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Lab no.07

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Lab no.07

  1. 1. Lab No.07 Analysis of LTI systems (Impulse response, convolution) Designed by : Dawar Awan dawar@cecos.edu.pk CECOS College of Engineering and IT March – July 2012
  2. 2. LTI systems  Systems that are both linear and time invariant are called LTI systems.  The behavior of LTI systems is completely characterized by their impulse response.  The input and output of an LTI system is related by convolution sum/integral. CECOS College of Engineering and IT March – July 2012
  3. 3. Impulse response  Impulse response ‘ h[n] ’, is the output of an LTI system, when the input is a unit impulse.  Given the impulse response, we can find the output for any input using convolution. CECOS College of Engineering and IT March – July 2012
  4. 4. Difference equation  A very common representation of LTI systems is in the form of difference equation  The general difference equation is K=M K=N K=0 K=0 ∑ ak y[n-k] = ∑ bk x[n-k]  example : for , y[n] – 5/6y[n-1] + 1/6y[n-2] = 1/3x[n-1] a0=1, a1=-5/6, a2=1/6, CECOS College of Engineering and IT and bo=0, b1=1/3 March – July 2012
  5. 5. Impulse response  Plot the impulse response of the following difference equation . y[n] – 5/6y[n-1] + 1/6y[n-2] = 1/3x[n-1] CECOS College of Engineering and IT March – July 2012
  6. 6. Impulse response CECOS College of Engineering and IT March – July 2012
  7. 7. Practice 1- Plot the impulse response of the following systems I. y[n] = x[n] + y[n-1] (Accumulator) II. y[n] – y[n-1] = 1/7 (x[n] + x[n-1] + x[n-2] + x[n-3] + x[n-4] + x[n-5] + x[n-6]) (Moving average system)
  8. 8. Convolution  y[n] = x[n] * h[n] =  here y[n] is the output signal, x[n] is the input signal, and h[n] is the impulse response of the LTI system.  in MATLAB use the instruction ‘ y=conv(x,h) ‘ to perform convolution. CECOS College of Engineering and IT March – July 2012
  9. 9. Convolution  Convolve the following two sequences in MATLAB. CECOS College of Engineering and IT March – July 2012
  10. 10. Convolution CECOS College of Engineering and IT March – July 2012
  11. 11. Convolution NOTE  MATLAB assumes that both the convolving signals are starting from zero index, hence the time/sample no. of the output signal is not correct always.  You have to adjust the time axis of the output signal keeping in view some rules related to convolution. CECOS College of Engineering and IT March – July 2012
  12. 12. Convolution Hint : length of y = length of (x) + length of (h) - 1 and the starting index for y will be the sum of starting indices of x and h CECOS College of Engineering and IT March – July 2012
  13. 13. Convolution (without time adjustment) * = CECOS College of Engineering and IT March – July 2012
  14. 14. Convolution (after adjusting time) * = CECOS College of Engineering and IT March – July 2012
  15. 15. Task 1- Convolution is associative. Given the three signals x1[n], x2[n], and x3[n] as: x1= [ 3,1,1] x2= [ 4,2,1] x3= [ 3,2,1,2,3] Show that (x1*x2)*x3 = x1*(x2*x3) 2- Convolution is commutative. Given x and as: x=[1,3,2,1] h=[1,1,2] Show that x* h = h*x 3- Determine the output of the LTI system with: h[n] = 2δ[n] + δ[n‐1] + 2δ[n‐2] + 4δ[n‐3] + 3δ[n‐4] x[n] = δ[n] + 4δ[n‐1] +3δ[n‐2] + 2δ[n‐3] CECOS College of Engineering and IT March – July 2012
  16. 16. Task 4- Convolve the signal x=[1,2,3,4,5,6] with an impulse delayed by two samples. Plot the original signal and the result of convolution. 5- Find the output signal y[n] for any range of ‘n’ using convolution, for the following cases. a) h[n] = 5(-1/2)n u[n] b) h[n] = (0.5)2n u[-n] c) h[n] = 2n u[-n-1] CECOS College of Engineering and IT , , , x[n] = (1/3)n u[n] x[n] = u[n] x[n] = u[n] – u[n - 10] March – July 2012

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