Power of CT Signals (Solved Problems) | Part 2

[@sayan486]

i) Average power=8 RMS=2.828 ii) Average power=18 RMS=4.242

[@thirumalai186]

best lectures for students

[@Spiros219]

5:49 he is such a rock star

[@study-kl8ur]

1q.pavg 8watts.rms 2 root2 2q.pavg 18watts. rms 3root2

[@slatetoscreenlife]

you are using very easy examples how can we get strong subject

[@vishwasjain3162]

For first one Average power:8 RMS:2√3 Second one Average power:18 RMS :3√2

[@vishwasjain3162]

Please confirm??

[@yogikkhurana]

Sir i have a confusion, U said dc signals are periodic so they has to be power signals...but in question 2 you are using the formula for non periodic signals .Why? Pls answer my ques😅

[@sathiyamoorthy8831]

fundamental time period undetermine

[@harryharika5443]

1).8,2root2 2)18,3root2

[@ValaHardik-x1q]

First -> 8 , sqrt(8) Second -> 18 , sqrt(18)

[@akashdhetarwal5138]

1. 8,2√2 2. 18,3√2

[@hemantkumarkalosiya5145]

1) 8 and 2√2 2) 18 and 3√2

[@meghana3773]

Sir, in previous lecture you have told that dc signal is periodic. Then why did you use formula of nonperiodic signal. If anyone knows please give a reply.

[@saurabhsinghbhadauria419]

Bcoz time period of this dc signal can not be determined

[@akashdhetarwal5138]

Can you also provide the presentation of Fourier series and Fourier transform please

[@ashishtiwary7947]

first power is 8 and rms is two root 2 for the later answer are as follows 18 and three root two

[@aravindhangk0066]

1) p =8, rms =√8 2) p = 18, rms = √18

[@saikumar7253]

sir u sai dc signal is a periodic signal, but y u used non-periodic formula for avg power calculations

[@aksharaviju6624]

Exactly my doubt

[@ThomasShelby-nc2xh]

The fundamental time period is undefined for DC value, that's why he applied aperiodic formula..

[@abhignathota9304]

Thank you so much Sir

[@thirumalai186]

thank you very much sir

[@eneskuzuoglu1422]

you are awesome

[@Indore..]

16, 4 and 36, 6

[@bipulboruah1236]

1) p=8 RMS=2√2 2) P= 18 RMS=3√2

[@gnquince]

Why do you use non periodic avg power formula for DC signal??

[@travelwithmbdelhiivaale3615]

because DC is periodic signal but the time period I.e T is undefined for DC.

[@jaikhatri4783]

When we use limit?

[@SanjayKumar-fu7vb]

dc signal is periodic!but u applied aperiodic power formula??

[@tejasriram1480]

dc signal is periodic but we can't tell fundamental time period so he used aperiodic power formula

[@ThomasShelby-nc2xh]

The fundamental time period is undefined for DC value, that's why he applied aperiodic formula..

[@sudheerjalli4707]

8, root8-1 18,root18

[@rohitdatwani8467]

Sir u said that dc signal is always periodic so we should apply the other formula of avg power

[@ThomasShelby-nc2xh]

The fundamental time period is undefined for DC value, that's why he applied aperiodic formula..

[@yashkhosla408]

but you said that DC signals are always periodic so why are we using avg power for non-periodic signal here for that dc.

[@RAVITEJAP-ss9rl]

For using the formula of period functions, fundamental time period should be known which is undefined for dc signal. So the other formula was used

[@pcid2251]

1st one and 2nd one have infinite area then how can we calculate power

[@uradiajay9202]

Sir signum function is ? Power signal or energy signal ? Or NENP? Plz rply

[@uradiajay9202]

Neso Academy :: thankyou sir

[@jancydas5569]

Power signal

[@bhanuprakashreddy]

Sir Can you please text book name which you referenced

[@nandapeela1080]

DC is periodic.....

[@raveenaravi3786]

If sgnl1 nd sgnl2 have their respective power as P1 nd P2...will the power of sgnl3=P1+P2..if sgnl3=sgnl1+sgnl2.??!?!

[@akashdhetarwal5138]

This is quite interesting but generally applied in orthogonal signals

[@PritamSoni]

For the first problem Average Power = 8 Joule R.M.S value=2√2 Joule For the second problem Average Power=18 Joule R.M.S value=3√2 Joule

[@pritomde2929]

Unit of power is watt not joule

[@zedlepplin9450]

why is everyone getting 3root2 for RMS for the second one? I got RMS=6root2

[@exodus5948]

Power is 18. For the second one .. there is 1 transition . So from -T/2 to 0 the value of amplitude is 0. And the integral of limit -T/2 to 0 is ZERO. and from 0 to +T/2 the value is |-6|^2 so it becomes integral (36dt) limit is 0 to T/2 . So Integral value is 36T/2 = 18T. And this T is cancelled by another T. So power is 18. Now rms is sqrt(18) = (9*2) Sqrt of 9 is 3. R comes outside the answer will be 3_/2

[@amutharamakrishnan6871]

Dc signal is periodic signal ryt? Then y u used the formula of non periodic signal in 2nd problem?.. Please anyone clarify my doubt

[@amutharamakrishnan6871]

May be in dc signal to is undefined so we cannot use periodic formula ah?

[@travelwithmbdelhiivaale3615]

because T is undefined for DC signal

[@evidencew]

why we use time period for a unit step function(nonperedic) as long as it is non peridic segnal i know that it has transition but still its no peredic so that it has no time period

[@monuramsunder8605]

For the first one Average power:8 RMS:2√2 Second one Average power:18 RMS:3√2

[@chidanandadatta4695]

1. 8,2√2 2. 18,3√2