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Electrical Circuits II - Non sinusoidal waveform — Transcript

by Lizandro Bitang · 6,147 words · 975 segments · language en · Watch on YouTube

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0:02So good day
0:04everyone. It's my pleasure to welcome
0:07you today in lecture on important and
0:11practical topic in electrical circuits
0:142. So this is the
0:17non-sinosoidal wave form. So as an
0:20electrical engineering student and
0:23future
0:24practitioners you will frequently
0:26encounter a variety of wave forms in
0:29your field work in the
0:31future. So understanding how these
0:35nonsenosodal weights behave and how we
0:38can analyze them will significantly
0:41strengthen your ability to design and
0:45troubleshooot electrical and
0:48electronic
0:52systems. So in electrical
0:55engineering, wave forms are fundamental
0:59to understand how circuits behave over
1:03time. Most of us are familiar with
1:06sinusoidal wave forms. They are
1:10smooth, periodic, and naturally occur in
1:15AC power system. But
1:18however, real world system rarely
1:21operate on ideal sign waves
1:26alone. Devices such as switching power
1:29supplies, logic circuits, and motor
1:32drivers produce wave forms that are far
1:36from sinosoidal.
1:39These are known as non-synosoidal wave
1:42forms and they are characterized by
1:45sharp edges, abrupt transitions and
1:49complex harmony content. So in today's
1:53session we explore what these wave forms
1:56are, why they matter and how we analyze
2:00them using powerful tools such as the
2:06four-year analysis.
2:12So nonsosodal wave forms include any
2:16repetitive wave forms that is not a pure
2:21sign wave. So common types include
2:24square
2:25waves, triangular waves, sooth waves,
2:30pulses, and even clip or restorted side
2:35wave.
2:37Unlike sign waves that have a single
2:40frequency, non-sinosoidal waves
2:43typically consist of a fundamental
2:46frequency combined with multiple
2:50harmonics. These wave forms are
2:53essential in digital
2:55circuits where binary signals transition
2:59rapidly between high and low states. And
3:02in power systems where converters and
3:05inverters generate complex voltage
3:12shapes. So why should we electrical
3:16engineers devote our time to
3:19understanding nonsenosoidal wave forms?
3:23This is because these wave forms are the
3:27backbone of modern technologies.
3:32In power electronics, they appear in
3:34switching converters and pulse with
3:38modulations signal. In digital
3:41electronic, logic gates and
3:43microcontrollers generate square and
3:46pulse signals. In signal processing, we
3:50shape and modify nonsenosidal signal to
3:54filter or extract information.
3:57Moreover, harmonics introduced by these
4:01wave forms can cause power quality
4:04issues such as overheating equipment or
4:09signal distortion.
4:12Hence understanding their impact and
4:15behavior is a crucial for designing
4:19reliable systems and complying with
4:21standard such as the
4:24IE 519.
4:32To handle the complexity of nonsosoidal
4:35wave
4:36forms engineers turn to 4 year analysis
4:41so this is a powerful tool this method
4:44allows us to decompose a complex
4:48periodic wave form into a series of s
4:52and cosine wav the beauty of this method
4:57lies in universality
5:00almost any periodic signal no matter how
5:04complex can beesented a sum of sinosoids
5:09this helps us analyze signals in the
5:13frequency domain making it easier to
5:16understand the system behavior design
5:19filters and mitigate
5:22harmonic okay so we will be using this
5:26equation
5:28the year equation
5:33so
5:35any complex signal can be written as sum
5:40of s and cosine function so including
5:43the a z0 the dc
5:46component
5:52so the general form of body here series.
5:57So the fouryear series is a mathematical
6:00representation of a periodic signal as
6:03an infinite okay infinite sum of s and
6:08cosine. So as you can see on our
6:11equation
6:13fx so
6:15this is composed of the sum of sine and
6:19cosine but they differ on its kato ang
6:24frequency.
6:26So sin x cos x is what you call the
6:31fundamental frequency x.
6:35So as you increase the number of term so
6:39just multiply it by 81 depending on what
6:44term you are finding for example a sub2
6:47so make the frequency
6:512 for also 4 cos up to nx or cos nx a
6:59sub n and the b sub n so omega guys the
7:05fundamental angular frequency so
7:08familiar na this is equivalent to 2π
7:11times the
7:13fundamental
7:17frequency. Okay. So our challenge here
7:21is to find or to solve for the 4year
7:25coefficients such as a sub0 a sub1 b
7:28sub1 a2 b2 so on and so forth.
7:37Okay.
7:40So the first term of sine and cosine
7:43series of the series analysis is what it
7:47called the fundamental component. No the
7:51first term
7:54so this is what you call the fundamental
7:57component. So the fundamental component
8:00represents the minimum frequency term
8:04required to represent a wave form and
8:09this fundamental component should always
8:13be present in any year representation.
8:18So the other term with higher order
8:20frequencies are what you call the
8:24harmonic
8:29terms. Okay.
8:31So determination of fouryear equation
8:34which specify a particular wave is
8:37called wave analysis.
8:39So as what I have mentioned a while ago
8:42that uh in representing a complex wave
8:45form into a 4 year series so we need to
8:48solve for or determine the coeffici of
8:51the coefficients a sub0 a sub 1 b etc.
8:56So the determination of coefficients can
8:58be accomplished by some operation on the
9:01equation that will eliminate all terms
9:05except the decided quantity.
9:11So we have I have here now the
9:14formula in solving for the coefficients
9:17of the series.
9:18So for a
9:21sub0 1/ 2π integral of y dx from 0 to
9:272π a sub 1 / π integral of 0 2π y sin x
9:32dx b sub 1 / π integral of y cos x dx 0
9:38to 2π for a always s b always cosine so
9:46if you are finding for a sub 1. So you
9:51will just multiply sin x and b sub 1 cos
9:54x. For
9:56b2 you will multiply cos 2x
10:01a3x and b3 cos 3x and of course a sub 2
10:07y sin 2x. So same limit 0 to
10:112π and the general
10:15formula we will use this a sub n / π
10:19integral of y sin nx dx 0 to 2π d sub n
10:26/ π integral of 0 to
10:282π cos n x dx so note so i have here
10:34note no even functions only cosine terms
10:38b sub n because cosine is even so if the
10:42function is even so the s component are
10:46all zero only cosine terms are present
10:51and for add function also only terms are
10:55present and all cosine terms are
10:59zer so for mixed functions both sine and
11:02cosine terms are
11:05needed okay
11:07So this topic even add and mixed
11:12functions just review no on your topic
11:15on functions no po during your
11:18differential calculus I think
11:20differential calculus this this topic is
11:25was
11:26introduced differential calculus just
11:30review so for
11:34example write the four series which
11:37represents the south wave form. So this
11:41is an example of a
11:43nonsenosoidal wave form. So we can we're
11:46going to write the fouryear series
11:49no of this
11:51given wave form. So two so it
11:54started
11:56π uh the one
11:58cycl complete cycle is from -0 to 2π. So
12:04this corresponds to the one cycle.
12:061 from -
12:22π 4π na siya di so 4π lang
12:28differensya okay so let's solve this
12:33problem so it will be observed that the
12:36wave form is simply a straight line
12:38variation so straight line lang kini
12:41from t to
12:43diha so our challenge here is to solve
12:46for the function y okay function y so
12:51that we can substitute to the formula.
12:54So it can be seen no as you can observe
12:57on this illustration here this
13:01one that point A no for example uh look
13:06at point A so point A has a coordinates
13:11of what 0 and -
13:16π point B as you can see here the X
13:20component is
13:222π and The y component is posi pi. So we
13:28have two points no so we can use we can
13:32compute for the slope of this line. So
13:36we will use the slope formula y sub2- y
13:41sub 1 / x sub2 - x sub 1. Okay? So
13:45substitute. So you will have slope
13:48equivalent to
13:501 and we will use the point slope form
13:55equation of a line point slope form or
13:58the slope intercept form y = mx +
14:06[Musika]
14:07b okay so substitute m =
14:111 for the
14:14slope at
14:170- π so that we can solve for the y
14:22intercept
14:24b okay so at
14:290-π meaning we're going to put the value
14:32for y π equ m our slope is
14:381 x component here
14:41is b is equ
14:45to uh + b so solving for b so we have
14:51b-
14:53π okay so finally we have now the
14:56equation of this
14:58line y x - π so this will be our f x
15:08okay or the equation of this line here
15:13from zero 0 to
15:172π. Okay.
15:21Next. Now we can solve for the
15:25coefficient of the 4 year I mean the 4
15:28year coefficient. So let's start with a
15:33sub0. So our fx is x- π. So this is good
15:40for uh from 0 to 2π. Okay. This
15:47function is from 0 to
15:512π. Now as you can see in our previous
15:53slide so we have from 0 to 2π one
15:57complete cycle.
16:01Okay.
16:03So the formula we use this formula a
16:06sub0 is 1/ 2π integral of fx dx from 0
16:12to 2π. So
16:15substitute fx as x - π
16:21integrate. So
16:24integrate o siya buwagon.
16:28So integral of x dx
16:31- integral of π dx. So integral of x is
16:36x² / 2 limits from 0 to 2π. So
16:41substitute the upper and lower limit. So
16:43we have 2π s for integral
16:48of x dx. And for the integral of π dx.
16:52So you we have 2π s.
16:57So combining the two and multiplied it
17:00with 1 / 2π. So the result for a sub0 is
17:05zer. So for your a sub z0 term
17:09is
17:10zero. Okay? Zero.
17:14Next uh let's solve for a sub one term.
17:20So for a sub of one
17:21term same
17:23fx for a sub of one term uh we have the
17:28formula 1/ integral of fx sin x dx from
17:340 to 2π
17:40substitute a sub 1 is equivalent to
17:44integral of x- π sin x dx from 0 to 2π
17:48F. So
17:50integrate we just separate the
17:56integral
17:58so x integral of x sin x dx - integral
18:02of sin x I mean π integral of sin x dx
18:08from 0 to 2π. So the integral of x sin x
18:13dx can be solved using integration by
18:17parts.
18:19So by introducing u = x and du dx and dv
18:26sin x dx and the integral of sin x dx is
18:30- cosine. So that formula UV so UV U V
18:38or equivalent to - X cos X minus
18:42integral of video - V RV is negative so
18:47negative times negative so
18:50positive integral of cos x dx so
18:56integral of cosine
18:59is sin so the integral of x sin x dx is
19:03- x cos x + sin x substitute the limit
19:08evaluate it from 0 to
19:112π result
19:14is
19:152π and
19:17also
19:19for the integral
19:22of sin
19:24x dx from 0 to 2π the result is so you
19:30can use your calculator
19:32para madali.
19:37Okay. So our a sub 1 di
19:41is a sub 1 will be equivalent
19:45to how
19:47much? Ah ne
19:51π evaluation zer. So for this quantity
19:56here is zero man siya and daman tay - 2π
20:03π
20:05-2 okay so a sub one term is -
20:112 so i-note lang na
20:15siya sa solbo ni iyang kuan next for b
20:19sub
20:20one o the formula hulihan lang na o
20:25cosine. Okay. So substitute.
20:29O so just follow the
20:32solution integral of x cos x can be
20:36solved again by integration by
20:40parts and evaluating the integral from 0
20:44to 2π the result is zero and of course
20:48the integral of π integral cos x 0
20:542π so zero nahapon
20:57siya So therefore B sub 1 is
21:020. Next we solve for a sub 2.
21:06Coefficient of a sub
21:092. B sub 1 is 0.
21:15So for a sub
21:192 same fx but the formula now will
21:25be multiplied by sin
21:292x since a sub 2 atolve so we will
21:33multiply multiply sin 2x 4a sub
21:392 okay
21:41so this quantity
21:44here integral of x sin 2x dx from 0 to
21:492π- π integal of sin 2x dx 0 to 2π
21:54and multiped by 1 π so
22:00nagsa this this equation here can be
22:04solved using again integration by parts
22:14So integration by parts and
22:16evaluate from 0 to 2π. So the result is
22:21-
22:24π integral of sin 2x dx from 0 to 2π the
22:29result is 0. Then substitute to the
22:33formula.
22:36Okay. 1 / π - π - 0 a sub 2
22:41is
22:441.
22:47So we now have a sub2 -1. Next we solve
22:51for b sub 2. Coefficient of b sub 2. So
22:56same process no. So we now going to
22:59multiply instead of cos x we multiply
23:03cos 2x kay sa b sub
23:072. So follow the
23:09steps. Same procedure as what we did on
23:12the previous uh
23:16solution or you can use your calculator
23:18to immediately solve for the
23:21coefficient. na mo just make sure that
23:25your calculator is a radian mode okay
23:29radian mode dapat inyoang
23:34calculator because if your calculator is
23:37in degree mode so it will arrive on a
23:41different
23:42answer. Okay take note on
23:45that. So our B sub 2 is zero.
23:51So zero next a sub
23:553. Same procedure or same process but
23:59instead of sin 2x
24:02multiply sin 3x same fx.
24:11So integrating the
24:14function integrate using a
24:17calculator. So the result
24:21is
24:2323.
24:25Okay. You can double check on your
24:28calculators - 2/3. So next we solve for
24:33v sub 3. Same process instead of
24:37multiplying it by cosine 2x multiply
24:40cosine
24:433x. Okay. So by calculator
24:48works the result is b sub 3 is 0.
24:56a sub 4. So a sub 4 o instead of sin 3x
25:02you will now going to multiply by sin
25:084x. Okay so calculator
25:13works. So you will arrive with the value
25:17of
25:211/2 B sub
25:244. So cos 4x na sad.
25:27P 4 cos
25:334x-multiply 0 4 i-calcule na lang na
25:36para
25:38mali a sub n or the general equation or
25:42the general
25:45coefficient use the formula sin
25:48nx from 0 to
25:502π and same procedure again use
25:55integration by parts no after you
25:58distribute the sign to the equation so x
26:03this will become f sin nx dx from 0 to
26:062π-
26:10sin nx dx from 0 to 2π so to solve this
26:14quantity here this integral we will be
26:17using integration by parts so the answer
26:21is -2π / n for the integral
26:27sin nx the answer
26:30is so over a sub n is ne 2
26:38n and b sub n will be equivalent
26:42to zer so we can now write our 4 years
26:46series representation of the sooth wave
26:50so this will be equivalent to imo na
26:53lang ipang ipang ang substitute na n mo
26:56ang sa a sub 0 a sub 1 b sub 1 and this
27:02will this this this will correspond also
27:05to the
27:08function gamitan for example a sub
27:121 dapat tapad siya sa sin x or sin omega
27:19t if your a iser so this term will be
27:22will not be included in the equation so
27:27as you can see here our a sub0er is zer
27:30so wala na
27:31siyael maatulag na mga value our b walay
27:36cosine so kasabot zero imong
27:41cosine okay so up to so the last term
27:47will be sin nx / n so ma itong a sub n
27:53'di ' ba na our n is - 2 / n so ato
27:59langigawas ang 2 common man ang 2 sa
28:03common man ang two sa
28:06tan i
28:08mean or formally pwede na siya ma-write
28:11in this form the series representation
28:16of our wave will be the summation of 2 n
28:21sin nx x and from 1 to infinity so pwede
28:29naon so k this is now enough enough
28:32enough enough enough na
28:33siya yung answer dapat nakaingani
28:37ngani
28:40ngrepresent okay so let's try another
28:45example okay so write the first four
28:48terms of the year series which will
28:52represent the wave form. So as you can
28:56see our wave form is a half wave no so
29:01from zero to
29:05π from 0 to π one
29:09cycle with a maximum value of i sub m.
29:13So this is a sinosoidal sinosoidal here.
29:16So after that from π to 2π the value is
29:22zer or the value has been
29:25clip cut and again same
29:31value from uh in the next cycle. So for
29:36one complete cycle half wave lang siya
29:38no meaning usa ka wave dira for the
29:42first half and the next half equivalent
29:44to
29:46zero. So it can be seen in the figure
29:49that the expression of current can be
29:52expressed between the limits of 0 to 2π.
29:57So du ka function na to makuha ni no
30:00from 0 to π no from 0 to
30:04π we can have equation of the current as
30:09im sin alpha or sin alpha.
30:14And for the next cycle or the next half
30:20cycle from π to
30:222π
30:24soer i i
30:28iser or we're going to i-replace ang a
30:31as omeg t we can use as we can use this
30:35equation for i in the first half im sin
30:39omega t from 0 to π and i0 4 π gand to
30:482π 0 so
30:50this are our
30:53equation im sin omega t and i = 0 so im
30:58omega sin omega t is good from 0 to π
31:04and i = 0 from π to 2π so atong ka
31:11functionunction so The piece wise
31:13defined wave form for one
31:16period from 0 to
31:182π i is im sin alpha or sin omega t 4
31:25alpha 0 to π and 0 for alpha π to 2π
31:32okay so we can also solve for the first
31:35term or the a0 term of the year
31:40series okay So a sub z0 instead of using
31:43y dx lah namanong variable so ano pa
31:48onong gamiton integral of π 1 /
31:532π i mean 1 / 2π integral of i alpha d
31:58alpha from 0 to 2π so our i alpha duha
32:03no duha so im sin alpha and zer so for
32:09one complete cycle
32:13for
32:16cycle so first for i alpha for the first
32:21half meaning from 0 to π so
32:26limit kung kong gamit equation sa so
32:31makita saong
32:32function ng 0 to π lang siyaang instead
32:36of 0 to
32:372π we just use 0 to π and kini nga sin
32:41alpha im sin
32:43alpha siya sa first half okay so dili ka
32:47magamit
32:48to
32:492π so im sin alpha d alpha 0 to π plus
32:54the next isong from π to 2π para usa
32:59siya
33:00ka-complete cycle sa wave so 0 d alpha
33:05from π to 2 to 2π so of course di it
33:08will be it will be zero zero. So
33:11integrating using your
33:13calculator na siya i-integrate using
33:27calcul a sub 0 is
33:34im a sub 1. So for a of one term same
33:39lang giapon ng piecewise
33:42function or outong function on gamiton
33:44kani for one complete
33:48cycle o multiply now by sin alpha for
33:54a1 so again this part will still be
33:59zero so p na
34:01lang so integrating this function or i
34:06meaning this
34:07function. So the result is I am over
34:132. Okay. Im 2. This is now our a sub 1
34:19value. So let's proceed
34:22to B sub one. So the problem is just
34:27asking for four terms only ha. O meaning
34:31up to four terms niya na ay value. So
34:34kung zero siya continue pa. until such
34:37makaarrive upat ka term sa atong 4 year
34:44series
34:46so b sub
34:49one so i alpha cosine
34:53alpha so k zero lang di
34:59zeroapon substitute i alpha from 0 to π
35:03im sin alpha cosine alpha d alpha
35:06So integrating this just follow the
35:09procedure here. So B1 is
35:140 so 0 B1.
35:17Next we solve for the A2
35:23coefficient. So just multiply by sin 2
35:28alpha. No, as you can see sin 2
35:34alpha from 0 to 2π.
35:42So I mean that this is not 0 to 2π this
35:46is 0
35:49to 0
35:52to
35:552π this
35:57function is applicable only
36:00on from 0 to
36:04π 0
36:062πunda
36:10k
36:12so integrating and
36:15substituting
36:17limits will arrive with a sub 2 again
36:23zero next b sub
36:252 so integrating b sub2 or i mean
36:29solving for b sub 2 so cos 2 alpha no
36:35cos 2
36:38alpha okay integrate so final
36:42answer will arrive with b sub 2 - 2 /
36:503π okay so pwede lang
36:54calchan calcul lang para
37:01madali a sub
37:043. So a sub 3 is
37:07multiply sin 3
37:11alpha. So after integrating the arrive
37:15with the value of a sub 3 which is
37:21zero b sub
37:243 sin cos 3 alpha m cos 3 alpha pigamit.
37:30So after integrating arrive naaponta
37:34zero.
37:37So next a sub 4.
37:41So a sub 4 sin 4
37:44alpha o zero
37:47ian we still continue to find another
37:54term and b sub 4 cos 4
37:59alpha so cos 4 alpha our b sub 4 is
38:05-2 im /
38:0915π okay so i think we now have four
38:12terms
38:14available so we can now
38:17stop here so our a sub0 will be the
38:22coefficients a sub0 a sub 1 b sub 2 and
38:26b sub 4 so we now have four coefficient
38:29so pwede na siya takut kay four terms
38:33naangita o so a sub0 im / 5 + a sub 1 is
38:40im / 2 sin alpha then
38:43our b
38:45sub uh this is b sub
38:482 so b sub 2 is - 2 / 3π cos 2 alpha
38:57base sub 4 is 2 15 cos 4
39:02al
39:03so this is now your 4 yearesentation of
39:08the half wave
39:10the given one cycle of half
39:14wave. So I alpha is I / π + im / 2 sin
39:19alpha - 2/ 3π cos 2 alpha and 2 / 15π
39:26cos of 4
39:30alpha. Okay. Ah another
39:34example so write the first four terms of
39:37the four year series.
39:39which will represent the wave form
39:41shown. So we have here a
39:44rectangular wave form no so extend from
39:490 to π we have a maximum value of 100
39:53and π to
39:542π0 and again repeat
39:57naag continue na siya from 2π to 4π 100
40:03thener
40:04na another so and so forth so the value
40:08of our e voltage
40:12100 from 0 to
40:16π and from π to
40:192π our iser this is similar to the prev
40:24example but ang nak
40:28is rectangular
40:32wave so let's solve first
40:41Okay. So our piece wise
40:46function e alpha is equivalent to 100
40:49for alpha 0 to π and 0 for alpha
40:56to
40:572π. Use the formula 1 / 2π integral of e
41:03alpha d alpha 0 to 2π.
41:07So our e alpha duha no 100 from 0 to
41:14π 100 d alpha + 0 d alpha from π to
41:222π sige so integrating the function so
41:26our a sub z0 will be
41:3250 next a sub one
41:36So again mar gagamit
41:38nga
41:40equation kan siya for one complete
41:43cycle a sub 1 man so sin alpha 100 sin
41:49alpha d alpha zero naman usa pwede na
41:53siya
41:53ibutang this is the limit is from 0 to
41:57lang
41:581 so integrating this function so arrive
42:04Sir
42:05nga a sub 1 is 200 /
42:10π.
42:12Okay. Na a sub 1 b sub 1
42:18na. So for b sub 1 cos
42:22alpha. So integrating from 0 to 0 to
42:27π b sub one will be equivalent to 0.
42:36Next is a sub
42:372. So for a sub 2, so sin 2 alpha na
42:42gamiton.
42:44ng 100 sin 100 2 alpha na d alpha 0 to π
42:51so integrating the function so we arrive
42:54with a sub ng zero so continue
42:58kayakaabot four
43:01terms b sub 2 so computing for b sub 2
43:06so cos 2 alpha mm cos of 2 alpha
43:12is Okay cos to alpha d alpha from 0 to π
43:16so integrating the function so the
43:19result answer for b2 is
43:250 next a sub
43:273 so a sub 3 sin 3 alpha pod 100 sin 3
43:34alpha from 0 to
43:37π so integrating the function so come
43:40upag answer sa a sub
43:42200 /
43:473π sub 3 cos 3 alpha so
43:53integrating zero na yan a sub 3 na
43:58to a sub
44:004 so sin 4 alpha so sin 4 alpha
44:05iintegrate na siya from 0 to
44:08π 100 sin 4 alpha
44:11d alpha from 0 to π then multip by 1 π
44:15so the resulting answer is
44:20zero dapat base of 4 so base of 4 then
44:24cosine 4 alpha po
44:26ka so you will have b of 4 equivalent to
44:32[Musika]
44:340 m so a sub 5 so for a sub of 5 sin 5
44:40alpha no 100 sin 5 alpha from 0 to π
44:47divided all divided by 5π divided by π
44:52so
44:53integrate just use your
44:55calculator so the result will be 40 over
45:00π so i think nakaabotag terms na present
45:04we can now write the year representation
45:07or the year series
45:10nakuha is a
45:11sub0 a sub 1 200 π a sub 3 200 3π a sub
45:184 a sub 5 i mean is 40 / π
45:23soay up ka term present so our e alpha
45:28or the 4 representation of the
45:31rectangular wave form half rectangular
45:33wave form will be 50 +
45:37200π 4 a sub 1 so sin alpha lang ta then
45:42coefficients sa a sub 3 is 200 / 3π a
45:45sub 3 then sin 3 alpha + 40 / π 4 sub 5
45:52so sin 5 alpha so ang 4 year
45:59seriesentation even
46:00given wave
46:03form this is the
46:07term and the
46:09fundamental term fundamental
46:15harmonic
46:17okay so another example
46:20okay now another example write the for
46:23series which will represent the
46:25triangular wave form shown so the
46:28triangular wave form shownya from 0 to
46:31point a then kana from diha to diha
46:372π. So this is the first half of the
46:42triangular wave form. Then the second
46:44half from point B, point C and D. So the
46:48first half is from 0 to π and the second
46:51half is from π to 2π.
46:56Okay. So point A is at π / 2, point B is
47:02at π and point C is at 3π/ 2 and point D
47:07is at 2π.
47:10Okay. So we can observe on this
47:13sketching sketch that the wave form is
47:16defined over from 0 to 2π and this
47:22consist of a three linear segments no 1
47:282 and
47:293. Okay. So pwede po tana from diha to
47:33diha dire to dire another segment line
47:36po
47:38dire then okay
47:41di isaito direito
47:46[Musika]
47:49direito plus di to 2π okay so from 0 to
47:54π / 2 ng
47:56interval increasing from 0 to 1 no one
48:02value then π / 2 to
48:07π so decreasing from 1 to 0 and π / π to
48:133π / 2 decreasing from 0 to
48:16-1 and 3π / 2 + 2π na-increase na po
48:20siya 2 from -1 to 0. So these are the
48:27corresponding coordinates nga at makuha
48:29ana. Okay we can now use this example is
48:33just similar to the example of first
48:36example the soft wave form. So atong
48:40nagsa equation no for the whole
48:45cycle. Okay. So we we're now going to
48:49define the piece wise function.
48:53So we can use the slope formula and the
48:57point slope form.
49:01Okay. From 0 to π/ 2 m iang points ba sa
49:08line is 0 then π/ 2 and 1. So we can
49:11have a slope
49:13of 2 / π and the y intercept equivalent
49:18to 0.
49:20So 2/ 2/ 5 x FX
49:27kin gikan sa ibalik from point A to
49:33point
49:34B so from point A to point B so
49:39naay points π/ 2 1 and
49:45π0 ang coordinates point A will
49:49tungod sa π / 2 then pos 1 ang
49:53coordinate sa point b naat ay 5 and zer
49:59okay
50:00so the slope of the line from point a to
50:05b is - 2 /
50:09π and we can have our f x equation as -
50:172 / πx +
50:222 kuan mo na gusto mo makahiba naani ah
50:27just apply what we did on example number
50:31one okay i-check ninyo kung sakto ba
50:36next from π to 3π/ 2 so ma
50:43coordinates of course expected na siya
50:46ng pareha na sil
50:47equation kay supp na no
50:51no and the last is from point c to d mga
50:56point c point c ba from point c to
51:04d from point
51:06c to point d coordinate than here is 3π/
51:112 and
51:12-1 is 2π0
51:15zero ato makuha na equation sa line
51:20diong so our equation sa line is 2π /
51:26x 2π x - 4 so finally we now have the
51:33wise function so kaning 2 3 pwede na
51:37siya usahon pero i-change ang limit
51:39instead of π/ 2π at π/ So 2 3π / 2. So
51:49the first is 2 / πx. So this is this is
51:53good from 0 to π /
51:592 and k - 2 / π + 2. So from π / 2 to 3π
52:08/ 2 para mausa na natin siya. And the
52:12last from 3π / 2 to 2π. So 2 / πx - 4.
52:19This is now the fx of our triangular
52:23wave form given. Okay. So magamit na
52:28formula. So first is zero atong
52:32kaon. So we will use this formula 1 / 2π
52:36integral of fx from 0 to 2π for the
52:41complete cycle. So in this example
52:46siya integral pwede since we're
52:50integrating the complete cyclean.
52:54So the first is the integral of 2π/ x
52:58from 0 to π/ 2. So as you can see in
53:03here 0 to π/ 2. So the result is π/ 4.
53:10And the next is the integral of -2/ πx +
53:142. So our limit will be from lower limit
53:18π 2 to 3π/ 2. The result is 0.
53:24And last is the integral of
53:272πx 2πx 2 / π x - 4. So our limit is
53:33from 3π/ 2 to 2π. So combining all the
53:38terms or all the result. So mura na po
53:42kag nagkuha sa integral of fx from 0 to
53:462π. So money combine the
53:51total come up will arrive in answer for
53:54a sub equivalent
53:56to
53:58zero zero lang
54:01siya. Next
54:04is a sub one. 1
54:07m I mean B A sub 1 A sub 1. So fx sin X.
54:15So same procedure buhaton
54:19butanganya sin
54:23x so integral of 2π x sin x 2π 2/ π
54:31answer next segment 2
54:34from π / 2 to 3π / 2 soay 4 / π segment
54:403 is from 3π / 2 to 2π so atay 2π
54:462
54:51π
54:53combine so after
54:55combining come up result 8
55:00π
55:031 so our a will be
55:078
55:11π B sub
55:131 so b sub 1 to
55:17cosine again from 0 to
55:212πagsaon na to i-sum total i-sum up na
55:25siya after
55:29integrating
55:30so the result will be zero zero
55:35result okay so given that the
55:39triangular given that the triangular
55:41wave form is unfunction or meaning the
55:43symmetric about the origin no all cosine
55:47coefficients are zero so pwede na to
55:52apply kaong kuan about sa and even
55:56function so just review ko sa when can
55:58we say that the function is add and when
56:01can we say that the function is even so
56:05that our resolution will not be kuan
56:07mataas sa kaay so in this case the
56:10triangular wave
56:12is a add function add na siya meaning
56:17symmetric about sa origin so review lang
56:21mo about an ha or pwedeapon mag-solve mo
56:24sa value sa b1 b2
56:28b3 b sub pwede rang pwede rapon para
56:32ma-check ninyo no kung kwan
56:36ah sakto ba no mag-check ma-dble check
56:39ninyo e pwede rin calculuh ninyo pwede
56:43nam mag-calcul magkuha sa iang mga term
56:46para medyo madali mo kay medyo taas
56:50solusyon so in this case ang triangular
56:54atong triangular wave is
56:56a is an add function so meaning all
56:59cosine terms will be zer ang present
57:03lang niya is kadong mga sign terms and
57:06even the dc term zer
57:10So no need to solve for B
57:13paraayo mataas ong
57:15solusyon ang mga A na natutahon.
57:20Okay. So for a sub 3 or I mean a sub 2.
57:25So mapuno naay
57:26atong
57:29e ah ganito na formula. So sin 2x ka na
57:34fx sin
57:372x saon
57:39mo then makuha mo mga value pwede i-add
57:43na or i-combine then ito multiplyan
57:471 /
57:49π so ang result is zer zero atong a sub
57:542 okay so pa ka makaingon ng uban ng
57:57term zero
57:59kay kwan man siya add function
58:02eh so
58:04[Musika]
58:06nad
58:10[Musika]
58:20presentate up to segment
58:253
58:28-8 / 9π π
58:32s 4 coefficient of a sub
58:373 next coefficient of a sub
58:444 so dapat note o since the wave form is
58:48add so moon characteristic na to again
58:50for an add function no wave form is add
58:54and composed of only add harmonics so
58:58kung atong wave form is add ang mga had
59:00harmonics po present meaning all even
59:03harmonics like a sub2 a sub 4 a sub 6
59:07etc are all zero so pwede na no need na
59:11to solve ani para mas madali no
59:14technique lang siya para mas madali pero
59:16kung
59:18na ganahanunaan ng mga
59:21techniqueshapon mo magamit sa formula na
59:24to na shortcut na lang niyo
59:29Then just review ha review about when
59:33can we say even the function is even and
59:37the function is
59:39add. Okay so let's now proceed to a sub
59:435. A sub 4. So a sub 5. So sin
59:505x. So same pang procedure same ang
59:53pamaagi buhaton.
59:56So the total integral for the whole
59:59cycle is 8/ 25 π multiply pala to 1/ π.
1:00:07So the result is a sub a sub 5 is 8 / 25
1:00:14π s. So sevh
1:00:21term the 7th term sin 7x
1:00:26ka o same procedure kahapon just
1:00:34integrate so the total integral for the
1:00:36seventh term will be
1:00:38-8 or the a sub 7 will be -8 /
1:00:4449 s.
1:00:47So excuse me the non coefficients man
1:00:50siya a sub 1 a sub 3 a sub 5 a7
1:00:55soag takut as long as na kuha na ong mga
1:00:59coefficient no medyo taas na so 8 / π²
1:01:04min ang a sub 3 is - 8 /
1:01:079π² and h of 5 is 8 / π² kahapon 25 a 7
1:01:14is 8 / -49 s as you can see makita an
1:01:21commonya so for a sub 1 nat 8 / π sare
1:01:26napay 8 / π s sa a sub 3 and up to a sub
1:01:317 so makita na to we can see that terms
1:01:35have common na equivalent to 8 / π s so
1:01:42pwede na to na siya i-factor out Then
1:01:45ang denominator sa kita that
1:01:48ipang-square siya no starting from 1 s 3
1:01:52s no 9 is 3 s 25 is 5 s 49 is 7 s
1:01:59unexpected the a sub9 term kaha what do
1:02:04you think so a sub9 term will
1:02:08be pos 8
1:02:12/ sunod sa 49 s 9 so 81 81
1:02:20π so mana siya man onong sunod na term
1:02:24is 9 so on and so forth and alternate
1:02:27lang iyang sign so starting from a sub
1:02:30one positive pagabot sa a sub 3 na minus
1:02:33so alterate po dire na plus na minus
1:02:36sunod plus na po na siya no sa ang sunod
1:02:39an na is 8 / 81 π² squ sin of 9x na san
1:02:46okay yan ano na so this is the 4 year
1:02:49series representation of the given
1:02:53triangular wave form okay
1:03:00so i think that would be all thank you
1:03:04for your time and attention so today we
1:03:08explore the fascinating world of a non
1:03:11sinosoidal wave form. So understanding
1:03:15their nature,
1:03:17importance and how we can
1:03:20apply we can analyze them using 4 year
1:03:24series.
1:03:25No. So as you advance in your
1:03:28engineering journey, these concepts will
1:03:30help you tackle real world challenges in
1:03:34power system, communication and digital
1:03:38electronic. So I hope this session
1:03:41dependation for signal analysis and
1:03:44equip you with tools to interpret the
1:03:48wave form that define modern technology.
1:03:51So on the next session so I will be
1:03:55discussing to you about the effective
1:03:57value of a
1:04:00nonsenosoidal wave and also the RMS the
1:04:06general expression of complex
1:04:08wave and power due to
1:04:12aensionidal voltage and
1:04:15current uh so then circuit analysis
1:04:20using I
1:04:23So just standby for the next video.
1:04:26Okay. So again, thank you everyone.

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