00:00in the last lecture we saw the use of
00:02convolution convolution is a
00:04mathematical tool which is used to
00:06calculate the output of an LTI system
00:08when the impulse response and input is
00:10available if you remember the initial
00:13lectures I told you there are five
00:15important operations we are required to
00:17study in this course the first operation
00:19was shifting operation both amplitude
00:22shifting and time shifting the second
00:24operation was a scaling operation both
00:27amplitude scaling and time scaling there
00:30was one operation known as reversal I am
00:33NOT considering reversal as a separate
00:35operation because it is an especial case
00:37of scaling operation amplitude reversal
00:40and time reversal both are special case
00:44of amplitude scaling and time scaling
00:47the third operation was differentiation
00:49operation we saw how to perform the
00:51graphical differentiation the fourth
00:53operation was integration operation and
00:55here also we saw how to perform the
00:58graphical integration but we did not
01:00move to our fifth operation that is the
01:03convolution operation because we know
01:05convolution operation is used to
01:07calculate the output of an LTI system
01:10when the impulse response and input is
01:12given so in this course the convolution
01:15is mainly used to calculate the output
01:17of an LTI system so it's good to discuss
01:20convolution once we know the basics of
01:22healthy eye systems now we know the
01:25basics of LTI systems it's important
01:27parameters what it is and now we can
01:30easily understand the convolution
01:33operation convolution operation can be
01:35used for different purposes but in our
01:38course it is mainly used to calculate
01:40the output of an LTI system so in this
01:43presentation we will see the definition
01:45of convolution we will see the
01:47mathematical formula of convolution we
01:50will see the number of steps involved in
01:52calculating the output using the
01:55convolution and once we are done with
01:57all these things we will solve one
01:59question which will explain all these
02:01steps involved in calculating the output
02:04using the operation known as convolution
02:07and after this before ending this
02:09lecture I will show you the animation of
02:12so this lecture is very important
02:15initially you will find the convolution
02:18a little bit confusing but if you know
02:20the integration well the integration I
02:22have already explained if you know it
02:24too well you can easily understand the
02:26convolution also so let's begin our
02:29discussion we will start with the
02:31definition of convolution a convolution
02:34is an integral that expresses the amount
02:39of overlap this is very important the
02:42amount of overlap of one function when
02:46it is shifted over another function so
02:52you can see five important points in
02:55this definition the first point is
02:57integral the second point is amount of
02:59overlap the third point is one function
03:02fourth point is shifted over and the
03:04fifth point is another function so there
03:07are two functions involved this is the
03:10first function this is the second
03:12function there is integration involved
03:15there is amount of overlap in a Walt
03:18when when the function is shifted over
03:21the another function this definition
03:24will be clear after solving one example
03:26now we will see the formula of
03:29convolution the impulse response HT of
03:34an LTI system is given and the input of
03:37the LTI system is also given and now we
03:41are interested in calculating the output
03:44YT so output YT is equal to convolution
03:48of HT and XT so this operator here is
03:53known as convolution operator
03:58convolution operator and the convolution
04:02operation between these two signals is
04:04equal to integration minus infinity to
04:08infinity X tau multiplied to H t minus
04:17tau D tau so this is the formula of
04:20convolution and I will explain all the
04:23steps involved in performing the
04:26using this formula so we are having two
04:29signals the first signal is XT the
04:32second signal is HT so we have two
04:34signals XT and HT here we are discussing
04:41the different steps involved in
04:43calculating the output using the
04:46convolution so in first step we are
04:49required to replace T by a dummy
04:53variable tau we will replace T by a
04:57dummy variable tau we already know the
05:00reason to replace T by dummy variable
05:02tau we are going to move our signal we
05:06are going to shift or move our signal
05:09waveform and for that we will vary the
05:13time T so we don't want to confuse
05:15between the particular instant of time T
05:18and the variable time T we are having in
05:22the signals so it's good to replace T by
05:24tau it will eliminate all the confusions
05:27while calculations so we will have X of
05:30tau and here we will have H of tau this
05:34is step number one now we want one
05:37signal to get fixed we want one signal
