Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2014

(i) Autocorrelation of sum of signals is equal to the sum of autocorrelation of the individual signal. Let $x(t) = A + B ……………… (1)$

Where, $A = 3 \cos t \text {and} B = 4 \cos 3t.$

Autocorrelation is given by,

$A = ∫\limits_{-∞}^∞ x(k)x(k-t)dk$

By changing the limits from $-∞$ to $∞$ to time period –T/2 to T/2 $$A = \dfrac 1T ∫\limits_{-T/2}^{T/2}3\cos(k) X\space \space 3\cos(k-t)dk$$

Since $f= \dfrac 1{2π} ∴T=2π \\ A= \dfrac 1{2π} ∫\limits_{-π}^π 9\cos(k)\space X \cos(k-t)dk \\ =\dfrac 1{2π} ∫\limits_{-π}^π \dfrac 92[\cos(2k-t)+ \cos(-t)dk ⋯⋯(2\cos A\cos B=\cos⁡(A+B)+\cos⁡(A-B)) \\ =\dfrac 9{4π}[∫\limits_{-π}^π\cos(2k-t)dk+ ∫\limits_{-π}^π\cos(-t)dk] \\ =\dfrac 9{4π} [\dfrac {\sin⁡(2k-t)}2+ \cos⁡(-t)(k)]^π_{-π} \\ =\dfrac 9{4π} [\dfrac{\sin⁡(2π-t)-\sin⁡(-2π-t)}2+2π\cos⁡(t)] \\ = \dfrac 9{8π} [\sin2π\cos t-\cos2π\sin t+\sin2π\cos t+\cos2π\sin t]+\dfrac92 \cos(t) \\ A = 0 + \dfrac92 \cos t = \dfrac92\cos t …………… (a) \\ \text {Similarly,} B = ∫\limits_{-∞}^∞ x(k)x(k-t)dk $

By changing the limits from $-∞$ to $∞$ to time period –T/2 to T/2. $$B = \dfrac 1T ∫\limits_{-T/2}^{T/2} 4\cos(3k) \space\space X 4\cos3(k-t)dk $$

$ \text {Since} \space \space f= \dfrac 3{2π} ∴T=2π/3 \\ =\dfrac 3{2π} ∫\limits_{-π/3}^{π/3} 16\cos(3k) \space \space X \cos3(k-t)dk \\ =\dfrac 3{2π} ∫\limits_{-π/3}^{π/3} 8[\cos(6k-3t)+ \cos(3t)dk] ⋯⋯(2 \cos A\cos B=\cos⁡(A+B)+\cos⁡(A-B)) \\ = \dfrac {12}π [∫\limits_{-π/3}^{π/3}\cos(6k-3t)dk+ ∫\limits_{-π/3}^{π/3}\cos(3t)dk] \\ = \dfrac {12}π [\dfrac {\sin⁡(6k-3t)}6+\cos⁡(3t)(k)]^{π/3}_{-π/3} \\ =\dfrac { 12}π [\dfrac {\sin⁡(2π-3t)-\sin⁡(-2π-3t)} 6 + \dfrac {2π}3 \cos⁡(3t) ] \\ B = \dfrac 2π x0+8 \cos⁡(3t)=8\cos⁡(3t) …………... (b) $

Put (a) and (b) in eqn. (1) we get,

$(R_{xx} (t))= A + B $

By putting t = 0 in the above equation, power can be obtained.

Power $= (R_{xx} (t))|t=0 \\ = 4.5 \cos 0 + 8 \cos 0 \\ = 4.5 + 8 \\ = 12.5W.$

Power spectral density is given by,

But $f{\cos wt} = 0.5[δ(f+f_0 )+δ(f-f_0)] \\ ∴PSD=4.5[δ(f+\dfrac 12π)+δ(f-\dfrac 12π)]+8[δ(f+\dfrac32π)+δ(f-\dfrac32π)]$