Power Signals

[@rahulkumarsaw1632]

Thank you so much for free lectures providing by Neso Academy all these lectures help me lot either for my gate exam or semester. Actually i tried all lectures such as Gate Academy , Kretryx , Unacademy , and few youtube lectures but i found Neso is best.

[@aishwaryabuyya3793]

avg power = (Ao^2)/3,false. The line equation to be taken is x(t)= - (2Ao/To)*t btw -To/2 to 0 and x(t)= +(2Ao/To)t btw o and +To/2. and solve...

[@amankumargiri3163]

How ?

[@fazilhamza1476]

@@amankumargiri3163 use the formula Y=Mx formula where M = (y2-y1)/(X2-X1). This will give yo x(t) as Aishwarya mentioned above.

[@sethusri0123]

@@fazilhamza1476 bro can you please send me the solved ans to me please to watsapp please bro please 9444556281

[@NocturnalCoder022]

But the avg power is (4Ao^2)/(To^2)

[@sudeepreddy5130]

Can anyone pls explain why it is false , we got p as finite value

[@mretyo]

power:(A^2)/3. if time is T it is periodic so every periodic signal is power signal but not true for vice versa. correct me if feel wrong thanks

[@soumabratabhattacharya3375]

avg.power=(A0^2)/3.False

[@gianlucarocca295]

bro he wrote a0^2/2! just by computing the integral of the finite area -t0/2 to t0/2

[@study-kl8ur]

@Neso Academy....please explain the hw question

[@nickthewinner2194]

its good to have straight A's.

[@rovibb2128]

You are a great teacher

[@rovibb2128]

By the way as feedback: The sound coming from outside sometimes distracts my focus from you maybe if you solve that problem your videos will be liked more. Just friendly feedback :D

[@ertugrul98835]

Amazing tutorials

[@uniqueversalunicorn2412]

Your explanation is super dude Keep rocking

[@madhavkhindri2817]

Why we are using the formula for Aperiodic power signal, while the signal is periodic? I calculated it with the periodic formula and got the same result...

[@AMITKUMAR-ug9lz]

thanks a lot sir because of you my basic are cleared

[@ritikrustagi]

i) P = (Ao^2)/3 RMS = Ao/((3)^1/2) ii) False

[@neamaessa4037]

What's the RMS

[@einsteinash2207]

@@neamaessa4037 root mean square value

[@princekonwar1672]

p average is Ao^2 and rms valuse is Ao and signal is power signal so true .sir reply its right or wrong

[@chaitanyakishore2150]

I got same answer 🙌 but don't know it is true or not may be you can tell as its been 3 year now so .

[@AshishVerma13yq]

​@@chaitanyakishore2150 I also got same answer pls tell it's true or not as it's been two years now so

[@nitinkamra8972]

avg= A^2/3 and rms is A/sq. root of 3 and signal is power signal so false 😊😊😊😊😊 thnku soo much neso

[@muhammadwaqas209]

the average power is Ao^2 while the rms value is A0. the signal is power signal

[@r2editx]

the avg power is p=to square divided by 4*Ao square RMS value is =to divided by 2 into AO

[@mbhagyarajareddy4639]

Square of Ao divided by 3 is average power of signal.

[@DeepakKumar-gf8mz]

P= A.A./3 RMS= A./√3 E=infinite

[@DeepakKumar-gf8mz]

@Boy Say thank u gentleman

[@mujtabakhurram8707]

How do you get A^2/3

[@Lucky-wi6or]

The answer is 4/3A0^2

[@monikayadav3554]

Sir u shld have made vd for discrete time signals also

[@blackboxtechnologies8841]

Thanks

[@saikiransivaji3378]

excellent lectures sir....if you are professor in my btech....i learn so many things from you....thank you so much sir

[@meghana3773]

Though it is a periodic signal ,why are we taking the limits as -t/2 to t/2 like non periodic signals ?.we need to check for amplitude transition only in non periodic right?. Can anyone tell me please.

[@nehabhusari8243]

We can think this -T/2 to T/2 as generalized formula where for periodic signal T will have some finite value hence no need of limit, and for aperiodic signal limit will be T--> inf. As given signal is periodic, it doesn't matter where we are considering the amplitude transition, whether between 0 to T or between -T/2 to T/2, both are same.

[@shiva0304]

thank you for the information @@nehabhusari8243

[@abhishek.p8744]

Bro can we get this notes as pdf all the lectures

[@annu_2341]

Sir, can u please provid ethe correct answer of the homework question ? My answer is Avg Power = (A^2). And its a power signal

[@nadiatahseen7100]

same here its A^2..

[@MrSamMrLonely]

In the second property, won't there be dimension error?

