Final answer:
To find the frequency response H(e^jω) of a given LTI system represented by the difference equation y[n]-1/2y[n-1]=x[n]+2x[n-1]+x[n-2], one must apply the z-transform to the equation, solve for H(z), and then substitute z with e^jω to obtain H(e^jω).
Step-by-step explanation:
To determine the frequency response H(ejω) of a linear time-invariant (LTI) system with a given difference equation, you first need to write the equation in terms of the z-transforms of the input x[n] and output y[n]. The original difference equation given is:
y[n] - \frac{1}{2}y[n-1] = x[n] + 2x[n-1] + x[n-2].
Applying the z-transform and using the fact that the z-transform of y[n-k] is z-kY(z), the transformed equation becomes:
Y(z) - \frac{1}{2}z-1Y(z) = X(z) + 2z-1X(z) + z-2X(z).
This can be rearranged to solve for the system function H(z):
H(z) = \frac{Y(z)}{X(z)} = \frac{1 + 2z-1 + z-2}{1 - \frac{1}{2}z-1}.
To find the frequency response H(ejω), we substitute z with ejω:
H(ejω) = \frac{1 + 2e-jω + e-2jω}{1 - \frac{1}{2}e-jω}.
By further simplifying the expression, we may write the frequency response in terms of sine and cosine functions using Euler's formula, eikx = cos kx + i sin kx, but the given question does not require this additional step.