Linear Time-invariant Systems And The Convolution Integral.docx

Lab Experiment Invariant Systems Convolution Integral 6.1 The

Impulse

Response

6: and and

the

Linear the

Time-

Convolution Integral

The

impulse response of a continuous-time system 𝑆, denoted by ℎ 𝑡 , is defined as ℎ𝑡≜𝑆𝛿𝑡. In words, if a continuous-time system is given a Dirac delta function as an input, the resulting output is known as the impulse response. It turns out that for LSI (Linear Shift-Invariant) systems, knowing the impulse response ℎ 𝑡 = 𝑥 ∗ ℎ and the input signal 𝑥 𝑡 is enough to find the output 𝑦 𝑡 . / In this case, the following equation holds: 𝑦𝑡 𝑡 ≜ 𝑥𝜏ℎ𝑡−𝜏

𝑑𝜏

0/

6.2 Symbolic Evaluation

of

the

Convolution Integral

with MATLAB

First of convolution ℎ𝑡.

will if

see we

how have

the for

all we integral

we an

evaluate expression

𝑥𝑡

and

>> syms t tau >> x = @(t) dirac(t); >> h = @(t) exp(-t) .* heaviside(t); >> y = @(t) int(x(tau) .* h(t - tau), tau, -inf, inf); >> y(t) ans = (exp(-t)*(sign(t) + 1))/2 >> tr = -1:0.01:5;

1

>> plot(tr, y(tr)) 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

0-1

0

However, the way integral integral,

1

2

this takes MATLAB at each and store

3

4

5

a long time to implements it, as point. An alternative it manually:

compute because it evaluates the is to evaluate

>> y = @(t) (exp(-t).*(sign(t) + 1))/2; >> plot(tr, y(tr)) You should notice that this evaluates much faster. One care of is you must replace all operators point-wise counterparts. Symbolic evaluation is very

6.3 Numerical

Evaluation

of

the

An alternative approach is to numerically. This may not be very needed when analytical solutions the property that if 𝑥𝑡 is 𝑡 , is nonzero only nonzero for 𝑡12 + 𝑡15 ≤ 𝑡 ≤ 𝑡42 + 𝑡45. >> >> >> >> >> >> >> >> >> >> >> >> >> >>

thing to by their accurate.

of the

take

Convolution Integral evaluate the accurate but cannot be found. nonzero only for 𝑡15 ≤ 𝑡 ≤ 𝑡45, Let’s put this

integral is sometimes Here, we use for 𝑡 then 𝑥 ∗ ℎ 𝑡 is into practice.

x = @(t) heaviside(t) - heaviside(t - 1); h = @(t) heaviside(t) - heaviside(t - 2); tx = 0:0.01:1; t1x = 0; t2x = 1; t1h = 0; t2h = 2; ts = 0.01; tx = t1x:ts:t2x; th = t1h:ts:t2h; t1y = t1x + t1h; t2y = t2x + t2h; ty = t1y:ts:t2y; xr = x(tx); 2

>> hr = h(th); >> yr = ts * conv(xr, hr); >> plot(ty, yr) 1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00

0.5

Here, ts calculates in the

is the next

6.4 Lab

Tasks

1

1.5

2

2.5

3

the sampling time and discrete-time convolution. We experiment.

6.4.1 Question 1 Calculate the symbolic pairs of functions. a. 𝑥1 𝑡 = 𝑢 𝑡 and ℎ1 𝑡 ANS— ans =

convolution

the will

between

conv function learn more about this

the

following

= cos 10𝑡 𝑒0>𝑢 𝑡 .

-((exp(-t)*(exp(-t*10i)*(1/2 - 5i) + exp(t*10i)*(1/2 + 5i)))/101 1/101)*(sign(t)/2 + 1/2)

3

>>

b. 𝑥4 𝑡

= sin 5𝑡

and

ℎ4 𝑡

= 𝑒01B>𝑢 𝑡 .

Ans ans =

(2*sin(5*t))/25 - cos(5*t)/25

4

>>

c. 𝑥C 𝑡

= 𝑡𝑢 𝑡 and

ℎC 𝑡

= D>D 𝛿 𝑡 .

ans =

int(tau*kroneckerDelta(t - tau, 0), tau, 0, Inf)

5

>>

For part Calculate Hint: Type You a

c, what does your the result manually help dirac if you

should submit a *.mat file handwritten sheet with the

intuition tell and submit it are stuck.

you as

is well.

correct?

that calculates answer to part

the c.

answers

and

6