Signals and Systems - Electrical Engineering

630 C H A P T E R 10: Fourier Analysis of Discrete-Time Signals and Systems

Table 10.2Properties of Discrete Fourier Series and Discrete Fourier Transform.

Fourier Series of Discrete-Time Periodic Signals

x[n], periodic signal of periodN X[k], periodic FS coefficients of periodN

Z-transform x 1 [n]=x[n](u[n]−u[n−N]) X[k]=N^1 Z(x 1 [n])

∣∣

z=ej^2 πk/N

DTFT x[n]=

∑

kX[k]e

j 2 πnk/N X(ejω)=∑

k^2 πX[k]δ(ω−^2 πk/N)

LTI response Input:x[n]=

∑

kX[k]e

j 2 πnk/N Output:y[n]=∑

kX[k]H(e

jkω (^0) )ej 2 πnk/N
H(ejω), frequency response of system
Time shift (circular shift) x[n−M] X[k]e−j^2 πkM/N
Modulation x[n]ej^2 πMn/N X[k−M]
Multiplication x[n]y[n]
∑N− 1
m= 0 X[m]Y[k−m], periodic convolution
Periodic convolution
∑N− 1
m= 0 x[m]y[n−m] NX[k]Y[n]
Discrete Fourier Transform
x[n], finite-lengthNaperiodic signal ̃x[n], periodic extension of periodL≥N
̃x[n]=^1 N
∑L− 1
k= 0 X ̃[k]e
j 2 πnk/L X ̃[k]=∑Ln−=^10 x ̃[n]e−j 2 πnk/L
IDFT/DFT x[n]=x ̃[n]W[k],W[n]=u[n]−u[n−N] X[k]=X ̃[k]W[k],W[k]=u[k]−u[k−N]
Circular convolution (x⊗Ly)[n] X[k]Y[k]
Circular and linear convolution (x⊗Ly)[n]=(x∗y)[n], L≥M+K− 1
M=length ofx[n],K=length ofy[n]
(a) The given filter is LTI, and as such the eigenfunction property applies. Obtain the magnitude
response|H(ejω)|of the filter using the eigenfunction property.
(b) Compute the magnitude response|H(ejω)|at discrete frequenciesω= 0 ,π/ 2 , andπradians. Show
that the magnitude response is constant for 0 ≤ω≤π, and as such this is an all-pass filter.
(c) Use the MATLAB functionfreqzto compute the frequency response (magnitude and phase) of this
filter and to plot them.
(d) Determine the transfer functionH(z)=Y(z)/X(z). Find its pole and zero and indicate how they are
related.
10.2. Frequency transformation of low-pass to high-pass filters—MATLAB
You have designed an IIR low-pass filter with an input–output relation given by the difference equation
y[n]=0.5y[n−1]+x[n]+x[n−1] n≥ 0
wherex[n]is the input andy[n]is the output. You are told that by changing the difference equation to
y[n]=−0.5y[n−1]+x[n]−x[n−1] n≥ 0
you obtain a high-pass filter.