Trigonometric Fourier Series

[@hafizuddinsyed3887]

"Symmterical abpout t-axis" it means that the positive and negative area is equal in magnitude.

[@mayankbarapatre7817]

Yes. Average value will become 0.

[@nickbeats9883]

oh

[@skmgamer8214]

can anybody plz tell me how will we know a signal is even or odd.

[@ayshaakter5676]

@@skmgamer8214 when the signal is symmetrical about y axis, it is even signal. But when its not then it is an odd signal.

[@skmgamer8214]

@@ayshaakter5676 ok thanks a lot🙏

[@animeshverma584]

This is the best you tube channel for system and signal, with great detail

[@MatheusSilva-dragon]

Homework answer: x(t) is not symmetrical, so it's dc/average value cannot be nule; x(t) is purely even which implies bn = 0 (there is no sines as components).

[@anikaitdash6519]

Yes. It is symmetrical about the time axis. So, an = 0.

[@rohithrs21]

no it is symmetrical about y axis therefore bn=0 and an and ao can't be zero@@anikaitdash6519

[@vidhiruparelia2491]

how did you figure its purely even?

[@sanjusolra6359]

you can check for it x(-t) = x(t) or it is a mirror image w r to y axis @@vidhiruparelia2491

[@vproductions7389]

@@vidhiruparelia2491 time reversal

[@adithyarajapakshe3791]

Sir, You have no idea how much this videos worth to thousands of students arround the world. Thank you!!!!

[@kavinmoorthi7108]

Finally youtube has recommended a good lecture series.

[@dembashow]

*x(t) is not symmetrical about t-axis ==>a0#0, *x(t) is symmetrical about y-axis==>x(t) is Even signal ==> bn = 0, *an#0

[@sheesh9210]

Y axis or x axis?

[@mayankmathurofficial]

​@@sheesh9210 x-axis

[@HipHop-xv2zn]

Bro how did say that it is symmetrical or not, please reply me

[@diags_1109]

@@HipHop-xv2zn If you go to -.5 on x it has the same corresponding y value as if you were to go to .5 on the x so therefore it is even. If odd then the values would be the opposite i.e (-.5,-1) ; (.5,1)

[@der__forschera3589]

Thank you very much for this video, your explanations are better than my profs's

[@MatheusSilva-dragon]

Same here

[@Fantasticsman]

For homework question, *a0* will be some DC or average value since it is not symmetrical to the t-axis. *an* will be some value since it is even signal. *bn* will be zero since it has no sine term.

[@prakhartiwari9893]

Sir u r doing great job u are the best teacher we students are blessed to have a teacher like u ,thanks very much

[@sarahhughman2651]

this really helps me with my online learning during covid19 quarantine. thank you

[@prashantsinghtomar5727]

What you pursuing???

[@tanushadithiya7136]

Offline too ✌️

[@rajkumari13484]

Tq tq tq so...much sir . Ur teaching is better than my faculty ... I can easily understood Ur class And once again tq sir...this is really fantastic KZbin channel for ECE studies...🤗🤗👍👍

[@pk_officl]

Bn =0, otherwise ≠0. Bcoz it's symmetry to y axis and not to x axis. The waveform doesn't going down. It's only on the positive region.

[@sheesh9210]

Then it's symmetry to x axis not y axis, y axis is for odd

[@Rajveersingh-mb8fi]

@@sheesh9210 if symmetry in y- axis the signal is even

[@ovidioescuderofranco233]

Sir, you truly do not know how much you are helping students, definitely great series.

[@snehdeep712]

Hello! At around 12:08, you say that the saw tooth function (Fn. (ii)) is symmetric about the 't' axis. Symmetry generally means reflective symmetry. However, the function doesn't have a reflective symmetry. But since the amplitude is the same on both sides of the axis, they yield an average value of 0. This is probably why a0=0. Please comment.

[@collegetv7986]

same doubt here

[@aravindsubhash3740]

Sneh Deep eey even signals are Signals which when folded around yaxis( time reversal) remains Same as that of original signal.. This does not hold true for fun (ii). This it is an even signal.

[@emanianjana1781]

i didn't get you can you please expand

[@aravindsubhash3740]

EMANI ANJANA Wat

[@socratesocrate3172]

The integration over the period will give you zero , if u calculate it u will get a°=zero

[@samuelyawtenkorang352]

Not symmetric about t-axis and so a0 ≠0, signal is symmetric about x(t) axis and this makes it Even, an≠0, bn=0

[@쥬마펠키키]

Isnt "symmetrical about the t axis" a little vague here...? I guess it just means if the area above and below the t axis is the same, a0 is 0?

[@arijitdey8419]

Yes I guess you are right... because clearly about time axis,not a single graph has a mirror image

[@nipunsenarath1561]

next video -144(example 1)

[@makingmycareer2608]

T-axis is nothing but the x-axis in these cases, I was also confused, but that's how you see it I guess.

[@asrithperuri629]

Even I have same doubt

[@gopikrishnamoodu6152]

Sir i don't understand how the signals are symmetric about time axis. Can you clarify my doubt.

[@sanketsingh8288]

here, the time axis means x axis.

[@eddyr978]

@@sanketsingh8288 lakin zese in even signal..about y axis it is reflection But x axis or time axis me to nhi hora asa plz explain

[@gauravnagamaheswarpattem7069]

Bn = 0 ; An and A0 are not equal to zero; because that is even signal (An != 0 ) and since there is no signal below the X-axis(time axis) so it is not symmetric about time axis. so it is clear...

[@sethusri0123]

A0=zero bro

[@techtalkasad]

(iii): In the given signal a0=0(because signal reflection is same or its magnitude is same), since the signal is an Even signal (because it is symmetric about y-axis) therefore "an" not equal to zero. Also bn=0

[@RitayanMitraggmu]

Thank you sir for resuming the lectures..hope this subject will be finished before GATE 2018 :)

[@apoorvojash1292]

Ritayan Mitra kya hua finish hua??

[@anshul4216]

how was your exam? XD

[@RitayanMitraggmu]

@@anshul4216 LOL!

[@GovindRajM16]

How much you scored in gate sir?

[@RitayanMitraggmu]

93.33

[@okoriejessica697]

How do you know when a signal is symmetrical to any axis?

[@raveenabharathi740]

Sir, At 11.10 is signal is symmetrical about X axis?? I can see only halfway symmetry..

[@MathEnth-rz3tt]

How is iii) symmetrical about the t-axis?

[@prabhakardas4261]

it is an even signal ,average,a^o=0,a^n not equal to 0 and b^n equal to 0....correct me if i am wrong

[@GurpreetSingh-zz3mj]

a^0 will exist for it.

[@mostafadx7236]

how can i know the signal from waveform whether it's even or odd??

[@tehreemsheikh1386]

please explain how the signal is symmetric about time axis,i am confused.

[@hemanth7804]

Take single time period of sin signal. Top and bottom parts areas are equal. So it's symmetrical signal.

[@eddyr978]

@@hemanth7804 will u plz explain me .. As we are saying that it is symmetric..than it must has reflection but it is not Because in even we are seeing reflection about y axis and in odd reflection about origin... Plz explain me

[@surajhimanshu1962]

How at 11:08 signal is symmetric

[@theraja1732]

It is a half wave symmetry then =a°not equal to 0 It is a even signal then=an not equal to 0 And for even signals bn=0 not required..

[@roushansingh9936]

13:19, an !=0, a0 != 0, and bn = 0

[@xXMaDGaMeR]

you are amazing, thank you so much for this series

[@dhritirajborah219]

why is a0 not equal to 0

[@virajpansuriya8605]

here , i am at right place for signal and systems course great playlist 😁

[@yashvisolanki4340]

Here waveform is symmetric along y-axis but not along x-axis that means bn=0 and we need to calculate ao&an

[@ahmernadeem3521]

How to write the harmonics in trigonometric and exponential foirier series.

[@awesomeboyztamil3651]

neso acadamy is awesome acadamy

[@saraj866]

x(t) is not symm.above the time so a0=0, becouse this signal is even then bn=0 . for that x(t)=a0+sum of(an cos nwt)

[@ankushrana2001]

ao not equal to zero ( unequal across y or t axis), an not equal to zero ( x(-t) = x(t) , thus even signal, therefore cosine part is not zero), bn = 0 (not odd part of signal is there)

[@akshatrailaddha5900]

hw ans: a0 , an not zero as not symmetrical about time axis and origin respectively and bn is 0 as function is even

[@inteluhdgaming1524]

Ao =/ 0 Bn = 0 (since it symmetrical along the y axis) An=/0

[@pagalavan7472]

Why 1/(T/2) which is equal to (2/T) is taken for an & bn but Ao=(1/T) Please clarify

[@subhraneelacharjee2616]

Keep doing the great work that you're doing

[@altuber99_athlete]

Do you (or someone else) know what is the trigonometric form of the *discrete-time* Fourier series? The one shown in this video is for the continuous-time. And I've only found the complex exponential form of the DTFS.

[@SIDDHARTHSHAH-m3i]

a0=0 an=o as there is no symmetry with y axis bn#0

[@Livedukk]

can anyone please explain how the signals are symmetrical around the X-axis, i understand it for Y axis but not for X.

[@Deepak-ip1se]

just take the mirror image about x axis if it comes same as the original signal then they are symmetrical about x axis

[@jijojohn3303]

sir, I had a doubt. Why do we multiply 2 to integrating factors for an and bn

[@rafiausman9955]

The signal is not symmetrical with y axis...and hence it is.a odd signal.so an should be eqaul.to zero.???

[@dalinsixtus7648]

Sir Iam a great fan of yours...sir please do kindly make lectures on DSP...it's only because of you to we understand Signals and system better... please sir

[@solidworksforbeginnersin2053]

As the third signal is symmetrical so a0=0 an is not equal to zero while bn is zero.

[@adoanmian6701]

graph 3: a0 not= 0, bn = 0, an not= 0 --> reason : this graph is an even function without symmetry to t-axis

[@benimadimeva]

a_0 ≠ 0 ---> there is a dc component in the signal because signal is not a symmetric about the x axis a_n ≠ 0 ---> the signal is include a cos component because the signal is a pure even signal. The signal is symmetric about y axis b_n=0 ---> the signal is not include a sin component because the signal is a pure even signal.There is no any odd component.

[@sudheerjalli4707]

Reflection along y axis Even function so bn=0

[@hailuu625]

I dont understand how we can check whether x(-t)=x(t) or -x(t). Can someone pls explain for me

[@adiljan1305]

This is the condition for the odd signals. You can verify it by first taking the time-reversal of the original signal and matches it with the amplitude reversal of the original signal. If R.H.S=L.H.S, then it is an odd signal otherwise not.

[@sureshpeddini6443]

@@adiljan1305 consider sint....if u take -t in place of t...then sin(-t) = -sint so it is odd signal....if u consider cost....if u take -t then cos(-t) = cost so it is even signal

[@PardeepLabVIEW]

Thanks sir ji

[@vsmemegenix3512]

Signal is even so We have to calculate a० and an in last question. Where bn = o

[@janh83]

very well explained

[@amarjitroy2308]

a0 is not zero as the signal is not symmetry over the time axis, bn is zero as the signal is even signal, so an with cosine term exist.

[@saisagardasari227]

Fourier series expansion is dc value + cosine terms +sine terms...instead of calculating dc value ...shall we calculate even component of signal????

[@utkarshshirbhate2938]

I think there is a formula mistake at 5:39 , instead of just a(not), it should be a(not)/2. correct me if I am wrong

[@ibnefayaz5107]

it is correct

[@shivijain2899]

Why an=!0?

[@thelsazzad]

Signal is identical, and even signal, so a0≠an≠0 and bn=0

[@itsrao5245]

sir a0 is not equal to zero and bn will be zero as signal is even signal and an will not be equal to zero. thanku so much for this signal and system playlist

[@afrinakhanam4696]

How to understand whether a signal is symmetrical to y-axis, I do not understand it? Can anybody help?

[@annyni6640]

Thank you! I'm passing my class because these videos help!

[@MrPotaatooo]

sir my teacher notes said a0 = 1/2T and an =1/T and bn =1/T but in your video you wrote a0 = 1/T and an = 2/T and bn = 1/T is it the same?

[@insafyouthwing8477]

a0 = 1/T

[@MrBooster.]

Sir, can you please upload the rest of the videos for signals and systems.

[@workerpowernow]

good explanation. thank you

[@kothojumadhuri5410]

Ur explanation fantastic 🔥 sir...

[@i_am_-bj6lx]

an != 0, bn = 0, an !=0 thank you very much for these lectures

[@wilsonkomla6521]

Looking at the waveform, it is not symmetric about the t-axis, thus making a_0 !=0; But from the look of the time reversal property of the signal, the output is reflective of the input and thus makes it an odd signal. That is, x(-t)= -x(t) and this is an odd signal. And we know that for odd signals b_n=0 and there a_n !=0. So in summary, a_0 !=0, a_n !=0 and b_n=0 respectively. Thank you

[@myonlynick]

My book says x(t) = ao/2 +..... bla bla. I do not see how ao comes to be valid. I can derive ao/2 from the an formula... But I do not see how to derive just ao.

[@gowthamsaikrishnatanikonda2594]

Hi sir,this signal can be represented in both sin and cos right?

[@manjulatasahoo3835]

How it is symmetrical about x axis?????

[@yaredmin5999]

a0=0 symmetry to y axis, and bn =0 even

[@muhammadmarufnurrifai1792]

Can someone tell me about the example signal (1) , why the bn = 0? how do we know the signal is symethrical with y axis?

[@rammayabaniya4445]

Just fold the graph keeping the corner at the y-axis..

[@ec039venom4]

how should i find symmetrical about y axis ?? anyone can explain plzz......

[@rakeshraj688]

thank you sir for uploaded this video...

[@sportsuper9489]

which is even signal bn =0 sir thanks

[@saketsamrat5038]

ao ≠ 0 an ≠ 0 bn = 0

[@degavenkatavedavyaswfkwbgd3162]

thank u so much bro

[@niashree]

can you explain what is symmetry about time axis?

[@ElboxD]

It means that if you analyse only one cycle, let's say, from - T/2 to T/2 (where T is the period of the signal) the area below the x axis (time axis, also known as horizontal axis) will be the same as the area above it so the areas will cancelled out so the total area is equal to zero. That's why Ao = 0. Hope I explained it well.

[@kautukraj]

@@ElboxD Yes, you did. Thank you!

[@ElboxD]

@@kautukraj Glad to know!

[@ashwanianand7424]

Why An and Bn divided by To/2 rather than To

[@vedanaikar5931]

How an is calculated? And how it is unequal to 0

[@PareeksAcademy10october2001]

#include int main() { printf ("I think your videos should be in Hinglish(Hindi+English)"); return 0; } //TRUE ENGINEER CAN UNDERSTAND IT

[@mr.sneakyff9133]

Thank u so much sir

[@poornasekhar20]

Thank you sir

[@emc2knowledge999]

Sir please continue Analog circuits

[@musadiqalammaniyar9546]

Why an and bn is taken as T/2 ?

[@atherawan1898]

Sir which book you follow ?

[@MDSHOAIBURRAHMAN-qz4bg]

a0 not equal to 0; an not equal to 0; bn equal to zeno;

[@roopachinnu4707]

Even signal

[@gangsterkhan125]

Good 🌟 sir

[@manishankarmukherjee9456]

13:16 bn=0

[@aakriti7373]

How can we say that a signal is symmetrical along t-axis?

[@abhishekagrawal8376]

by observing if a signal have exact mirror image of itself along x axis

[@pavanajsridhar939]

why in some texts its written a0/2

[@BEEMonikaBishnoi]

Nyc sir

[@lilianpatience8969]

Ao is not equal to zero because the signal is not symmetrical to the time axis An is not equal to zero as the signal is the reversal of the signal does not yield the same value BN is equal to zero

[@dotilaxmaiah4779]

a(n)=0,b(n)=o

[@santoshsharma8445]

Nice one sir ji

[@captainkite]

For the assignment: a0 != 0 an != 0 bn = 0

[@warsnewz4094]

Sir actually the last one is not symmetrical it is a dc so a=/= not equal to zero and also odd an not equal to zero and also an even bn =/=/ not equals to zero