Symmetricities in Fourier Series (Part 1)

[@urjasolanki631]

Why x(t)= Cn1 and x(t-t0)= Cn1 e^-jnw0t0

[@sidaina1434]

As usual grt lecture....sir we are highly relying on ur lectures and as gate exam is close pls provide ur lectures asap....we will be extremely grateful to u

[@sidaina1434]

Thanks again sir

[@im.meer-asif]

Neso Academy yess sir...plz complete...only dependent on u in this subject....

[@SanatanHiSatyaha-]

@@sidaina1434 bhai ho gya gate clear

[@souravgorai9481]

Didn't understand the part where x(t) got time shifted to x(t-t°)...how come Cn1 changed to (Cn1)e^(-jnw°t°) sir please help

[@myonlynick]

e^(something) represents the shift in time in fourier... You need to solve the formula to see how it appears.

[@NishantKumar-tx7sk]

i am confused too

[@ziyancheng8122]

at 7:17, x(t) should not be considered a constant of Cn1, this is confusing, instead, try to say x(t) is a canonical complex number represented by polar form

[@ziyancheng8122]

to be specific, let us assume x(t) = c exp[inwt]; then x(t+t0/2) should be c exp[inw(t+t0/2) ]. We have x(t) = -x(t + T0/2), so c exp[inwt] = -c exp[inw(t+t0/2) ] ==> 1 = -exp[inw T0/2].

[@StCham]

Thank you 🙏

[@nasar8480]

Didn't understand the part where x(t) got time shifted to x(t-t°)...how come Cn1 changed to (Cn1)e^(-jnw°t°)

[@phanindrareddy4885]

Nasar Ahmad Zafar Substitute in complex Fourier expansion equation Reply me if clear

[@bhanu8391]

Phanindra Reddy, I didn't understand please let me know how?

[@phanindrareddy4885]

If u substitute t minus to In Fourier equation And then separating exponential parts U will get it

[@bhanu8391]

Phanindra Reddy , ohh thank you bro

[@NishantKumar-tx7sk]

i have the same confusion....

[@biswajitsamal5641]

Add lectures on control system also

[@akibjawed5076]

in case of odd symmetry how can you say that avg. value is zero, an average value is zero only when the signal is symmetric about the x-axis. in case if it is not symmetric about x-axis then avg value will not be zero even the signal has odd symmetry.

[@ankitabarman4380]

If a signal is odd then it will be symmetric about the x-axis because x(-t) = -x(t) which means the amplitude is reversed along the x-axis.

[@chandran-youtube]

Brilliant 💛

[@exile341]

Can you proof half wave symmetry with a Fourier series expansions that all cos terms 0 out?

[@aditipandey4724]

sir could you explain how to write function for any given waveform?? if you have already explained then let me know about that video please sir if you can then please explain me

[@NishantKumar-tx7sk]

if x(t) having the fourier coefficient C1 then how we can write the F.C. of x(t-t0) = c1* e^-jnw0t0 please i need clearification

[@gauravnagamaheswarpattem7069]

yes! this is my doubt also... please, somebody, clarify!

[@user-rf5um7wt2l]

with time shifting the magnitude won't change. The phase will be changed only.

[@user-rf5um7wt2l]

if we take any func lets say,x(t)= c1*exp(jnw0t) now after time shifting ,ie,by t0, we have x(t-t0)= c1*exp(jnw0(t-t0)) now we have x(t-t0)=c1*exp(jnw0t)*exp(-jnw0t0) =x(t)*exp(-jnw0t0)

[@animeshgautam6197]

@@user-rf5um7wt2l but x(t-t0) is cn1exp^(-jnw0t0) But you got x(t)exp^(-jnw0t0) How x(t) is equal to cn1

[@kunalaggarwal1928]

@@animeshgautam6197 see it like this in x(t-t0) = c1*exp(jnw0t0)*exp(jnw0t) , the whole coefficient is c1*exp(jnw0t0) , thats only you need here , in the video sir has written only coefficients in both cases

[@EE-SANTHIYAB]

Why average value is zero for odd signal?

[@nesoacademy]

Don’t skip lectures, the explanation is already there in the earlier chapter.

[@EE-SANTHIYAB]

@@nesoacademy sure ji thank you

[@im.meer-asif]

nicee

[@ch.kaviya9151]

Sir your video is not clear

[@quantised1703]

why do you speak like a robot? anyways, good video!