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Please go through the pictures...... Solution(1)(a) E( Energy )=int_(-oo)^(oo)|x(t)|^(2)dt : for bo th periodicP={[1//T_(0)int_(T0)^(-oo)|x|t||^(2)dt;],[lim_(T rarr oo)(1)/(T)int_(-T//2)^(T//2)|x(d)|^(2)dt;" Periodic Signals "],["; Signals "]:}(i) x∣t1=e^(-2t)u(t)quad u(t)={[0;,t < 0],[1;,t⩾0]:}So, E=int_(-oo)^(oo)|x(t)|^(2)dt quad So, e^(-2t)u(t)={[0;,t < 0],[e^(-2t);,t⩾0]:} E=int_(-oo)^(0)0dt+int_(0)^(oo)|e^(-2t)|^(2)dt=>E=0+int_(0)^(oo)e^(-4t)dt E=[(e^(-4t))/(-4)]_(0)^(oo)=>E=(e^(-oo)-e^(0))/(-4)=>E=(0-1)/(-4)E=1//4=> Finite so it is an exergy signal. (if E= finite the Av ... See the full answer