Discrete Time Energy Signals (Solved Problems)

Signal & System: Solved Questions on Discrete-Time Energy Signals
Topics discussed:
1. Calculation of total energy of discrete-time signals.
2. Effect of time-shifting on the total energy of a discrete-time signal.
3. Effect of time scaling on the total energy of a discrete-time signal.
4. Effect of time-reversal on the total energy of a discrete-time signal.
5. Effect of amplitude reversal on the total energy of a discrete-time signal.

Music:
Axol x Alex Skrindo - You [NCS Release]

#SignalAndSystemByNeso #Signal&System #DiscreteTimeSignal #DiscreteEnergySignal

[dyuthig]

Your dedication can be seen with the fact that... You tell us to solve the HW problems but never to subscribe your channel like the other people....

[Official-tknc]

I am from nit and u teach better then my professors

[pkofficl]

The Answer is 12 J.
Explanation:
|1+j|^2
First take |1+j|
Formula for modulus of complex number is,
|a+jb|= root ((a)^2 + (b)^2)
Similarly, |1+j| = root ((1)^2 +( -1)^2)
Therefore we got, =root (1 + 1)
=root (2).
And after square the root (2),
Final answer is = 2
Apply above steps to next value (1-j).
Finally the solution is = 2+2+4+4=12J

[huzzikhan]

Ex=(1+j) ^2+(1-j)^2+(-2)^2+(2)^2
=(1+2j+j^2)+(1-2j+j^2)+4+4
=(1+2j-1)+(1-2j-1)+8
=2-2+2j-2j+8
=8 Joules to your question sir

[pratyushkesharwani]

superb teaching but tell the answer of h.w also please

[prakashsharma]

To simplify the expression (1−j)2+(1+j)2+8(1 - j)^2 + (1 + j)^2 + 8(1−j)2+(1+j)2+8, we first need to expand the squares and then combine the terms.

Step 1: Expand (1−j)2(1 - j)^2(1−j)2

(1−j)2=(1−j)(1−j)=1−2j+j2(1 - j)^2 = (1 - j)(1 - j) = 1 - 2j + j^2(1−j)2=(1−j)(1−j)=1−2j+j2

Since j2=−1j^2 = -1j2=−1:

1−2j+(−1)=1−2j−1=−2j1 - 2j + (-1) = 1 - 2j - 1 = -2j1−2j+(−1)=1−2j−1=−2j

Step 2: Expand (1+j)2(1 + j)^2(1+j)2

(1+j)2=(1+j)(1+j)=1+2j+j2(1 + j)^2 = (1 + j)(1 + j) = 1 + 2j + j^2(1+j)2=(1+j)(1+j)=1+2j+j2

Again, since j2=−1j^2 = -1j2=−1:

1+2j+(−1)=1+2j−1=2j1 + 2j + (-1) = 1 + 2j - 1 = 2j1+2j+(−1)=1+2j−1=2j

Step 3: Combine the Results

Now we sum the two expanded terms and the constant 8:

(1−j)2+(1+j)2+8=−2j+2j+8(1 - j)^2 + (1 + j)^2 + 8 = -2j + 2j + 8(1−j)2+(1+j)2+8=−2j+2j+8

Notice that −2j+2j=0-2j + 2j = 0−2j+2j=0, so we are left with:

0+8=80 + 8 = 80+8=8

Thus, the simplified result is:

(1−j)2+(1+j)2+8=8(1 - j)^2 + (1 + j)^2 + 8 = 8(1−j)2+(1+j)2+8=8

[z_o_d]

[√2]^2 + [√2]^2 + 4 + 4 = 12 joules

[awesomearchit]

Sir you said from this Monday you'll upload lectures on networks

[eegauravshrivas]

8 was incorrect due to blindly squaring but it is actually modulus square hence correct answer is 12

[rishabhbahoria]

sir HW problem page for SS and AE is not accessible ! know, how to check for the correct solution ?

[maheedharkolli]

answer to homework problem is 8J and if u guys have confusion on how x[3n] 's energy is 17 here is the solution
{1, 4} are the samples of x[3n] so E=17

[anushrigupta]

Why energy changed in case of compression, not in expansion?

[anujsharma]

sir, in CTS you told that impulse signal are NENP and here said that impulse signal are energy signal

[Svenkatesh]

my doubt why an impulse is not an NENP signal as its mag is infinite at t=0

[shikhaal-ali]

The answer to Question 3 is ( 8 Joules )

[kritikabhateja]

How is x[n-1] energy 55J.
As x[n-1] = {0, 1, 2, 3, 4} It should be 30J.