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11. Difference Equations—Inverse Z Transforms—Using Partial Fractions -21MAB201T — Transcript

by Dr. E. Suresh · 2,409 words · 294 segments · language en · Watch on YouTube

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0:00Dear students, welcome to the problem
0:01solving session. In today's session, we
0:03are going to solve difference equation
0:06using Z-transform.
0:09A difference equation is a mathematical
0:12expression that relates the value of a
0:14discrete function at different point in
0:17time or space. It involves the sequence
0:19of values typically represented by YN,
0:22where N is integer. Sometime, it may be
0:25expressed as Y of N also. Recurrence
0:28relation, it express the value in terms
0:31of its previous value. For example, you
0:34can see YN is equal to function of Y N
0:38minus 1. There are two types of
0:40difference equation. Linear difference
0:42equation, you can see the example. All
0:45my AN, AN minus 1, etc., A naught and GN
0:49are constant, then it is going to be
0:52linear difference equation. Otherwise,
0:54it is going to be a non-linear
0:56difference equation. It is widely used
0:59in computer science engineering for
1:01algorithms, codes, biology, etc. Let us
1:04take a simple example. Everyone knows
1:07the Fibonacci numbers, where the
1:09relation is FN is equal to FN minus 1
1:13plus FN minus 2 with the starting value
1:15F naught equal to 0 and F1 equal to 1.
1:19Now, let us go into the problem solving
1:22point of view. Solving difference
1:24equation using Z-transform. Here, we are
1:26going to consider Y of N is my basic
1:30unit. Z of Y of N is F of Z, then Z of
1:34YN plus 1 is going to be Z into F of Z,
1:39and we need one initial condition Y of 0
1:42into Z. Now, Z of YN plus 2 is going to
1:47be very easy. It needs two conditions,
1:50that is two initial condition. So, now
1:52multiply this by Z. We'll be getting
1:55small Z squared F of z minus z squared y
1:58naught and subtract with z into y one.
2:02And similarly, z of y n plus three, now
2:06you would have guessed, just multiply by
2:08z in the previous formula, z cubed f of
2:11z minus z cubed y naught minus z squared
2:14y one and subtract z into y of two. Hope
2:18you understand. You can see the typed
2:20version. So, we are going to apply z
2:22transform on the difference equation and
2:25then inverse, find the solution for the
2:27difference equation. Let us go into the
2:30problem. Now, it is very important in
2:32exams, they give u n, u of n, y n, y of
2:37n, z of n, omega of n, etc., etc. in
2:41various notations. We are going to
2:43follow the standard notation y of n. If
2:47u naught is given, how can I convert
2:49this? I can write it as y of zero. u one
2:52means y of one. Hope you understand.
2:55Now, let us write the given difference
2:57equation and write in our format. Let us
3:00write in our format. y of n plus two
3:02minus five times y of n plus one plus
3:05six times y of n is equal to four power
3:08n with the conditions y naught is equal
3:11to zero, y of one is equal to one. Once
3:14you rewrite the problem in our format,
3:17write the formulas. Here, we need z of y
3:20n, z of y n plus one and z of y n plus
3:23two. Let us write the formula one by
3:25one. This is the very important
3:27assumption. z of y n is f of z. Then, z
3:31of y n plus one is z into f of z minus z
3:34into y of zero. Similarly, z of y n plus
3:37two is z squared f of z minus z squared
3:40y naught minus z into y of one. Now, let
3:44us take this equation as one. Applying z
3:48transform on both sides, we'll be
3:50getting the new data. Now, we know the
3:53formula. Just we have to write it. Z
3:56squared capital F of Z minus Z squared Y
4:00of zero minus Z into Y of one minus five
4:04times Z into F of Z minus Z into Y
4:07naught then plus six times Z of Y N is
4:10simply capital F of Z and we know that Z
4:14of A power N is going to be Z by Z minus
4:18A. So we are getting now Z by Z minus
4:21four. Now the condition is given to us Y
4:23of zero is zero, Y of one is one. Now Y
4:27of zero is zero, this term is going to
4:29be zero and this is going to be zero and
4:31this is equal to one. Now substituting
4:34and simplifying we need to isolate
4:37capital F of Z alone. So first let us
4:39take this terms Z squared F of Z minus
4:42five Z F of Z plus six F of Z. So Z
4:45squared minus five Z plus six F of Z.
