1
00:00:00,560 --> 00:00:02,760
PROFESSOR: Convolution
is a tricky concept

2
00:00:02,760 --> 00:00:04,420
and hard to understand.

3
00:00:04,420 --> 00:00:06,800
But it gives so much
insight that it's worthwhile

4
00:00:06,800 --> 00:00:08,710
coming to grips with it.

5
00:00:08,710 --> 00:00:10,840
The Convolution:
Accumulation mathlet

6
00:00:10,840 --> 00:00:14,710
provides one perspective on
the convolution integral.

7
00:00:14,710 --> 00:00:18,080
I'll explain the
applet using a story.

8
00:00:18,080 --> 00:00:20,890
There's a lake in Iowa with
a stream running into it

9
00:00:20,890 --> 00:00:23,000
and a stream running out of it.

10
00:00:23,000 --> 00:00:26,560
Farm runoff puts
phosphate into the lake.

11
00:00:26,560 --> 00:00:30,150
The rate of deposition of the
phosphate varies over the year.

12
00:00:30,150 --> 00:00:33,550
Most in the summer, none
in the dead of winter.

13
00:00:33,550 --> 00:00:37,330
Suppose the farm is new
and that t equals 0,

14
00:00:37,330 --> 00:00:39,550
there is no phosphate
in the lake.

15
00:00:39,550 --> 00:00:42,780
This is called rest
initial conditions.

16
00:00:42,780 --> 00:00:45,980
The question is, how
much phosphate is there

17
00:00:45,980 --> 00:00:49,860
in the lake at some
later time, say t.

18
00:00:49,860 --> 00:00:53,290
This will be an accumulation
of contributions from earlier

19
00:00:53,290 --> 00:00:55,770
times, but the
earlier contributions

20
00:00:55,770 --> 00:00:59,650
will have decayed
somewhat by time t.

21
00:00:59,650 --> 00:01:02,710
We can visualize this process
using the Convolution:

22
00:01:02,710 --> 00:01:04,780
Accumulation mathlet.

23
00:01:04,780 --> 00:01:08,100
I'll set it up to model
what we're looking at.

24
00:01:08,100 --> 00:01:13,010
The signal f of t is the
rate of input of phosphate

25
00:01:13,010 --> 00:01:14,290
into the lake.

26
00:01:14,290 --> 00:01:20,260
And I will model it by the menu
item f of t is 1 plus cosine

27
00:01:20,260 --> 00:01:21,810
b*t.

28
00:01:21,810 --> 00:01:24,765
You can see it drawn in the
lower graphing window here.

29
00:01:29,300 --> 00:01:32,700
A certain proportion of
the phosphate in the lake

30
00:01:32,700 --> 00:01:36,200
leaves it during
each unit of time.

31
00:01:36,200 --> 00:01:40,710
This results in an exponential
decay of each contribution.

32
00:01:40,710 --> 00:01:44,690
This decay profile is called the
weight function, denoted here

33
00:01:44,690 --> 00:01:46,390
by g of t.

34
00:01:46,390 --> 00:01:48,755
And I will select
the weight function,

35
00:01:48,755 --> 00:01:52,320
giving exponential decay
e to the minus a*t.

36
00:01:56,082 --> 00:01:57,540
Now let's see what
happens if I let

37
00:01:57,540 --> 00:01:59,620
a small amount of time elapse.

38
00:01:59,620 --> 00:02:00,990
How small?

39
00:02:00,990 --> 00:02:04,630
Well I can select that
by picking the step size.

40
00:02:04,630 --> 00:02:08,479
And I will select a
step size of one eighth.

41
00:02:08,479 --> 00:02:11,810
So now let's let a little
bit of time elapse.

42
00:02:11,810 --> 00:02:17,060
Say, amount of
time of one eighth.

43
00:02:17,060 --> 00:02:20,650
If I click on the time slider at
one eighth, I get this picture.

44
00:02:20,650 --> 00:02:23,570
t equals 0 happens to
occur at mid-summer,

45
00:02:23,570 --> 00:02:25,980
so that the signal
has a value of 2

46
00:02:25,980 --> 00:02:29,090
in kilograms per unit time.

47
00:02:29,090 --> 00:02:35,780
This phosphate decays away
as water comes into the lake

48
00:02:35,780 --> 00:02:38,360
and carries it off according
to the weight function.

49
00:02:38,360 --> 00:02:41,927
And this is indicated by the
exponential decay function

50
00:02:41,927 --> 00:02:43,010
here in the bottom window.

51
00:02:46,790 --> 00:02:48,940
The weight function is a rate.

52
00:02:48,940 --> 00:02:51,540
It needs to get multiplied
by the step size

53
00:02:51,540 --> 00:02:55,450
before it gives an actual
amount of phosphate in the lake.

54
00:02:55,450 --> 00:02:59,270
The step size is one
eighth, and the product

55
00:02:59,270 --> 00:03:01,980
is the signal, the
value of the signal at t

56
00:03:01,980 --> 00:03:06,840
equals 0, that's 2, times
the weight function, that's

57
00:03:06,840 --> 00:03:12,360
the exponential decay, times the
step size, which is one eighth.

