Digital Signal Processing

Digital Signal Processing - DSP -1 Home Work # 1 Sriram...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: DSP -1 Home Work # 1 Sriram Yarlagadda sy19@uakron.edu sid # 2196944 PROBLEM-1 Given second order differential equation d 2 f ( t ) dt 2 2 o df ( t ) dt o 2 f ( t ) o 2 ( t ) ------(1) Applying Laplace Transforms on both sides L { d 2 f ( t ) dt 2 } s 2 f ( s ) sf (0) f (0) L { df ( t ) dt 2 } sf ( s ) f (0) L { f ( t )} f ( s ) Since ( t ) represents an impulse function on applying Laplace transform to ( t ) a unit function is obtained L { ( t )} 1 Given initial conditions to be zero Therefore Therefore the above equations transform as follows L { d 2 f ( t ) dt 2 } s 2 f ( s ) L { df ( t ) dt } sf ( s ) L { f ( t )} f ( s ) Substituting these obtained Laplace Transforms in equation 1 i.e. second order differential equation s 2 f ( s ) 2 o sf ( s ) o 2 f ( s ) o 2 1 Taking f(s) common on LHS we have f ( s ){ s 2 2 o s o 2 } o 2 f (0) f (0) f ( s ) o 2 s 2 2 o s o 2--------(2) finding the roots for denominator equation in denominator is similar to ax 2 bx c roots for ax 2 bx c 0 are b b 2 4 ac 2 a Similarly the roots for equation 2 are 2 1 o o j The equation 2 can be written as follows ) 1 ( ) ( ) ( 2 2 2 2 o o o s s f Applying Inverse Laplace Transforms on both sides } ) 1 ( ) ( { )} ( { 2 2 2 2 1 1 o o o s L s f L----------(3) L 1 { f ( s )} f ( t ) RHS of equation 3 is of form at a s a L sin } { 2 2 1 Multiplying and Dividing equation 3 by 2 2 ) 1 ( } ) 1 ( ) ( ) 1 ( ) 1 ( 1 { )} ( { 2 2 2 2 2 2 1 1 o o o s L s f L Here s s o and a= ) 1 ( 2 o and From Shifting property L 1 { 1 s a } e at Therefore t e t f o t o o ) 1 sin( 1 ) ( 2 2 Hence the Laplace transform of second order differential equation d 2 f ( t ) dt 2 2 o df ( t ) dt o 2 f ( t ) o 2 ( t ) is t e t f o t o o ) 1 sin( 1 ) ( 2 2 ---------(4) (II) MATLAB is used to plot the f(t) obtained above and the values of and o are chosen such that the system is stable i.e. the poles should be in the left hand side of the S plane MATLAB PROGRAM clc clear all; close all; W=2.4; E=0.4; t=0:0.01:10; a=sqrt(1-(E^2)); b=(W/a)*exp(-1*W*E.*t).*sin(W*a.*t); plot(t,b) title('Response for Second order differential equation'); xlabel('time period --t'); ylabel('amplitudeof response --f(t)'); Plot (III) To plot the normalized power spectrum (0 to -60 db scale) and phase spectrum, we compute the Fourier transform of f(t) ....
View Full Document

Page1 / 23

Digital Signal Processing - DSP -1 Home Work # 1 Sriram...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online