05:41to remain on its own position and we
05:46want X tau to remain on its own position
05:48because this another function here will
05:51remain on its own position and this
05:54function here will shift and calculate
05:56the total amount of overlap so we are
05:59interested in moving H T which is now H
06:03tau and for this we will first perform
06:05the time reversal operation the time
06:11reversal operation or you can say this
06:15scaling operation and this will give us
06:18H minus tau actually we are implementing
06:22the formula written here because using
06:25this formula we can easily perform the
06:27convolution and we will have our output
06:30YT so we have X of tau we have X of tau
06:34and we are trying to get H t minus tau
06:37we have obtained hat
06:39- Tao now it's time to perform the
06:46we are performing the time shifting
06:52operation and we know one important
06:55point related to multiple operations the
06:58time scaling operation and the time
07:01shifting operation is only performed
07:03against the variable we are having the
07:07variable is tau so we cannot perform the
07:10time shifting against minus tau so it is
07:13important to take the negative sign out
07:16we are having tau and here we will
07:20subtract T and when you simplify this
07:25you will have at t minus tau the same
07:30thing we are having here but this is the
07:32exact process we need to follow we will
07:35perform the time shifting against tau
07:38because it is our independent variable
07:40now it is a dummy variable which is
07:43acting as an independent variable so we
07:45are done with time shifting we have HT -
07:48tau we already had X tau and now we will
07:51multiply them we will multiply them so
07:55we will have X tau multiplied to H t
07:59minus tau and once you have the result
08:02of multiplication you need to perform
08:04the integration so from here we will
08:07have integration minus infinity to
08:10infinity X tau multiplied to H t minus
08:15tau D tau so we are done with all these
08:18steps this step number one is to replace
08:22T by tau we will have X tau we will fix
08:25it and then in step number two we will
08:28perform the time reversal we will have H
08:30minus tau in step number three we will
08:34perform time shifting against tau this
08:37will give us HT minus tau the step
08:40number four is to multiply X tau and HT
08:43minus tau and the fifth step the last
08:46step is to perform the integration so
08:48these are the five steps involved in the
08:51convolution operation
08:53now we will move to our example in this
08:56example we have XT the input of the LTI
09:06system and we also have HT the impulse
09:14response of the LTI system and we are
09:19interested in calculating the output of
09:22the LTI system so I will quickly draw
09:26the waveforms of input and the impulse
09:29response we are performing the graphical
09:32convolution here following all the five
09:35steps so let's see how the waveforms
09:37look the pink waveform is the wave form
09:40of input XT and the green waveform is
09:43the wave form of the impulse response at
09:46T both are square wave forms but the
09:49width of the first waveform waveform of
09:52XT is more as compared to the width of
09:54second waveform the waveform of HD and
09:57it is important to notice their widths
10:00because we will obtain one important
10:02conclusion after obtaining the output YT
10:04now we will perform our step number one
10:07in which we will replace T by tau in
10:10both the waveforms so let's quickly
10:14remove T and replace it by tau here we
10:22will have X tau tau H tau tau so we are
10:31done with step number one now in step
10:33number two we will perform the time
10:35reversal that is the special case of
10:37time scaling in the signal X tau we will
10:41keep X tau as it is and we will perform
10:44the time scaling which is the time
10:49reversal to be exact so this is how the
10:53waveform of H minus tau will look we
10:56have simply obtained the mirror image
10:58about devices and this is the waveform
11:02now we will move to step number three in
11:04which we will perform the time shifting
11:06and we know the process of performing
11:08the time shifting I will directly plot
11:11the wave form of H t minus tau but first
11:15let us mention that we have performed
11:20the time scaling or time reversal to
11:26obtain this waveform and now we will
11:28we will perform the next operation that
11:32is time shifting operation I will
11:38quickly plot the waveform after
11:41performing the time shifting so this is
11:44the waveform of HT - tau now don't
11:48confuse yourself that the signal is
11:50shifted towards the left Y signal is not
11:53shifted towards the right
11:54because T can be positive or it can be
11:57negative we will start performing the
12:00integration by moving our signal from
12:03infinity towards the right our signal X
12:06tau is nonzero from 0 to 2 and we will
12:10start moving our signal HT - tau from
12:14infinity and we will calculate all the
12:17possible values of this integration but
