Complex symmetry Signals and Systems Fall 2015

Transcription

1 18-90 Signals and Systems Fall 015 Complex symmetry 1. Complex symmetry This section deals with the complex symmetry property. As an example I will use the DTFT for a aperiodic discrete-time signal. The notions can be applied to all the other categories (FS, DTFS, FT, DTFT). First we will talk what happens to the frequency representation X(e jω ) when x[n] has different properties. (a) Prove x[n] is real X (e jω ) X(e jω ) If x[n] is real then x [n] x[n]. X(e jω ) x[n]e jωn X (e jω ) X(e jω ) n ( n n n n n n x[n]e jωn ) ( x[n]e jωn ) x [n] ( e jωn) x [n]e jωn, we know that x [n] x[n] x[n]e jωn x[n]e jωn X (e jω ) X(e jω ) x[n] is real X (e jω ) X(e jω )

2 Complex symmetry (b) Prove x[n] is pure imaginary X (e jω ) X(e jω ) If x[n] is pure imaginary then x [n] x[n]. X(e jω ) X (e jω ) X(e jω ) n n n n n x[n]e jωn, we know that x [n] x[n] x[n] is pure imaginary (c) Prove x[n] is even X(e jω ) X(e jω ) If x[n] is even then x[n] x[ n]. X(e jω ) x[n]e jωn n n x[n]e jω( n) we can do a variable change m n x[ m]e jωm, we know that x[n] x[ n] m m x[m]e jωm X(e jω ) x[n] is even X(e jω ) X(e jω )

3 Complex symmetry 3 (d) Prove x[n] is odd X(e jω ) X(e jω ) If x[n] is odd then x[n] x[ n]. X(e jω ) x[n]e jωn x[n] is odd n n we can do a variable change m n, we know that x[n] x[ n] m m Time domain x[n] is real x [n] x[n] x[n] is pure imaginary x [n] x[n] x[n] is even x[n] x[ n] x[n] is odd x[n] x[ n] Frequency domain X (e jω ) X(e jω ) X(e jω ) X(e jω ) We will next mix the properties. In other words, we want to see what happens when the signal is real and even or complex and odd. (a) Prove x[n] is real and even X (e jω ) X(e jω ) X(e jω ) x[n] is real X (e jω ) X(e jω ) x[n] is even X(e jω ) X(e jω ) x[n] is real and even X (e jω ) X(e jω ) X(e jω )

4 4 Complex symmetry X (e jω ) X(e jω ) X(e jω ) is real, therefore ImX(e jω )} 0 X(e jω ) X(e jω ) X(e jω ) is even (b) Prove x[n] is real and odd X (e jω ) X(e jω ) X(e jω ) x[n] is real X (e jω ) X(e jω ) x[n] is odd X(e jω ) X(e jω ) x[n] is real and odd X (e jω ) X(e jω ) X(e jω ) X (e jω ) X(e jω ) X(e jω ) is pure imaginary, therefore ReX(e jω )} 0 X(e jω ) X(e jω ) X(e jω ) is even (c) Prove x[n] is pure imaginary and even x[n] is pure imaginary x[n] is even x[n] is pure imaginary and even (d) Prove x[n] is pure imaginary and odd x[n] is pure imaginary x[n] is odd x[n] is pure imaginary and odd Time domain Frequency domain x[n] is real and even ImX(e jω )} 0 x[n] is real and odd ReX(e jω )} 0 x[n] is pure imaginary and even x[n] is pure imaginary and odd

5 Complex symmetry 5 Thinking about the duality property, if we know information about the frequency representation X(e jω ) we can determine properties for the signal x[n]. Rule All of these properties can be applied to the rest of the Fourier friends - FS, DTFS, FT and DTFT.

6 6 Complex symmetry Problem 0 Given the following time domain signals, determine whether their corresponding Fourier transform is real/imaginary and even/odd. (a) x(t) cos( π 4 t) (b) x[n] j sin( π 8 n) (c) x(t) ( 1 (d) Solution: ) t ( 1 ) n, n 1 x[n] 0, n 0 n, n 1 (a) x(t) cos( π 4 t) Step 0: Compute the complex conjugate x (t) x (t) cos( π 4 t) x(t) x (t) x(t) X [k] X[ k] Step 1: Determine whether the signal is even or odd x( t) cos( π 4 t) cos( π 4 t) x(t) x( t) x(t) X[k] X[ k] Step : Draw the conclusion x (t) x(t) and x( t) x(t) X [k] X[ k] X[k] From this we can extract that the FS of x(t) is real and even. Alternate route: Compute the FS coefficients x(t) cos( π 4 t) 1 ej π 4 t + 1 e j π 4 t The FS coefficients are real and even. X[ 1] X[1] 1 for one period T 0 8