58
00:03:12,360 --> 00:03:15,010
So that gives all
together one quarter

59
00:03:15,010 --> 00:03:17,380
times that exponential decay.

60
00:03:17,380 --> 00:03:20,090
And you see that drawn in
the top graphing window here,

61
00:03:20,090 --> 00:03:23,480
the value near t
equals 0 is a quarter

62
00:03:23,480 --> 00:03:26,880
and it decays away
exponentially.

63
00:03:26,880 --> 00:03:30,370
Now let's move on to
the next time interval.

64
00:03:30,370 --> 00:03:36,450
I'll click here, and
you see what happens.

65
00:03:36,450 --> 00:03:39,970
The signal is now
a little bit less

66
00:03:39,970 --> 00:03:44,180
and the time is a
little bit later.

67
00:03:44,180 --> 00:03:47,580
So the weight function is
scaled and shifted a little bit

68
00:03:47,580 --> 00:03:50,980
differently in this
bottom graphing window.

69
00:03:50,980 --> 00:03:54,620
It gets multiplied by the
step size one eighth again,

70
00:03:54,620 --> 00:03:59,690
and laid down on top of
the graph of the phosphate

71
00:03:59,690 --> 00:04:02,450
arising from the
first time interval.

72
00:04:02,450 --> 00:04:06,870
So the sum of those
two is a record

73
00:04:06,870 --> 00:04:09,610
of the phosphate in the lake
arising from these first two

74
00:04:09,610 --> 00:04:14,070
time intervals as it
decays away in later time.

75
00:04:14,070 --> 00:04:16,050
This process continues.

76
00:04:16,050 --> 00:04:19,060
You can see the next few
steps by clicking them up.

77
00:04:19,060 --> 00:04:21,490
Here's the effect of
the third time interval

78
00:04:21,490 --> 00:04:25,190
as it decays away, but laid
down on top of the preceding

79
00:04:25,190 --> 00:04:29,530
contributions to
the lake, and so on.

80
00:04:29,530 --> 00:04:32,070
Or I can animate
the entire process

81
00:04:32,070 --> 00:04:34,925
and watch the thing
evolve as time increases.

82
00:04:44,310 --> 00:04:47,470
So you can see, the
effect is a steady state

83
00:04:47,470 --> 00:04:53,210
is reached after a while, you
get a sinusoidal total amount

84
00:04:53,210 --> 00:04:56,950
of phosphate in the lake.

85
00:04:56,950 --> 00:05:00,040
It's delayed a little
bit from the peak.

86
00:05:00,040 --> 00:05:02,286
The maximum amount of
phosphate in the lake

87
00:05:02,286 --> 00:05:03,660
occurs a little
after mid-summer.

88
00:05:07,670 --> 00:05:10,820
Now imagine shrinking
the step size.

89
00:05:10,820 --> 00:05:14,340
In the limit, as the
step size goes to 0,

90
00:05:14,340 --> 00:05:17,940
this process is described
by the convolution integral.

91
00:05:17,940 --> 00:05:22,970
At time t, all the contributions
from time 0 to time t

92
00:05:22,970 --> 00:05:24,560
are present.

93
00:05:24,560 --> 00:05:27,800
And so we're going to take an
integral from u equals 0 to u

94
00:05:27,800 --> 00:05:30,230
equals t.

95
00:05:30,230 --> 00:05:35,650
The contribution from time u
decays by the factor of g of t

96
00:05:35,650 --> 00:05:39,970
minus u by time t.

97
00:05:39,970 --> 00:05:45,930
So the contribution from between
time u and time u plus du

98
00:05:45,930 --> 00:05:51,690
is given by the product f of u,
the signal of time u, times g

99
00:05:51,690 --> 00:05:54,700
of t minus u, the weight
function evaluated

100
00:05:54,700 --> 00:05:57,010
at t minus u, times du.

101
00:05:59,890 --> 00:06:01,720
And the integral of
this differential,

102
00:06:01,720 --> 00:06:04,790
from u equals 0
to u equals t, is

103
00:06:04,790 --> 00:06:07,570
the amount of phosphate
in the lake at time t.

104
00:06:07,570 --> 00:06:11,810
And it's given by the
convolution integral.

105
00:06:11,810 --> 00:06:14,820
This system is modeled by
a differential equation,

106
00:06:14,820 --> 00:06:20,600
namely dx/dt plus
a*x equals f of t.

107
00:06:20,600 --> 00:06:26,940
The weight function of the
operator dx/dt plus a*x is e

108
00:06:26,940 --> 00:06:29,290
to the minus a*t.

109
00:06:29,290 --> 00:06:31,400
And the solution to the
differential equation

110
00:06:31,400 --> 00:06:33,220
with rest initial
conditions is given

111
00:06:33,220 --> 00:06:36,290
by the convolution integral
of the input signal

112
00:06:36,290 --> 00:06:38,260
with the weight function.

113
00:06:38,260 --> 00:06:41,740
This statement is a general
fact for LTI operators,

114
00:06:41,740 --> 00:06:44,590
not just first-order operators.