12:20it is smart choice to look for the time
12:23instants at which the signal value is
12:25changing for example when tau is equal
12:28to 0 the signal value is changing from 0
12:32to 1 and when tau is equal to 2 the
12:34signal value is changing from 1 to 0 so
12:37these points or time instants are very
12:39important to perform the convolution
12:41because we don't want to perform the
12:44convolution for each and every small
12:47increment in time I will explain this in
12:50the next steps but for now keep one
12:52thing in your mind after performing the
12:55time shifting this waveform can shift
12:57towards the right also depending on the
13:00value of T but we want to move our
13:02signal from minus infinity to worse the
13:04right so we are considering the value of
13:07T such that the signal is shifted
13:09towards the left so this is all for step
13:12number three now we will move to step
13:15number four and step number five we will
13:17perform step number four and step number
13:19five multiple times it is repeated one
13:23two and three are the steps we are
13:24required to do one time but four and
13:27five will repeat more than one time so
13:30let's see how we can calculate the
13:32integration by obtaining this
13:35multiplication for this I will make
13:38another plot which is the plot of X tau
13:41replied to HT minus tau in this plot
13:45this waveform is the waveform of hex tau
13:50and this waveform here is the waveform
13:53of h t minus tau and we are moving HT
13:57minus tau from left this arrow
13:59represents the movement of XT minus tau
14:02by the variation of T here and you can
14:05see one important thing in the waveform
14:09when we shifted this waveform the
14:12instant at which the signal value is
14:14changing from 0 to 1 becomes t minus 1
14:17because initially it was minus 1 and
14:20when minus tau is replaced by t minus
14:23tau we will have t minus 1 and this
14:27instant here which is the instant at
14:30which signal value is changing from 1 to
14:320 is 0 that's why we have t here 0 plus
14:36T is equal to T and this is important
14:38because we will use it while calculating
14:40the multiplication and hence the
14:43integration let's say this case is case
14:46number 1 and in this particular case we
14:51are considering time T to be less than 0
14:54here time T is less than 0 so you can
14:58see there is no overlap between the two
15:00signals HT minus tau and X tau and as
15:04there is no overlap the multiplication
15:06is going to be 0 so we will have X tau
15:10into HT minus tau equal to 0 and when
15:13you integrate it from minus infinity to
15:15infinity you will get 0 so I can write
15:20YT is equal to 0 when T is less than 0
15:25we are integrating X tau multiplied to
15:31when you multiply X tau 2 HT minus tau
15:33you will get 0 because h t minus tau is
15:37equal to 1 from t minus 1 to t but in
15:40this time interval signal X tau is equal
15:43to 0 so the multiplication is also equal
15:45to 0 and X tau is nonzero from 0 to 2 it
15:49is equal to 1 but in this time interval
15:51HT minus tau is equal to 0 so again they
15:55is equal to zero and elsewhere both
15:57these signals are equal to zero so
16:00multiplication is again equal to zero I
16:02hope you understand what I mean to say
16:04here so the multiplication is equal to
16:07zero and the integration of zero is
16:09equal to zero and the integration here
16:11is nothing but the output of the LTI
16:15system so why it is equal to zero in
16:17case number one when T is less than zero
16:19now we will move to our next case which
16:23is case number two and in case number
16:25two we will consider time T to be
16:30greater than zero but less than one this
16:33one here so time T is less than 1 but
16:36greater than zero this means this edge
16:39of the signal will exist between 0 and 1
16:42somewhere in this region so let's
16:44quickly plot the waveform so here you
16:47can see there is overlapping between X
16:50tau and HT minus tau and as we are
16:53multiplying them we will have some
16:54non-zero value between this interval
16:57from minus infinity to zero you can see
17:01the two signals are not overlapping HT
17:05minus tau the green waveform h t minus
17:09tau is non zero from t minus 1 to 0 but
17:13X tau is equal to 0 so the
17:15multiplication in this interval is equal
17:18to 0 in this interval from 0 to T the
17:22multiplication is equal to 1 because X
17:25tau is equal to 1 + h t minus tau is
17:28also equal to 1 after this from t to 1
17:31you can see x tau is one but HT minus
17:36tau is 0 so the multiplication is again
17:38equal to 0 and from 1 to 2 again
17:41multiplication is 0 and from 2 to
17:44infinity multiplication is equal to 0 so
17:47the multiplication is nonzero only
17:49between 0 to T so we will integrate from
17:530 to t there is no need to integrate
17:55from minus infinity to infinity because
17:57you will get the same result so we don't
18:00want to waste our time writing this
18:04toughs which will give us the result
