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.
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).
@@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)
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.
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.
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 TH-cam channel for ECE studies...🤗🤗👍👍
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...
(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
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.
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)
@@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
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
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
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.
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.
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
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
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.
@@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
This is the best you tube channel for system and signal, with great detail
Sir, You have no idea how much this videos worth to thousands of students arround the world. Thank you!!!!
No, sir have the idea by the likes and views.
Finally youtube has recommended a good lecture series.
"Symmterical abpout t-axis" it means that the positive and negative area is equal in magnitude.
Yes. Average value will become 0.
oh
can anybody plz tell me how will we know a signal is even or odd.
@@skmgamer8214 when the signal is symmetrical about y axis, it is even signal.
But when its not then it is an odd signal.
@@ayshaakter5676 ok thanks a lot🙏
Thank you very much for this video, your explanations are better than my profs's
Same here
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.
Sir u r doing great job u are the best teacher we students are blessed to have a teacher like u ,thanks very much
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).
Yes. It is symmetrical about the time axis. So, an = 0.
no it is symmetrical about y axis therefore bn=0 and an and ao can't be zero@@anikaitdash6519
how did you figure its purely even?
you can check for it x(-t) = x(t) or it is a mirror image w r to y axis
@@vidhiruparelia2491
@@vidhiruparelia2491 time reversal
*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
Y axis or x axis?
@@sheesh9210 x-axis
Bro how did say that it is symmetrical or not, please reply me
@@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)
this really helps me with my online learning during covid19 quarantine. thank you
What you pursuing???
Offline too ✌️
13:19, an !=0, a0 != 0, and bn = 0
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.
same doubt here
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.
i didn't get you
can you please expand
EMANI ANJANA Wat
The integration over the period will give you zero , if u calculate it u will get a°=zero
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.
Then it's symmetry to x axis not y axis, y axis is for odd
@@sheesh9210 if symmetry in y- axis the signal is even
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 TH-cam channel for ECE studies...🤗🤗👍👍
Sir, you truly do not know how much you are helping students, definitely great series.
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...
A0=zero bro
you are amazing, thank you so much for this series
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
(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
Thank you sir for resuming the lectures..hope this subject will be finished before GATE 2018 :)
Ritayan Mitra kya hua finish hua??
how was your exam? XD
@@anshul4216 LOL!
How much you scored in gate sir?
93.33
here , i am at right place for signal and systems course
great playlist 😁
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?
Yes I guess you are right... because clearly about time axis,not a single graph has a mirror image
next video -144(example 1)
T-axis is nothing but the x-axis in these cases, I was also confused, but that's how you see it I guess.
Even I have same doubt
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Sir i don't understand how the signals are symmetric about time axis. Can you clarify my doubt.
here, the time axis means x axis.
@@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
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.
Thank you! I'm passing my class because these videos help!
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..
Reflection along y axis Even function so bn=0
neso acadamy is awesome acadamy
hw ans: a0 , an not zero as not symmetrical about time axis and origin respectively and bn is 0 as function is even
How at 11:08 signal is symmetric
How do you know when a signal is symmetrical to any axis?
an != 0, bn = 0, an !=0
thank you very much for these lectures
How is iii) symmetrical about the t-axis?
Ao =/ 0
Bn = 0 (since it symmetrical along the y axis)
An=/0
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)
please explain how the signal is symmetric about time axis,i am confused.
Take single time period of sin signal. Top and bottom parts areas are equal. So it's symmetrical signal.
@@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
Signal is even so We have to calculate a० and an in last question. Where bn = o
Keep doing the great work that you're doing
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
a^0 will exist for it.
Signal is identical, and even signal, so a0≠an≠0 and bn=0
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
13:16 bn=0
a0=0 an=o as there is no symmetry with y axis bn#0
Here waveform is symmetric along y-axis but not along x-axis that means bn=0 and we need to calculate ao&an
graph 3: a0 not= 0, bn = 0, an not= 0 --> reason : this graph is an even function without symmetry to t-axis
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.
An=0,ao#0,bn#0
Are ye network theory me kam ayegi????
a0=0 symmetry to y axis, and bn =0 even
very well explained
Thanks sir ji
which book are using
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)
good explanation. thank you
How to write the harmonics in trigonometric and exponential foirier series.
Good 🌟 sir
Ur explanation fantastic 🔥 sir...
how can i know the signal from waveform whether it's even or odd??
As the third signal is symmetrical so a0=0
an is not equal to zero while bn is zero.
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
it is correct
For the assignment:
a0 != 0
an != 0
bn = 0
Sir, can you please upload the rest of the videos for signals and systems.
thank u so much bro
Sir, At 11.10 is signal is symmetrical about X axis?? I can see only halfway symmetry..
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
can you explain what is symmetry about time axis?
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.
@@ElboxD Yes, you did. Thank you!
@@kautukraj Glad to know!
a0 not equal to 0;
an not equal to 0;
bn equal to zeno;
why is a0 not equal to 0
Sir which book you follow ?
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.
thank you sir for uploaded this video...
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
Nyc sir
Bro can u send the pdf u are explaining on
can anyone please explain how the signals are symmetrical around the X-axis, i understand it for Y axis but not for X.
just take the mirror image about x axis if it comes same as the original signal then they are symmetrical about x axis
Sir please continue Analog circuits
may Allah bless u brother for making these videos ❤❤
Thank you sir
add the remaining videos on fourier series sir plzzzzzzzz
which is even signal bn =0 sir thanks
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
bn,an=0
Nd Ao=!0
Correct me if I'm wrong 🥺
Why an=!0?
I dont understand how we can check whether x(-t)=x(t) or -x(t). Can someone pls explain for me
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.
@@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
Thank u so much sir
Nice one sir ji
I got interest in SS subject only bcz of ur videos
Why 1/(T/2) which is equal to (2/T) is taken for an & bn but Ao=(1/T)
Please clarify
Fourier series expansion is dc value + cosine terms +sine terms...instead of calculating dc value ...shall we calculate even component of signal????
y axis symmetry=like cosine
How it is symmetrical about x axis?????
a(n)=0,b(n)=o
Hi sir,this signal can be represented in both sin and cos right?
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?
Even signal
ao ≠ 0
an ≠ 0
bn = 0
a0 and an not equal to zero
bn equal to zero
sir your are so talented...
In video at time 12.00 u said that ao=0 if it symmetric about t axis but it example is not symmetric about t-axis
PARAMESWARA RAO modda gudu
@@raviarumilli2028 lol.
How to understand whether a signal is symmetrical to y-axis, I do not understand it? Can anybody help?