EE123 Discussion Section 3 - University of California ...

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EE123 Discussion Section 3

Jon Tamir (based on slides by Frank Ong)

February 8, 2016

Jon Tamir EE123 Discussion Section 3

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Announcements

I Submit Lab 1 by Thursday, Feb. 11 on bCourses

I Submit Homework 2 self-grade by Friday, Feb. 12 on bCourses

I Submit Homework 3 by Friday, Feb. 12 on bCourses

I Midterm 1 on Monday, Feb. 22 in class (2 hours)

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I DFT problems

I DCT demo and problem

I FFT demo

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The Discrete Fourier Transform

X [k] =N−1∑n=0

x [n]W knN [Analysis]

x [n] =1

N−1∑n=0

X [k]W−knN [Synthesis]

Equivalent interpretations of the DFT:

I Sampling the DTFT at ω = 2πN k

I The DTFT of periodically extended signalI Sampling the z-transform at z = e j

2πN k

I Direct implementation: O(N2)

I Fast implementation: O(N logN)

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X [k] =N−1∑n=0

x [n] =1

N−1∑n=0

2πN k

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X [k] =N−1∑n=0

x [n] =1

N−1∑n=0

2πN k

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Question 1

Given a 10-point sequence x [n], we wish to find equally spacedsamples of its Z transform X (z) on the contour as shown. Showhow to achieve this using the DFT.

Circle withradius = 1/2

��

z-plane

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Solution 1

��

z-plane

I Want to find X (z)|z= 12e j(2πk/10+π/10)

I X (z)|z= 12e j(2πk/10+π/10) =

∑n x [n]0.5−ne−j πn

10 e−j 2πkn10

I =⇒ DFT{x [n]0.5−ne−j πn10 }

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Solution 1

��

z-plane

I X (z)|z= 12e j(2πk/10+π/10) =

10 e−j 2πkn10

I =⇒ DFT{x [n]0.5−ne−j πn10 }

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Solution 1

��

z-plane

I X (z)|z= 12e j(2πk/10+π/10) =

10 e−j 2πkn10

I =⇒ DFT{x [n]0.5−ne−j πn10 }

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Question 2

x [n] = {−3, 5, 4,−1,−9,−6,−8, 2}

I a) Evaluate∑7

k=0(−1)kX [k]

I b) Evaluate∑7

k=0 |X [k]|2

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Solution 2

I part a)

(−1)k = e−jπk = e−j 2π4k8 = W 4k

x [4] =1

7∑k=0

X [k]W 4k8

⇒7∑

(−1)kX [k] = 8x [4] = −72

I part b) By Parseval’s Theorem,∑7k=0 |X [k]|2 = 8

∑7n=0 |x [n]|2 = 1888.

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Solution 2

I part a)

(−1)k = e−jπk = e−j 2π4k8 = W 4k

x [4] =1

7∑k=0

X [k]W 4k8

⇒7∑

(−1)kX [k] = 8x [4] = −72

∑7n=0 |x [n]|2 = 1888.

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Solution 2

I part a)

(−1)k = e−jπk = e−j 2π4k8 = W 4k

x [4] =1

7∑k=0

X [k]W 4k8

⇒7∑

(−1)kX [k] = 8x [4] = −72

∑7n=0 |x [n]|2 = 1888.

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Solution 2

I part a)

(−1)k = e−jπk = e−j 2π4k8 = W 4k

x [4] =1

7∑k=0

X [k]W 4k8

⇒7∑

(−1)kX [k] = 8x [4] = −72

∑7n=0 |x [n]|2 = 1888.

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Question 3

X [k] is the 9-point DFT of x [n]

I Given:

X [0] = −j ,X [2] = 1−j ,X [3] = 2−j ,X [5] = 3−j ,X [8] = 4−j

If possible, determine the remaining 4 samples with thefollowing assumptions:

I a) x [n] is real

I b) x [n] is conjugate symmetric

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Solution 3

I a) x [n] is real, then X [k] is conjugate symmetric:

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Solution 3

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Solution 3

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Solution 3

X [1] = X ∗[9−1] = 4+j ,X [4] = 3+j ,X [6] = 2+j ,X [7] = 1+j

BUT we also need X [0] = X ∗[0]. Clearly X [0] is not real, sothis first condition cannot be met.Not Possible.

I b) x [n] is conjugate symmetric, then X [k] is real. BUT this isnot the case.Not possible

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Solution 3

X [1] = X ∗[9−1] = 4+j ,X [4] = 3+j ,X [6] = 2+j ,X [7] = 1+j

BUT we also need X [0] = X ∗[0]. Clearly X [0] is not real, sothis first condition cannot be met.Not Possible.

