当前位置:首页 >> 信息与通信 >>

DSP实验答案Solution_lab5

Name: Section:

Laboratory Exercise 5
LINEAR, TIME-INVARIANT DISCRETE-TIME SYSTEMS: FREQUENCY-DOMAIN REPRESENTATIONS
5.1 TRANSFER FUNCTION AND FREQUENCY RESPONSE Transfer Function Analysis

Project 5.1 Answers: Q5.2

The plot of the frequency response of the causal LTI discrete-time system of Question Q5.2 obtained using the modified program is given below:

The type of filter represented by this transfer function is - BF

5.2

TYPES OF TRANSFER FUNCTIONS Filters

Project 5.2

According to Programme P4_1. Answers: Q5.8 The required modifications to Program P4_1 to compute and plot the impulse response of the FIR lowpass filter of Project 4.2 with a length of 20 and a cutoff frequency of ωc = 0.45 are as indicated below: clf; fc = 0.45; n = [-10:1:10]; y = 2*fc*sinc(2*fc*n);k = n+10; stem(k,y);title('N = 20');axis([0 20 -0.2 0.6]); xlabel('Time index n');ylabel('Amplitude');grid; The plot generated by running the modified program is given below:

Q5.20

The plots of the impulse responses of the four FIR filters generated by running Program P4_3 are given below:

From the plots we make the following observations: Filter #1 is of length ____9___ with a __ symmetrical____ impulse response and is therefore a Type _1_ linear-phase FIR filter. Filter #2 is of length ____10__ with a __ antisymmetric____ impulse response and is therefore a Type _2_ linear-phase FIR filter. Filter #3 is of length ___9_____ with a ___ symmetrical ____ impulse response and is therefore a Type _3_ linear-phase FIR filter. Filter #4 is of length __10_____ with a __ antisymmetric___ impulse response and is therefore a Type _4_ linear-phase FIR filter. From the zeros of these filters generated by Program P4_3 we observe that: Filter #1 has zeros at z = 2.3273 + 2.0140i

2.3273 - 2.0140i -1.2659 - 2.0135i -0.2238 - 0.3559i 0.2457 - 0.2126i 2.5270 - 2.0392i

-1.2659 + 2.0135i -0.2238 + 0.3559i 0.2457 + 0.2126i
Filter #2 has zeros at z

= 2.5270 + 2.0392i

-1.0101 + 2.1930i -1.0000

-1.0101 - 2.1930i

-0.1733 + 0.3762i 0.2397 + 0.1934i
Filter #3 has zeros at z

-0.1733 - 0.3762i 0.2397 - 0.1934i 0.2602 + 1.2263i 1.0000 0.6576 - 0.7534i 0.1655 - 0.7803i 2.0841 - 2.0565i

= -1.0000

0.2602 - 1.2263i 0.6576 + 0.7534i 0.1655 + 0.7803i
Filter #4 has zeros at z

= 2.0841 + 2.0565i

-1.5032 + 1.9960i 1.0000 -0.2408 + 0.3197i 0.2431 + 0.2399i

-1.5032 - 1.9960i

-0.2408 - 0.3197i 0.2431 - 0.2399i

Plots of the phase response of each of these filters obtained using MATLAB are shown below:

From these plots we conclude that each of these filters have ____ linear ___ phase.

The group delay of Filter # 1 is - 4 The group delay of Filter # 2 is – 4.5 The group delay of Filter # 3 is - 4 The group delay of Filter # 4 is – 4.5

5.3

STABILITY TEST According to Programme P4_4.