4:49Since this terms are zero, only Y of one
4:52exists. This is nothing but Z into one
4:55that is equal to Z by Z minus four. Now
4:58simplifying this will be getting Z minus
5:01two, Z minus three. Taking LCM on the
5:04right hand side and we'll be getting Z
5:06squared minus three Z and we take Z
5:09outside we'll be getting Z into Z minus
5:12three divided by Z minus four. As I said
5:15I want capital F of Z. Push the all the
5:18other things to the other side. Now Z
5:20minus three Z minus three get cancelled.
5:22I'll be getting Z by Z minus two into Z
5:26minus four. If you are going to solve by
5:28partial fraction method, I need to
5:30always reserve one Z. Otherwise I cannot
5:33apply inverse. Therefore whatever the
5:36problem comes, reserve one Z from the
5:38numerator and move to the denominator of
5:40F of Z like this. So now F of Z is equal
5:44to one by Z minus two into Z minus 4.
5:48I'm going to apply partial fraction. See
5:50the highest denominator power. It's
5:53going to be two because here we have Z
5:56into Z. And the numerator power is going
5:58to be zero. So, I can apply partial
6:01fraction because the rule is highest
6:03denominator power should be greater than
6:06highest numerator power. Now, I can
6:08split this fraction as A by Z minus 2
6:11plus B by Z minus 4. I'm just reserving
6:15Z here. Now, taking LCM, we'll be
6:17getting A into Z minus 2 plus B into Z
6:21minus 4. Both sides denominator get
6:23cancelled. We'll be getting one. Now, I
6:25want to find the values of A and B. It
6:28is very easy. If you put Z equal to 4, A
6:32term will become zero. So, B into 4
6:35minus 2 is equal to 1. Therefore, B
6:38equal to 1 by 2. Now, when you put Z
6:40equal to 2, then B into 0, B term will
6:44get eliminated. We have 2 minus 4 is
6:47equal to 1. Therefore, A equal to minus
6:501 by 2. We found both A and B. Next, now
6:54substitute the value of A and B in a
6:57legible way. Write like f of Z by Z is
7:00going to be A value is minus 1 by 2 1 by
7:04Z minus 2 plus 1 by 2 1 by Z minus 4. We
7:09know that all our Z transform formulas
7:12numerator has an Z. That's why we
7:15reserve this Z in the denominator. So,
7:17now taking the Z to the other side,
7:20we'll be getting capital f of Z is minus
7:231 by 2 Z by Z minus 2 plus 1 by 2 Z by Z
7:29minus 4. From the formula of Z transform
7:32of the difference equation, we know that
7:34Z of Y of N is capital f of Z. I said Y
7:38of N is my solution. So, if you apply Z
7:42inverse on both sides, we'll be getting
7:45Y of N is equal to Z inverse of f of Z.
7:49Therefore, applying Z inverse on both
7:52sides of this equation will give the
7:55final solution. So, now we know Z
7:58inverse of Z by Z minus A is A power N.
8:02So, Z by Z minus 2 is going to be 2
8:05power N. Z inverse of Z by Z minus 4 is
8:084 power N. Therefore, my answer is minus
8:121 by 2 2 power N plus 1 by 2 4 power N.
8:16If you take this 1 by 2 commonly out,
8:18we'll be getting 4 power N minus 2 power
8:22N. This is my final answer. Next, we go
8:25for a different problem
8:28where both initial conditions are zero.
8:31First, let us write in our proper format
8:34and then write the Z transform formula.
8:37Z of YN, Z of YN plus 1, and Z of YN
8:40plus 2. Now, take this as equation 1.
8:43Apply Z transform on both sides of the
8:46equation. Now, substituting the value of
8:50Z of YN plus 2, YN plus 1, YN etc.,
8:54we'll be getting like this. And the
8:55right hand side is Z of 2 power N. And
8:58the right hand side is Z of 2 power N.
9:00We know that Z of 2 power N is going to
9:03be Z by Z minus 2. Since all my initial
9:07conditions are zero, everything is going
9:09to be zero. We'll be getting Z squared
9:11plus 6 Z plus 9 into capital f of Z is Z
9:16by Z minus 2. Now, keep f of Z on left
9:19hand side, push all the things to the
9:21right hand side. As I said, if you want
9:24to do by partial fraction, if you want
9:26to solve the problem by partial
9:28fraction, then we have to reserve this Z
9:30to the denominator of f of Z. So, f of Z
9:34by Z is going to be 1 by Z minus 2 into
9:37z squared plus 6 z plus 9. Now, we know
9:40that z squared plus 6 z plus 9 is
9:43nothing but z plus 3 whole squared.
