数字信号处理与应用作业答案

This is an open book exam. You must work independently without consulting anyone.

Problem 1 (40-pts)

All pass filters are useful in audio signal processing. Fig. 2 shows an all-pass filter. Find (a) transfer function (b) impulse response (c) frequency response (d) plot its impulse response (0 to 50) and list y(8), y(16), and y(24), magnitude and phase responses e) Is this a stable filter? Justify your answer. where

(1.1)

(1.2)

Fig. 1. An all-pass filter.

Solution:

As Fig.1 shows,)(')()('M n x a n x n x M --=,)(')(')(y 0n x b M n x n +-=,so we can get

M M

M z b z X z Y z H z a z X z --+==+==

0)

(')

()('',11)((z)X')('H , then M

M M z

a z

b z H z H z X z X z X z Y z X z Y --++====1)(')('')()

(')(')()()(H(z)0 (1.1.a) :

we have so the transfer funciton is 8

8

01.011.01H(z)----++=++=z

z z a z b M M M (1.1.b) :

The impulse response h(n) is the inverse of z-transform of H(z).

Because H(z) is not the common form that can be solved by hand computation,so we can use Matlab software to solve this probelm.The Matlab codes and results are as followed.

codes: syms z

00.1, delay 8M b a M ===010, delay 8

M b a M ===00.1, delay 8

M b a M ===

H(z)=(0.1+z^(-8))/(1+0.1*z^(-8)) h=iztrans(H) results:

H(z) =(1/z^8 + 1/10)/(1/(10*z^8) + 1)

h(n) = 10*kroneckerDelta(n, 0) + (99*sum(-r3^n/8, r3 in RootOf(z1^8 + 1/10, z1)))/10

that is,we can get impulse response:

h(n) = 10*kroneckerDelta(n, 0) + (99*sum(-r3^n/8, r3 in RootOf(z1^8 + 1/10, z1)))/10

(1.1.c) :

By replacing z with jw

e

in H(z),we can get frequency response )(jw e H .so

w

8j w

8j jw

e

1.01e 1.0)H(e --++= (1.1.d) :

Matlab codes: figure;stem(h(0:50)) title('impulse response')

impulse response

y(8)=0.99, y(16)=-0.099, and y(24)=0.0099

(1.1.e) :

This is a stable filter.The zero-pole mapof the filter is as followed.

-1.5-1

-0.500.51 1.5

Real Part

I m a g i n a r y P a r t

We can see from the picture that all the poles are inside the unit circle,so we can draw the conclusion that it ’s stable. Matlab codes:

a=[1 0 0 0 0 0 0 0.1]; b=[.1 0 0 0 0 0 0 1]; figure;zplane(b,a);

(1.2.a) :

we have so the transfer funciton is 8

0101101H(z)--++=++=z

z z a z b M M (1.2.b) :

The impulse response h(n) is the inverse of z-transform of H(z).

Because H(z) is not the common form that can be solved by hand computation,so we can use Matlab software to solve this probelm.The Matlab codes and results are as followed. codes: syms z

H(z)=(10+z^(-8))/(1+10*z^(-8)) h=iztrans(H)

results:

H(z) =(1/z^8 + 10)/(10/z^8 + 1)

h(n) =kroneckerDelta(n, 0)/10 - 99*sum(-r4^n/80, r4 in RootOf(z3^8 + 10, z3)) that is,we can get impulse response:

h(n) =kroneckerDelta(n, 0)/10 - 99*sum(-r4^n/80, r4 in RootOf(z3^8 + 10, z3))

010, delay 8

M b a M ===

(1.2.c) :

By replacing z with jw

e

in H(z),we can get frequency response

)(jw e H .so

w

8j w

8j jw

e

101e 10)H(e --++= (1.2.d):

Matlab codes: figure;stem(h(0:50)) title('impulse response')

6

impulse response

y(8)=-99, y(16)=990, and y(24)=-990

(1.2.e) :

This is not a stable filter.The zero-pole mapof the filter is as followed.

-1

-0.500.51Real Part

I m a g i n a r y P a r t

We can see from the picture that not all the poles are inside the unit circle,so we can draw the conclusion that it ’s not stable.

Matlab codes:

a=[1 0 0 0 0 0 0 10]; b=[10 0 0 0 0 0 0 1]; figure;zplane(b,a);

Problem 2 (30-pts)

Assume that you need to implement a filter which has two sub-filters shown in below.