Answers: Q5.23 The pole-zero plots of H1(z) and H2(z) obtained using zplane are shown below:

From the above pole-zero plots we observe that Q5.24 Using Program P4_4 we tested the stability of H1(z) and arrive at the following stability test parameters {ki}: -0.9989 0.8500 From these parameters we conclude that H1(z) is _____stable________ . Using Program P4_4 we tested the stability of H2(z) and arrive at the following stability test parameters {ki}: -1.0005 0.8500 From these parameters we conclude that H2(z) is

____unstable_________ .

5.4

spectrograms

Background: A spectrogram of a time signal is a two-dimension representation that displays time in its horizontal axis and frequency in its vertical axis. A gray scale is typically used to indicate the energy at each point. “white”: low energy, “black”: high energy.
Tasks: Produce a spectrogram in Matlab and then compare it with specgram.m in Matlab to assure what you did is right. Outline: To produce a spectrogram, you will need to assemble many individual spectral slices produced via the Fourier transform - from successive segments of a speech signal. The

basic operations required are ? loading a speech signal ? selecting individual time segments of the speech signal ? using the FFT to transform from the time domain to the frequency domain ? extracting the log magnitude spectrum ? assembling the sequence of spectra into a 2D time-frequency matrix ? displaying the matrix as an image

You will learn how to write MATLAB scripts and functions to produce spectrograms from signals. You will also use MATLAB's plotting facilities to display signals as a means of checking that your code is doing the right thing. Q 5.25 a. Load the supplied speech signal into MATLAB using the command: [y,fs,nbits] = wavread('oilyrag.wav');
b. Design a programme to divide this speech into several frames, and compare it with the

command ‘buffer’.
x=wavread('oilyrag.wav'); %调用自己的分帧函数 w=256; ov=128; y=fenzhen(x,w,ov); %调用matlab的分帧函数 z=buffer(x,w,ov);

function y=fenzhen(x,w,ov) %===分帧函数 if (size(x,1) > size(x,2)) x = x';

end s = length(x); if nargin < 2 w = 256; end if nargin < 3 ov = w/2; end h=w-ov; c=1; ncols=1+fix((s-w)/h); d=zeros(w,ncols); for b=0:h:(s-w) d(:,c)=x((b+1):(b+w))'; c=c+1; end y=d;

c. Design a programme to synthesize the framed speech to a section of speech.
function y=hecheng(x,ov) %==合成语音 [framleng framnum] = size(x); if nargin < 2 ov = framleng/2; end h=framleng-ov; sigleng=h*(framnum-1)+framleng; d=zeros(1,sigleng); for i=1:framnum d(((i-1)*h+1):((i-1)*h+framleng))= d(((i-1)*h+1):((i-1)*h+framleng))+x(:,i)';

end y=d; d. Design a programme to represent the spectrogram of a section of speech, and compare it with the command ‘specgram’.

[x,sr]=wavread('Tea_10kHz_16bit.wav'); %调用自己的语谱程序 y=my_specgram(x,sr,256,128); %调用matlab的语谱程序 figure,

specgram(x)

function y=my_specgram(x,sr,w,ov) %====语谱程序 if (size(x,1) > size(x,2)) x = x'; end s = length(x); if nargin < 2 sr = 1; end if nargin < 3 w = 256; end if nargin < 4 ov = w/2; end h=w-ov; win=hanning(w)'; c=1; ncols=1+fix((s-w)/h); d=zeros(1+w/2,ncols); for b=0:h:(s-w) u=win.*x((b+1):(b+w)); t=fft(u); %???ò???? d(:,c)=t(1:(1+w/2))'; c=c+1; end tt=[0:h:(s-w)]/sr; ff=[0:(w/2)]*sr/w; if nargout < 1 axis xy xlabel('Time/s'); ylabel('Frequency/Hz') ; else y = d; imagesc(tt,ff,20*log10(abs(d))); axis xy xlabel('Time/s'); ylabel('Frequency/Hz') end imagesc(tt,ff,20*log10(abs(d)));

Date:

Signature:


相关文章:
DSP实验答案Solution_lab5(精)_图文.doc
DSP实验答案Solution_lab5(精) - Name : Section
DSP实验答案Solution_lab6.pdf
DSP实验答案Solution_lab6_信息与通信_工程科技_专业资料。武汉大
DSP实验答案Solution_lab4.pdf
DSP实验答案Solution_lab4_信息与通信_工程科技_专业资料。武汉大
DSP实验答案Solution_lab3.pdf
DSP实验答案Solution_lab3_信息与通信_工程科技_专业资料。武汉大
DSP实验答案Solution_lab8.pdf
DSP实验答案Solution_lab8_信息与通信_工程科技_专业资料。武汉大
DSP实验答案Solution_lab1.pdf
DSP实验答案Solution_lab1_信息与通信_工程科技_专业资料。武汉大
DSP实验答案Solution_lab2.pdf
DSP实验答案Solution_lab2_信息与通信_工程科技_专业资料。武汉大
dsp实验5_图文.doc
四.实验步骤工程文件:Lab308-EPWM\lab0307-pwm 目录中的“Example_2833xEPwmUp...DSP实验1-5 暂无评价 3页 1下载券 DSP实验答案Solution_la... 11页 免费...
lab7参考答案_图文.pdf
lab7 Open Shortest Pat... 12页 免费 C_lab7-2_结构体 5页 1下载券 Lab7 多域OSPF配置 28页 1下载券 喜欢此文档的还喜欢 DSP实验答案Solution_la...
DSP实验报告指示灯.doc
,将找到 C:\ DSP281x_examples\ Lab02-UseCMD 目录, 将文件类型改为“...实验 5.3 :快速傅立叶变换(FFT)算法试验结果: 64 点 输入: 滤波: 输出: ...
DSP实验报告 .doc
4 DSP 技术应用实验报告 图 5 CCS3.3 平台主界面 2、试验步骤 (1) 实验准备:给实验板加电,启动 CCS,打开工程,连接目标板。 (2) 通过键盘设置输入信号。 ...
DSP实验报告定时器2.doc
DSP实验报告定时器2_电子/电路_工程科技_专业资料。有关DSP的实验,仅供参
DSPLab4实验报告.pdf
DSPLab4实验报告_理学_高等教育_教育专区。数字信号处理实验报告 ...0
DSP实验报告(完美版)_图文.doc
8 5.实验体会 通过本次实验,我了解了 DSP 开发系统的组成和结构,熟悉了 DSP 开发系 统的连接,熟悉了 DSP 的开发界面,了解了 C54x 常用的寻址方式,了解了 ...
DSP实验报告.doc
打开工程,浏览程序,工程目录为 C:\ICETEK\VC5416AES61\VC5416AES61\Lab0502-...DSP实验报告 (5) 4页 2下载券 DSP第二次实验报告 9页 免费 ...
dsp实验报告 3.doc
dsp实验报告 3_数学_自然科学_专业资料。实验三 PWM 实验:一、实验目的了解 ...命令,在文件 lab24-PWM 下加载 Debug 目录下的.out 可执行文件; (5) 在 ...
dsp实验报告4.doc
2.打开工程文件: C:\ICETEK\VC5509AES60\VC5509AE\Lab0304-XINT\v60\XINT....DSP实验报告1 14页 免费 DSP实验报告8 5页 免费 电子科大通信学院DSP数字....
DSP实验.doc
DSP 原理与应用 实验报告姓专班学成名:业:级:号:绩: 2013 年 5 月 13 ...项,选择 D:\2407EDULab\Lab6-IOPin 目录中的 IOPin.out 文件,通过仿真器将...
东南大学DSP实验报告..doc
东南大学DSP实验报告._职业技术培训_职业教育_教育专区。第三章 DSP 芯片系统...C:\ICETEK-F2812-A-EDUlab\DSP281x_examples\Lab0201-Memory\Memory.pjt 5....
DSP实验报告6.doc
DSP_CCS\\Labs\\09lab\\lab56\\DSP54X-28-T...DSP实验六实验报告 暂无评价 5页 5下载券 DSP...答案无对错之分,我们会筛选出您提出的宝贵建议和...