I b) x [n] is conjugate symmetric, then X [k] is real. BUT this isnot the case.Not possible

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Question 4

Let x [n] be some length-N sequence. Let X [k] = DFT{x [n]}I Express x2[n] = DFT{X [k]} in terms of x[n]

I Hint: Try to match DFT{X [k]} with the equation

x [n] =1

N−1∑k=0

X [k]W−knN

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Solution 4

x2[n] = DFT{X [k]}

=N−1∑k=0

X [k]W knN

=N−1∑k=0

(X [−k])NW−knN

= 8IDFT{(X [−k])N}= 8(x [−n])N

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Solution 4

x2[n] = DFT{X [k]}

=N−1∑k=0

X [k]W knN

=N−1∑k=0

(X [−k])NW−knN

= 8IDFT{(X [−k])N}= 8(x [−n])N

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Solution 4

x2[n] = DFT{X [k]}

=N−1∑k=0

X [k]W knN

=N−1∑k=0

(X [−k])NW−knN

= 8IDFT{(X [−k])N}= 8(x [−n])N

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Solution 4

x2[n] = DFT{X [k]}

=N−1∑k=0

X [k]W knN

=N−1∑k=0

(X [−k])NW−knN

= 8IDFT{(X [−k])N}

= 8(x [−n])N

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Solution 4

x2[n] = DFT{X [k]}

=N−1∑k=0

X [k]W knN

=N−1∑k=0

(X [−k])NW−knN

= 8IDFT{(X [−k])N}= 8(x [−n])N

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Question 5

Let x [n] be some length-N sequence. Let X [k] = DFT{x [n]}I Express x3[k] = DFT{DFT{X [k]}} in terms of X[k]

I Express x4[n] = DFT{DFT{DFT{X [k]}}} in terms of x[n]

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Solution 5

I Using the DFT properties:

x2[n] = 8(x [−n])N

x3[k] = DFT{x2[n]} = 8(X [−k])N

x4[n] = DFT{x3[n]} = 64x [n]

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Solution 5 continued

I Ignoring the scaling factors:

x0[n] = DFT 0{x [n]} = x [n]

x1[k] = DFT 1{x [n]} = X [k]

x2[n] = DFT 2{x [n]} = (x [−n])N

x3[k] = DFT 3{x [n]} = (X [−k])N

x4[n] = DFT 4{x [n]} = x [n]

I The usual DFT operator is four-periodic. The nth power ofthe Fourier transform can be generalized as the fractionalFourier transform, which is used in optics.

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Question 6: DCT

I The Discrete Cosine Transform (DCT) is a DFT-relatedtransform that decomposes a finite signal in terms of a sum ofcosine functions

I The DCT is often used in compression schemes, such as MP3,JPEG and MPEG.

I One of the reasons is its energy compactness

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DCT Demo

I DCT Demo

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Question 6: DCT

Wanting to know more why DCT performs much better thanDFT, you decide to look closer at the definition of DCT

Given that the definition of the DCT (Type 2) is

Xc [k] = 2N−1∑n=0

x [n] cos(πk(2n + 1)

Express Xc [k] in terms of X [k], the 2N-point DFT of x [n]

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Solution 6: DCT

Xc [k] = 2N−1∑n=0

=N−1∑n=0

x [n](e−j πk(2n+1)2N + e j

πk(2n+1)2N )

=N−1∑n=0

x [n]e−j 2πkn2N e−j πk

2N +N−1∑n=0

x [n]e j2πkn2N e j

= X [k]e−j πk2N + X [−k]e j

= X [k]e−j πk2N + X ∗[k]e j

= 2Real(X [k]e−j πk2N )

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Solution 6: DCT

Xc [k] = 2N−1∑n=0

=N−1∑n=0

x [n](e−j πk(2n+1)2N + e j

πk(2n+1)2N )

=N−1∑n=0

2N +N−1∑n=0

x [n]e j2πkn2N e j

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Solution 6: DCT

Xc [k] = 2N−1∑n=0

=N−1∑n=0

x [n](e−j πk(2n+1)2N + e j

πk(2n+1)2N )

=N−1∑n=0

2N +N−1∑n=0

x [n]e j2πkn2N e j

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Solution 6: DCT

Xc [k] = 2N−1∑n=0

=N−1∑n=0

x [n](e−j πk(2n+1)2N + e j

πk(2n+1)2N )

=N−1∑n=0

2N +N−1∑n=0

x [n]e j2πkn2N e j

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Solution 6: DCT

Xc [k] = 2N−1∑n=0

=N−1∑n=0

x [n](e−j πk(2n+1)2N + e j

πk(2n+1)2N )

=N−1∑n=0

2N +N−1∑n=0

x [n]e j2πkn2N e j

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Solution 6: DCT

Xc [k] = 2N−1∑n=0

=N−1∑n=0

x [n](e−j πk(2n+1)2N + e j

πk(2n+1)2N )

=N−1∑n=0

2N +N−1∑n=0

x [n]e j2πkn2N e j

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Solution 6: DCT

Key point is:

Xc [k] = X [k]e−j πk2N + X [−k]e j

Xc [k] = e−j πk2N (X [k] + X [−k]e j

2πk2N )

xc [n] = Shift 12{x [((n))2N ] + x [((−n − 1))2N ]}

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Solution 6: DCT

Periodicity assumed by DFT

Periodicity assumed by DCT

DCT symmetric extension is better because sharp transitionsrequire many coefficients to represent

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FFT Demo

I FFT Demo

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