)

The overall impulse response of the filter is

The impulse response of the high pass filter

= [-0.014137 -0.10433 0.062933 0.091263 -0.046729 -0.31389 0.54717 -0.31389 -0.046729 0.091263 0.062933 -0.10433 -0.014137]

(a) What is the impulse response of 1()h k ? (b) What is the frequency response of ()h k ? (c) Plot the magnitude response in dB.

(d) What is the type of the overall filter ()h k ?

Solutions:

(a)：

We can learn from the picture above that ))(()()(1k h k h k h high -+= so )6(6

when 1

6when 0))((-)()(1-=??

?=≠=-=k k k k h k h k h high δ

(b)：

))((DTFT )(k h e H jw =,DTFT means Discrete-time Fourier Transform.so we can get

(), when 6()()1, when 6

high high h k k h k h k k -≠??=?-+=??

]

45283.0)cos(62778.0)2cos(093458.0)3cos(182526.0-)4cos(125866.0-)

5cos(20866.0)6cos(028274.0)[()()(612

++++==--=∑w w w w w w e e k h e H w j jkw k jw

which is the frequency response of h(k).

(c)：

We can learn from (b) that w

j e

6- is phase response,which means the magnitude response

]

45283.0)cos(62778.0)2cos(093458.0)3cos(182526.0-)

4cos(125866.0-)5cos(20866.0)6cos(028274.0[Amp(w)++++=w w w w w w

so the magnitude response in dB is as followed.

00.5

1

1.52

the magnitude response in dB

w(pi)

a m p (d B )

Matlab codes:

clear; close all;

w=0:2*pi/13:2*pi-2*pi/13;

amp=0.028274*cos(6*w)+0.20866*cos(5*w)-0.125866*cos(4*w)-0.182526*cos(3*w) ... +0.093458*cos(2*w)+0.62778*cos(w)+0.45283;

figure;plot(w/pi,20*log(abs(amp))) title('the magnitude response in dB') xlabel('w(pi)') ylabel('amp(dB)')

We can also use these codes to get magnitude response,the results of both ways are same.

Another way:

close all; clear

H =[0.014137 0.10433 -0.062933 -0.091263 0.046729 0.31389 0.45283 ... 0.31389 0.046729 -0.091263 -0.062933 0.10433 0.014137]; FFT_H=fft(H);

Amp_dB=20*log(abs(FFT_H)); w=0:2*pi/13:2*pi-2*pi/13; figure;plot(w/pi,Amp_dB);

title('the magnitude response in dB') xlabel('w(pi)') ylabel('amp(dB)')

(d):

Because the length of impulse response is limited,so it is a FIR filter and FIR filter must be stable. Besides,

]

45283.0)cos(62778.0)2cos(093458.0)3cos(182526.0-)4cos(125866.0-)

5cos(20866.0)6cos(028274.0)[()()(612

++++==--=∑w w w w w w e e k h e H w j jkw k jw

we can know from the formula that the phase responsed of H(e jw ) are linear. Problem 3 (30-pts)

Given the following sequences,

(a) ][98765432100b b b b b b b b b b x = (b) ]0

[98765432102b b b b b b b b b b x =

(c) ]0[9

8765432103b b b b b b b b b b x =

Assume that 9...,1,0},{)(},{)(3322===m x DFT m X x DFT m X Find: (1)The mathematical relationship between X 2(m) and X 3(m). (2) Interpret your result in part (1)

Solutions:

(1)：

The mathematical relationship are as follows.

)()(31122m X

e m X m j π-=

(2)：

we can interpret the result in two ways as follows.

The view of Mathematics:

n m

j

n n n m

j

n e

b e

n x m X 11

29

011

210

033)()(ππ-=-=∑∑==

n m

j

n n m j

n m

j

n n n m

j

n n n m

j

n e

b e

e

b e

b e

n x m X 11290

11

2)1(11

29

11

210

1

111

210

22)()(πππππ-=-+-=-=--=∑∑∑∑====

so we can get )()(31122m X

e m X m j

π-=

The perspective of physical significance:

From ]0

[98765432102b b b b b b b b b b x =,]0[9

8765432103b b b b b b b b b b x =,we can say that x 2

is delayed for an unit time than x 3,according to the properties of Fourier transform and characteristics of delay,we can draw the conclusion that )()(31122m X

e m X m j π-=

which means

X 2(m) and X 3(m) share the same magnitude response,but have the different phase response.X 2(m)

delayed 11/2m π than X 3(m) in phase.