1 Length = 2 Length = 1+2 = 3

Example 4:

Example 5:

Solution:

• Graph x(t ) and h(t )

• Replace t with τ in x(t) and h(t)

Answer:0 t 02 0 t 16 1 t 2

y(t) x(t)*h(t) 12 2 t 310 3 t 46 4 t 50 5 t

• Graph x(t- τ)

I. For t < 0: Two functions do not overlap - Area under the product of the functions is zero, then y(t)=0

II. For t ≥0: - Part of x(t- τ) overlaps part of h(τ) - Area under the product of the functions is Result of convolution:

Example 6.

There is one critical time point defined by t-1 = 0, i.e. t = 1; we have 2 intervals to consider for t:

t < 1 : y(t) = 0

Example : p2.4.11

b-

2( 3) ( )

0

2 6 )

0

6

0

6

6 6 2

( )

( )

( )

1( )1

( ) the zero state response

tt

tt

tt

tt

t t

y t e e d

y t e d

y t e e d

ey t e

y t e e

τ τ

τ τ

τ

τ

τ

τ

− − − −

− + − +

− −

−−

− −

=

=

=

−= − = − →

∫

∫

∫

d-

[ ] ( )0

( ) ( )

0

0 1

( 1)

( ) ( ) ( 1)

( ) ( ) ( 1)

( )

( ) 1

thus the zero state response is y(t)=

tt

tt t

t tt t

t t t t

t t

y t u u e d

y t e u e u d

y t e e d e e d

y t e e e e e

e e

τ

τ τ

τ τ

τ τ τ

τ τ τ

τ τ

− −

− − − −

− −

− −

− − −

= − −

= − −

= −

= − − −

−

∫

∫

∫ ∫

P: 2.4.17

-2π+t=0 -2π+t=2π t=0 t=2π

t=2π,0, 4π

For 0