[@alaminhossain6146]

avg p = (4/3)*Ao^2 RMS = 2Ao/sqrt(3)

[@alaminhossain6146]

No...it will be Pavg = Ao^2/3 RMS = Ao/sqrt(3)

[@HELLOWORLD-wz1uj]

Super guruji

[@Yuvraj191996]

A0/√3

[@havanahou517]

Thank you for your amazing lecture. BTW, are you writing by a mouse or an electronic pen?

[@GauravKumar-bx7ry]

P = 4/3A.^2 Rms = 2A/3^1/2 False

[@keerthanajospin2874]

Me getting the same

[@sudeepreddy5130]

Power is finite naa, why false

[@SudarshanreddyChaganti123]

Power=A°^2 & rmsA°

[@mohammadtawfiq7115]

Can't figure out the answer! Somebody Explain please !

[@Atulkumar-bl3sv]

(A^2)*T/3

[@manishbeniwal01]

1) A/sqrt 3 2) True

[@manishbeniwal01]

A^2 by 3 and False

[@santanubhowmik9128]

In homework problem ,this is a periodic signal. So it's a power signal??

[@manishsolanki174]

p=square of Ao divided by 3

[@palletinavanthreddy2826]

Pls post solution of homework problem

[@VedantBadukale]

hw = { "avg_power": [" Ao^2/3," "false" ] }

[@happyclub7509]

every periodic signal is a power signal??? x=tan (t)??

[@AshishKumar-bs7um]

nice explained sir ji

[@rohit-kt1qq]

p= Ao^2/2 and RMS = Ao/√2

[@Dynamite-bc2jk]

area of triangle -- A^2*T/2 power=area/T --> A^2/2??

[@adekoyataiye7088]

P = Aö²/2 It's a power signal

[@prachimane5716]

m not getting that answer will you please explain...

[@karimadel1691]

can you explain why false?

[@renixden9849]

All periodic signals are power signals. The signal in the exercise is periodic and thus a power signal. The question says if the signal is NOT a power signal.

[@akashdhetarwal5138]

It is a periodic signal Av. power=Ao^2 divided by To And it is a power signal so second statement is false.

[@saisumanth8064]

howwwwww sirrrrrrrrrrrr can u explain

[@samrafarooq9402]

P=(A0)^2/3 Is it correct?

[@excommunicado3955]

Could you please explain how?

[@samrafarooq9402]

I used slope method m1(-ve)= -2Ao/To and m2(+ve)= 2Ao/To y1=-2Ao/To +C and y2= 2Ao/To +C here c=0 in both equations, then just put these two equations in formula.

[@excommunicado3955]

But how you got positive valves while performing limits from -T/2 to 0? Sorry I still little bit confuse.

[@excommunicado3955]

If u don't mine, can you please write the steps or snap photo of the solved copy.

[@dkkhalid3454]

Till here, I got the same results but can you explain the steps further when you evaluate the integral? Thanks

[@KISHORKUMAR-cm2fv]

Please give me detail solution ........

[@reshmaslukose6424]

Please write the answer....

[@Mehraj_IITKGP]

Average power=A_{0}^{2}/3, RMS=A_{0}/sqrt3, false

[@panostef]

Why a periodic signal is a power signal?

[@srimoyeeghosh4575]

As it's avg power is finite

[@suhassumanth4801]

Sir please pin crt answers in comment section

[@MrSiddusivva]

Sir it's not comfortable to note down your writings on left bottom side....

[@chidanandadatta4695]

(To^2)/3, false

[@PoluBalaji]

Broo how to do..??

[@rprvk]

P=(2/3)*Ao^2

[@reficeismail6520]

P=A₀²*4/9 RMS= A₀*2*3

[@ashishn.c.7913]

P=(A0)^2/3 false,correct?

[@asabu80]

ASHISH N.C. ....power= (4/3)Ao^2 RMS= (2*Ao)/√3 so...the given signal is power signal.

[@hitabratanath305]

Ao*Ao/3

[@abhinavkishor2469]

@@asabu80 can u explain how to calculate average power

[@_DEBANJANDHAR]

Its true

[@AshishKumar-bs7um]

A/2

[@shortsaftershots]

can anyone hear the backgrd construction noice. If not play it on 3x

[@AshishKumar-bs7um]

power signal

[@sayan486]

P=Ao^2 /3 False

[@dharmeshparmani8473]

How ??

[@karimadel1691]

why false?

[@abby.studios]

homework sum answer.?

[@emeraldcreations3196]

1.Average power = A^2/ 3, rms=A/root3 2. Statement is false.

[@study-kl8ur]

how avg power come explain me

[@avinashmaurya6087]

False

[@Atulkumar-bl3sv]

false

[@worldwide6626]

its good to have straight A's.