18:08integrate from zero to T X tau
18:12multiplied to HT - tau is equal to one
18:15so we will have integration of 1 with
18:18respect to tau we already know
18:20integration of 1 is equal to tau the
18:23lower limit is 0 the upper limit is T so
18:26finally we have T so YT is equal to T
18:30when T is less than 1 and greater than 0
18:34now we will move to our next case
18:36case number 3 in this case we will move
18:40our signal HT - tau further you can see
18:44we have moved signal XT minus tau from
18:47this position to this position and now
18:48we will move it further and this time
18:51our time T will be greater than 1 but it
18:56is less than 2 so let's see how the
18:59waveform will look so this is how the
19:02waveform will look and as we have
19:03already seen the two cases we will
19:06quickly perform the calculations in our
19:08third case the two signals are non zero
19:11from t minus 1 to t you can see the
19:15signal value is equal to non zero the
19:17two signals are overlapping and the
19:19multiplication x tau HT - tau is equal
19:23to 1 so we will perform our integration
19:28from t minus 1 to t and the result of
19:33multiplication is 1 so we have 1 D tau
19:36and this will give us tau the lower
19:40limit is t minus 1 the upper limit is t
19:43let's simplify this it will give us t
19:46minus t plus 1 this is equal to 1 so the
19:50output YT is equal to 1 when T is less
19:53than 2 but greater than 1 now we will
19:56move to our fourth case in the fourth
20:00case we will move our signal further we
20:04will move it further and this time T is
20:07greater than 2 but less than 3 it is
20:11greater than 2 but it is less than 3 I
20:14will quickly plot the waveform so this
20:17is how the waveform will look in
20:19number four you can see T is greater
20:22than two but it is less than three and
20:25you can see the overlapping interval it
20:28is from tau equal to t minus one to tau
20:31equal to two so we will integrate in
20:34this interval only the output YT is
20:37equal to integration from t minus 1 to 2
20:41and the integration of 1 we have to
20:45perform because the multiplication is
20:47equal to 1 if you have some other result
20:50as the multiplication you have to
20:52integrate that result so here we will
20:54have tau the lower limit is t minus 1
20:59the upper limit is 2 let's put them 2
21:01minus t plus 1 so it is equal to 3 minus
21:07T now we will move to our fifth and last
21:10case in the 5th case I will plot the
21:14waveform but here you need to notice
21:16that T is greater than 3 so you can
21:20visualize what will happen to the
21:22overlapping when you further move the
21:25green waveform towards the right there
21:27will be no overlapping when T is greater
21:30than 3 I will show this by the help of
21:33the waveform so you can see there is no
21:35overlapping between the two signals hex
21:38tau and hatch t minus tau so the output
21:41variety is simply equal to 0 so we have
21:44the output and now we can easily obtain
21:47the output waveform so let's quickly
21:50define our output I will define the
21:54output here that is not a good place to
21:57write but I want to write the output
22:00here only so the output variety is equal
22:04is equal to we will start from our first
22:09case in the first case the output YT is
22:12equal to 0 so it is 0 and when it is 0
22:16that is 0 when T is less than 0 it is
22:19less than 0 from second case we can see
22:22the output YT is equal to T so it is T
22:28and it is T when T is less than 1 but
22:33zero so T is less than one but it is
22:36greater than zero let's move to our case
22:39number three from here we have output YT
22:42equal to 1 and it is 1 when P is less
22:46than 2 but greater than 1 less than 2
22:49but greater than 1 let's move to our
22:52case number 4 from case number 4 we have
22:56YT equal to 3 minus T 3 minus T when T
23:01is less than 3 but greater than 2 and
23:05from the last case YT 0 and it will
23:10happen when T is greater than 3 so this
23:14is your answer you can write this as
23:16your answer and if you want to find out
23:19the wave form you can easily plot it
23:21using this definition and you will see
23:25the wave form of the output is
23:26trapezoidal because we have two
23:30rectangular pulses with unequal width so
23:33whenever you have two rectangular pulses
23:35with unequal width you will get the
23:38trapezoidal waveform as the output and
23:40when you convolute the equal width
23:42rectangular pulses you will get the
23:45output has a triangular signal so this
23:48is one important point because sometimes
23:50while choosing the four options given
23:53you can use these small small properties
23:55so I will repeat it once more when you
23:58have two rectangular pulses with unequal
24:00width you will get the trapezoidal
24:01signal as the output and when you have
24:04the two signals which are rectangular
24:07but having the same width you will get
24:09the triangular output waveform so for
24:12now we will quickly plot the waveform of
24:14the output we have obtained so I will
24:16plot the waveform and once we are done
24:19with the waveform we will obtain the
24:22expression of the waveform in terms of
24:25ramp signals I am telling you we will
24:28get the expression in terms of ramp
24:31signal because we have the continuous
24:33waveform the trapezoid is a continuous
24:37waveform from the definition of YT you
24:39can see when T is less than 0 we have 0
24:42when T is from 0 to 1 we have T
24:45so we will quickly plot this part of the
24:48waveform when time T is less than zero
24:53why it is also equal to zero so we will
24:56have the waveform like this it will
25:01reach to minus infinity having the
25:03magnitude equal to zero and after this
25:06when T is between 0 and 1 we have Y T
25:12so before plotting the waveform I will
25:15first mark down all the important time
25:21and now we will plot the remaining part
25:24of the waveform from 0 to 1 YT is equal
25:29to T so it will increase linearly like
25:34this and from the definition you can see
25:36when T is between 1 & 2 it is equal to 1
25:42it is constant so we will have the
25:46waveform like this from 1 to 2 and after
25:52this when T is between 2 & 3 we have y T
25:58equal to 3 minus T it is equation of
26:01straight line with a negative slope and
26:07intercept equal to 3 so this is how the
26:11waveform will look and when T is greater
26:14than 3 again why it is equal to 0 till
26:18infinity so this is how the waveform
26:21will look but I am interested in finding
26:23out the expression of Y T in terms of
26:29ramp signals and for this we will use
26:38the knowledge we have obtained from the
26:41lecture mathematical representations of
26:43signal waveforms initially we will
26:46follow the signal flow it is like this
26:49so YT is equal to 0 but when T is equal
26:54to 0 you can see the upward turn and
26:58whenever there is upward turn we take
27:01positive sign and after taking the
27:03upward turn the flow of the signal will
27:06change it will become like this and when
27:09you take the upward turn you need to
27:11write down the magnitude of the slope
27:14the slope magnitude you can calculate it
27:17is 1 divided by 1 so the magnitude is 1
27:20and the ramp signal will be our t minus
27:250 because the line is passing through 0
27:28and once we have T equal to 1 again the
27:32signal who will take the turn but this
27:35time the turn is downward and the flow
27:38of this signal will become like this for
27:40downward turn we take the negative sign
27:42and again we are required to write down
27:45the magnitude of the slope these two
27:47angles are same so the slope will also
27:50remain same 1 and the ramp signal will
27:53be our t minus 1 because t is equal to 1
27:57here and after this when T is equal to 2
28:01there is downward turn again the flow of
28:04this signal will change it will become
28:07like this for downward turn again we
28:10will take negative sign and the
28:11magnitude of the slope is equal to 1
28:13again because 1 divided by 1 is equal to
28:171 and the ramp signal will be our t
28:22minus 2 and after this when we are at t
28:27again there is turn which is upward turn
28:30so plus 1 the magnitude is 1 because
28:35these two angles are same so the
28:37magnitude of the slope is equal to 1 and
28:39after this after the support turn the
28:42flow of the signal is like this so we
28:44will have our t minus 3 so we are done
28:50our analysis and now we can simplify the
28:54obtained results from here we have our
28:57tea from here we have minus RT minus 1
29:02and from here we have minus R t minus 2
29:06plus R T minus 3 so in this way you can
29:10represent the output waveform in terms
29:12of ramp signals so this is all for this
29:16lecture you can obtain the same result
29:18using the Laplace transform which I will
29:21explain in the coming presentations but
29:23from next lecture we will start the
29:25properties of convolution now we will
29:27see the animation of the convolution we
29:30have performed in this lecture you can
29:33see two windows the first window is
29:35having two signals one signal is
29:37stationary and it is X tau the other
29:40signal is moving and it is HT minus tau
29:42when HT minus tau is moving towards the
29:45right the amount of overlap is
29:47increasing and convolution is an
29:50integral which gives us the amount of
29:53overlap between the two signals when one
29:56signal is shifting towards the other
29:58signal and you can see that in the
30:01second window here you can see the
30:04convolution is initially equal to 0 and
30:07then it is increasing linearly and after
30:10some time it becomes constant and when
30:13this signal XT minus tau is moving
30:15further the amount of overlap is
30:17decreasing and the convolution is also
30:20decreasing and finally becomes 0 so I
30:23hope this animation is clear to you so
30:26if you have any doubt you may ask in the
30:27comment section I will end this lecture
30:29here see you in the next one
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