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Determine whether the signals are power or energy signals. Calculate energy/power accordingly:
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(i) x(t)=0.9e3tu(t)

E=|x(t)|2dt

=|0.9e3tu(t)|2dt=0.810e6tdt=0.81[e6t6]0=0.81/6E=0.135J

Since the signal is not periodic hence it’s not a power signal but it is a energy signal.

(ii) x[n]=u[n]

The signal is periodic because u(n) repeats after every sample and of infinite duration.

Hence it is a Power Signal

P=limN12N+1NN|x[n]|2=limN12N+1NN|u[n]|2=limN12N+1N0(1)2

Since N0(1)2=1+1+1+=(N+1)P=limN12N+1(N+1)=limNN(1+1N)N(2+1N)=0.5W.P=0.5W

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