In summary If x[n] is a finite-length sequence (n 0 only when |n|<N) , its DTFT X(e jw ) shall be a periodic continuous function with period 2 . The DFT of x[n], denoted by X(k), is also of length N. where , and W n are the the roots of W n = 1.
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If x[n] is a finite-length sequence (n0 only when |n|<N) , its DTFT X(ejw) shall be a periodic continuous function with period 2.
The DFT of x[n], denoted by X(k), is also of length N.
where , and Wn are the the roots of Wn = 1.
(is a linear system)
when |x[n]| Bx.
h[n] = 0 for all n <0.
S = 1 +|a| +|a|2 +…+ |a|n+…… =1/(1|a|) is bounded.
The term y[n] y[n1] suggests the implementation can be cascaded with an accumulator.
when x[n] is available, y, y, … y[n], … can be computed recursively.
y[n] = ay[n-1] + x[n].
y = a1 y = a2 c, …, and y[n] = an+1c for n<1.
y[n] = an+1c+Kanu[n],
y[n] =x[n nd],
Hence, the frequency response is
By the principle of superposition, the output is
(magnitude and phase)
M1= 0 and M2 = 4
y[n](1/2) y[n1] = x[n] (1/4)x[n1]
To find the impulse response, we set x[n] = [n]. Then the above equation becomes
h[n](1/2) h[n1] = [n] (1/4)[n1]
Applying the Fourier transform, we obtain
H(ejw) (1/2)e-jwH(ejw) = 1 (1/4) e-jw
So H(ejw) = (1 (1/4) e-jw) / (1 (1/2) e-jw)
thus, (1/2)nu[n] 1 / (1 (1/2) e-jw)
By the shifting property,
(1/4)(1/2)n1u[n1] (1/4) e-jw / (1 (1/2) e-jw)
h[n] = (1/2)nu[n] (1/4)(1/2)n1u[n1]
x[n] = ejwn u[n]
Illustration for the FIR case by convolution
Illustration for the IIR case by convolution