Filtering in the Frequency Domain

Transcription

1 Filtering in the Frequency Domain Dr. Praveen Sankaran Department of ECE NIT Calicut January 11, 2013

2 Outline 1 Preliminary Concepts 2

3 Signal A measurable phenomenon that changes over time or throughout space. Case in hand for this class an image. Signals undulate variation in pixel values. Frequency-domain representation An exact description of a signal in terms of its undulations. Any Real Signal has a Frequency-Domain Representation.

4 Fourier Series Any periodic function can be expressed as the sum of sines and/or cosines of dierent frequencies, each multiplied by a dierent coecient. The sinusoids are called basis functions. The multipliers are called Fourier coecients.

5 Fourier Series T = period f (t) = n= c n e j 2πn T t (1) n = 0,±1, c n = 1 T T /2 2πn j f (t) e T t dt (2) T/2 Note e jθ = cosθ + jsinθ

6 Non-periodic Functions? Any Real Signal has a Frequency-Domain Representation. So what about non-periodic functions? Fourier Transform µ = frequency. I{f (t)} = F (µ) = f (t) e j2πµt dt (3) f (t) = F (µ) e j2πµt d µ (4) Fourier transform pair. Note integral of the square of f (t) =nite.

7 Fourier Transform Example

8 Impulses A unit impulse of a continuous variable t located at t = 0, is dened as, { if t = 0 δ (t) = 0 if t 0 (5) and, δ (t) dt = 1 (6) Unit Discrete Impulse δ [x] = { 1 if x = 0 0 if x 0 (7)

9 Sifting Property Sifting Property f (t)δ (t) dt = f (0) (8) if f (t) is continuous at t = 0 f (t)δ (t t 0 ) dt = f (t 0 ) (9) Discrete Form f [x]δ [x] = f [x] (10) x= f [x]δ [x x 0 ] = f [x 0 ] (11) x=

10 Impulse Train s (t) = n= δ (t n T )

11 Convolution Convolution f (t) h (t) = f (τ) h (t τ) (12) Convolution Theorem I(f (t) h (t)) = H (µ) F (µ) (13) I(f (t) h (t)) = H (µ) F (µ) (14)

12 Outline 1 Preliminary Concepts 2

13 Sampling 1 Sampling theorem? 2 Aliasing?

14 Fourier Transform of an Impulse F (µ) = f (t) e j2πµt dt f (t) = F (µ) ej2πµt d µ Note that the dierence is in the sign of the exponential. Thus if, f (t) F (µ), then F (t) f ( µ) I(δ (t t 0 )) = e j2πµt 0 Then, e j2πtt 0 (notice the variable replacement), will have a transform, δ ( µ t 0 ). If we put t 0 = a, transform of e j2πat is δ (a µ) or δ (µ a). ( ) So we have, I e j2πnt T = δ ( ) µ n T

15 Fourier Series of an Impulse Train Let f [t] represent the product of f (t) and an impulse train. (sampled function). } F (µ) = I { f [t] = F (µ) S (µ) (15) This is obtained from equation 14, convolution theorem. So now we need to nd the transform of an impulse train, S (µ) = I(s (t)) = I ( n= δ (t n T ) ), which is periodic. We rst nd the Fourier series expansion for the impulse train as, s (t) = c n = 1 T s (t) = 1 n= T T 2 T 2 c n e j 2πnt T n= 2πnt j s (t) e T dt = 1 T e j 2πnt T

16 ( ) So now, S (µ) = I S (µ) = 1 T I( n= 1 T n= e j2πnt T ) e j2πnt T, the term inside the function summation we recognize as being that of impulse! So now, S (µ) = 1 T ( δ µ n ) n= T Remember - F (µ) = F (µ) S (µ), leading to, F (µ) = 1 ( F µ n ) T n= T (16) (17) Note notice that F (µ) is a continuous function. And since F (µ) consists of copies of F (µ), this is also continuous.

17 Fourier Transform - Illustration

18 Reconstruction Problem: Illustration

19 Reconstruction Problem 1 Generate two sinusoids with frequencies F 0 = 1 Hz and 8 F 1 = 7 Hz. Sample both at 8 F s = 1Hz. Generate stem plots for each, with a dierent color for each. Plot the original continuous signals over the sampled one, again each color coded. What do you observe? Determine the sampling rate at which you can avoid a problem here? Repeat the procedure at this new sampling frequency. What do you observe? 2 You have done such a process in your DSP lab. 1 You may try this out in Matlab.

20 Illustration

21 Reconstruction Function? f (t) = I 1 {F (µ)} = I 1 { H (µ) F (µ) } = h (t) f [t]

22 Outline 1 Preliminary Concepts 2

23 Continuous Transform Revisited F (µ) = = = = f [t] e j2πµt dt (18) n= n= n= Remember F (µ) = 1 f (t)δ (t n T ) e j2πµt dt (19) f (t)δ (t n T ) e j2πµt dt (20) f [n] e j2πµn T (21) T F n= f [n] is discrete, F (µ) is continuous. ( µ n ). T

24 Illustration - FT We have F (µ) is continuous, repeating itself with period 1 T. Consider one period µ = 0 to 1 T.

25 Getting the DFT What we want What we will do - K equally spaced samples of F (µ) between µ = 0 and µ = 1 T. Take samples at following frequencies - µ = k K T k = 0,1, K 1 (22) Substituting above µ to F (µ) = K 1 F [k] = n=0 n= f [n] e j2πµt, f [n] e j2πnk /K k = 0,1, K 1 (23)

26 Getting the DFT What we want What we will do - K equally spaced samples of F (µ) between µ = 0 and µ = 1 T. Take samples at following frequencies - µ = k K T k = 0,1, K 1 (22) Substituting above µ to F (µ) = K 1 F [k] = n=0 n= f [n] e j2πµt, f [n] e j2πnk /K k = 0,1, K 1 (23)

27 Getting the DFT What we want What we will do - K equally spaced samples of F (µ) between µ = 0 and µ = 1 T. Take samples at following frequencies - µ = k K T k = 0,1, K 1 (22) Substituting above µ to F (µ) = K 1 F [k] = n=0 n= f [n] e j2πµt, f [n] e j2πnk /K k = 0,1, K 1 (23)

28 DFT Pair DFT K 1 F [k] = n=0 f [n] e j2πnk /K k = 0,1, K 1 Inverse DFT f [n] = 1 K K 1 F [k] e j2πkn /K n = 0,1, K 1 (24) k=0

29 Convolution K 1 f [n] h [n] = k=0 f [m] h [n m] n = 0,1, K 1 (25) Circular convolution.

30 Sample Calculation In terms of variable x (f [x]). x = 0, 1, 2, 3. Calculate, F [0], F [1], F [2], F [3] f [0], f [1], f [2], f [3] from the above.

31 Summary Fourier series. Fourier transforms. Sampling, sampling theorem. Fourier transform of sampled functions. DFT, IDFT, Circular convolution.

32 Assignment? 1 Perform histogram equalization of image provided. Save the input image histogram and the output image histogram as.pgm images. 2 Repeat histogram equalization on the processed output of previous question. Save the histogram as.pgm. Repeat this process a few times. 1 Do you notice any dierence over successive histograms? 3 Using images provided, perform Histogram Matching. Save histograms of both input image and the reference image. Save histogram of processed image. (All in.pgm format).

33 References Dr. Richard Alan Peters II: