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Determine power and energy for the following signals $(i) x (t) = 3 \cos 5 \Omega t.$ $(ii) x[n] = (\dfrac 14)^n u[n]$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 05

Year : MAY 2014

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We know that, Energy

$= \lim\limits_{T→∞} ∫\limits_{-T}^T |x(t)|^2 dt$ and

Power $= \lim\limits_{T→∞} \dfrac 1{2T}⁡∫\limits_{-T}^T |x(t)|^2 dt,$ so let us first find

$∫\limits_{-T}^T|x(t)|^2 dt \\ ∴ ∫\limits_{-T}^T|x(t)|^2 dt =∫\limits_{-T}^T|(3\cos5Ωt)|^2 dt \\ = ∫\limits_{-T}^T(3\cos5Ωt)^2 dt \\ = ∫\limits_{-T}^T9\cos^2 (5Ωt)dt \\ = 9∫\limits_{-T}^T[\dfrac {1+\cos10Ωt}2]dt = \dfrac 92 ∫\limits_{-T}^T[1+\cos10Ωt]dt = 4.5 [t+ \dfrac {\sin10Ωt}{10Ω}]_T^{-T}$

Since $\sin (-t) = -\sin (t)$ and $Ω = 2π/T \\ ∴ =4.5[2T+ \dfrac {2\sin10ΩT}{10Ω}] \\ =4.5[2T+\dfrac {2\sin10T*( 2π/T)}{10*( 2π/T)}] \\ =4.5[2T+ \dfrac T{10π} sin20π]$

As we know $\sin Nπ=0$ for integer N,

$$= 4.5*2T = 9T$$

Now, Energy = $\lim\limits_{T→∞}⁡∫\limits_{-T}^T|x(t)|^2 dt = \lim\limits_{T→∞} 9T= ∞.$

And Power $= \lim\limits_{T→∞} 1/2T⁡∫\limits_{-T}^T|x(t)|^2 dt = \lim\limits_{T→∞} \dfrac 1{2T}*9T = 4.5 W.$

Since energy is infinite and power is constant, the signal is power signal.

$(ii) x[n] = (\dfrac 14)^n u[n]$

Energy for DT signal is $E = ∑\limits_{-∞}^∞|x(n)|^2 \\ ∴ E = ∑\limits_{-∞}^∞|(\dfrac 14)^n u[n]|^2 \\ = ∑\limits_0^∞(\dfrac 14)^{2n} ……………… (since\space \space u[n] = 1 for n = 0 to ∞ ) \\ = ∑\limits_0^∞(\dfrac 1{16})^n \\ = ∑\limits_0^∞(0.0625)^n$

As we know $∑\limits_0^∞a^n= \dfrac 1{1-a} |a|\lt1 \\ ∴ E = \dfrac 1{1-0.0625} \\ E = 1.067 J$

Power for DT signal is $P = \lim\limits_{N→∞} \dfrac 1{2N+1} ∑\limits_{-N}^N|x(n)|^2 \\ ∴ P = \lim\limits_{N→∞} \dfrac 1{2N+1} ∑\limits_{-N}^N|(\dfrac 14)^n u[n]|^2 \\ =\lim \limits_{N→∞} \dfrac 1{2N+1} ∑\limits_0^N(\dfrac 14)^{2n} ……………… (since \space \space u[n] = 1 for n = 0 to ∞ ) \\ = \lim\limits_{N→∞} \dfrac 1{2N+1} ∑_0^N (\dfrac 1{16})^n \\ =\lim\limits_{N→∞} \dfrac 1{2N+1} ∑_0^N (0.0625)^n$

As we know $∑_0^N a^n=\dfrac {a^{n+1}-1}{a-1} |a|\lt1 \\ = \lim\limits_{N→∞} \dfrac 1{2N+1} \dfrac {0.0625^{n+1}-1}{0.0625-1} \\ =\dfrac 1∞ \dfrac {0.0625^{n+1}-1}{0.0625-1} = 0.$ Power of the signal is zero and energy of the signal is constant, hence it is a energy signal.