Question: Given x [ n ] = s i n ( 1 1 2 0 n ) s i n ( 2 0 n ) l

Given
x[n]=sin(1120n)sin(20n)longleftrightarrowDTFS;10x[k],
evaluate the time signal y[n] with the following DTFS coefficients, using only DTFS properties:
(a)Y[k]=x[k-5]+x[k+5]
(b)Y[k]=cos(k5)x[k]
(c)Y[k]=x[k]x[k]
(d)Y[k]=Re{x[k]}
Given x [ n ] = s i n ( 1 1 2 0 n ) s i n ( 2 0 n

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