7 Complex symmetry 7 (b) x[n] j sin( π 8 n) Step 0: Compute the complex conjugate x [n] x [n] j sin( π 8 n) x[n] x [n] x[n] X [k] X[ k] Step 1: Determine whether the signal is even or odd x[ n] j sin( π 8 ( n)) j sin( π 8 n) x[n] x[ n] x[n] X[k] X[ k] Step : Draw the conclusion x [n] x[n] and x[ n] x[n] X [k] X[ k] X[k] From this we can extract that the DTFS of x[n] is real and odd. Alternate route: Compute the DTFS coefficients x[n] j sin( π 8 n) j j ej π 8 n j j e j π 8 n 1 ej π 8 n 1 e j π 8 n X[ 1] X[1] 1 for one period N 0 16 The FS coefficients are real and odd. ) t (c) x(t) ( 1 Step 0: Compute the complex conjugate x (t) x (t) ( ) t 1 x(t) x (t) x(t) X (jω) X( jω)

8 8 Complex symmetry Step 1: Determine whether the signal is even or odd Step : Draw the conclusion ( 1 x( t) ( 1 x(t) ) t ) t x( t) x(t) X(jω) X( jω) x (t) x(t) and x( t) x(t) X (jω) X( jω) X(jω) From this we can extract that the FT of x(t) is real and even. Alternate route: Compute the DTFS coefficients ( ) t 1 x(t) ( ) t 1 X(jω) e jωt dt 0 ( ) t 1 ( ) t 1 e jωt dt + e jωt dt 0 ( ) t 1 ( ) t 1 e jωt dt + e jωt dt 0 0 ( ) e jω t ( ) e jω t dt + dt 0 0 ( ( e jω ) t ln( ejω ) ln( ejω The FT is real and even. e jω ) t ln( e jω ) 0 ln( e jω ) ) jω ln() + 1 jω ln() 1 ln() jω + 1 ln() (ln()) + ω ln() + jω

9 Complex symmetry 9 (d) ( 1 ) n, n 1 x[n] 0, n 0 n, n 1 Step 0: Compute the complex conjugate x [n] x [n] x[n] X (e jω ) X(e jω ) Step 1: Determine whether the signal is even or odd ( 1 ) n, n 1 x[ n] 0, n 0 n, n 1 n, n 1 x[ n] 0, n 0 ( 1 n ), n 1 ( 1 n ), n 1 x[ n] 0, n 0 n, n 1 x[ n] x[n] X(jω) X( jω) Step : Draw the conclusion x [n] x[n] and x[ n] x[n] X (jω) X( jω) X(jω) From this we can extract that the FT of x(t) is pure imaginary and odd. Alternate route: Compute the DTFS coefficients ( 1 ) n, n 1 x[n] 0, n 0 n, n 1 X(e jω ) x[n]e jωn k k1 ( ) n 1 e jωn + 1 k ( n ) e jωn

10 10 Complex symmetry ( ) n 1 ( ) n 1 e jωn e jωn k1 ( ) e jω n ( ) e jω n k1 k1 k e jω 1 1 ejω 1 1 ejω ( e jω 1 1 e jω) ( 1 1ejΩ) j sin(ω) 1 cos(ω) 1 4 j sin(ω) 3 cos(ω) 4 j sin(ω) cos(ω) 3 4 The FT is pure imaginary and odd. Problem 1 You are given the following magnitude and phase of a DTFT of a discrete time aperiodic signal x[n]. X(jω) X(jω) π 1 π ω ω 1 π π Determine: (a) whether the signal is real or pure imaginary (b) whether the signal is even or odd

11 Complex symmetry 11 Solution: 1e j π, 4 ω X(jω) 1e j π, ω 4 e j π, 4 ω e j π, ω 4 j, 4 ω j, ω 4 (a) Step 0: Compute the complex conjugate X (jω) j, 4 ω j, ω 4 X (jω) X(jω) (b) Step 1: Determine whether the FT is even or odd j, 4 ω X( jω) j, ω 4 j, ω 4 j, 4 ω j, 4 ω j, ω 4 X( jω) X(jω) (c) Step : Draw conclusions based on X( jω) X(jω) X (jω). x(t) is real and odd. Problem Determine wheter the signal that has the follwing frequency representation is real/pure imaginary and even/odd. X(jω) jω cos(ω) Solution:

12 1 Complex symmetry (a) i. Step 0 is to determine X (jω) X (jω) jω cos(ω) ii. Step 1 is to determine X( jω) X( jω) j( ω) cos( ω), cos is even jω cos(ω) X(jω) iii. Step is to determine the relationship between X (jω) and X( jω) X (jω) X( jω) This implies that the signal x(t) is pure imaginary. (b) X(jω) X( jω), therefore the signal is even.