9:46Therefore, next we have to do partial
9:49fraction. Since my highest denominator
9:52power is going to be 3 and highest
9:54numerator power is going to be 0. So,
9:57highest denominator power is greater
9:58than highest numerator power, I can
10:00apply partial fraction. And we know how
10:03to apply partial fraction. z minus 2 z
10:06plus 3 whole squared. For z minus 2, it
10:09is unique, so I'm writing a by z minus
10:122. For z plus 3 whole squared, I'm
10:14writing b by z plus 3 plus c by z plus 3
10:19whole squared. Now, just take the LCM
10:21and don't write the denominator like
10:23your school days. Just write the
10:24numerator alone, denominator
10:26automatically get cancelled. So, a into
10:29z plus 3 whole squared because it has
10:31only z minus 2 term. b into z minus 2 z
10:36plus 3. c into z minus 2 alone. Next, we
10:40have to find the values of a b c. As I
10:43said, we have to substitute some z value
10:46such that it make any one of the term at
10:48least 0. Now, put z equal to minus 3.
10:51When you substitute z equal to minus 3,
10:54this term will get 0, this term will get
10:560. I'll be getting c directly. So, minus
10:593 minus 2 is going to be minus 5.
11:02Therefore, c equal to minus 1 by 5.
11:06Next, we have to put z equal to 2. When
11:09you put z equal to 2, this both terms
11:11will get 0. So, we'll be getting a into
11:142 plus 3 whole squared is 1. So, a equal
11:18to 1 by 25. Next, don't go for a random
11:22value or don't go for 0 because if you
11:25put the 0 or random value, you'll be
11:27getting equation in terms of a b c. So,
11:30the
11:31takes a
11:32time. So, in this situation, we use our
11:34handy tool. Just compare the
11:37coefficients of either
11:39power or the constant. Highest power is
11:42the wisest decision here. So,
11:43coefficient of z squared, the highest
11:45power is z squared. So, the coefficient
11:47of z squared is going to be A in this
11:50term. Similarly, coefficient of highest
11:53term is z into z squared into B. Here,
11:55it is going to be B. And for the third
11:58term, there is no z squared. So, it is
12:00going to be zero. And right hand side is
12:02also there is no z squared term. That is
12:04also zero. We know A already, so B equal
12:08to -1/25.
12:10Immediately, we found all the three
12:12values, A, B, and C. Now, it's our job
12:16is to substitute this ABC in a legible
12:19way, like this. Next, we have to push
12:22the z to the numerator to apply z
12:25inverse. We know that z inverse of z by
12:28z minus A is A power n. Therefore,
12:32answer for this is going to be 2 power
12:34n. On the other side, we know the
12:36formula z inverse of z by z plus A is
12:41going to be minus A whole power n.
12:44Therefore, the answer for the second
12:46term is minus three whole power n. Now,
12:49I can write the answer directly for the
12:52first two terms. But for the third term,
12:55I don't know the formula. Now, just
12:57recall the z inverse formula. z inverse
13:00of Az divided by z minus A whole squared
13:04is going to be n into A power n. But
13:07here, we have plus term. So, I can
13:10rewrite the formula z inverse of z plus
13:13A whole squared is going to be minus A
13:16into z. That is equal to n into minus A
13:20whole power n. So, for this problem, I
13:22need this minus three in the numerator.
13:24-1/5, I multiply and divide by minus
13:28three, you'll be getting Z inverse of
13:30minus three into Z divided by Z plus
13:34three whole square. Now, I can apply
13:36this formula directly and I'll be
13:39getting the answer as N into minus three
13:42whole power N because just now I said Z
13:44inverse of minus A Z divided by Z plus A
13:48whole square as N into minus A whole
13:52power N. This is my solution. And we
13:55know that Z inverse of F of Z is nothing
13:57but Y of N. That is my answer. Hope you
14:00understand. We see some more problems on
14:02the next session.
14:04Thanks for watching. Subscribe our
14:06channel and share to your friends. See
14:09you in the next video. Bye-bye.

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